microphone amplification with OPA134?

Nigel Goodwin

Super Moderator

Most Helpful Member

• Jan 30, 2007

• #21

It looks exactly the same as it did when it left the electret mike - except it's now centered at half the supply rail - so if the electret ouput is swinging +/-10mV, and the supply is 12V, then it's swinging from 6.01V to 5.99V.

I think Audioguru confused you by the term 'modulated', which while technically correct doesn't really help you to understand.

 

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whiz115

Member

• Feb 7, 2007

• #22

i bring back this subject....because i want to ask
if this 5.99V to 6.01 has anything to do with DC offset...
(basically i want to know what DC offset is.... )

also this 5.99V to 6.01 i think it isn't an AC signal...it's just a DC
with our signal like a small ripple (noise)... the real AC comes from the electret which has both positive and negative period right??

 

audioguru

Well-Known Member

Most Helpful Member

• Feb 7, 2007

• #23

When you speak into the mic, its output through the coupling capacitor causes the 6.0VDC at pin 3 of the opamp to swing down to 5.99V and up to 6.01V. The swing is 20mV peak-to-peak of AC signal. Then the opamp amplifies it.

 

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whiz115

Member

• Feb 8, 2007

• #24

i can't understand it...check this photo and
tell me if this is the case...

**broken link removed**

also i would like to know if it has anything to do with DC offset

 

Last edited: Feb 8, 2007

audioguru

Well-Known Member

Most Helpful Member

• Feb 8, 2007

• #25

No.
The electret mic does not idle at 0VDC. It is powered by a 10k resistor from +6V and draws current so its DC output voltage is about +2VDC. Then the signal causes the 2V to rise to 2.01V and to drop to 1.99V.

Your sketch has the polarity of the signals reversed. They should be exactly the same, except the coupling capacitor allows them to have different DC voltages.

The OPA134 has a typical input offset voltage of only 0.5mV. This circuit has a DC gain of 1 so if the input DC voltage is exactly 6.0000V then the output DC voltage could be anywhere from 5.9995VDC to 6.0005VDC.

 

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whiz115

Member

• Feb 8, 2007

• #26

sorry if if look odd...but i have to understand those things...and web sites
can't help much because you can't make questions to the special things you don't understand.

is this the case?

**broken link removed**

 

audioguru

Well-Known Member

Most Helpful Member

• Feb 8, 2007

• #27

Yes, the two signals in your sketch have the same polarity. It is correct.

You show 1.0VDC for the microphone. But it is probably about 2.0VDC.

 

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whiz115

Member

• Feb 8, 2007

• #28

audioguru said:

You show 1.0VDC for the microphone. But it is probably about 2.0VDC.

Click to expand...

yes...sorry typographical mistake!
thank you for helping me understand this!!!!

one last question on this topic please?

you said that the output DC voltage could be anywhere from 5.9995VDC to 6.0005VDC i'm guessing that the waveform looks like as the one my last post...

if this is true what we have succeed?

 

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whiz115

Member

• Feb 9, 2007

• #29

i know that i maybe asking excessive informations possibly difficult for
someone to answer etc...but it's more easy for me at the moment to source the specific things from the forum than from my school or from friends out of the school.

so please someone help me because i'm confused with the op amp....

 

audioguru

Well-Known Member

Most Helpful Member

• Feb 9, 2007

• #30

The opamp circuit is designed to amplify the small AC output from the microphone, not amplify the opamp's DC voltage.