Unknown circuit

Discussion in 'Homework Help' started by wave runner, Jun 1, 2009.

1. wave runner Thread Starter New Member

May 21, 2009
4
0
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

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Apr 5, 2008
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3. beenthere Retired Moderator

Apr 20, 2004
15,815
282
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

4. wave runner Thread Starter New Member

May 21, 2009
4
0
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

5. wave runner Thread Starter New Member

May 21, 2009
4
0
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??

6. Audioguru New Member

Dec 20, 2007
9,411
895
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.

7. wave runner Thread Starter New Member

May 21, 2009
4
0
Thanks all the members for their answers..