Difference-Amplifier

dark said:

Hi,

I see a circuit of difference amplifier where the reference is set to 10V (the rest of the circuit require a bipolar supply , therefore the reference) . I know the transfer function of the difference amplifier when referenced to GND , but what is the ckt behaviour in the attached image . When we reference it to 10V , is it the same as before Vout=((V2-V1)(Rf/Rin) , hope V2,V1,Rf,Rin are understood . And this time the ckt is expected to get a VCC of 14V? , is this correct.


Thanks
-Adi

Click to expand...

ericgibbs said:

hi,
How do you get a value of 14Vout.?

Click to expand...

audioguru said:

When the AC input signal is less than +10V then the output of the opamp will be less than +10V.
When the AC input signal is exactly +10V then the output of the opamp is exactly +10V.
When the AC input signal is more than +10V then the output of the opamp is more than +10V.

R3 and R1 attenuate the input to 0.0909 times.
The gain of the opamp is (150k/10k) + 1= 16. The total gain from input to output is
0.0909 x 16= 1.45 times.

Click to expand...

dark said:

Hi audioguru ,Thanks ;

How do you get ' 0.0909 times.' figure?

It seems to be reasonable what you explain the 10V becomes the center point of the either side swing .

With (V2-v1)*Gain I end up with a -ve sign , what transfer function holdes now.

Regards
-Adi

Click to expand...