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Resistors and?

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The microphone needs a preamp circuit with a voltage gain of about 30.
Then you can use one jack on the input of your circuit for the output of the mic preamp and use a second jack for the output of your MP3 player.

Use a switch to switch between the two inputs or make an audio mixer circuit.
 
In response to your PM request, here's a circuit for a preamp. Audioguru is, as his name says, the audio guru here. I used his advice for the nominal gain of 30, with adjustment over a range of ≈13 to ≈65. He might be able to suggest improvements, including but not limited to a better op amp.
 

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Thanks Ron,
I would do it like this:
 

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Thank you both for all your suggesions.
I have to use 2 jacks for both mic and the mp3 player with your circuit right?

What about using a dynamic mic? I can use the above circuit for a dynamic mic or need to
change the parameters?

You know that my circuit uses a volume to generate dc offset for the modulator so my MAIN problem is how to connect circuits like your above circuit to it?
 
Oh I forgot to say, I do not need a mixer. just want to remove the mic jack and then insert the mp3player to the modulator if nesesary.
 
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R1 in the preamp circuit powers an electret mic. A dynamic mic generates a signal and does not need to be powered so don't use R1 with a dynamic mic.

The input resistance of your modulator is 1k (the inverting opamp) which might be too low for the line output of an MP3 player to drive. It is also very low for the opamp in the mic preamp to drive. An input resistance of 10k will be better.
 
R1 in the preamp circuit powers an electret mic. A dynamic mic generates a signal and does not need to be powered so don't use R1 with a dynamic mic.

The input resistance of your modulator is 1k (the inverting opamp) which might be too low for the line output of an MP3 player to drive. It is also very low for the opamp in the mic preamp to drive. An input resistance of 10k will be better.

agu,
Just in case you dont have it already [ for LTSpice]
 

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agu,
Just in case you dont have it already [ for LTSpice]
Thanks Eric. I needed that sim for an electret mic.
I think the 2.2k resistor value is too low.

I simulated Colin's "hearing aid" circuit with the 100nF muffler capacitor across its mic. Its response at 3kHz is down almost -10dB.
 

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R1 in the preamp circuit powers an electret mic. A dynamic mic generates a signal and does not need to be powered so don't use R1 with a dynamic mic.

The input resistance of your modulator is 1k (the inverting opamp) which might be too low for the line output of an MP3 player to drive. It is also very low for the opamp in the mic preamp to drive. An input resistance of 10k will be better.

Yea, but what about impedance matching? An electret mic is a high impedance component while the dynaimic one is the low (600 ohms) one?

Is the last scheme for me?

How did you calculated the modulator's resistance?

Anyway I have changed the Ron's first circuit (the modulator) so that it does work with 14V single ended power supply, I did modify the input like the below pic (please consider that I used 4047 as oscillator/divider instead of 4013 too):
 

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Yea, but what about impedance matching? An electret mic is a high impedance component while the dynamic one is the low (600 ohms) one?
Impedances are never matched in audio circuits. If they are matched then half the signal's level is thrown away. A 600 ohm mic is actually 150 ohms and a "600 ohm" input is actually 1k ohms or 1.2k ohms. The input impedance of the input is low to reduce noise.

How did you calculated the modulator's resistance?
The input impedance of an inverting opamp circuit is the value of its input resistor.
 

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Impedances are never matched in audio circuits. If they are matched then half the signal's level is thrown away. A 600 ohm mic is actually 150 ohms and a "600 ohm" input is actually 1k ohms or 1.2k ohms. The input impedance of the input is low to reduce noise.


The input impedance of an inverting opamp circuit is the value of its input resistor.

Can you explain the first paragraph please? I would like to learn more about your saying.
So you are saying that we can handle both a dynamic and an electret mic through just the same ciruitry (except of the feeding resistor for the electret mic)? Are you meaining that the impedance is not important for an input? What about the input of a transistor?)?


I used a 10k resistor for it before (please take a look at Ron's modifiction). But why this is true ""The input impedance of an inverting opamp circuit is the value of its input resistor""? Is that due to this fact that an op amp has a very high impedance at its input then we can use a resistor at its input and consider it as a the input impedance of the opamp?

Now that I am thinking I reach to another question, you know that I am using a mp3 player for the input too, we know that we can exchange the output of the mp3 player with a voltage source and a resistor in series with it. So do we need to consider that resistor as part of the input impedance too?
 
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Can you explain the first paragraph please? I would like to learn more about your saying.
Impedances are never matched in audio circuits because then you would have a voltage divider and the output level would be reduced to half.
A preamp with an input impedance of 1.2k ohms is used for for a 150 ohm dynamic mic.

So you are saying that we can handle both a dynamic and an electret mic through just the same ciruitry (except of the feeding resistor for the electret mic)?
When a preamp has a low input impedance then it has less noise when nothing is connected to it. An electret mic has an impedance of about 3.3k ohms which is in parallel with the 10k resistor powering it so its preamp should have an input impedance of about 20k ohms.

What about the input of a transistor?)?
The input impedance of a transistor is more when its current is less and when it has a resistor in series with its emitter to ground.

I used a 10k resistor for it before (please take a look at Ron's modifiction).
I know. Ron's inverting opamp has an input impedance of 10k.

But why this is true ""The input impedance of an inverting opamp circuit is the value of its input resistor""?
That is how an inverting opamp behaves.

Is that due to this fact that an op amp has a very high impedance at its input then we can use a resistor at its input and consider it as a the input impedance of the opamp?
No.
The input impedance of a non-inverting opamp is extremely high.
The negative feedback to the inverting opamp's input causes it to have an input impedance that is the value of its input resistor.

Now that I am thinking I reach to another question, you know that I am using a mp3 player for the input too, we know that we can exchange the output of the mp3 player with a voltage source and a resistor in series with it. So do we need to consider that resistor as part of the input impedance too?
Yes.
The input to your modulator circuit is either an MP3 player or the output of the mic preamp.
The output impedance of the mic preamp's opamp is extremely low so the gain of the inverting part of your modulator is 1.
But you don't know what is the output impedance of the MP3 player. Then the gain of the inverting input of the modulator circuit changes with a different output impedance of an MP3 player.
 
Agin need more help please

Hello again.

Finally I decided to amplify the output signal of this modulator by factor 4 or 5 or even more.
The power supply is 14V single ended.
Do you suggest a transistor or an Op-amp to do so?
Any suggestion for a reasonable and low noise amplifier would be appreciated.
 

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The two opamps in your modulator can have any amount of gain that you want.
But amplify the AC, not the DC.
 
The two opamps in your modulator can have any amount of gain that you want.
But amplify the AC, not the DC.

Well I was talking about amplifying the output of the modulator.
That opamp you are talking about is not helpful I think.
Anyway I could not amplify the output of the modulator by changing the resistor values for the said opamp.
 
Hello Ron, audioguru, and other guys.
Thanks for your guides.

I noticed that Ron's modification circuit uses an Op-amp with the gain of just 1. So I decided to modify it again to just get more gain out of the op-amps, maybe by factor 10 or 11. So here it is.

Please let me know if I have done it right?
Does it work like what the original circuit did (but just with a higher gain)?
The symmetrically is very important for the modulator so please check it out and see If I have done it just right or there might be a problem??

To let you compare both circuits I will put Ron's modification too.

Thanks a lot for any guide.
 

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