preamp op amp circuit question

[walid1]

Hi
For the circuit shown below I want:
to calculate the o/p gain
to know what the function of C1

I noticed that there are something strange in it, the feedback is connected to the noninverting i/p and the voltage divider to the inverting i/p

thank u in advance

View attachment 40177

 

[aProgrammer]

Clearly there are a mistake,
the + and - inputs are inverted
no doubt is possible !

 

[audioguru2]

When you swap the inputs of the opamp you will notice that the input impedance of the preamp circuit is only 2.2k ohms which is too low for an electret microphone. A non-inverting opamp circuit would have a much higher input impedance so that an electret mic is not loaded down.

For a very low resistance source, the gain at low audio frequencies of the inverting preamp circuit is R2/R1= 100. The impedance of the mic is in series with R1 so reduces the gain.

You will also notice that the lousy old 741 is noisy and cuts off frequencies above 9kHz. A more modern TL071 single, TL072 dual or TL074 quad audio opamp IC costs about the same, is low noise and has a much wider bandwidth.

 

[walid1]

Hi aProgrammer
thank u for fast reply

Yes I agree, there are a mistake, the + and - inputs are inverted

If they are
Av = 1 + (R2/R1)

but why they connect the (+) i/p  [the -ve in the figure] to adivider that half of VCC applied to it, why not simply connect R4 to ground and omit R3?
thanks

To guru whose reply reached me just now

Are u agree that there are a mistake, the + and - inputs are inverted
I read your reply and I'll analyze it now:

When you swap the inputs of the opamp you will notice that the input impedance of the preamp circuit is only 2.2k ohms which is too low for an electret microphone. A non-inverting opamp circuit would have a much higher input impedance so that an electret mic is not loaded down.

from this i deduce that someone can swap the inputs of the opamp to get low input impedance if he want, and that the circuit as its work.

For a very low resistance source, the gain at low audio frequencies of the inverting preamp circuit is R2/R1= 100. The impedance of the mic is in series with R1 so reduces the gain.

thank u guru

 

[audioguru2]

Hi Walid,
The opamp has a single supply so R3 and R4 form a voltage divider to provide it with a half-supply reference voltage. Then the output of the opamp can swing symmetrically. If it had only R4 then its output would be saturated near 0V and it would be a rectifier.

I just noticed that the volume control is only 1k. Most opamps have a minimum load of 2k. I would use 10k.
I also notice that the volume control has DC in it which produces noise as it is adjusted. It should be fed only AC through a coupling capacitor from the output of the opamp.

 

[walid1]

very good guru, u r really a professor

but please
is this with the reverse of the figure
and
the 10u farad cap why?
thanks

 

[Kevin Weddle]

It's possible that it might oscillate. Mono meaning "one frequency" I think. Could be just a rudimentary oscillator. Does not appear to be a linear sort of amplifier.

 

[audioguru2]

Hi Walid,
I fixed the backwards inputs. The 10uF capacitor is a filter so that any noise from the supply isn't amplified 50 times.

Hi Kevin,
The opamp was drawn as a comparator with hysteresis. It won't oscillate, it will just latch with its output either high or low.

View attachment 40181

 

[walid1]

Hi guru

The 10uF capacitor is a filter so that any noise from the supply isn't amplified 50 times.

Av = R2/R1 = 220/2.2 = 100 and not 50

correct me

 

[audioguru2]

BEEP!
Hee, hee. ;D

R3 and R4 are a voltage divider. They reduce noise from the supply to half. Then it is amplified 100 times so the output is 50 times as noisy as the supply.

 

[walid1]

hi guru

IF i apply audio sif to the input cap of 10uV, then I'll fet it from the o/p (before the volume) as 100*10u = 1mV
or 50* 10u = 0.5mV????????????

 

[audioguru2]

Hi Walid,
We were talking about the 10uF filter capacitor. If it isn't there and the supply has 40mV of mains hum then the R3 and R4 divider would reduce it to 20mV and the opamp would amplify it by 101 to 2.02V at the output.

Signals at the input need to have a very low impedance or they will be divided by the 2.2k input resistance of the circuit. Then they will be amplified 100 times.

 

[Kevin Weddle]

Just wanted to touch on supply ripple noise. I applied a signal to the bases of two transistor, collectors connected together, for a gain of 1000, whatever it's called, and the 60 Hz supply ripple never made it to the output. The signal was there, gain of 1000, but the supply ripple disappeared. Could there have been a cancellation effect? The circuit no longer exists.

 

[audioguru2]

I applied a signal to the bases of two transistor, collectors connected together, for a gain of 1000, whatever it's called, and the 60 Hz supply ripple never made it to the output. The signal was there, gain of 1000, but the supply ripple disappeared. Could there have been a cancellation effect? The circuit no longer exists.

A differential amplifier cancels the same signal on both inputs but its emitters are connected together, not its collectors. This one is the differential input of an opamp:View attachment 40184