Hello to everyone.
Im new in the forum and i need your help.
Can you explain me the operation of the circuit below ?
I tried but failed to give explanation.
Is it a peak detector??
How do I calculate the output,the capacitors kai the resistors?

Thanks,
waverunner

 

Hello,

As far as I can see it is a "precision peak detector".
Take a look at the PDF in the link.
www.site.uottawa.ca/~rhabash/ELG4135L8.pdf

Greetings,
Bertus

 

The diode will have some effect on the output. The capacitors will have some reactance at 4 KHz that will also affect circuit operation.

Use Xc = 1/2PI*F*C

 

Thanks for your answers.
I did that:
R3\\Xc4=9,2KΩ , R4+Xc3=497,88Ω , Vpeak=0,54V

Α=(9,2ΚΩ/497,88Ω)+1=19,47

Vdc=Vpeak/PI=0.172V

Vout=(0,172V*19,47)+2,5V=5,84V ... but, that is not true.

Where i do wrong??
Thanks

 

If : Vin=0,212Vrms --> Vout=2,8Vdc
Vin=0,382Vrms --> Vout=3,03Vdc
Vin=0,743Vrms --> Vout=3,528Vdc

If I do : Vout = Vpeak+Voffset
for example Vout=0,382/0,707 + 2,5V=3.04V
or 0,212/0,707 + 2,5 = 2,799V

Is that way true??

 

Yes the circuit is a peak detector.
The opamp has an internal gain of about 200,000 that reduces the 0.7V forward voltage of the diode to almost nothing since the diode is inside the negative feedback loop.
Then the output voltage is the 2.5V input bias voltage plus the peak of the input sine-wave which is 1.414 times its RMS value.

 

Thanks all the members for their answers..