# Differential mode gain

## Homework Statement

http://i59.tinypic.com/1smqlu.png

I'm not sure how to calculate differential mode gain

## Homework Equations

I know formuals for CMRR and Acm but not Adm

Rb/Ra = (1-ε)Rd/Rc

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gneill
Mentor
Hi bnosam. There's lots of information on differential and common mode gain on the web. A quick search on "differential mode gain" will turn up some excellent material. For example, this site explains the derivations quite clearly.

Essentially the idea is to define a couple of parameters that represent the differential and common mode inputs in terms of the voltage sources $v_a$ and $v_b$ in your diagram. To wit:

$V_d = v_b - v_a$ is defined to be the differential mode input, and
$V_{cm} = \frac{v_b + v_a}{2}$ is the common mode input

If you add those two equations together you can find a "new" expression for $v_a$ in terms of $V_d$ and $V_{cm}$, and if you subtract the first from the second you arrive at a new expression for $v_b$, also in terms of $V_d$ and $V_{cm}$. Plug these in for the sources in your circuit and solve your circuit for the resulting output voltage. The site I gave the link for above has more details.

So using:

Ad = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2]

1/2[1000/(1000+1000)] * [(24000 + 25000)/1000 + 24000/25000] = 12.49

So it's exactly half of what the answer should be according to the answers up there? 24.98? Why does it have the 1/2 in front of it? Without it the answer would be 24.98 which would be correct...

gneill
Mentor
So using:

Ad = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2]

1/2[1000/(1000+1000)] * [(24000 + 25000)/1000 + 24000/25000] = 12.49

So it's exactly half of what the answer should be according to the answers up there? 24.98? Why does it have the 1/2 in front of it? Without it the answer would be 24.98 which would be correct...
Well let's see. You've used 1000 for R3 when the circuit says it's 24000. You've used 24000 for R4 when your circuit says it's 25000, and 25000 for R2 when your circuit gives it as 1000.

I should mention that just copying a formula and plugging in values is a dangerous way to go. You won't increase your understanding and you'll be left hanging on an exam. You should be able to analyze the circuit and derive the gain from scratch. I don't think you would have made the above mismatch if you laid the groundwork yourself.

I was looking at the wrong diagram online and used this formula for it, sorry.

1/2(24000/(1000+24000))* ((25000 + 1000)/1000 + 25000/1000) = 24.48

I must have messed up here or something. I remember my prof deriving this for us but he told us we don't need to know how to derive it?

gneill
Mentor
I remember my prof deriving this for us but he told us we don't need to know how to derive it?
Lucky you! I never had it so easy :)

My answer in post #5 seems to be off by .5. I can't spot anything that's wrong

LvW
My answer in post #5 seems to be off by .5. I can't spot anything that's wrong

EDIT: Sorry, I have to correct myself (calculatiuon error): The result is, in fact: Ad=24.98

Last edited:

EDIT: Sorry, I have to correct myself (calculatiuon error): The result is, in fact: Ad=24.98
Where did I mess up? I keep putting it in my calculator and getting it wrong haha

gneill
Mentor
Where did I mess up? I keep putting it in my calculator and getting it wrong haha
You and your calculator must have a difference of opinion on the order operations implied by a given key sequence.

Try breaking down the calculation into smaller units and use memories to store these intermediate values. Then combine the unit results.

Even when I do it bit by bit I get the wrong answer heh

1/2(24000/(1000+24000))* ((25000 + 1000)/1000 + 25000/1000)

24000/(1000+24000) = 24000/25000 = .96

(25000+1000)/1000 = 26000/1000 = 26

25000/1000 = 25

1/2 * .96 * (25+26) = 24.48

Drives me insane a little.

gneill
Mentor
Hmm. I think perhaps the given formula may have suffered from careless algebra. I derived the expression myself and used that, obtaining the correct result. The expression I arrived at was:

$$A_{dm} = \frac{1}{2} \left[ \left( \frac{R3}{R1 + R3} \right) \left( \frac{R2 + R4}{R2} \right) + \frac{R4}{R2} \right]$$

Hmm. I think perhaps the given formula may have suffered from careless algebra. I derived the expression myself and used that, obtaining the correct result. The expression I arrived at was:

$$A_{dm} = \frac{1}{2} \left[ \left( \frac{R3}{R1 + R3} \right) \left( \frac{R2 + R4}{R2} \right) + \frac{R4}{R2} \right]$$
Yeah that was the issue, thank you :)