# Differential amplifier design

Thread Starter

#### boks

Joined Oct 10, 2008
218
1. The problem statement

http://i36.tinypic.com/hugkyb.jpg

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

#### boks

Joined Oct 10, 2008
218
R4=3.3K

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

thus

5=R2/R1=R2/3.3

thus R2=5*3.3=15.15K

R3=R2=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?

Thread Starter

#### boks

Joined Oct 10, 2008
218
It's 1.2 Mega ohms, not milli ohms.

Thread Starter

#### boks

Joined Oct 10, 2008
218
Thread Starter

#### boks

Joined Oct 10, 2008
218
Then I get negative resistance.

#### mik3

Joined Feb 4, 2008
4,843
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.

#### mik3

Joined Feb 4, 2008
4,843
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

5=((1+R3/R4))/(1+R1/R2))=((1+16.5/3.3))/(1+1200/R2))=6/(1+1200/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

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