Opamp Feedback problem, Urgent !!

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jaiminajmeri

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1) At resistor R3, Voltage at R3 is 5V initially.
But, After the feedback this voltage changes to 500mV. Can somebody explain this with equations ?

see the ckt here : **broken link removed**

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2) If I replace Voltage@ R3(5V) with 5V battery, then after feedback the voltage remains 5V itself, Any reason ?

see the ckt here : **broken link removed**
 

On first circuit You have postive feedback. It can not work as amplifier this way.
 

Could you please share the Application.

As in 1st fig. its +ve feed back and one of the input is 12V (Greater then VCC,VEE) circuit will saturate to -10V. We can apply superposition theorem on this circuit.
i) V(R3) due to OpAmp (10V supply is shorted) will be -3.33V.
ii) V(R3) due 10V supply need to consider the equivalent resistance looking at R3 node. and use Res-div theorem over it.
iii) Addition of both will provide final node voltage.


In 2nd fig. you are forcing the -ve terminal to 5V. Here also output will saturate but to +10V.
Please correct me if I am wrong.
 

In both circuits, the OP will be destroyed by applying an input voltage 2V beyond the supply rails.

In so far there is no feedback problem.
 

Hello Every1 !
I am not designing a amplifier, Its a schmitt trigger. The output will oscillate between +Vsat and -Vsat.
I just want to know how V(R3) (i.e 5V) changes to 500mV.

- - - Updated - - -


Hi skamthey,
This is just for selfstudy on schmitt trigger.Have a look at its datasheet.
Can u please prove the (ii) point(I mean with equations) ? I tried I had no luck on it.
 

I am not designing a amplifier, Its a schmitt trigger. The output will oscillate between +Vsat and -Vsat.
I just want to know how V(R3) (i.e 5V) changes to 500mV.
Click to expand...
It's a simple resistor network calculation, not specific to OPs. You get 500 mV, if the OP output voltage is -8.5 V ("-Vsat" in your terms).

Did you realize in the meantime that the feedback resistor in circuit 2 is meaningless, because the input is driven by a voltage source?
 
The Eq is:

V(R3) or Vref = Vout * {(R2||R3)/(R1+(R2||R3)) } + V2 * {(R1||R3)/(R2+(R1||R3)) }

Its Theoretical Formula and we are considering the R1 as just R1. But practically ( I think ) there will be some resistance at the output node of OPAMP which will add with R1 and that's why V(R3) will come around 0.5V.

Please correct me if I am wrong.
 

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