# Design an op-amp circuit.

Thread Starter

#### TL314

Joined Mar 29, 2015
64
Design an op-amp circuit to produce this transfer function, Vout=3*(Vb1+Vb2). Vb1 and Vb2 come from two different batteries.

If the transfer function is = to Vout/Vin then how would I get a Vout=3*(Vb1+Vb2)?
I dont really understand op-amp circuits or how to get a correct transfer function.

Thread Starter

#### TL314

Joined Mar 29, 2015
64
will Vin just be V1+V2?

edit: oh I see -Vout=Rf/Rin(V1+V2)

so my Rf/Rin is = to 3 which makes Vout=3(V1+V2) right?

#### shteii01

Joined Feb 19, 2010
4,644
Notice the minus sign. The summing amp in that tutorial uses inverting configuration. Which means that positive goies in, negative comes out. You don't want this.

You want positive goies in, positive comes out. The simple way to do your equation is to use two inverting amps. First amp is the summing amp: Vout1=(-Rf1/Rin1)(V1+V2), where the gain=Rf1/Rin1=1. Second amp provides the gain of 3: Vout2=(-Rf2/Rin2)(-Vout1), where gain=Rf2/Rin2=3, this way Vout2=-3(-1)(V1+V2)=3(V1+V2)

Thread Starter

#### TL314

Joined Mar 29, 2015
64
Notice the minus sign. The summing amp in that tutorial uses inverting configuration. Which means that positive goies in, negative comes out. You don't want this.

You want positive goies in, positive comes out. The simple way to do your equation is to use two inverting amps. First amp is the summing amp: Vout1=(-Rf1/Rin1)(V1+V2), where the gain=Rf1/Rin1=1. Second amp provides the gain of 3: Vout2=(-Rf2/Rin2)(-Vout1), where gain=Rf2/Rin2=3, this way Vout2=-3(-1)(V1+V2)=3(V1+V2)
So will my second op amp just be a regular not a summing if i connected them together on a schematic?

#### shteii01

Joined Feb 19, 2010
4,644
So will my second op amp just be a regular not a summing if i connected them together on a schematic?
That is correct. Second op amp act as simple amplifier. It does two things:
1) provide gain of 3
2) invert the signal coming out of first op amp
Work through the example. Op amp 1 (summing amp) has V1 is 2 volt, V2 is 3 volt, Rin1 is 1 kOhm, Rf1 is 1 kOhm, the output will be -5 volts. Why? Because 2+3=5, Gain is 1, 5*1=5, and the Op amp 1 is configured as inverting so 5*(-1)=-5. Now we feed -5 to Op amp 2, Rin2 is 1 kOhm, Rf2 is 3 kOhm (three 1 kOhm resistors in series), the output of Op amp 2 will be 15 volts. Why? Because Gain is 3, -5*3=-15, it is configured as inverting amp, -15*(-1)=15. You have: Vout=3(V1+V2)=3(2+3)=15. The math matches the output of the circuit.
All this assumes that both op amp are used in inverting configuration.

You don't have to do it this way. You can use one op amp in none inverting configuration, in this case positive goes in, positive comes out. Here is article about it: http://masteringelectronicsdesign.c...f-the-summing-amplifier-with-n-input-signals/
Figure 1 has what you need, you just need to figure out Rf1, Rf2, R1, R2 so that you have gain of 3.

Last edited:
Thread Starter

#### TL314

Joined Mar 29, 2015
64
if R1 and R2 = 1 and Rf2=2 and Rf1 = 1 that would give a gain of 3(V1+V2)? or can I not have the same value resistors as R1 and R2?

Thread Starter

#### TL314

Joined Mar 29, 2015
64
oh wait no i dont think that correct

what if R1 and R2 = 1 but Rf2=5 Rf1=1?

#### shteii01

Joined Feb 19, 2010
4,644
if you are talking none inverting summing amp, then you should pick r1 and r2, say 1 kOhm, then you have
Vout=(1+rf2/rf1)(1*v1/2+1*v2/2)
factor out 1/2
Vout=(1+rf2/rf1)(0.5)(v1+v2)
so... we want (1+rf2/rf1)(0.5)=3
all you have to do now is pick either rf2 or rf1. if you pick rf2, then solve for rf1.

The above is demonstration of design process. You, the user/designer, have to make choices. In the abov e example we had to make choice twice. First time we chose what values to use for r1 and r2. Second time we chose what value to use for rf2. If you don't make those choices, then obtaining the working design is impossible.

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