# BJT Differential Pair Amplifier Design

#### jegues

1. The problem statement, all variables and given/known data

See first figure attached.

2. Relevant equations

3. The attempt at a solution

See 2nd figure attached for my work so far at attempting 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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#### uart

Science Advisor
The resistor bias (as opposed to current source) wont change things too much.

- Start with a bias point. Does it have to be DC coupled? If so your bias point is approx zero volts at the bases. I'd start with that as a first attempt in any case.

- So now you've got about -0.6 volts at the emitters, work out the value of RS to get 2mA bias current (1mA for each BJT).

- Now choose the Rc's to establish a bias point to enable +/- 3.5 volts collector swing. I'd be looking at about 4 to 4.3 volts Vc for this.

- Now choose the emitter ballast Re for the required gain and you've got a good starting point to analyze and perhaps tweak.

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#### jegues

The resistor bias (as opposed to current source) wont change things too much.

- Start with a bias point. Does it have to be DC coupled? If so your bias point is approx zero volts at the bases. I'd start with that as a first attempt in any case.

- So now you've got about -0.6 volts at the emitters, work out the value of RS to get 2mA bias current (1mA for each BJT).

- Now choose the Rc's to establish a bias point to enable +/- 3.5 volts collector swing. I'd be looking at about 4 to 4.3 volts Vc for this.

- Now choose the emitter ballast Re for the required gain and you've got a good starting point to analyze and perhaps tweak.

I'm not really sure whether is has to be DC coupled or not, I'm a little confused about what that means. (The question I posted is all the information we are told)

Did I indicate the -0.6V in the correct place? How did you obtain this value of -0.6V?

I tried solving for Rs but I wasn't sure if I should have negative Vcc or ground at the bottom so I did the analysis for both cases. Which is correct?

How did you determine a Vc of 4-4.3V for a +/- 3.5 V swing at the collector? Anyways, I used the range of Vc you provided to determine Rc. I did two values, 1 for 4V and one for 4.3V.

I determined $$r_{e}$$ using $$V_{T} = 25mV$$. I also did a few different calculations for Re, using Rc = 5k and Ad = 20. Then again with the averages of the values, Rc = 4.85k, Ad = 22.5.

What am I doing wrong? Are the values I'm getting reasonable?

Thanks again!

#### uart

Science Advisor
How did you obtain this value of -0.6V?
The Vbe of the bjt will be about 600 to 650 mV. This is approx, but it's precise value doesn't make much difference to your Rs calculation.

I tried solving for Rs but I wasn't sure if I should have negative Vcc or ground at the bottom so I did the analysis for both cases. Which is correct?
The question specifies a dual supply voltage of +/9 Volts, so you should assume that the negative rail is -9V.

Your collector resistance calculation looks good. What value of Re did you determined to get the correct gain?

#### jegues

The Vbe of the bjt will be about 600 to 650 mV. This is approx, but it's precise value doesn't make much difference to your Rs calculation.

The question specifies a dual supply voltage of +/9 Volts, so you should assume that the negative rail is -9V.

Your collector resistance calculation looks good. What value of Re did you determined to get the correct gain?
Whoops. I had scanned an image of my work, but it seems as though I forgot to attach it.

I'll edit this post and attach it when I get home so you can see the rest of my workings.

#### jegues

Here I started my design attempt from scratch again.

I'm going to simulate the circuit right now and see if it agrees with the values I calculated in 2).

How's it look?

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#### jegues

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.

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?

Edit: See the 3rd figure attached for another, better, attempt at my simulation.

What do I need to do in the simulation in order to confirm the gains found in part 2 of my design?

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#### uart

Science Advisor
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)
Yes the input is differential but it cant be floating, you need a path for bias currents to flow in the base. Place a resistor (say 10k) from base to ground for each of Q1 and Q2. Apart from that the circuit looks good.

Some tips on the schematic.

- You don't need separate voltages sources V1 and V2. Nodes 2 and 3 should be the same Vcc node.

- It would be easier to read if V3 had a value of +9 instead of -9, but where simply reversed 180 degrees (as in placed upside down).

- In the simulation, if you probe the voltage between nodes 1 and 4 (the two collectors) you'll see your true differential output.

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