A simple microphone preamp can someone explain this circuit to me?

[Whiteblooded]

Hi guys.

Firstly, sorry if this question is far to easy for this section of the forums.

Basically, I was bored and decided to take apart a microphone I that I had just lying around.

It came with a box that you have to put the microphone through, which had the following circuit in it:

http://img16.imageshack.us/img16/2475/labspreampcircuit.jpg

I'm just wondering if anyone could talk me through this circuit? Basically, the whole thing is powered by a 1.5V battery, and the signal comes from the microphone from the two inputs from the left on the diagram. The potentiometer controls the volume.. which I understand. However, can anyone explain:
- why the two outputs are connected via a resistor?
- why there is a need for a capacator (I'm guessing some kind of filter?.. but why?)
- why there is a second resistor separating the battery and the microphone signal in?
- why this circuit is required in the first place.

Any comments on this circuit would be a great help.. I really don't understand electronics =/ again, sorry for spamming the forum up with this easy stuff :S

 

[vk6kro]

If you include the microphone it looks a lot more friendly:

http://dl.dropbox.com/u/4222062/Mic%20Amp.PNG

The microphone is an electret type which requires a DC voltage to operate. This is what the 1.5 volts is for, although a higher voltage is usually used.

The capacitor is used to stop the DC getting to the volume control pot. Capacitors block DC but allow AC signals to pass.

I don't understand what the 2.2 K resistor on the right is doing there or why the diagram showed a switch in series with the pot. Their effects are predictable but I can't guess why these effects would be needed.

 

[Whiteblooded]

Ah, thanks for that diagram.. that's looking much better. Thats explained a few things in itself..

Hmm. Yeah, I really have no clue about the resistor on the right either. I have no idea why it's there. would it have anything to do with DC voltage mixing with an AC voltage?... that's just a guess.

 

[N.Saravanan]

1.The output signal level of any transducer(microphone here) also depends on the input impedance of the circuit to be coupled with the output say an amplifier. So if you are not sure about the input impedance of the circuit you would be adding it is always better to keep the output impedance of transducer as low as possible. This is when you connect the low impedance output with a relatively high impedance input the effective impedance always takes the value nearest to the lowest impedance value coupled which is the output impedance of the transducer. In other words the output impedance of the transducer remains same even after connecting a circuit to its output. So the transducer will be able to provide signal with same amplitude. On the other hand if you connect a circuit with further lower impedance at the output the effective impedance at the output of transducer further drops down. This results in a demand for greater current flow which the battery cannot provide. So the signal level drops down.
2.When the switch is open and you vary the potentiometer the effective resistance across the output varies from 2.2k to 12.2k. When the switch is closed the same effective resistance varies between 2.2k to 10k. So I guess the switch may be used for fine tuning the gain of the microphone if required.

 

[leodpenrose]

Not 100% sure but i believe that The other resistor acts as a pad that attenuates the signal in case the source is really loud resulting in higher voltage you would use it, for instance if your were recording a kick drum which generates high spl and you can turn it off when recording soft sources like vocals.

also there is a capacitor there i believe because the type of microphone it is may be a Condenser microphone also know as a Capacitor microphone. It stores voltage from an internal or external power supply (Phantom Power 48V). As the microphones diaphragm gets closer to the back plate a result from speaking into the microphone it releases the voltage creating a analog of the acoustic energy.