Differential amplifier design

Thread Starter


Joined Oct 10, 2008
1. The problem statement


2. Relevant equations

According to my book, if R3 = R2 and R4 = R1, I can use the following formula:

Vo = (R2/R1)(V2 - V1).

3. The attempt at a solution

We want R2/R1 to equal 5. This means that R3 and R2 must be 5*3,3 = 16.5 k(ohms).

However, I don't know what to do with the inner resistance of V2 and Vo. Any suggestions?

Thread Starter


Joined Oct 10, 2008

thus R1 must equal 3.3K as for the formula given to you to work



thus R2=5*3.3=15.15K

R3 = 5*3.3 = 16.5 k(ohms)

But how about the inner resistance of V2, and the 22k(ohms) resistance? Don't I need to take them into consideration?


Joined Feb 4, 2008
Forgot all we said. Your calculations are right but on your final answer (5*(V1-V2)) you have o subtract the voltage drop across the 1.2M resistor due to the current through R1+R2.

If you want i can tell you in a few hours the complicated answer because now i am busy.


Joined Feb 4, 2008
Ok before i was busy and my mind was on my job so i wrote some stupid things and i deleted my posts.

According to your circuit diagram, the output voltage will equal

Vo=(V2-V1)*(R3/R4) (1)

if R4=R3 and if R2=R1 and not as you said.

Thus 5=R3/R4=R3/3.3K thus R3=16.5K

But now R3 is not equal to R4 anymore so we cant use equation 1 to calculate the output voltage.

We will use the general equation for the output voltage of the differential amplifier which is :

Vo=V2((1+R3/R4))/(1+R1/R2))-V2(R3/R4) (2), according to your schematic

Now, to take the desired equation for the output voltage we need to equate ((1+R3/R4))/(1+R1/R2))=5 because (R3/R4) equals 5 (we calculated it before) and to be able to take 5 as a common factor.

Thus, R4=3.3K , R3=16.5K
lets say that R1=0 but because V2 has an inner resistance of 1200K then we can say that R1=1200K

Thus, we only need to calculate R2


thus R2=5*1200=6000K=6M

If we apply our resistor values to equation (2) we get:

Vo=5*V2-5*V1=5*(V2-V1) as desired