# I have problem in op-amp can any help me please

#### ksalatawi

Joined Nov 29, 2004
21
question in this attachment

#### pebe

Joined Oct 11, 2004
626
Originally posted by ksalatawi@Apr 19 2005, 06:31 PM
question in this attachment
[post=7067]Quoted post[/post]​
This looks like homework. If so, you should have been taught about op-amps before being set this excercise. So which part don't you understand?

#### David Bridgen

Joined Feb 10, 2005
278
The two opamps on the left need not contribute any gain. They need to provide only an inversion of sign.

The right hand one is the summer.

Remember that the ratio of the feedbak R to the input R gives the voltage gain from an input to the output. This is true for eachof the three stages.
You need to do the simple arithmetic for each of the three inputs.

#### ksalatawi

Joined Nov 29, 2004
21
thank you but
how can i find the value of the resistors
If i find that R2R8/R1R3=3 and
R5R8/R4R6=2 and R8/R7=4

#### dragan733

Joined Dec 12, 2004
152
Hi ksalatawi,
1. The solution is that:
R2R8/R1R3=3
R5R8/R4R6=2
R8/R7=4
We have 8 unknown resistors and only three equations.

2. Each op -amp. amplifies and inverts the input signal

3.

3V1+2V2-4V3=15
2V1+2*3-4*1=15
V1=4,33V
3V1+2*3-4*1=-15
V1=-5,67V
Then the range of V1 for linear operation is: -5,67V until 4,33V

#### ksalatawi

Joined Nov 29, 2004
21
thaaaaaaaaank you very much dragan i missed you

#### ksalatawi

Joined Nov 29, 2004
21
can i find the value of each resistor

#### dragan733

Joined Dec 12, 2004
152
Thank you ksalatawi. The answer I gave you also on your E-mail.
1. The solution is that:
R2R8/R1R3=3
R5R8/R4R6=2
R8/R7=4
We have 8 unknown resistors and only three equations. That means we can select for example:
R7=1K Ω , R1=1K Ω , R3=1K Ω , R4=1K Ω , R6=1K Ω then R8=4K Ω , R5=500 Ω, R2=750 Ω
or for example:
R8=4KΩ, R4=1K Ω , R6=2K Ω , R2=3K Ω ; R3=2K Ω then: R7=1K Ω , R5=1K Ω , R1=2K Ω
Therefore there are infinity combinations