Engineersphere.com » Electronics 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 Smith Charts Explained http://engineersphere.com/linear-systems/smith-charts-explained.html http://engineersphere.com/linear-systems/smith-charts-explained.html#comments Mon, 28 Feb 2011 21:35:56 +0000 Safa http://engineersphere.com/?p=2025

Why We Need Smith Charts

In the world of RF (Radio Frequency) electronics, normal “bench-top” circuit components cease to operate the way they were designed to.  This means a normal resistor can become a capacitor, a capacitor can become an inductor, and a normal wire can become a distributed network of inductors and capacitors.  This highly non-ideal behavior occurs because, in reality, no true resistor, capacitor, inductor, or wire exists; rather they are all processed and manufactured to operate within a certain frequency range – at frequencies where the real-world effects are quantitatively insignificant.  However, as one approaches RF frequencies, these real-world effects become much more pronounced in cheap components.   Eventually, the frequency of operation can become so high that the transmission line itself – no longer a simple wire – will exhibit significant signal-loss.  But even with lossless transmission lines, it is important in communications to “match impedances,” i.e. attach an antenna whose impedance matches that of the signal source – this maximizes the transmitting-antenna’s power dissipation (and “reflects” back zero power).  Indeed, being able to calculate and measure the impedances of antennas, transmission lines, etc is very important within RF design, which are almost always complex numbers.  Another reason determining load impedances is important is because of the 1:1 mapping between a value of load impedance and a corresponding value of \Gamma , the reflection coefficient (a ratio of how much a signal is reflected versus how much a signal is radiated for a given load).

How To Read a Smith Chart

One way of simplifying the analytical problems communication engineers typically face is by using a Smith Chart.  Smith Charts provide a graphical representation of the impedance of any load – whether that load be an antenna or simply an open-circuited transmission line, such as a coax cable.  Because these impedances may very well be complex in nature, a Smith Chart is designed such that each point on it represents both the real and imaginary parts of the load’s impedance.   To begin, observe the basic “format” of any Smith Chart:

 

smith chart

Generic Smith Chart

What is seen here is the generic, normalized Smith Chart.  Each point on the chart represents an impedance, and the numbers marked on the chart represent different coefficients needed to multiply by the original load value  (from which the chart was normalized from – If this is confusing, don’t worry about it – the “original load value” will almost always be known).  The chart consists of yellow circles and red arcs.  The yellow circles represent contours of where the Real part of the impedance magnitude is the same, e.g. for any point along the yellow circle marked “2.0,” the real part of the impedance is:

where  is the impedance the chart was normalized from.

Similarly, the red arcs represent contours where the Imaginary part of the impedance magnitude is the same, e.g. for any point along the red arc marked “0.5,” the imaginary part of the impedance is:

This is very handy for displaying complex impedance values.  For instance, notice the following Smith Chart:

 

Smith Chart for Z0 = 50 Ohms

What is the impedance for the load represented on the Smith Chart by the blue dot?  This is easy to determine, because:

  1. The blue dot is along the yellow circle marked “2.0,” so the real part of the impedance must be:
  2. The blue dot is also along the red arc marked “0.5,” so the imaginary part of the impedance must be:

So, from this, we have determined that the impedance of the “load” represented by the blue dot is:

By using this method, it is simple to find the impedance represented by any point on the Smith Chart!

Useful Smith Chart Relations

As mentioned before, a Smith Chart is really just a 1:1 mapping between a value of load impedance and a value of \Gamma, the reflection coefficient of a load.  The reflection coefficient is defined as:

\Gamma =\frac{V_{reflected}}{V_{incident}} = \frac{V_{refl}}{V_{inc}}

The reflection coefficient is a very important metric.  For antennas, a reflection coefficient expresses how much signal voltage is used in exciting the antenna and how much signal voltage is reflected back to the source.  For an ideal antenna, \Gamma would be zero – corresponding to Z = Z_0 \Omega , which is the origin of the Smith Chart!  In fact, simple relations exist between Z and \Gamma :

and

Useful Notes

Simple observations allow for a more “intuitive” approach when using Smith Charts.  Note the following:

  • The straight red line in the center is an “arc” representing all points where the imaginary part of the impedance is zero
  • The furthest-left point on the straight red line represents where the impedance is zero (or, a short circuit), and the furthest-right point on the straight red line represents where the impedance is \infty (or, an open circuit)
  • The very top point of the Smith Chart is where the impedance is +j
    • For this reason, the top half of the Smith Chart represents inductive loads
  • The very bottom point of the Smith Chart is where the impedance is -j
    • Similarly, the bottom half of the Smith Chart represents capacitive loads

____

Safa Khamis

Questions?  Comments?  safa@ksu.edu – Or please leave a comment below for the author!

 

 

]]> http://engineersphere.com/linear-systems/smith-charts-explained.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 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 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

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