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Op-Amp subtractor strange problem

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dr.power

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Hi guys,

I have made a difference amplifier (subtractor) based upon OP-AMPs as below pic.
The whole 4 resistors which I used were 10K. I have just a problem by it. When I connect a sine input to the V1 the said sine wave just drops in volatge by say a factor of 2!! But when I connect the said sine signal to the V2, The signal does not drops at all!
By the way When I increasing the feedback resistor to a higher one the signal at V1 is less dropped!?
What is the problem please and how to solve it?
I myself guess that the input impedances at V1 and V2 are not the same, even by using the same resistors for them though It is very strange and should not be correct!? But I am not sure?!
 

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When I connect a sine input to the V1 the said sine wave just drops in volatge by say a factor of 2!! But when I connect the said sine signal to the V2, The signal does not drops at all!
Are you sure you've got that the right way round? If your circuit is wired as shown the sine input to V2 will be halved by R3 and R4 acting as a potential divider, so only half will reach the non-inverting input of the amplifier.
With the sine input at V1 the amplifier gain is -1, i.e. the output is the same magnitude as the input sine but is inverted.
Increasing the feedback resistor increases the amplifier gain so will increase the output voltage.
 
The circuit is correct. It just requires a low (ideally zero) source impedance for both inputs. Thus when you input the voltage to V2, V1 must also be connected to a low impedance source (ground if you don't have a source connected). Otherwise the non-inverting gain from V2 is 1/2 as you observed, not 1 as it should be.

If you calculate the gains you will see that, for the proper connections, you have an inverting gain of 1 from the V1 input and a non-inverting gain of 1 from the V2 input, thus it acts like a differential amplifier with a gain of one.
 
Are you sure you've got that the right way round? If your circuit is wired as shown the sine input to V2 will be halved by R3 and R4 acting as a potential divider, so only half will reach the non-inverting input of the amplifier.
With the sine input at V1 the amplifier gain is -1, i.e. the output is the same magnitude as the input sine but is inverted.
Increasing the feedback resistor increases the amplifier gain so will increase the output voltage.

I think the Pic shows the right circuitry. I guess the reason of using a resistor divider in V2 is the gain of a non inverting op-amp which is 2 if R1 and R2 happen to have similar values.
 
The circuit is correct. It just requires a low (ideally zero) source impedance for both inputs. Thus when you input the voltage to V2, V1 must also be connected to a low impedance source (ground if you don't have a source connected). Otherwise the non-inverting gain from V2 is 1/2 as you observed, not 1 as it should be.

Sorry But If I got it right, You are saying that the output gain from noninverting input while the Ri and Rf are simillar is just 1/2,right??!! I think it is 2 not 1/2 and thats why a subtractor has resitive divider at noninverting input (to reduce the incommong volatge to 1/2). Am I right?

,In a subtractor whole resistors are simillar, so If you calculate the gains you will see that, for the proper connections, you have an inverting gain of 1 from the V1 input and a non-inverting gain of 1 from the V2 input, thus it acts like a differential amplifier with a gain of one.

Sorry as I said I think the gain for the non inverting input should be 2 (1+Rf/Ri), What the volatge divider at noninverting input will cause to have an propper output.
 
crutschow,
Can you please kindly let me know what would be the final output for a circuit like what I posted (Suppose the while resistors are similar), If the input signals happen to be
V1=Vmsinwt
V2= -Vmsinwt
and if :
V1= -Vmsinwt

Thanks
V2= Vmsinwt
 
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I think and almost sure that here is the formula for the above subtractor circuit
 

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crutschow,
Can you please kindly let me know what would be the final output for a circuit like what I posted (Suppose the while resistors are similar), If the input signals happen to be
V1=Vmsinwt
V2= -Vmsinwt
and if :
V1= -Vmsinwt
Since the V1 input has a gain of minus 1 and the V2 input has a gain of plus 1 the output for the first case will be
-(Vmsinwt) + (-Vmsinwt) = -2Vmsinwt.

Edit: For the second case with V1 = -Vmsinwt, the output will be
-(-Vmsinwt) + (-Vmsinwt) = 0
 
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Since the V1 input has a gain of minus 1 and the V2 input has a gain of plus 1 the output for the first case will be
-(Vmsinwt) + (-Vmsinwt) = -2Vmsinwt.

Edit: For the second case with V1 = -Vmsinwt, the output will be
-(-Vmsinwt) + (-Vmsinwt) = 0

Thanks,
And what if the onw input happens to be Vm sinwt and the other happens to be -Vm coswt?
 
For those just starting out with OP amps, the need to drive with a LOW Z source isn;t stressed enough for many circuits. The second issie is that you have to provide a place for the input bias current to go and in some cases you have to provide a resistance for the output to drop across. It will get you every time.
 
Thanks,
And what if the onw input happens to be Vm sinwt and the other happens to be -Vm coswt?
The math is very simple. I'm sure you can do it.
 
Since the V1 input has a gain of minus 1 and the V2 input has a gain of plus 1 the output for the first case will be
-(Vmsinwt) + (-Vmsinwt) = -2Vmsinwt.

Edit: For the second case with V1 = -Vmsinwt, the output will be
-(-Vmsinwt) + (-Vmsinwt) = 0

For the second one V2 was V2= Vmsinwt .
So the response is 2Vmsinwt.
crutschow, So giving 2 signals just having 180 degrees output phases causes just having one signal on the output by just a voltage as high as 2 times of each input.

The last question is that can you guys please tell me if the input ipedances are the same in the above subtractor Pic while whole resistors are the same?

Thanks
 
The math is very simple. I'm sure you can do it.

Actually I am not cure if I can do it for whiles that one input is sine and the other is cosine!
Ok I will try my best and plz corrcet me.
for whiles that:
V1= -Vm coswt
V2= Vm sinwt
the output is V2 - V1:
Vm sinwt + Vmcoswt which can be simplified as Vm(sinwt + coswt). Sorry but I do not know what to do whith the term of (sinwt + coswt)?! Is there any Trigonometry formula for it?

If:
V1= Vm coswt
V2= -Vm sinwt
we'll have an output as:
-Vm sinwt - Vm coswt which can be simplified as -Vm(sinwt + coswt). As Above any Trigonometry formula?
 
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