Engineersphere.com » Basic Electrical Engineering Concepts 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 Finding the Inverse of a Matrix http://engineersphere.com/math/finding-the-inverse-of-a-matrix.html http://engineersphere.com/math/finding-the-inverse-of-a-matrix.html#comments Sun, 06 Mar 2011 03:26:46 +0000 Luke http://engineersphere.com/?p=2103

Matrix manipulations and properties

Finding the inverse of a matrix is much more complex than finding the inverse of a number. All real numbers have an inverse (i.e. 6^{-1}= \frac{1}{6} ). However, not all matrices have an inverse. There are several characteristics that allow us to visibly determine whether a matrix has an inverse but we will only focus on one. A matrix must be square (i.e. 2×2, 3×3, etc.) to have an inverse. Performing the following manipulations will be a waste of time if a matrix is not square. It is also important to know the inverse matrix property. Using my example above, 6^{-1} * \frac{1}{6} = 1 and similarly with matrices, A * A^{-1} = In where In is the identity matrix (diagonal from top left to bottom right contains all 1′s, and everything else is 0) . We take advantage of this property when solving systems of matrices.

In words, the general algorithm for determining the existence of an inverse matrix is to manipulate the matrix into row reduced echelon form (rref). If the rref matrix is an identity matrix, then the inverse matrix exists. Hang on now, earlier I mentioned that there were other, visible characteristics that allow us to determine the existence of an inverse matrix, but now I’m asking you to perform a tedious process (without a calculator) with the same goal? Wouldn’t it be easier to first determine if finding the rref of the matrix is worthwhile? You’re right, except we are going to make a simple manipulation, and at the same time that we finish our rref process and determine that an inverse matrix exists, we will have found the inverse matrix! How do we do that? We will create an augmented matrix between our matrix in question, A, and the appropriate identity matrix where the size of matrix A is equal to the size of matrix In . We will perform the same rref process to the augmented matrix | A In | . If the portion of our augmented matrix previously belonging to matrix A reduces to an identity matrix (indicating the existence of A^{-1} ), then the portion previously belonging to the identity matrix, will equal A^{-1} .

Some matrix math

Now, for the math…

Suppose we are asked to find the inverse of the following matrix:

\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}

First, we must set up the augmented matrix discussed above. Notice that I have simply placed the identity matrix (of the same size as A ) on the right of matrix A .

\begin{bmatrix}1&3&3&1&0&0\\1&4&3&0&1&0\\1&3&4&0&0&1\end{bmatrix}

Finding the rref of an augmented matrix

Next, we will attempt to find the rref of the augmented matrix. If the portion of the augmented matrix previously belonging to A yields an identity matrix, A is invertible.

rref \begin{bmatrix}1&3&3&1&0&0\\1&4&3&0&1&0\\1&3&4&0&0&1\end{bmatrix} = \begin{bmatrix}1&0&0&7&-3&-3\\0&1&0&-1&1&0\\0&0&1&-1&0&1\end{bmatrix}

Ok great! The left half of our augmented matrix reduced to an identity matrix. That means two things to us: the matrix has an inverse and we’ve already found the inverse. If you recall from above, A^{-1} is the right half of the augmented matrix (after finding it’s rref, of course). So we can conclude:

\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}^{-1} = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}

If our rref of the augmented matrix had yielded anything other than an identity matrix, we would conclude that A^{-1} does not exist. This method will simply allow us to determine the existence of and entries to A^{-1} for any size matrix.

]]> http://engineersphere.com/math/finding-the-inverse-of-a-matrix.html/feed 1 Solving a Linear System Using the Inverse Matrix http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html#comments Thu, 03 Mar 2011 22:49:47 +0000 Luke http://engineersphere.com/?p=2009

Describing the process of solving a linear system using the adjacent matrix is best done while performing an example. Suppose we have a system A*x = B where A is the coefficient matrix of our system, x is the column vector containing our variables, and B is the solution column vector. We are asked to solve for the column vector x made up of variables x_1, x_2, and x_3.

\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 12\\-10\\16\end{bmatrix}

Typically, we would divide B by A to solve for x , however there is no method for performing division between matrices. By taking advantage of the inverse matrix property A^{-1}*A = 1 , we can simply the formula to solve for the column vector x . The commutative property does not apply in matrix multiplication so A^{-1}*B \not= B*A^{-1}.  Therefore we have have to be aware of the ‘order’ in which we multiply:

(A^{-1}) * A * x = (A^{-1}) * B       simplifies to      x = A^{-1}*B

Notice that since we multiplied by A^{-1} ‘first’ on the left side of the equation, we also multiply ‘first’ on the right side. Now, multiplying the inverse of matrix A by matrix B will yield a column vector matching our x_1, x_2, and x_3. Below, I have used the equation x = A^{-1}*B and plugged the values for A^{-1} into the equation. The product between A^{-1} and B is shown on the far right. Note: This article assumes you know how to find the inverse of a matrix. This process is described in my article Finding The Inverse of a Matrix.

\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix} \begin{bmatrix} 12\\-10\\16\end{bmatrix} = \begin{bmatrix} 66\\-22\\4\end{bmatrix}

Therefore, x_1=66, x_2=-22, and x_3=4. Simple systems (i.e. this 3×3 system) are much easier to solve with algebra instead of finding the inverse of the coefficient matrix and performing matrix multiplication. This application is more practical for larger systems or while working on Matrix Theory homework.

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]]> http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html/feed 0 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 BJT Circuit and Symbol Conventions http://engineersphere.com/basic-electrical-concepts/bjt-circuit-and-symbol-conventions.html http://engineersphere.com/basic-electrical-concepts/bjt-circuit-and-symbol-conventions.html#comments Mon, 19 Apr 2010 18:13:39 +0000 Luke http://engineersphere.com/?p=1484

The following is an explanation of symbol conventions , voltage polarities and current directions for npn and pnp BJTs. The goal is to help understand these characteristics but not on the physical level of electrons and holes. The following figure shows practical operation of each BJT in the active mode.

pnp-and-npn-bjts

npn or pnp

When looking at a BJT, the easiest way to decide whether it is npn or pnp is to look at the emitter, which is always modeled as the arrow. If you remember that the arrow tail is always at a ‘p’ node and the tip is at an ‘n’ node, you can easily decide whether the BJT is npn or pnp. Remember that the collector and emitter are always either both ‘n’ or both ‘p’.

Determining Voltage Polarities

It is important to know which direction the voltage’s will appear positive when we begin using nodal analysis to solve BJT circuits. Typically, there will be a voltage drop of .7 V over the |V_{BE}| = |V_{EB}| nodes that will be used in these calculations. Whether |V_{BE}| or |V_{EB}| is positive is decided by the type of BJT. The voltage polarities are flipped between pnp and npn BJTs. Obviously, the only difference in the symbols between the two types of BJTS is the arrow, which is the emitter. If we remember the tip of the arrow is the lower voltage, we are able to deduce that V_{BE} = .7V for an npn BJT and V_{EB} = .7 for pnp.

To be in the active mode, a BJT’s collector-emitter voltage must be above approximately .3 V. As above, this voltage polarity is reversed between npn and pnp BJTs. To determine, whether V_{CE} or V_{EC} should be positive, we can use our deduction of the base-emitter voltage polarity. The voltages, in active mode, drop from collector to base to emitter in npn BJTs and from emitter to base to collector in pnp BJTs. So, if we have figured out that we are using an npn BJT because V_{BE} was a positive .7V, we know that the base voltage is higher than the emitter voltage. From here we know the collector must be higher than the base, and therefore, higher than the emitter. We have just figured out that V_{CE} must be greater than the .3V to be working in active mode. Using the same logic, V_{EC} must be greater than .3V for a pnp BJT to remain in active mode.

Current Flow Directions

Current directions are very simple to figure out. Just use the arrow. The collector and emitter currents always go in the direction of the arrow in active mode. The base current is a little more tricky to figure out, but is also fairly obvious when using the arrow as a reference. As you can see in the above npn circuit, where the arrow is ‘pointing’ away from the base, the base current flows towards the BJT, in the direction the arrow is pointing. Oppositely in the pnp circuit, the base current flows away from the BJT, in the direction the arrow is pointing. There is a table of basic equations listed in my post titled “BJT Transistor Nodal Analysis” which would allow us to calculate each current using a different current, but using Kirchhoff’s Current Law, knowing two currents, we could calculate the third. For a npn BJT, I_E - I_B - I_C = 0 and for a pnp BJT I_C + I_B -I_E = 0 . Note that both of these equations evaluate to I_E = I_C + I_B .

]]> http://engineersphere.com/basic-electrical-concepts/bjt-circuit-and-symbol-conventions.html/feed 0 BJT Transistor Nodal Analysis http://engineersphere.com/basic-electrical-concepts/bjt-transistor-nodal-analysis.html http://engineersphere.com/basic-electrical-concepts/bjt-transistor-nodal-analysis.html#comments Fri, 16 Apr 2010 20:53:33 +0000 Luke http://engineersphere.com/?p=1361

Basic BJT Equations:

i_C = \alpha i_E \hspace{20 mm} i_B = (1- \alpha ) i_E = \frac{i_E}{\beta + 1}

i_C = \beta i_B \hspace{20 mm} i_E = ( \beta + 1) i_B

\beta = \frac{ \alpha }{ 1 - \alpha } \hspace{21mm} \alpha = \frac{ \beta }{ \beta + 1 }

It is also important to know that |V_{EB}| = |V_{BE}| can be modeled as .7V .

These equations are not very informative by themselves so a few examples are demonstrated below. In both examples we will assume \beta is very large. What this means for our calculations is i_B \approx 0 . Since i_B \approx 0 we also assume that i_C \approx i_E .

Finding missing voltages in a BJT circuit

Example 1. Solve for V3:

bjt-voltagesThere are several ways to find V_3 . The more “difficult” way is to first find the emitter current, i_E , and then use Ohm’s Law. Since we know i_C \approx i_E , we can find the collector current, i_C , and then solve for V_3 .

\frac{-4-(-10)}{2.4k } = i_C = 2.5mA

12 - (i_E)(5.6k) = 12 - (i_C)(5.6k) = V_3

V_3 = 12 - (2.5mA)(5.6k) = -2 V

The easier way to find V_3 is to recall that |V_{EB}| behaves like a diode. For this pnp BJT: V_{EB} = V_E - V_B = .7 .

We know that V_B = -2.7V so V_E = V_3 = -2 V

\beta may not always be a very large number. Had that been the case here, we would have started by finding the collector current (since it’s voltage drop and resistance are given) and since i_B \neq 0 anymore, we would use the formulas above to the find the base and collector current.

Finding BJT Bias Voltages and Currents

Example 2 Solve for V2 and I1:

bjt-bias-current

Here we will want to start by finding I_1 . I_1 also equals I_E which approximately equals I_C and this collector current will allow us to find V_2 .

I_1 = \frac{10.7-.7}{10k} = 1mA

V_2 = (1mA)(10k) - 10.7 = -.7 V

Notice that V_E was given as .7 V .  If this had not been given, we would have been able to find it because V_{EB} = V_E - V_B = .7 V and V_B = 0 V .

Similar to the previous example, if \beta was not a very large number. We would first find the emitter current and then use the equations in the table to find the other branch currents.

Note that both of these examples used pnp BJTs. The difference in an npn BJT is the base-emitter voltage is reveresed. You would use V_{BE} = V_B - V_E = .7 V.

General Rule of Thumb

Most of these problems are very simple to solve. Typically \beta is given and you will need to use Ohm’s Law to identify one of the currents. After one of the currents is found you will be able to solve for the other currents using the basic equations listed above. If one of the currents is not immediately obvious, the base-emitter voltage is likely needed. Most problems have you deduce the emitter voltage from the base, but it is easily possible to find the base voltage from the emitter voltage and then use that to find the base current.

]]> http://engineersphere.com/basic-electrical-concepts/bjt-transistor-nodal-analysis.html/feed 0 Time Shifting and Scaling of Functions http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html#comments Wed, 07 Apr 2010 21:19:17 +0000 Luke http://engineersphere.com/?p=1317

We’ll begin with a square function, f(t), that has a an amplitude of 1, a start time of 2 seconds and an end time of 4 seconds.

square-wave

Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph.

positive-shifted-square-wave

In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal.

time-shifted-square-wave
time-shifted-square-wave2

Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2.

taller-square-wave
truncated-square-wave

Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative.

mirrored-square-wave


Example:
Here we will attempt to convert f(t) into 2*f(.5t+3). The graph of f(t) is shown below.

triangle-wave

The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2*f(t). Only the peaks are changed here (by a factor of 2).

triangle-wave2

Next, I convert 2*f(t) into 2*f( \frac{1}{2} t). Notice how the \frac{1}{2} actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration).

shifted-triangle-wave

Finally, we move from 2*f( \frac{1}{2} t) to 2*f( \frac{1}{2} t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course).

shifted-triangle-wave2

]]> http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html/feed 2 Calculating Electron and Hole Concentrations in a p-n Junction http://engineersphere.com/basic-electrical-concepts/acceptorsdonors-and-holeselectrons.html http://engineersphere.com/basic-electrical-concepts/acceptorsdonors-and-holeselectrons.html#comments Wed, 24 Mar 2010 21:47:30 +0000 Luke http://engineersphere.com/?p=1249

Calculating hole and electron concentrations

Sometimes it can be complicated understanding and calculating hole and electron concentrations. My intent in this article is to briefly, but thoroughly describe what the variables used in these calculations mean and how to use them.

To begin I will introduce our variables

n = concentration of free electrons (donors)
p = concentration of holes (acceptors)
n_i = number of free electrons and holes in a unit volume

In thermal equilibrium(or no doping)

n=p=n_i and, therefore n \cdot p=n_i^2

However, doping is common in most examples. To increase the concentration of free electrons, an element with 5 valence electrons is used (i.e. Phosphorous). The resultant material is said to be n-type. To increase the number of holes, an element with 3 valence electrons is used (i.e. Boron). The resultant material is said to be p-type.

This introduces subscript n’s and p’s along with our concentration of free electron and hole variables.

n-type silicon:

n_n = concentration of free electrons (in n-type silicon)
p_n = concentration of holes (in n-type silicon)

p-type silicon:

n_p = concentration of free electrons (in p-type silicon)
p_p = concentration of holes (in p-type silicon)

Note: The subscript indicates whether the material is n-type or p-type.

Calculations

Typically you first want to identify whether the material you are working with is p-type or n-type. This introduces two new variables. N_D which refers to the concentration of donor atoms and N_A which refers to the concentration of acceptor atoms.

n-type silicon:

Here you will use the variables n_n, p_n, n_i^2, and N_D.

n_n \approx N_D \quad \quad n_n \cdot p_n = n_i^2 \quad \quad p_n = \frac{n_{i}^{2}}{N_{D}}

p-type silicon:

Here you will use the variables n_p, p_p, n_i^2, and N_A.

p_p \approx N_A \quad \quad p_p \cdot n_p = n_i^2 \quad \quad n_p = \frac{n_{i}^{2}}{N_{A}}

In most cases n_i^2 and N_D or N_A will be given and you will be able to find n_n or p_p. Then you will find p_n or n_p from the equations above.

]]> http://engineersphere.com/basic-electrical-concepts/acceptorsdonors-and-holeselectrons.html/feed 0 Amplifiers – Part II http://engineersphere.com/basic-electrical-concepts/amplifiers-part-ii.html http://engineersphere.com/basic-electrical-concepts/amplifiers-part-ii.html#comments Mon, 22 Mar 2010 17:59:54 +0000 Jeff http://engineersphere.com/?p=1188

Important Amplifier Properties

Isolation

  • For the biological subject, against electric shock from the amplifier’s power source(s)–or, how to not kill your patient while taking measurements!
  • For the signal-to-noise ratio (SNR): isolation keeps the (60-Hz mains and other) noise out of the sensor pickup–and out of the amplifier input.

One simple way to isolate the input is to use an LED/phototransistor pair:

led-pairThe sensor is connected to an LED, which outputs light of intensity proportional to signal voltage.

The light from the LED falls on a photodiode or phototransistor, which is biased in such a way that current only flows when light hits the device, and the current (and thus measured voltage) is proportional to light intensity.

Phototransistor output is to the amplifier.

The power circuitry for the amplifier is completely isolated from the sensor (electrodes etc.) Power supplies for the sensor use a transformer or battery, and the “ground” for the sensor side floats relative to the amplifier side (which should have ground connected to earth eventually.)

These devices are manufactured in one piece, so the frequency of light output from the LED is matched to the ideal absorption frequency of the phototransistor. However, there must be no electrical connection between the two sides of the circuit. More exactly, the impedance between the two circuits should be near a Teraohm (10^{12} \Omega ).

Single-Ended vs. Differential Input

  • Referenced Single-Ended: two leads of sensor are + signal and earth (computer) ground. Note: to avoid ground loops, it is best to avoid connecting sensor earth to computer earth.
  • Non-Referenced Single-Ended: two leads of sensor are + signal and – signal, which is connected through a bias resistor to earth ground.
  • Differential: three sensor leads are available: + and – signal and (earth) ground.

The 4-BNC -to- DAQ-card lead you’ve probably used in a lab before is referenced single-ended: each of 4 sensors has available a signal wire, and the 4 sensors share a common ground wire. If the DAQ card is configured in Differential mode using this input lead, the signal gradually rises to a saturated level, as the DAQ card assumes a signal on the – differential input, which is in reality floating.

Problems with Single-Ended Input:

  • Cross-talk: Because multiple signals share a ground wire, they also share some portion of the signal, so if only one input is connected that signal shows up, somewhat weakened, on the second input trace.
  • Lack of isolation: using the ground wire as the negative signal lead means the subject and the amplifier are strongly coupled–and strong electric signals from the amplifier may travel to the subject. Also, stray noise (such as from the power mains) easily couples into the input.

Differential Input:

I’m talking a lot about DAQ cards, these are very important when working with A/D systems and analog and digital filters.  If you are not familiar with these, go do a little light reading.  Hopefully this is making enough sense to get the concepts across.  The DAQ card comes with a finite number (16, in our case) of analog input ports (channels), each with paired input pins: a signal (ACH#) and a ground (AIGND). All the analog input ground pins are tied together on the DAQ card. In differential mode, the analog input ports are paired, so channel 1 uses AICH0 as +, AICH8 as -, and the ground wires for the pair are tied together as the reference ground. Thus a 16-channel DAQ card has only 8 channels in differential mode.

  • Because no channels share leads, cross-talk is reduced. The DAQ card has high common-mode rejection, so if the – input leads are tied to ground (simplest configuration), channels will not interfere because the cross-talk is common to both + and – inputs.
  • The ground lead cannot be isolated from the input electrodes without causing the DAQ circuitry to saturate, so the DAQ card itself does not provide isolation.

Wiring a DAQ card in Differential Mode:

daq-card-internal-wiringThe figure is adapted from Fig. 4.6, p. 4-15, of the NI 6024E (DAQ card) User Manual, and illustrates the appropriate connections for differential input to the card. Recommended values of R+ and R- depend on the impedance Rs (in series with Vs) and coupling of the source signal Vs:

source-voltages-table

R+ and R- provide bias current return paths. Bias currents result from not-quite infinite input impedance to the DAQ, and if not balanced the noise they represent will not be common to both + and – inputs, and thus won’t be rejected. However, R+ and R- load down the source with an equivalent 2R+, which will decrease gain if R+, R- are too low. If R+, R- are too large, they will produce a DC offset at the DAQ input.

Bandwidth

Bandwidth is a critical parameter in determining the type of amplifier needed. Often a bioamp has adjustable bandwidth; typical applications have both low-pass and high-pass adjustable filters.

Typical Bandwidths of Biological Signals: (Table from Webster Fig 6.16, p. 259.)

bio-signal-bandwidth-table

(Electro-oculogram: signal from electrodes placed either side or above and below the eye. Linearly proportional to angle of gaze; DC signal)

Noise Reduction

Can be either or both analog and/or digital. If digital, it is done by post-processing collected data. If analog, wiring of the printed-circuit board, handling of wires on the bench, and internal circuitry are all details to consider carefully.

Protective Shielding

For equipment, against transient large signal sources–includes grounding the outer case of the equipment and surge protection, as well as isolation.

Power Supplies

AC (mains) with rectification and usually transformer isolation. Subject to power failures, risk of electrocution.
Battery: provides its own isolation. Limited lifetime and undesirable behavior just before failure.

Specialized Amplifiers

lock-in-amplifier-schematicLock-In Amplifier Block Diagram

Lock-in Amplifier

(www.lockin.de)If the signal source is mostly at a single frequency but is very weak and/or subject to a great deal of noise, a lock-in amplifier can be used to extract the signal. In biological cases the single-frequency property of the signal is most-often externally generated, by applying a single-frequency excitation in one way or another.

Theory:

Signal source S(t) = A cos(ω1t + θ1) + B cos(ω2t + θ2)

Reference signal R(t) = C cos(ω1t)

Product S(t)R(t) = AC cos(ω1t) cos(ω1t + θ1) + other terms in cos(ω1t) and cos(ω2t)

= AC/2 (cos(2ω1t + θ1) + cos(-θ1) ) + other terms in cos(ω1t) and cos(ω2t)

Integrating over an even number of cycles of the reference signal reduces all terms in ω1 to zero, so the displayed signal is proportional only to the amplitude of the component of the signal source at the reference frequency.

To get the reference frequency into the signal source various means are employed: one may have to excite the subject (nerve, membrane…) at the reference frequency, or one may already know the source has a dominant resonance; alternatively one may look at each of several spectral components in the source signal piecemeal, by tuning the reference frequency.

Pre-Amplifier

high input impedance, moderate gain, high CMRR. Often differential input and isolation. May include DC offset control, gain control switches, and/or calibration signal.

Chopper-stabilized Amplifier

removes thermal DC drift from a (very-low-frequency or DC) signal by using negative feedback and chopping the low-frequency signal at a frequency above the amplifier’s high-pass limit (effectively, this is frequency modulation). The signal can be reconstructed by demodulation after amplification; noise signals at both high frequencies and those below the chopper frequency are rejected.

This post was made using some old class notes, a DAQ card user manual and just some good old knowledge.  I know this is not a traditional engineersphere.com lesson, it is more of an ‘informative read’ for the avid electrical engineer interested in biomedical applications.  Enjoy :)

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