# Homework Help: Differential Amplifier find currents, and Gain

1. Apr 18, 2014

### DODGEVIPER13

1. The problem statement, all variables and given/known data
For the differential amplifier circuit given above, for all transistors except Q1 and Q2 we have:
β=100, VA=80 V, RC =20 k, VT=0.026 (Thermal voltage), VBE(on)=0.7 V

a-Determine the collector currents IC1, IC2, IC3 and IC6
b-Calculate the Differential mode voltage gain Vo2/Vd.
c-What is the output resistance of the three transistor current source that is biasing the differential pair? Find the numerical value. Does this effect the common mode voltage gain? Explain briefly.
d-calculate the overall Gain

2. Relevant equations

3. The attempt at a solution
So the first thing I did was find I1 = V-Vbe-V(-)/R1 = 1 mA and Ic3=I1=1 mA then Ic1=Ic2=Ic3/2=0.5 mA and I have V02=V-Ic2(Rc) = 15-0.5(2) = 5 Volts then Ic6 = 15-5-.7/6 =1.55 mA. Next I found differntial mode gain rpi1= Vtβ/Ic = 0.026*100/0.5 = 5.2 kΩ and Rib6= rpi6+(1+β)Re = ?? rpi6= 100*0.026/1.55 = 1.67 mA so Rib6=1.67+(101)(4.3) = 435.97 and gain is β(R||Rib6)/2(rpi1+Rb) = 132.8. For part c Ro= Va/Ic3 = 80/1 = 80 kΩ and yes because if Ro changes so does Av when Av=something/(something)*(Ro). Calculate the overall gain well I would first need to find A2 and A3 and then plug into the formula Ad=Ad1*A2*A3 = ?? However I am unsure of how to proceed what should I do now

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2. Apr 18, 2014

### DODGEVIPER13

Ok I uploaded a pdf of my work to make it cleared and I also made an attempt at part d.

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3. Apr 20, 2014

### DODGEVIPER13

Hey is there anything I can do to make the question more clear? ? I know I tend to write poorly worded questions, so just let me know if it is too confusing.

4. Apr 22, 2014

### rude man

DodgeV, I think one reason for no-shows is your attempt description is too run-in. Split the sections up into paragraphs would help a lot. On top of that, there's a lot of detail here that I suppose many of us just don't have the time/energy/patience/inclination/whatever to sift thru all your work.

It would be good if you concentrated on one of the parts you're least sure about. We're here to give you hints if you're stuck, not necessarily proofread all your work (if I'm out of line here I'm sure I'll hear about it soon ....).

Anyway, for part (d) the overall gain is the gain from either input to Q2-c and then from that to Q6-c. I would consider Q2-c unloaded but maybe your teach wouldn't agree.

How are you modeling the Early voltage? By a resistor = Va/Ic between emitter and collector?

Last edited: Apr 22, 2014
5. Apr 22, 2014

### DODGEVIPER13

Ok man yah you are probably right well I found my error on part (a) but I am stuck on the differentail mode gain I had one of my friends assist me Adm = -gmRC/2 = -IC1RC/2VT = -(0.5)(20)/2(.026) = -192.3075 is that ok?

6. Apr 22, 2014

### rude man

Do you define the differential gain as Vo2/(v1-v2)? If so, why the 'divide by 2'? You're already figuring the Q2 current as 0.5 mA.

You're ignoring the 2K base resistors but frankly I would too.

7. Apr 23, 2014

### DODGEVIPER13

Ok so in your opinion which looks better because I got that equation -Ic1RC/2VT from a friend and im defining it as Vo2/Vd. Honestly as far as dividing by 2 I am not sure what does the division by 2 mean? I get lost on the gain stuff I can handle the DC analysis.
This is my equation:
Beta(R||Rib6)/2(rpi1+RB)

8. Apr 23, 2014

### rude man

The differential gain IS a dc computation.
I don't see any Vd on the schematic. I guess Vd = v1 - v2?
You "equation" is the differential gain? I don't think so. What is Rib6? What is R? Try to stick to symbols on the schematic where possible. You were a lot closer before. I don't have time right now but I think I goofed and I believe the divide by 2 is correct. Right now looks like the differential gain is indeed IC1*RC/2VT but i want to double-check.
More later.

9. Apr 24, 2014

### Jony130

Because we have 2K resistors at differential pair input the voltage gain will not be equal to RC/2re = IC1*RC/2VT = 192.
The gain will drop to

$\Large Av = \frac{Rc}{2re + \frac{2*2K}{\beta+1}} = 139V/V$

And if we include Early voltage rce ≈ 80V/500μA = 160kΩ and Q6 input impedance we have

rce = 160kΩ and RinQ6 ≈ (β + 1)*5.3kΩ = 535kΩ. The gain will drop to

$Av =\Large \frac{Rc||rce||RinQ6}{2re + \frac{2*2K}{\beta+1}} = 120V/V$

10. Apr 24, 2014

### The Electrician

Q

Hi, Jony130,

I noticed that in post #1, the OP gives us:

"For the differential amplifier circuit given above, for all transistors except Q1 and Q2 we have: β=100, VA=80 V, RC =20 k, VT=0.026 (Thermal voltage), VBE(on)=0.7 V"

You have done what I would, which is to assume that Q1 and Q2 have the same specs, but I wonder why the problem excepted Q1 and Q2? Does the problem want the student to use some other parameters for them?

11. Apr 24, 2014

### DODGEVIPER13

Hmmm just read that I think it means they have different parameters. would that affect Ic1 and 2 I guess not because I never used parameters to find them. On another note if I where to try to find the gain like jony how would I do that

12. Apr 24, 2014

### DODGEVIPER13

Also im in an area where I cant really give you Rib6 or the other values requested but I will later on

13. Apr 24, 2014

### The Electrician

It won't affect Ic1 and Ic2, but if the Early voltage and β is different for Q1 and Q2, the gain will be affected.

14. Apr 24, 2014

### DODGEVIPER13

ok so how would I go about starting to find gain? I mean I know what I see in the book and what my buddies show me but I really dont get it.

15. Apr 24, 2014

### DODGEVIPER13

Rib6=rpi6+(1+beta)Re rpi6=beta*Vt/Ic6 rpi1=beta*Vt/Ic1 and Rc= 20 k

16. Apr 24, 2014

### The Electrician

It would make it easier for those of us who are helping you if you wouldn't put everything on one line; it's hard to read. Do it like this instead:

Rib6=rpi6+(1+beta)Re
rpi6=beta*Vt/Ic6
rpi1=beta*Vt/Ic1
Rc= 20 k

To get the voltage gain of the differential pair you have take into account that the load on Q1 and Q2 is not just Rc.

The Early effect creates an apparent resistance (sometimes denoted "ro"; Jony130 called it rce) from collector to emitter of a transistor. The ro for Q1 and Q2 is 160000 ohms; you have to include that in the load on Q2, as well as Rpi6.

Look at what Jony130 did in post #9. That should give you some guidance.

17. Apr 24, 2014

### DODGEVIPER13

Ok, I am starting to get it. Where exactly is rpi located? I now know ro is over the collector to emitter. Also would it help for me to draw the small-signal equivalent? Furthermore what is K and re is his formulation I assume re means Re which is 5.3 K

Last edited: Apr 24, 2014
18. Apr 24, 2014

### The Electrician

Where is rpi located? I guess you could say that it's inside the transistor, just like ro and re are. Yes, it might help to draw the small-signal model, because then Rpi, ro and re will be shown explicitly.

K means 1000. Where Jony130 has 2*2K, that means 4000 ohms.

re means the internal emitter resistance, given by .026/Ic for each transistor.

Your schematic shows a resistor RE of 5300 ohms connected from the emitter of Q6 to ground; that's not the same as Re.

19. Apr 25, 2014

### Jony130

Last edited: Apr 25, 2014
20. Apr 26, 2014

### DODGEVIPER13

I get how to do small-signal equivalents for single transistor circuits, but multistage circuits are giving me problems. Are there any example multistages, that break it doen step by step? I have uploaded my first attempt.

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21. Apr 26, 2014

### The Electrician

That's a start, but you have a few problems.

The emitter of Q6 is not connected to the emitters of Q1 and Q2. The emitter of Q6 goes to ground through a resistor, RE.

The emitters of Q1 and Q2 are connected together, but then they go to ground through the collector of Q3. I wouldn't bother showing Q3 as a model; just replace it with an 80k resistor to ground (for small signal purposes that's its equivalent; it's really the ro of Q3).

Q6 has a different emitter current than Q1 and Q2 (I think you calculated something like 1.55 mA), so its r∏ and ro are different. I called them r∏6 and ro6.

Once you get all this right, then you will have to write some nodal equations for the whole thing.

This is actually a fairly complicated problem if you include all the various items such as ro and r∏ for all the transistors, different values for load resistors and RE6, etc.

22. Apr 26, 2014

### DODGEVIPER13

Ok so is my second attempt improved over my first?

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23. Apr 27, 2014

### The Electrician

The top end of ro6 should not be connected to the r∏ just to its left.

The bottom end of ro6 should not go to ground, but to the emitter of Q6.

The end of Rc6 that you have grounded should go to V+ just for consistency, even though V+ is also AC ground.

The bottom end of r∏6 should go to the emitter of Q6, not to ground.

The top ends of all your r∏ resistor should NOT be connected to the top of the gm sources.

The collector of Q2 (the top end of its gm source) should go to the top end or r∏6.

The top end of Rc6 and the top end of r∏6 should not be connected.

Check out the layout and connections of the Q6 model.

24. Apr 27, 2014

### DODGEVIPER13

Ok the rpi appears to be tripping me up so it is the resistance between the base and emitter correct? If that is so then I probably still have mistakes however I feel that this is improved on my 3rd attempt

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25. Apr 27, 2014

### The Electrician

You've still got the emitter of Q6 connected to the emitters of Q1 and Q2. Remove that connection and add an 80k ohm resistor from the emitters of Q1 and Q2 to ground.

You should probably give the two rpi resistors associated with Q1 and Q2 different names. Let rpi1 be the Q1 resistor, and rpi2 the Q2 resistor. Then the gm sources need to be also particularized to the appropriate transistor. You'll need to be sure that the voltage applied to the base of Q1 controls the collector current of Q1, the Q2 base voltage controls Q2 collector current, the Q6 base voltage controls the Q6 collector current, etc.