BJT Differential Pair Design Question

Thread Starter

jegues

Joined Sep 13, 2010
733
See the first figure attached for the design we are attempting to achieve.

The second figure attached is my work so far on the design.

The last figure is what my textbook gives as the small signal analysis of a BJT differential amplifier with emitter resistances. However, this may be a little different than our design because we replace the current source with a single resistor.

I'm little overwhelmed and I'm not really sure where to start with this design.

I know that for a differential amplifier with resistances in the emitter leads the differential gain is given by,

\(A_{d} = \frac{\alpha(2R_{c})}{2r_{e} + 2R_{e}} \approx \frac{R_{c}}{r_{e} + R_{e}}\)

Are there any other parameters in my design that I can currently establish, so that the number of things I have to solve for is diminished? What should I work with first? What other equations do I need to concern myself with?

Thanks again!
 

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Jony130

Joined Feb 17, 2009
5,176
You could start as usual set the collector voltage equal to Vc = 0.5Vcc.
Rc = 4.5V/1mA = 4.5K
Re = Rc/Ad = 4.5K/22 = 200Ω
Rs = (Vee - Vbe - VRe) / 2Ic = (9V - 0.65V - 0.2V)/2mA = 4KΩ


So now you know all resistance in the circuit.
Now you need to check whether using these values ​​satisfy our requirements.
 

Thread Starter

jegues

Joined Sep 13, 2010
733
We're now asked to simulate the circuit we've designed.

See figure attached for the question, as well as the circuit I built in multisim.

Is this correct?

I'm not entirely what peak to peak voltage I should use at the input, the problem only specifies the frequency.

I'm not the best at multisim so I'm going to see if I can figure out how to properly simulate this circuit.

Edit: When I do a transient simulation my Vc isn't sitting at 4.5V it sits at Vcc = 9V.
Why's that?

Edit: I tried connecting the base of Q2 to ground and the negative lead of the voltage source to ground and here is the graph from the transient analysis of the voltage from the source, and the voltage at the collector of Q1. (The red curve is input, and blue curve is output)

Is my input voltage of 2Vpp too large?
 

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Last edited:

t_n_k

Joined Mar 6, 2009
5,455
You do need to tie the common point of the source and Q2 base to ground.

Yes 2V p-p is too large. Try 0.2Vp-p
 

Ron H

Joined Apr 14, 2005
7,014
Your input voltage needs a path to ground to satisfy base current requirements. I would split the input into two voltages, of equal amplitude but opposite polarity.
The input voltage has to be low enough to avoid saturation or cutoff on the transistors. Do you think 2V pk multiplied by a gain of 22 will satisfy this requirement?

EDIT: t_n_k made essentially the same comments, with considerably less verbosity.:rolleyes:
 

Thread Starter

jegues

Joined Sep 13, 2010
733
Alrighty here's my 2nd attempt at the simulation.

How do things look?

What should I look at in my analyses in order to confirm the characteristics I calculated in step 2?

I can see from the graph in this post I will be able to maintain a swing of 7Vpp at the output.

The single ended, differential and common mode gains really depend on my resistance values, so how would I confirm these in a simulation?
 

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Ron H

Joined Apr 14, 2005
7,014
Alrighty here's my 2nd attempt at the simulation.

How do things look?

What should I look at in my analyses in order to confirm the characteristics I calculated in step 2?

I can see from the graph in this post I will be able to maintain a swing of 7Vpp at the output.

The single ended, differential and common mode gains really depend on my resistance values, so how would I confirm these in a simulation?
Applying all the input voltage to one input while grounding the other will give you exactly the same differential gain, but you will get a little more signal on the Q1 collector, and a little less on the Q2 collector. This is due to the resistor that acts as the tail current source.
The common mode gain is measured by connecting the bases together, then applying a signal between that junction and ground. You should get essentially zero common mode gain when you look at the outputs differentially, but each collector individually will have a lot of signal. This could screw up any following stages, depending on how much they depend on the common mode level at the output of the first stage. This is why a current source is generally used for the tail current in the first stage.

EDIT: I guess I didn't answer your first question.:(
 

Thread Starter

jegues

Joined Sep 13, 2010
733
Applying all the input voltage to one input while grounding the other will give you exactly the same differential gain, but you will get a little more signal on the Q1 collector, and a little less on the Q2 collector. This is due to the resistor that acts as the tail current source.
The common mode gain is measured by connecting the bases together, then applying a signal between that junction and ground. You should get essentially zero common mode gain when you look at the outputs differentially, but each collector individually will have a lot of signal. This could screw up any following stages, depending on how much they depend on the common mode level at the output of the first stage. This is why a current source is generally used for the tail current in the first stage.

EDIT: I guess I didn't answer your first question.:(
The information you provided for measuring the differential gain and the common mode gain was useful

Why did you write this,

I guess I didn't answer your first question.
That comment has me confused? Am I still doing something wrong?

Attached is the graph I get when I preform transient analysis on an expression that should be the differential voltage gain.

I.e. [V(4) - V(1)]/V(10)
 

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Ron H

Joined Apr 14, 2005
7,014
The information you provided for measuring the differential gain and the common mode gain was useful

Why did you write this,
I guess I didn't answer your first question.
That comment has me confused? Am I still doing something wrong?
I wrote that because I didn't understand this question:
What should I look at in my analyses in order to confirm the characteristics I calculated in step 2?
Attached is the graph I get when I preform transient analysis on an expression that should be the differential voltage gain.

I.e. [V(4) - V(1)]/V(10)
You get a graph whose scale is astronomical (notice the G's on the vertical axis) because v(10) passes through zero. Numerator divided by denominator then equals infinity at these points.
To display gain directly, you need to use AC analysis. Transient analysis is necessary to see voltage swings, Dc levels, distortion, etc., but with transient analysis you can only calculate gain by measuring Vout(p-p) and dividing by Vin(p-p). AC analysis is superior for gain and bandwidth measurements, and you don't even have to do waveform arithmetic. It is done for you. Just set V(in) to 1 and the output will be the gain (in dB or linear - your choice). You don't have to worry about device nonlinearities (saturation or cutoff), because spice calculates the bias point and then linearizes all components at that quiescent point. You can set the input at 1 microvolt, 1 volt, or 1 megavolt, and the resulting gain will be the same.
 

Jony130

Joined Feb 17, 2009
5,176
I build this circuit in the LTspice



And I use typical standard E24 values for resistors.

The voltage gain is equal to 20.7V/V

Vin = 0.4Vp and Vout = 8V

 

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Ron H

Joined Apr 14, 2005
7,014
I build this circuit in the LTspice



And I use typical standard E24 values for resistors.

The voltage gain is equal to 20.7V/V

Vin = 0.4Vp and Vout = 8V

The Bode plot from the AC analysis is, as I mentioned, the easy way to measure gain. I'm pretty sure that's how you got the gain value, Jony130.
 

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