Engineersphere.com » Circuit Theory http://engineersphere.com doesn't have to make sense to you, but it's probably great for your health. Fri, 28 Oct 2011 03:05:49 +0000 en hourly 1 http://wordpress.org/?v=3.3 DC Biasing & AC Performance Analysis of BJT & FET Differential Amplifiers http://engineersphere.com/basic-electrical-concepts/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers.html http://engineersphere.com/basic-electrical-concepts/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers.html#comments Sun, 20 Mar 2011 19:55:23 +0000 Safa http://engineersphere.com/?p=2340

DC Biasing & AC Performance Analysis of BJT and FET Differential Amplifier Sub-circuits with Active Loads

Any op-amp worth its salt has a differential amplifier at its front end, and you’re nobody if you can’t design one yourself.  So, this article presents a general method for biasing and analyzing the performance characteristics of single-stage BJT and MOSFET differential amplifier circuits.  The following images show the general schematic for both kinds of differential amplifiers, often referred to as a differential input stage when used in designing op-amps.  Notice that these types of differential amplifiers use active loads to achieve wide swing and high gain.

differential amplifiers with active loads

Figure 1. BJT and MOSFET differential amplifiers with active loads

Due to design processes and the nature of the devices involved, BJT circuits are “simpler” to analyze than their FET counterparts, whose circuits require a few extra steps when calculating performance parameters.  For this reason, this tutorial will begin by biasing and analyzing a BJT differential amplifier circuit, and then will move on to do the same for a FET differential amplifier.  But it should be noted that the procedures to analyze these types of differential amplifiers are virtually the same.

BJT Differential Amplifier

The first thing needed is to configure the DC biasing.  To accomplish this, a practical implementation of I_{BIAS} must be developed.  A very popular method is to use a current mirror.   A simple current mirror is shown below:

BJT Current Mirror

bjt current mirror

Figure 2. BJT Current Mirror

It is easy to understand how a current mirror works.  Observe the equation governing the amount of collector current in a BJT, denoted I_C:

I_C = I_S(e^{\frac{V{BE}}{nV_T}}-1)(1+\frac{V_{CB}}{V_A})

where:

  • I_C is the collector current
  • I_S is the scale current
  • V_{BE} is the DC voltage across the base-emitter junction
  • V_T is the thermal voltage, typically 25 mV
  • n is the quality factor, typically between  1- 2 and is frequently assumed to be 1
  • V_{CB} is the voltage across the collector-base junction
  • V_A is the early voltage

Note: [This equation may look intimidating at first, but what is important to understand is that the point of designing "by hand" is to get close. One should aim simply to get a good estimation of such parameters as necessary bias current, gain, input impedance, etc.  In this way, computer simulations can analyze the hand-designed circuit in much closer detail, which greatly aids in the process of designing a real-life differential amplifier.  Knowing this, the equations to be used in this tutorial will be rough estimates, but are still invaluable when it comes to designing these types of circuits.]

By assuming a very large equivalent resistance, one can estimate that the collector current through any BJT can be described by:

I_C \approx I_S e^{V_{BE}/V_T}

What can be noticed here is that the only controllable variable in that equation is V_{BE}.  All the other terms in the equation are constants that depend on either the environment or the actual physical size of the device.  This means that for any two same-sized transistors, the currents through their collectors will be the same as long as the voltage across their base-emitter junctions is the same. By tying their bases and emitters together, we can mirror the currents between them!  In order to implement a successful current mirror, one transistor (here, Q_5) must have a current induced in it to mirror it to the differential amplifier’s current source (here, Q_6).  After adding this current mirror to our BJT differential amplifier, the resulting schematic is:

bjt diff amp with current mirror

Figure 3. BJT differential amp with current mirror biasing

In order to properly bias this circuit, it is necessary to include R_{BIAS}.  Two things are accomplished by including R_{BIAS} in our circuit.  One of them is that we can induce the current in Q_5, and thus, the current in Q_6.  The other important thing this resistor does is drop a majority of the available voltage across itself, so that Q_5 doesn’t have the entire voltage difference between the supplies across it!  To bias this circuit, the first thing one must do is determine what the desired magnitude of the current source will be.  This parameter depends on how you want the circuit to operate, and is usually a known value.  In this tutorial, we will assume we want an I_{BIAS} of 1mA.  In order to determine the necessary size of R_{BIAS}, we analyze the loop that consists of:

 

VCC \rightarrow I_{BIAS} \cdot R_{BIAS} \rightarrow V_{BE5} \rightarrow VEE

 

Kirchoff’s Voltage Law (KVL) around this loop reveals:

 

0 = -VCC + I_{BIAS} \cdot R_{BIAS} + V_{BE5} + VEE

 

These kinds of circuits are typically supplied rails of \pm 10 to 15 V.  So, this tutorial will assume:


VCC = - VEE = 10 V.

 

For a given technology, all of the BJT transistors are designed to have the same turn-on voltage. This tutorial will assume .7 V for each BJT.  That being the case, and rearranging the above equation, results in:

 

R_{BIAS} = \frac{VCC - VEE - V_{BE5}}{I_{BIAS}} = \frac{10V - (-10V) - .7V}{1 mA} = 19.3 k\Omega

 

By introducing a resistor R_{BIAS} of 19.3k\Omega to the above schematic, the bias current is now established at 1 mA.  Due to symmetry, the currents through transistors Q_1 and Q_2 are each half of the bias current, described by:

 

I_{C1} = I_{C2} = \frac{I_{BIAS}}{2} =\frac{1 mA}{2} = .5 mA

 

Now that we know the collector currents through Q_1 and Q_2, characterizing the performance of this differential amplifier is a breeze.  Since the parameters we are interested in (gain, CMRR, etc) are small-signal parameters, the small-signal model of this circuit is needed.  To obtain this, a nice trick is to “cut the amplifier in half” (lengthwise, such that you only analyze the output side of the amplifier) to obtain:

bjt small signal model

Figure 4. Small-signal model for above differential amplifer

Note: [even though the output signal is single-ended here, the output is still a result of the entire input signal, and not just half of it.  This is because the small-signal changes in the currents flowing through Q_{2,4} are impeded from traveling down the branches controlled by current sources Q_{2,6}.  Also note that the connections between R_{\pi} and the voltage-controlled current source (VCCS) indicate that the voltage that controls the VCCS is the voltage across R_{\pi}.  This is because the resistance in the emitter of these transistors has been omitted, due to its typically small value (10 to 25 \Omega).  In addition to this, Q_6 is assumed to be a small signal (AC) open-circuit.  The frequency response has also been omitted, and the amplifier is assumed to be unilateral.]

Differential Mode Gain

It is simple to see that v_{out} (the small-signal output voltage) is equal to the current across the parallel combination of the resistors r_{o2} and r_{o1} multiplied by the size of the same parallel combination.  Since we know the value of the current through this combination is equal to the input voltage multiplied by g_m (the transconductance parameter):

v_{out} =- g_mv_{in} \cdot (r_{o2} \| r_{o4})

The transconductance parameter is a ratio of output current to input voltage. It is described mathematically as:

g_m = \frac{\partial i_c}{\partial v_{be}} = \frac{\partial i_{out}}{\partial v_{in}}

and can be solved for thusly:

\frac{\partial I_C}{\partial V_{BE}} = \frac{\partial (I_Se^{\frac{V_{BE}}{V_T}})}{\partial V_{BE}} = \frac{I_Se^{\frac{V_{BE}}{V_T}}}{V_T} = \frac {I_C}{V_T}

In this example, I_C is .5 mA and V_T is 25 mV.  With these values, we compute:

g_m = \frac{I_C}{V_T} = \frac{.5 mA}{25 mV} = 20 \frac{mA}{V}

Now that the transconductance parameter is known, the only other values needed to compute the differential mode gain are r_{o2} and r_{o4}.  Q_2 is an npn transistor, while Q_4 is a pnp transistor, so they will not have the same small-signal resistance, but the procedure to find these two values are nearly identical.  The following equation describes the small-signal output resistance of any BJT:

r_{o_{n,p}} = \frac{|V_{A_{n,p}}| + V_{CE}}{I_C} \approx \frac{|V_{A_{n,p}}|}{I_C}

The parameter V_{A_{n,p}} is typically given, and in this tutorial:

V_{A_n} = 130 V

V_{A_p} = 50 V

Which would result in:

r_{o2} = \frac{V_{A_n}}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega and

r_{o4} = \frac{V_{A_p}}{I_C} = \frac{50 V}{.5 mA} = 100 k\Omega

Now that the small-signal resistances are known, along with the transconductance parameter, the differential mode gain (A_{v,DM}) may be calculated:

A_{v,DM} =- g_m \cdot (r_{o2} \| r_{o4}) =- 20 \frac{mA}{V} \cdot \frac{260 k\Omega \cdot 100 k\Omega}{(260+100) k\Omega} = -1444.4 \frac{v}{v}

or, in decibels (dB):

A_{v,DM(dB)} = 20log(|A_{v,DM}|) = 63.2 dB

Differential Input Impedance

The differential input impedance of a differential amplifier is the impedance a “seen” by any “differential” signal. A “differential signal” is any and all signals that aren’t shared by V_{in-} and V_{in+}.  For instance, if:

V_{in-} = (2 + sin(2 \pi ft)) V and

V_{in+} = (2 + cos(2 \pi ft)) V

then the common mode signal and differential mode signals are:

V_{in,CM} = 2V and

V_{in,DM} = cos(2 \pi ft) - sin(2 \pi ft)

To find the differential input impedance, begin by following the loop consisting of:

V_{in-} \rightarrow V_{BE1} \rightarrow -V_{BE2} \rightarrow V_{in+}, as illustrated below:

bjt diff amp Rin

Figure 5. Loop analyzed in order to determine Rin(DM)

We see that, in the differential signal mode, the path to ground only consists of r_{\pi} of each input transistor. Since this is the case, the differential mode input impedance of any BJT diff-amp may be expressed as (omitting emitter resistance and assuming Q_{1,2} matched):

R_{in,DM} = r_{\pi 1}+r_{\pi 2} = 2r_{\pi}

where: r_{\pi} = \frac{\beta}{g_m}

\beta = \frac{i_c}{i_b} (current gain factor)

A typical value for \beta is 100, and knowing g_m allows one to compute:

R_{in,DM} = 2 \cdot \frac{\beta}{g_m} = 2 \cdot \frac{100}{20 \frac{mA}{V}} = 10 k \Omega

So, for the BJT differential amplifier in this tutorial, the differential mode input impedance is:

R_{in,DM} = 10 k \Omega (what impact will this have?)

Common Mode Gain

The CM gain (A_{v,CM}) is the “gain” that common mode signals “see,” or rather, is the attenuation applied to signals present on both differential inputs. A good op amp attempts to eliminate all common mode signals, but this is obviously not possible in the real world.  However, one may compute the common mode gain by “cutting the amplifier in half” by observing one of the loops in the following diagram.  The path differs from that of differential signals because common mode signals make it so that the two signal sources don’t “see” each other.  Notice:

 

common mode voltage gain

Figure 6. Loop(s) analyzed to determine common mode voltage gain and input impedance

We choose a loop and draw the small-signal model to obtain:

common mode gain BJT diff amp

Figure 7. Small-signal model for common mode input signals

Similar to the output voltage of the differential mode small signal model, we can see that V_o is the voltage across r_{o4}.  We also know the current running through this resistance, and may equate the output voltage to:

V_o = - g_mv_{\pi} \cdot r_{o4}

This time, though, v_{in,CM} isn’t distributed entirely over the resistances at the base.  Instead, a fraction of the input common mode input signal is across the base-emitter junction.  Referring back to the small signal model, we see that the loop composed of:

V_{in} \rightarrow v_{\pi2} \rightarrow (i_b + g_mv_{\pi 2}) \cdot 2 \cdot r_{o6}

reveals that:

V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot (i_b + g_mv_{\pi2})

but i_b is negligible compared to the current supplied by the collector, so we say:

V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot g_mv_{\pi2} = v_{\pi2} \cdot (1 + 2r_{o6}g_m)

which we use to solve for v_{\pi2}:

v_{\pi2} =\frac{ v_{in}}{1 + 2 \cdot r_{o6} \cdot g_m}

Which we then plug back into the equation for V_o:

V_o = - g_mv_{\pi} \cdot r_{o4} = - \frac{r_{o4}g_m}{1+2 \cdot r_{o6} \cdot g_m} \cdot V_{in}

From this we can solve directly for the common mode gain:

A_{v,CM} = \frac{V_o}{V_{in}} = -\frac{r_{o4}g_m}{1 + 2r_{o6}g_m}

Here, the common mode gain is:

A_{v,CM} = -.3845 = -8.3 dB

Common Mode Input Impedance

The common-mode input impedance is the impedance that common-mode input signals “see.” One can analyze the common mode input impedance (R_{in_{CM}}) by, again, “cutting the differential amplifier in half” and analyzing one side the resulting schematic, assuming a common mode signal.  This can be found by observing the figure 6, above.

Choosing one of these paths, we construct the corresponding small-signal model for common mode signals (assuming r_{o2} = \infty), which is shown in figure 7.  From this figure, deriving R_{in,CM} is simple.  Notice the currents flowing in the loop that consists of:

V_{in} \rightarrow i_b \cdot r_{\pi2} \rightarrow (i_b + g_mv_{\pi2}) \cdot 2 \cdot r_{o6}

from this loop, one may compute:

0 = -V_{in} + i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6}

which is used to find an equation for V_{in}

V_{in} = i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6}

and since:

g_mv_{\pi 2} = \beta \cdot i_{b}

and i_b = i_{in}

So:

V_{in} = i_b \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1))

which is the same as:

V_{in} = i_{in} \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1))

which can be rearranged for:

R_{in,CM} = \frac{V_{in}}{i_{in}} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1)

where: r_{\pi} = \frac{\beta}{g_m}

Which, in this tutorial, results in:

R_{in,CM} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1) = \frac{1 mA}{25 mV} + 2 \cdot \frac{130 V}{1 mA} \cdot (100 + 1) = 26.26 M\Omega

Common Mode Rejection Ratio (CMRR)

The common mode rejection ratio (CMRR) is simply a ratio of the differential mode gain to the common mode gain, and is defined as:

CMRR = \frac{A_{v,DM}}{A_{v,CM}}

Here, the CMRR is:

CMRR = \frac{-1444.4 v/v}{-.384 v/v} = 3761.46 = 71.5 dB

Analysis of FET Differential Amplifiers

As stated before, the analysis of these performance parameters are done virtually the same for FET diff amps as they are for BJT diff amps.  There are, however, a few key differences.  For one, all BJT transistors are typically built to be the same size on a given IC device.  But for an IC device that uses FETs, this is not the case.  Each FET has an adjustable length and width that affects how much current it will pass for a given voltage-drop across the device.  In fact, observe the equation for the drain current in a FET:

I_D = \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2

From this, the gate-source voltage is:

V_{GS} = \sqrt{\frac{2I_DL}{kW}} - V_{th_{n,p}}

where:

  • k is the process conductivity parameter, and is equal to:

    k = \mu_{n,p} C_{ox} , which is the electron mobility multiplied by the oxide capacitance

  • W, L are the width and length of the device, respectively
  • V_{GS} is the gate-to-source voltage
  • V_{th} is the threshold voltage of the FET

Analyzing BJTs in a circuit is more simple because all base-emitter voltages are assumed to be equal.  But this is not the case for mosfets, and one must analyze the above equation (or others) to find device voltages.  But there is the threshold voltage – the minimum gate-to-source voltage that will allow for any conduction whatsoever.  The threshold voltage is a result of the FET fabrication process, and is typically provided on datasheets for each FET gender.

For a differential amplifier composed of FETs to work, it is imperative that all the FETs be in saturation mode.  For a FET to be in saturation implies:

|V_{DS}| \ge |V_{GS}| - |V_{th}|

So this must be checked when analyzing these types of circuits.

Another important difference is the derivation of the transconductance parameter, g_m.  When analyzed for a BJT, it was defined as the ratio of the change in collector current to the change in the base-emitter voltage.  For a FET there is a similar procedure, as the transconductance is defined as the ratio of the change in drain current to the change in gate-source voltage.  Mathematically, the transconductance parameter is:

g_m = \frac{\partial{i_D}}{\partial{v_{GS}}} =\frac{\partial( \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2)}{\partial{v_{GS}}} = \sqrt{2I_Dk\frac{W}{L}}

The last notable difference is the computation for a FET’s small-signal resistance.  The equation describing r_{o} is:

r_o = \frac{1}{\lambda I_D}

where \lambda is the channel-length modulation parameter.

From this little discussion, you should be able to apply the principles used to analyze the BJT differential amplifier to the analysis of a FET-based differential amplifier.  But, of course, if you would like to see a FET differential amplifier explained in more detail, do not hesitate to ask a question!

Credit & Acknowledgment

This post was created in March 2011 by Kansas State University Electrical Engineering student Safa Khamis.  A million thank yous extended to Safa for taking the time to document this important process for everyone else to learn from.  Please leave questions, comments, or ask a question in the questions section of the website.


V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega

V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega

]]> http://engineersphere.com/basic-electrical-concepts/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers.html/feed 0 The Convolution Integral Explained http://engineersphere.com/math/the-convolution-integral-explained.html http://engineersphere.com/math/the-convolution-integral-explained.html#comments Thu, 10 Mar 2011 06:57:38 +0000 Safa http://engineersphere.com/?p=2179

Introduction to the convolution

Amongst the concepts that cause the most confusion to electrical engineering students, the Convolution Integral stands as a repeat offender.  As such, the point of this article is to explain what a convolution integral is, why engineers need it, and the math behind it.

In essence, the “convolution” of two functions (over the same variable, e.g. f_1(t) and f_2(t)) is an operation that produces a separate third function that describes how the first function “modifies” the second one.  Conversely, the resulting function can be seen as how the second function “modifies” the first function.  Sometimes the result is used to describe how much the first two functions “have in common.”  In all honesty, the concept of the convolution of two functions is quite abstract, but the frequency at which it appears in nature grants its importance to scientists and engineers.  Ultimately the aim here is to identify its use to electrical engineers – so for now do not dwell solely on its mathematical significance.

A convolution of two functions is denoted with the operator “* ”, and is written as:

convolution integral

Convolution of f1(t) and f2(t)

Where \tau is used as a “dummy variable.”  To aid in understanding this equation, observe the following graphic:

Convolution of 2 square pulses

Convolution of two square pulses, resulting in a triangular pulse

Before diving any further into the math, let us first discuss the relevance of this equation to the realm of electrical engineering.

Why is the convolution integral relevant?

Most electrical circuits are designed to be linear, time-invariant (LTI) systems.  Being “linear” implies that the magnitude of a circuit’s output signal is a scaled version of the input signal’s magnitude.  Further, an LTI system that is excited by two independent signal sources will output the sum of the scaled versions of each signal.  This is extended for an infinite number of independent signal sources, and gives rise to the concept of superposition.  Put in another way, if a function x_1(t) causes an LTI system to output y_1(t), then:

a_1 \cdot x_1(t) \to a_1 \cdot y_1(t)

Where a_1 is a multiplicative constant.  In addition to this, superposition allows us to say:

a_1 \cdot x_1(t) + a_2 \cdot x_2(t) + \ldots \to a_1 \cdot y_1(t) + a_2 \cdot y_2(t) + \ldots

Being a “time-invariant” system means it does not matter when the input signal is applied – a specific input signal will always result in the same output signal for a given LTI system.  Put mathematically, time-invariance can be expressed as:

x_1(t) \to y_1(t) \Leftrightarrow x_1(t+\tau) \to y_1(t+\tau)

where \tau can be viewed as a time delay when dealing with signals through time (i.e. “time-domain signals”).  Though not directly, this concept also signifies that an output signal cannot contain frequency components not inherent in the input signal (causality).

The vast majority of circuits are LTI systems, each with a specific impulse response. The “impulse response” of a system is a system’s output when its input is fed with an impulse signal – a signal of infinitesimally short duration.  A real-world “impulse signal” would be something like a lightning bolt – or any form of ESD (electro-static dischage).   Basically, any voltage or current that spikes in magnitude for a relatively short period of time may be viewed as an impulse signal.  The impulse response of a circuit will always be a time-domain signal, and exists because no signal can propagate through a circuit in zero time; each individual electron involved can only move so quickly through each component.  Typically, real-world electronic LTI systems exhibit an impulse response that consists of an initial spike in magnitude, followed by an everlasting and ever-decreasing exponential relationship in signal magnitude.  The following image describes this graphically.

Typical Unit Impulse Response

So, here’s the big deal: the fact that each LTI circuit has a specific impulse response function (here, referred to as h(t)) is very useful in predicting its behavior given a particular input signal (here, referred to as x(t)).  This is because the input signal itself may be viewed as an impulse train – a stream of continuous impulse functions, with infinitesimally short durations of time between each impulse.  This fact, along with superposition, allows one to find the output of an LTI system given an arbitrary input signal by summing the LTI system’s impulse response to each impulse function that make up the input signal. By allowing the time between each “impulse” of the input signal to go to zero, this approach can be used to determine the output time-domain signal of an LTI system for any time-domain input signal.  For example, the following graphic shows the output of an RC circuit when fed with a square pulse:

Convolution of RC network impulse response and square wave input to find the output signal.

What is seen here is the integral of the impulse response and the input square wave as the square wave is stepped through time. In the above convolution equation, it is seen that the operation is done with respect to \tau, a dummy variable.  In reality, we are taking an input signal, flipping it vertically through the origin (not evident with a square wave), and determining what the integral is at each value of \tau, which here is delay through time. Since the output of any LTI system is non-causal (meaning it cannot exist until the signal that excites the output has been applied), we must mathematically step through time to see how each impulse signal of the input affects the LTI system’s impulse response – again, achieved by stepping through \tau – the “time-delay” dummy variable.

A Convolution Example

To see how the convolution integral can be used to predict the output of an LTI circuit, observe the following example:

For an LTI system with an impulse response of h(t) = e^{-2t}u(t) , calculate the output, y(t), given the input of:

f(t) = e^{-t}u(t)

The output of this system is found by solving:

y(t) = h(t)*f(t) = \int_{0}^{\infty} h(\tau) \cdot f(t-\tau) d\tau

We only integrate between 0 and +\infty because, if we define t = 0 as the time that the input signal f(t) is applied, then both h(t) and f(t) have zero magnitude at any time t<0.

From there, we calculate:

y(t) = \int_{0}^{\infty} e^{-2\tau}u(\tau) \cdot e^{-\tau} d\tau= \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau

Next, we can simplify and compute the integral:

y(t) = \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau = e^{-2t} \int_{0}^{t}e^{\tau}d\tau = e^{-2t}(e^t-1) = e^{-t} - e^{-2t}

Since y(t) = 0 for all t < 0, we can write the output y(t) as:

y(t) = (e^{-t}-e^{-2t})u(t)

This result y(t) describes the output function for an LTI system with an impulse response h(t) when fed the input signal f(t).

5 Steps to perform mathematical convolution

Often, one may wish to compute the convolution of two signals that can’t be described with one function of time alone.  For arbitrary signals, such as pulse trains or PCM signals, the convolution at any time t can be computed graphically.  For signals whose individual “sections” can be described mathematically, follow these steps to perform a convolution:

1.) Choose one of the two funtions (h(\tau) or f(\tau)), and leave it fixed in \tau-space.

2.) Flip the other function vertically across the origin, so that it is time-inverted.

3.) Shift the inverted signal through the \tau axis by t_0 seconds.  Choose to shift the signal to the first “section” of the fixed function that is described by the same equation.  The inverted signal (say, f(- \tau) ), now shifted, represents f(t_0 - \tau) , which is basically a “freeze frame” of the output after the input signal has been fed to the LTI system for t_0 seconds.

4.) The integral of the two functions, after shifting the inverted function by t_0 seconds, is the value of the convolution integral (i.e. output signal) at t = t_0.

5.) Repeat this procedure through all “sections” of the function fixed in \tau-space.  By doing this, you can compute the value of the output at any time t!

Useful Properties

 

The following is a list of useful properties of the convolution integral that can help in developing an intuitive approach to solving problems:

1.) Commutative Property:

f_1(t)*f_2(t) = f_2(t)*f_1(t)

2.) Distributive Property:

f_1(t)*[f_2(t)+f_3(t)] = f_1(t)*f_2(t) + f_1(t)*f_3(t)

3.) Associative Property:

f_1(t)*[f_2(t)*f_3(t)] = [f_1(t)*f_2(t)]*f_3(t)

4.) Shift Property:

if f_1(t)*f_2(t) = c(t)

then f_1(t-T_1)*f_2(t-T_2) = c(t-T_1-T_2)

5.) Convolution with an Impulse results in the original function:

f(t)* \delta (t) = f(t) where \delta (t) is the unit impulse function

6.) Width Property:

The convolution of a signal of duration T_1 and a signal of duration T_2 will result in a signal of duration T_3 = T_1 + T_2

Convolution Table

Finally, here is a Convolution Table that can greatly reduce the difficulty in solving convolution integrals.

Thank you so much to Safa Khamis @ Kansas State University for taking the time to write this tutorial for Engineersphere and the electrical engineering community.

 

 

]]> http://engineersphere.com/math/the-convolution-integral-explained.html/feed 2 Frequency Response for MOSFET/BJT http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html#comments Sun, 30 May 2010 01:48:25 +0000 Riley http://engineersphere.com/?p=926

The frequency response of a BJT or MOSFET can be found using nearly the exact same process, with the only variations being caused by a single resistor and simple naming conventions that differ between the two devices.

Before we start let’s think a little bit about what we’re doing:
Our goal is going to be to find the pole(s) of the circuit.
Okay? What is a pole and why do I care where it is?
A pole is a frequency at which the gain of the device rolls off. (remember that when it rolls off , it will be at the -3dB frequency with a slope of -20dB/decade)

We care because if the gain of a device rolls off at a certain frequency, then we won’t be able to amplify a signal above that frequency very well because the gain will be decreasing by 20dB/decade.

The procedure is nearly identical whether we are using a BJT of a MOSFET, but we will work each of them side by side just in case there might be any confusion, and we’ll follow these steps as we go through.  (we will also use some values that came from the output file when running a simulation of this circuit in Cadence (or PSPICE) )mosfet-amplifier

bjt-amplifier
1. Take a look at one of the circuits and see what you notice, how about the MOSFET.  This step is just to help us with our knowledge understanding of the circuit.
- At a glance it just looks just like another MOSFET right? Sure is, but let’s take a look at a few things just for kicks. Notice that it is using a bypass capacitor at the source so we don’t have to worry about R_s (at when working with high frequency).  Since the capacitor C_s bypasses R_s to ground, you should notice that this is a common-source amplifier.  You could notice the Values for R_1 and R_2 and start to think about what the Gate voltage is and how that may affect the circuit.
2. We are talking about frequency response so that means we are probably going to want to draw the small signal equivalent circuit.
Remember that the capacitors C_1 and C_2 will act like short circuits at high frequencies so we will ignore them, but we will have to account for some of the capacitance internal to the device.

Both devices have internal capacitances that are very similar.  As you can see from the small signal models for a MOSFET (above) and BJT (below), the only significant difference is that the BJT has an additional resistance Rpi between the Base and Emitter.

Most of the analysis we will do is based on the small signal model. Note that small signal models are not typically used in PSPICE so this picture may look a bit odd, especially the controlled source but for our purpose it is good to have a visual reference. To start we will point out what everything is. Cgs is an internal capacitance betwemosfet-small-signal-model

en the gate and source. The

values for Cgs was similar to one the a PSPICE simulation may give.  CM1 and CM2 are Miller capacitances which we will find values for laterbjt-small-signal-model.  ro is a Norton equivalent resistance that makes the model more ideal.  And just pretend that the G2 looks more like a voltage controlled current source and that their gains are gm*Vgs and gm*Vpi. For the BJT CM1 and CM2 are both Miller capacitances, Cpi is similar to Cgs and Rpi the additional component used for BJTs but not MOSFETs. The other part should look familiar from the other figures.

ON TO THE ANALYSIS!!!

We will find the device gain, overall gain, equivalent input and output capacitances, and the input and output poles. The process for both is essentially the same.

Device Gain: This is the gain from the control source to the output so we are looking for Vout/Vgs (or Vout/Vpi for a BJT). We will ignore CM2 for this process. Notice the resistances ro, RD, and RL are in parallel. Vout should be given by that equivalent resistance times the current though it which is gm*Vgs from the control source. So the equation for device gain is,

V_{out} / V_{gs} = gm*(r_o//R_D//R_L)   (MOSFET)

V_{out} / V_{\pi} = gm*(r_o//R_C//R_L)   (BJT)

Overall Gain: This will be the gain from the source (Vs) to the output (Vout). We already know what Vout/Vgs is so if we find Vgs/Vs, we can multiply them to get Vout/Vs = (Vout/Vgs) * (Vgs/Vs).  Vgs/Vs is a simple voltage divider. Hopefully you can see this from the small signal model (remember that we are ignoring the capacitors for now but they will play a part later).  The equations we will get for Vgs/Vs and the overall gain are.

V_{gs} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s}  (MOSFET)

Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s} * gm(r_o//R_D//R_L)  (MOSFET)

V_{gs} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s}  (BJT)

Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s} * gm(r_o//R_C//R_L)  (BJT)

Now we will find the input and output poles.  For this we will need to look at the capacitances and use a formula to find the Miller capacitances, CM1 and CM2.  Any explanation for the miller capacitance will have to wait for another post or check out your Electronics Book, Wikipedia, Google, etc. but we will need to use a couple of special equations.  Overall we will need to find the input resistance and input capacitance for the input pole and the output resistance and output capacitance for the output pole.

Each pole will be at a frequency w=1/RC where the R and C are the equivalent R and C at that point, so to find the input pole, we will need to find the input resistance and the input capacitance.  These are found by looking into the input (the left side of the small signal model).  The voltage source will  act like a short so we see Rs in parallel with R1//R2 for the MOSFET (the BJT will have Rpi in parallel also).  The input capacitance will be Cgs in parallel with CM1 (the BJT will be the same).  The output resistance and capacitance are found the same way only looking in from the output (the right side of the small signal model).

\omega_{IN} = \frac{1}{R_{IN}C_{IN}}  \omega_{OUT} = \frac{1}{R_{OUT}C_{OUT}}     (MOSFET or BJT)

So the input pole will be: (MOSFET)

R_{IN} = R_S//R_1//R_2  =  950                                     R_{OUT} = r_o//R_D//R_L =

C_{IN} = C_{gs} + C_{M1}  =                                               C_{OUT} = C_{M2} =

\omega_{IN} =                                                                          \omega_{OUT} =

(BJT)

and the output pole will be: (MOSFET)

(BJT)

R_{IN} = R_S//R_1//R_2//r_\pi =                                  R_{OUT} = r_o//R_D//R_L =

C_{IN} = C_{BE} + C_{M1} =                                                 C_{OUT} = C_{M2} =

\omega_{IN} = \frac{}{}                                       \omega_{OUT} =\frac{}{}

To Do:

finish input & ouput R, input C, Pole (& calculate answers)

]]> http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html/feed 0 Amplifiers – Part I http://engineersphere.com/basic-electrical-concepts/amplifiers-part-i.html http://engineersphere.com/basic-electrical-concepts/amplifiers-part-i.html#comments Wed, 17 Mar 2010 19:54:47 +0000 Jeff http://engineersphere.com/?p=1165

This post is about amplifiers, how they work, and common applications. I will cover several operational amplifier configurations, and situations where each might be useful. This is part I of II for general discussion about amplifiers. Enjoy!

Definition of an amplifier

Definition (for Bioinstrumentation): Circuit that makes a small signal, usually voltage but occasionally current or power, big enough to do something useful–including excite an output mechanism.

Common amplifier uses:

• Biological measurements of small signals
• Audio engineering: a large current is needed to drive speakers
• Wireless communications: far from originating antenna, signal is very weak and must be
amplified to be useful.

Secondary amplifier applications:

Many amplifiers are also filters, preferentially amplifying some frequencies over others

• Don’t want to amplify noise along with signal
• Only interested in low- or high-frequency portion of signal
• Active filter provides amplification as added bonus

General Amplifier Characteristics

Common Mode Rejection Ratio (CMRR)

ratio (usually in dB) of the amplifier’s common-mode gain to its differential-mode gain. Common-mode signals are input signals common to both + and – inputs and are usually unwanted noise–60-Hz, thermal, etc; differential signals are applied to only one input.

Gain

voltage out over voltage in, or current out over current in. May be given in dB.  Bioamp requirement: often adjustable; should be 1000 or greater, should be calibrated.

Input Impedance

what the input source sees as its load working into the amplifier: if the entire amplifier circuit were modelled as a resistor, what would be the value of the resistor? Typical bioamp: Rin = 10MΩ–signal source need not provide much current.

Output Impedance

same as input impedance but from the output end: model the entire amplifier as a source, and this is its internal impedance. Bioamp requirement: Ro << Rload.

Frequency Response

over what range of frequencies is the gain constant? Graphically illustrated with a Bode plot of gain vs. frequency.

DC offset

usually an amplifier has an operator-adjustable DC offset knob, to null out any offset associated with non-ideal amplifier or sensor behavior. A DC offset signal results in an incorrect reading unless removed or filtered out (high-pass filter).

Operational Amplifier:

basis for most instrumentation-related amplifiers, cheap, readily available, easy to work with. “Operational” = good for mathematical operations (+, -, log, …)

op-ampMeanings and advantages:

• equal input voltages –> within the limits of external power supplies, an op amp outputs whatever current is needed to drive the two input voltages equal. Result is that the output voltage follows the input, scaled by a large gain.
• infinite input resistance means the op amp never loads down the source, even if the source cannot supply much power.
• zero output resistance means the op amp is an ideal voltage source, with output voltage independent of whatever load impedance it must work into.
• infinite open-loop gain means the amplification properties of a circuit containing an op amp are independent of the op amp internal properties.

Carr and Brown go through several common op-amp configurations and show how to derive their voltage gains. Suffice it for now to know that if you want to build an amplifier, an op amp is a good place to start.

Common op-amp circuit configurations:

• Inverting and non-inverting amplifiers
• Summing and difference amplifiers
• Integrating and differentiating amplifiers
• Log and anti-log amplifiers
• Instrumentation amplifier
• Low-pass filter
• High-pass filter
• Band-pass and notch filters
• Buffer (voltage follower, or unity-gain buffer)

Op-Amp Equivalent Circuit

This schematic illustrates the important properties of the op amp, and of any amplifier. It can also make it easier to understand circuit operation.

op-amp-equivalent

Note the open-circuit inputs– Rin = infinity. The
output voltage supply is a dependent voltage
source. Also, since the gain A is infinite, v2 – v1
must be zero to get a finite output.

 

 

 

 

Difference (Differential) Amplifier

Example: derive the gain relationship for the basic differential amplifier shown, assuming U1 is ideal and Vin = V2 – V1.

differential-amplifier

To get equal gain of both V1 and V2, set R2/R1 = R4/R3. Then Vo = R2/R1(V2-V1).

To get a high gain, R2 >> R1, but to get high input impedance R1 (and/or R3) should be large, making R2 and R4 even larger…Result: high gain and high input impedance are difficult to achieve together.

 

 

 

 

 

Instrumentation Amplifier

A difference amp with input buffer/gain stages to increase input impedance and gain. To analyze, realize that the same current must flow in R5, R6 and R5 (since no current flows into the op amps). Set R1=R3, R2 = R4; then Vo = G1* (v3(U1) -v2(U1)), where G1 = R2/R1 = gain of second (differential) stage.

instrumentation-amplifierGain of input stage is 1 + 2*R5/R6 = G2. Overall gain is G1*G2. Making R6 a potentiometer allows compensation for inequalities in the two R5s, as well as for variable gain of the entire circuit.

Overall Gain: A practical difference amp can have a gain of 100, so it is not hard to get an overall gain of 10,000 from an instrumentation amp.
Input Impedance: equal to that of the op amps U1 and U2–very large. Use FET-based amps for extremely high input impedance
Output Impedance: close to that of the op amp U1–very small: the amp will provide whatever current is needed to maintain the output voltage regardless of load impedance.

Equal resistors: in practice one cannot buy matched discrete resistors; however it is fairly easy to manufacture them within an integrated circuit. Monolithic diff-amps are available.

Non-idealities of amplifiers

Gain:  TANSTAAFL–you cannot have gain without a power supply to provide it. Real gain is limited by the external power  supplies (+/- 12 or 15 V, for op amp circuits) Exceeding the limits of the power supply results in Saturation, or “hitting the rail”.

Output impedance: a zero output impedance means the circuit will provide whatever current is needed to maintain the requested output voltage. Practically, however, an op amp can only provide some 20mA, meaning RO is negligible only for RL>>15V/20mA = 750 Ω.

Frequency dependence: to avoid oscillation or saturation, circuitry must often be added that limits the bandwidth of an amplifier.
• To keep DC offset signals (from polarizing electrodes, for example) out of the amplifier, a high-pass filter is used to cut off DC (and lower-frequency ac) signals.
• If the load to be driven contains substantial capacitance, the current output limit again becomes a problem, limiting gain at high frequencies, where capacitors look like shorts.

Input bias current: real op amps do have non-zero input currents, which produce voltage drops at the input–another source of DC offset. This source can be minimized by using FET op amps.

Impedance Bridge

Often the measurand is the relation between voltage and current (one applied, the other a response) rather than a biologically generated source. An example in Carr and Brown uses a wire heated by an applied current as an airflow sensor:  air flow from a breathing patient cools the wire, changing its resistance. Similarly, a voltage applied to a membrane induces a current flow; the ratio of voltage to current is a resistance. Such relations are best measured using a Bridge, and if the bridge is made solely of resistors it is called a Wheatstone Bridge.

wheatstone-bridgeUsually drawn as a diamond, this configuration of resistors is “balanced” when V+ – V- = 0. If Rtest then varies a little, a differential amplifier across V+ and V- will register a potential difference proportional to
the change in Rtest.

The impedances can have capacitance and/or inductance associated with them, in which case the bridge can measure both energy storage and  resistive loss in an element.

A return path to ground for (DC) bias currents is automatically provided by this circuit to prevent saturation.

Well there you have it, a few common amplifier configurations and some useful terms pertaining to them.  Remember important concepts such as amplifier saturation, Input Impedance, Output Impedance, and Gain.  A solid understanding of these concepts is sure to impress somebody!  Amplifiers  part II will continue to elaborate on more fun amplifier concepts.

References: References: Carr and Brown ch. 7; Webster chs. 3, 6; Neamen, Electronic Circuit Analysis and
Design (McGraw Hill, 2001) ch. 9

]]> http://engineersphere.com/basic-electrical-concepts/amplifiers-part-i.html/feed 1 Laplace Transforms http://engineersphere.com/math/laplace-transforms.html http://engineersphere.com/math/laplace-transforms.html#comments Sun, 06 Sep 2009 18:06:20 +0000 Luke http://engineersphere.com/?p=701

What is the Laplace Transform method?

The Laplace Transform is a method that simplifies integral and differential equations into algebraic equations. This practice is commonly used to solve for a function out of a differential equation, which otherwise may have been unsolvable or very difficult. The following integrals can be used to transform between F(s) \Leftrightarrow f(t) where \mathcal{L} denotes Laplace and \mathcal{L}^{-1} denotes Inverse Laplace:

\mathcal{L} [f(t)] = F(s) = \int\limits_0^\infty {f(t)*e ^ {-st}d} t

\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2 \pi j} \int\limits_{c - j \omega}^{c + j \omega} {F(s)*e ^ {-st}d}s

Table of Laplace Transforms

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

A Laplace Transform example

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for y(t) .

\frac{d^2y}{dt^2} + 7 \frac{dy}{dt} + 12y = 10 \quad y(0)=3, y'(0)

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

s^2 Y(s) - s y(0) - y'(0) + 7 (s Y(s) - y(0)) + 12 Y(s) = \frac{10}{s}

Inputting our initial conditions:

s^2 Y(s) - 3s - 0 + 7 (s Y(s) - 3) + 12 Y(s) = \frac{10}{s}

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have Y(s) in common and factor it out. Our goal is to find Y(s) = F(s) because, after all, \mathcal{L}^{-1}[Y(s)] = y(t) :

(s^2 +7s + 12)Y(s) - 3s -21 = \frac{10}{s}

After solving for Y(s) and factoring the denominator:

Y(s) = \frac{3s^2 + 21s + 10}{s(s+4)(s+3)}

Taking the Inverse Laplace Transform

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of Y(s) to find y(t) . If our function Y(s) does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the expand function on your TI-89. Just go to catalog \rightarrow expand and enter your function in parenthesis. Using this function:

Y(s) = \frac{\frac{5}{6}}{s} - \frac{\frac{13}{2}}{s+4} + \frac{\frac{26}{3}}{s+3}

Noting \mathcal{L} [f_1(t) + f_2(t)] = \mathcal{L} [f_1(t)] + \mathcal{L} [f_2(t)] and \mathcal{L} [k*f(t)] = k*\mathcal{L} [f(t)] , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

y(t) = \frac{5}{6} - \frac{13}{2}e^{-4t} + \frac{26}{3}e^{-3t}

To check your work you can plug y(t), \frac{dy}{dt}, \frac{d^2y}{dt^2} into the original, differential equation and at t=0 we find that 10=10.

]]> http://engineersphere.com/math/laplace-transforms.html/feed 3 Finding Thevenin’s Equivalent http://engineersphere.com/basic-electrical-concepts/thevenin-equivalent.html http://engineersphere.com/basic-electrical-concepts/thevenin-equivalent.html#comments Mon, 17 Aug 2009 20:44:58 +0000 Jeff http://engineersphere.com/?p=662

Thevenin’s Theorem

Thevenin’s theorem states that a two terminal circuit containing voltage sources, current sources, and resistors can be modeled as a voltage source in series with a resistor.  The benefit of using a Thevenin equivalent is that it makes analyzing how a circuit interacts with other circuits a much simpler process.  Consider the circuit below.  Suppose you want to know the loaded voltage of the circuit (VL) for three different loads connected to nodes a and b.  The three loads are 200 Ω,  2 kΩ, and 20 kΩ.  How fast could you find each Vab?

not-thevenin-equivalent

The best method for quickly assessing the performance of this circuit at a variety of loads is to find the circuit’s Thevenin equivalent.

thevenin-equivalent

Once we have the equivalent circuit, the problem can be solved using three simple voltage dividers.

thevenin-equivalent-loading

V_{L1} = V_{th} (\frac{R_{L1}}{R_{L1} + R_{th}}) = 2 (\frac{200}{200 + 6k}) = 0.065V       V_{L2} = 0.5V    V_{L3} = 1.54V

Finding Thevenin Equivalents in Practice

By using the Thevenin equivalent model, the problem was solved without the need to perform a full circuit analysis each time the load changed.  However, you must learn how to acquire Thevenin equivalents before you can take advantage of them.  In practice, finding the Thevenin equivalent of a circuit is simple.

  • Find Vth by measuring the open circuit voltage with a multimeter.
  • Find Rth by connecting a current meter to the two terminals and dividing Vth by the measured current (called a short-circuit current).  Note that this isn’t the safest method and should never be used in practice! A safer and more proper method for finding Rth is outlined in Lab 3.  That is, find a load resistance that produces a noticeable drop in the load voltage.  Then find Rth by applying the voltage divider formula.

V_{L} = V_{th} (\frac{R_{L}}{R_{L} + R_{th}}) \rightarrow R_{th} = R_{L} (\frac{V_{th} - V_{L}}{V_{L}})

Thevenin Equivalent Resistance

In theory, finding the Thevenin equivalent is more difficult because we can’t rely on lab equipment to do the work for us.  On paper, the Thevenin equivalent resistance (Rth) is easier to find.

remove-load

1. If there is a load, remove it – Remember that Thevenin’s theorem applies to two terminal circuits.  This implies that the circuit is unloaded (see Figure 4).  After all, Vth is equal to the open circuit voltage, as you saw above.

2. Short out any voltage sources and open up any current sources – All sources must be deactivated to find the Thevenin equivalent resistance.

example-circuit

3. Find the equivalent resistance looking into the two nodes – In Figure 4, the two nodes would be a and b.  The easiest way to find the equivalent resistance is to start at node a and end at node b.  If there are multiple paths from a to b, then there are parallel resistors somewhere in the circuit.

Example 1: Find the Thevenin equivalent resistance with respect to nodes a and b for the circuit in Figure 5.

Solution: Follow the steps listed above.  There is no load to remove because nodes a and b are already open.  Next, deactivate the sources by shorting the voltage source and opening the current source.  Shorting the voltage source effectively shorts out the 40 Ω and 15 Ω resistors, meaning they have no bearing on the value of Rth.  The last circuit in Figure 6 shows the two paths current can flow from a to b.  Rth is given by the equation below.

R_{th} = (4 + 26) || 10 = 7.5 \Omega

thevenin-current-flowthevenin-equivalent-voltage

Thevenin Equivalent Voltage

Any of the circuit analysis techniques learned so far can be used to find the Thevenin equivalent voltage (Vth).  Voltage dividers, branch currents, node voltage, superposition, and source transformations (the last two will be covered soon) are all legitimate methods for finding Vth.  Of the two Thevenin equivalent values, Vth tends to be the harder one to find because there is no one best way to work all problems.  Node voltage may work well on one circuit, but may be weighed down by multiple variables on another.  Voltage division isn’t always easy, either.

Until you can recognize which method to use, perhaps the best approach is to try one method and see if it works.  If the math becomes too overwhelming, move on to a different method.  Also, realize that the equivalent voltage is measured across two nodes, as shown in Figure 7.  Be certain Vb is connected to ground before declaring that Va is equal to Vth. In this case Vth is equal to Va – Vb. Finally, if there is a load, remember that you must remove it before working the problem! Once again, a Thevenin equivalent is derived from an unloaded circuit.  A loaded circuit alters the equivalent values.

Example 2: Find the Thevenin equivalent voltage with respect to nodes a and b for the circuit in Figure 5.

Solution: The circuit has no ground marked.  This must be placed first.  In general, it’s most convenient to place the ground at the negative terminal of a voltage source.  Placing the ground at the negative terminal of the 17.4 V makes the voltage at the node connected to the source’s positive terminal 17.4 V.

example-circuit2

Note: This node would not be at 17.4 V had the ground been connected elsewhere.

The next step is to find the two node voltages, Va and Vb.  Because there is a resistor between node b and ground, Vb must be found in addition to Va.  The current source combined with the arrangement of resistors make voltage division all but impossible.  Instead, try the node voltage method for node a.

I_{1} + I_{2}+ I_{3} = 0 \rightarrow \frac{V_{a} - 17.4}{26} - 0.1 + \frac{V_{a}}{14} = 0 \rightarrow V_{a} = 7V

thevenin-equivalent2

Notice that the last term used 14 instead of 10.  This is allowable because the 4 Ω and 10 Ω resistors are in series.  Find Vb by using either node voltage or voltage division.  In this case, it’s faster to use voltage division.  If this isn’t obvious, redraw the branch containing node b as shown in Figure 9.

V_{b} = 7(\frac{4}{4+10}) = 2V

Now solve for Vth using the equation from Figure 7 and draw the Thevenin equivalent.

V_{th} = V_{a} - V_{b} = 7 - 2 = 5V

See, Thevenin’s Equivalent is a lot of fun!  Get good at this and you have mastered quite a few important engineering concepts!  Thanks to Ryan Eatinger (reatinge@ksu.edu) for contribution of this lesson. :)

]]> http://engineersphere.com/basic-electrical-concepts/thevenin-equivalent.html/feed 4 Node Voltage http://engineersphere.com/basic-electrical-concepts/node-voltage.html http://engineersphere.com/basic-electrical-concepts/node-voltage.html#comments Tue, 11 Aug 2009 01:45:22 +0000 Jeff http://engineersphere.com/?p=639

As its name implies, the node voltage method is used to find a node’s voltage with respect to ground.  While a voltage divider can be used for the same purpose, the primary purpose of a voltage divider is to find voltage drops across resistances rather than with respect to ground.  One disadvantage of using a voltage divider is that a circuit must be simplified to a two resistor series circuit before the equation can be applied.  The larger the circuit, the more difficult it becomes to simplify the circuit, especially when there are multiple sources.  With the node voltage method, no simplification is necessary.  It can be applied to the circuit as is.

node-voltage-currents

The circuit in Figure 1 shows three resistors meeting at a node.  We’ll use this node to show how node voltage works in the most general sense (i.e. without using any numbers).  As you’ll see, the node voltage method is basically Ohm’s law applied to Kirchoff’s current law.

Note: All of the voltages in Figure 1 (and other figures throughout this article) are node voltages, or voltages at a node, which are taken with respect to ground.

Steps for solving a circuit using node voltage

1. Pick a node – This can be any node.  In the problems you’ll see, you’re told which node to use.  In reality, node voltage can be applied to any node because KCL can be applied to any node.  Let’s pick the Vx node for this example.

2. Apply KCL to the node – There are three paths or branches connected to the Vx node and each path has its own current.  These currents are labeled as I1, I2, and I3.  Rather than trying to guess which way these currents flow, assume that all currents leave the node.  If we assume that all currents leave the node, then we’re also assuming no currents enter the node, simplifying the KCL equation as shown below.

\Sigma i_{leaving} = \Sigma i_{entering} \rightarrow \Sigma i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0

Note that it doesn’t matter if our assumption is wrong.  If one of the currents is actually flowing into the node, it will show up later as a negative in the KCL equation (and we’re not afraid of negative numbers).

3. Write equations for each current – Unless one of the branches is a current source, you’ll have to place each current in terms of voltages and resistances according to Ohm’s law.

I_{1} = \frac{V_{x} - V_{1}}{R_{1}}               I_{2} = \frac{V_{x} - V_{2}}{R_{2}}             I_{3} = \frac{V_{x} - V_{3}}{R_{3}}

4. Derive the node voltage equation – Plug the equations you found in step 3 back into the KCL equation in step 2 to derive the node voltage equation for Vx.

i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{x} - V_{1}}{R_{1}} + \frac{V_{x} - V_{2}}{R_{2}} + \frac{V_{x} - V_{3}}{R_{3}}

5. Solve for any variables – In most problems you’ll encounter, all of the variables will be known except the voltage at the node.  In this instance, solve for Vx and you’re finished.  A node voltage equation can be written for any node in a circuit.  For multiple variables, you’ll need multiple equations (more on this later).

kirchoffs-current-law

Example 1: Find the voltage V1 for the circuit in Figure 2.

Solution: The first step has already been taken care of in the problem statement.  The next step is to write the KCL equation for the V1 node.  Mark the currents as shown in Figure 3, assuming that they all flow away from the node.  The KCL equation becomes:

\Sigma i_{leaving} = \Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

Next, write equations for each current.  The two voltage sources are connected to ground, making it easy to find the voltages at the nodes on the upper left and right corners of the circuit.  Notice that Va’s positive terminal is connected to ground, making the voltage at the negative terminal -8 V.

I_{1} = \frac{V_{1} - (-V_{a} )}{R_{1}} = \frac{V_{1} - (-8)}{150}            I_{2} = \frac{V_{1} - 0}{R_{2}} = \frac{V_{1}}{30}             I_{3} = \frac{V_{1} - V_{b}}{R_{3}} = \frac{V_{1} - 10}{20}

Using these equations, derive the node voltage equation for V1.  Solving for V1 gives the voltage with respect to ground at that node.

\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} + 8}{150} + \frac{V_{1}}{30} + \frac{V_{1} - 10}{20} = 0

300(\frac{V_{1} + 8}{150} + \frac{V_{1}}{30} + \frac{V_{1} - 10}{20} = 0) = 0(300) \rightarrow 2V_{1} + 16 + 10V_{1} + 15V_{1} - 150 = 0 \rightarrow V_{1} = 4.96V

Working with current sources

kirchoffs-current-law2

Example 2: Find the voltage V1 for the circuit in Figure 4.

Solution: The approach doesn’t change when there are current sources in the circuit.  Current sources can actually simplify the math.  Once again, the node has been chosen by the problem statement.  Now write the KCL equation and solve the problem the same way as before.  Note that the current source points into the node and therefore needs to be added as a negative current.

\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

I_{1} = \frac{V_{1} - 6}{3k}           I_{2} = \frac{V_{1}}{1k}          I_{3} = -0.01

\Sigma i_{leaving} = 0 \rightarrow I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} - 6}{3k} + \frac{V_{1}}{1k} - 0.01 = 0 \rightarrow V_{1} = 9V

Working with Multiple Variables

example-schematic3

Example 3: Find the voltages V1 and V2 for the circuit in Figure 6.

Solution: Approach every node voltage problem in the same manner: choose a node and derive its node voltage equation.  If the equation has more than one variable, then move to the next node and write another node voltage equation.  Continue this process until the number of equations equals the number of variables.  In this example, there will be two variables (V1 and V2).  This means two node voltage equations are required to find the two voltages.

This time there are two nodes to choose from.  Start with V1 first.  As always, assume the currents flow away from the node.

node-voltage-currents2\Sigma i_{leaving} =\Sigma i_{entering} \rightarrow I_{1} + I_{2} + I_{3} = 0

Deriving equations for I1 and I2 is the same as before, but I3 is slightly different.  The voltage V2 is unknown, but it can still be placed into the equation.  It will be treated as a variable just like V1.

I_{1} = \frac{V_{1} - V_{a}}{R_{1}} = \frac{V_{1} - 150}{20}                I_{2} = \frac{V_{1} - 0}{R_{2}} = \frac{V_{1}}{80}               I_{3} = \frac{V_{1} - V_{2}}{R_{3}} = \frac{V_{1} - V_{2}}{40}

The node voltage equation for V1 becomes

I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{1} - 150}{20} + \frac{V_{1}}{80} + \frac{V_{1} - V_{2}}{40} = 0 \rightarrow 7V_{1} - 2V_{2} = 600

Now follow the same procedure to find the node voltage equation for V2.

node-voltage-currents3I_{1} = \frac{V_{2} - V_{1}}{R_{1}} = \frac{V_{2} - V_{1}}{40}

I_{2} = -11.25 A

I_{3} = \frac{V_{2} - 0}{R_{4}} = \frac{V_{2}}{4}

I_{1} + I_{2} + I_{3} = 0 \rightarrow \frac{V_{2} - V_{1}}{40} -11.25 + \frac{V_{2}}{4} = 0 \rightarrow -V_{1} + 11V_{2} = 450

Finally, solve the system of equations to find V1 and V2.

V_{1} = 100V                            V_{2} = 50V

That should just about cover quite a few node-voltage examples and concepts.  If you have any questions feel free to ask them in the questions section of the website, or simply leave a comment below!  Many thanks to Ryan Eatinger (reatinge@ksu.edu) for contribution of this post. :)

]]> http://engineersphere.com/basic-electrical-concepts/node-voltage.html/feed 0 Current Divider http://engineersphere.com/basic-electrical-concepts/current-divider.html http://engineersphere.com/basic-electrical-concepts/current-divider.html#comments Mon, 10 Aug 2009 23:49:28 +0000 Jeff http://engineersphere.com/?p=618

Current Dividers

Current dividers are the inverse of voltage dividers.  Voltage dividers work with series circuits where current remains constant; any parallel components must be combined before the voltage divider equation works.  In contrast, current dividers work with parallel circuits where the voltage is the same across all components and any series components must be combined before applying the current divider equation.

current-flow

The general equation for a current divider is given below and illustrated in Figure 1.  Io is the measured current, Is is the source current (or the current entering the node, Ro is the resistance of the path Io flows through, and RT is the equivalent resistance of the circuit.

General Current Divider Equation:

I_{0} = I_{s} ( \frac{R_{t}}{R_{0}})

Notice that this equation is nearly identical to the voltage divider equation, except that RT and Ro are inverted.  RT is in the numerator because current dividers work exclusively with parallel resistors.  Therefore, RT is always smaller than any of the individual resistors because of the nature of the parallel resistance formula.  The result is that Io is always less than or equal to Is.

parallel-current-flow

The current divider equation is derived in the same fashion as the voltage divider equation.  Consider Figure 2.  All three components are connected in parallel, meaning all three have the same voltage across them.

V_{s} = V_{1} = V_{2}

Use Ohm’s law to place the equation above in terms of resistance and current.

Ohm’s Law: V = IR

\rightarrow I_{s} = (R_{1}||R_{2}) = I_{1}R_{1} = I_{2}R_{2}

Now use these equations to find I1 and I2.

I_{1} = I_{s} ( \frac{ R_{1} || R_{2} } {R_{1} } )                   I_{2} = I_{s} ( \frac{R_{1} || R_{2} }{ R_{2} } )

Current Divider Example

Example: Find the current Io for the circuit in Figure 3.

example-current-divider-circuit

Solution: This problem requires two current dividers to solve.  The current first splits between R1 and the rest of the circuit and then again between R3 and R4.  Because Io is part of the second split, the current through R2 must be found first.  Remember, though, that the current through R2 also passes through R3 and R4.  The current divider equation cannot be applied without taking the resistance of the entire path into consideration.

Combine R2, R3, and R4 before applying the first current divider equation.

R_{234} = R_{2} + R_{3} || R_{4} = 10 + \frac{6 \cdot 3}{6 + 3} = 12 \Omega

I_{2} = I_{s} ( \frac{ R_{1} || R_{234} } {R_{234}} ) = 4( \frac{ \frac{36 \cdot 12}{36 + 12}}{12} ) = 4( \frac{9}{12}) = 3A

example-current-divider-circuit-current-flow

The equation above shows that the 36 Ω resistor takes 1 A of the 4 A available to the circuit.  Therefore, the second current divider must use 3 A as the source current because that is the amount of current flowing into the node.  Also, do not try to incorporate R2 into the second equation.  The current splits between R3 and R4; R2 has nothing to do with where the current flows once it reaches the node.

I_{0} = I_{s} ( \frac{ R_{3} || R_{4} } {R_{4}} ) = 3( \frac{ \frac{6 \cdot 3}{6 + 3}}{3} ) = 3( \frac{2}{3}) = 2A

Thanks to Ryan Eatinger (reatinge@ksu.edu) for the contribution of this post.

]]> http://engineersphere.com/basic-electrical-concepts/current-divider.html/feed 2 Ohm’s Law http://engineersphere.com/basic-electrical-concepts/ohms-law.html http://engineersphere.com/basic-electrical-concepts/ohms-law.html#comments Sun, 26 Jul 2009 23:14:11 +0000 Jeff http://engineersphere.com/?p=390

Current Flow

What is Ohm’s Law?

Ohm’s law defines the relationship between voltage, current, and resistance by the equation below.  The second equation better represents voltage as the difference between two electric potentials.

V = IR

V_{1} - V_{2} = IR

Note that V1 and V2 are voltages measured with respect to ground and V is the voltage potential measured between them.

The equation derived from Ohm’s law is incredibly useful for an electrical engineer.  Ohm’s law allows many circuits to be fully analyzed with the aid of just a few measurements.  Many circuit analysis techniques you will learn involve some combination of KCL, KVL, and/or Ohm’s law.

Ohm’s law states that there must be a voltage drop (or voltage difference) across a resistor in order for any current to flow.  If V1 and V2 are the same, no current flows through the resistor.  In terms of currents, a current produces a voltage drop across a resistor; if there is no current, there is no voltage drop across the resistor.

A circuit component that follows Ohm’s law has a constant resistance.  Increasing the current through the component will produce a proportional increase in the voltage drop across it.  The plot of the current through the component versus the voltage drop across it will be linear, with the slope of the line determining the resistance of the component.

I-V Plot to describe Ohm’s Law

Figure 2 shows I-V plot of a 1 Ω and a 2 Ω resistor, both of which follow Ohm’s law.  Figure 3 shows the I-V curve of a component not following Ohm’s law.  You will learn about two components that don’t follow Ohm’s law later in the course (diodes and MOSFETs).

Linear I-V Relationship

Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!

]]> http://engineersphere.com/basic-electrical-concepts/ohms-law.html/feed 0 Mesh Current http://engineersphere.com/circuit-theory/mesh-current.html http://engineersphere.com/circuit-theory/mesh-current.html#comments Thu, 23 Jul 2009 03:43:17 +0000 Jeff http://engineersphere.com/?p=144

Here I will show you how to calculate the different currents in each loop of the figure below using the mesh current method.

This is my favorite approach to a problem like this one:Mesh Current Loops

1) Identify meshes in a planar circuit.

2) Identify currents unknown.

3) Write KVL for each mesh.

4) Simplify and Solve.

Easier said than done I suppose, so let’s do it:

It is important to understand how many equations you are going to need.  Here is how you do that:

Using Node-Voltage:   #Equations = Ne – 1

Using Mesh-Current: #Equations = Be – (Ne – 1)

Where Ne = Number of Node,  Be = Number of circuit components.

Here we have:

Ne = 4

Be = 6

Be – (Ne – 1) = 3 equations.

Now you are ready to start writing your mesh equations.  You should only do one loop at a time.  This is simply KVL at work here, adding up the voltages around the loop and making sure they sum up to zero or to the source voltage.

Mesh 1: 5 ( i1 – i3) + 26 (i1 – i2) = 80

Mesh 2: 26 ( i2 – i1) + 90 (i2 – i3) + 8 (i2) = 0

Mesh 3: 30 (i3) + 90 (i3 – i2) + 5 (i3 – i1) = 0

We now have 3 equations and 3 unknowns, this is what we want and we should go ahead and move the “80″ in the Mesh 1 equation to the left side so that all equations sum to zero.

Confused by this? “( i2 – i1)”

That is normal, and probably the most confusing concept.  All 3 of the currents in all 3 loops are heading in the same direction, but in relation to each circuit component (such as 90Ω resistor), they are opposite.  If you are in loop number 1, this current (i1) is your reference current for this given loop, and the current through a resistor in that loop is equal to that reference current minus the opposing current on that given component.  If the currents were heading in the same direction, they would simply add instead of subtract.  This way we have summed up the total current through each component and can therefore calculate the voltage.

You can either use a TI-89 to solve for these unknowns…

OR

\begin{pmatrix}31&-26&-5\\-26&124&-90\\-5&-90&125\end{pmatrix}\begin{pmatrix}i1\\i2\\i3\end{pmatrix} = \begin{pmatrix}80\\0\\0\end{pmatrix}

Solving the 3×3 system

I1 = 5 A

I2 = 2.5 A

I3 = 2 A


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