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Electret preamp layout question

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Torben

Well-Known Member
Hi all,

I've breadboarded a simple electret preamp based on the commonly-seen schematic presented at **broken link removed**, and after adding a resistor divider to drive the electret mic, it appears to work quite nicely in initial tests. (Not going for ultra quality off the bat here; just looking for design technique help).

I've been laying out the PCB in eagle for it, and I just have a couple of questions about ground and line lengths:

1) Would I likely benefit from running traces from each ground point to a single point at/near the power supply 0V input (i.e. star ground using traces)? This is as opposed to either using a ground pour on the bottom (solder-side) layer or using a ground plane on the top layer.

2) The electret element (ye olde radio shack cheapo) has a FET built in, right? Will this suffice as a line driver so I can mount the electret at the end of, say, a 30-60 cm line? Or would I need to further buffer it?

Pointers to reading are as welcome as flat-out responses. :) I don't mind terribly just testing all this, but I would like to avoid etching the boards just to find out...and the theory always helps understand the "why"s instead of just the "what"s.


Thanks all!

Torben
 
The currents in the mic preamp are very low so the grounding doesn't make much difference. It is important for making a power amplifier.

The electret mic needs a single resistor to power it, not a voltage divider. Usually 10k from a 9V supply. It needs 0.5mA.

The output impedance of the FET in the electret mic is high. The capacitance of 60cm of shielded audio cable is about 50pF so it will cut high frequencies above 320kHz.
 
audioguru said:
The currents in the mic preamp are very low so the grounding doesn't make much difference. It is important for making a power amplifier.

OK, thanks! Makes the layout easier.

The electret mic needs a single resistor to power it, not a voltage divider. Usually 10k from a 9V supply. It needs 0.5mA.

Alright. The reason I went with the divider was because the docs for the electret state that wants a 2.2K resistor to hot, and it can use 1V to 10V with 4.5V optimum--so I put the divider from the 9V supply to get 5V and then ran the 2.2K from that to the electret.

I just removed the divider and 2.2K and just stuck a 10K to 9V...and the response got better. I'm now getting ~5V to the electret, so I might try to tune that closer to 4.5V while I'm watching the canucks/oilers game.

What is the downside of the divider to achieve the required voltage as opposed to the single resistor? Just part count? I'm not just imagining the better response here, am I? (I'm going to do a simple freq. analysis using a basic soundcard scope during the game too...)

The output impedance of the FET in the electret mic is high. The capacitance of 60cm of shielded audio cable is about 50pF so it will cut high frequencies above 320kHz.

OK. I don't think my guitar gets much going on up that high. It might but I can't hear it with my bat detector. :)

Thanks for the help!


Torben
 
Torben said:
Alright. The reason I went with the divider was because the docs for the electret state that wants a 2.2K resistor to hot, and it can use 1V to 10V with 4.5V optimum--so I put the divider from the 9V supply to get 5V and then ran the 2.2K from that to the electret.

I just removed the divider and 2.2K and just stuck a 10K to 9V...and the response got better. I'm now getting ~5V to the electret, so I might try to tune that closer to 4.5V while I'm watching the canucks/oilers game.

What is the downside of the divider to achieve the required voltage as opposed to the single resistor? Just part count? I'm not just imagining the better response here, am I? (I'm going to do a simple freq. analysis using a basic soundcard scope during the game too...)
The resistor that powers the mic is also its load. A 2.2k resistor is lower than the output resistance of trhe FET in the mic so it reduces its output level. A 10k resistor feeding it results in a higher output level if it has a high enough source DC voltage so it can provide about 0.5mA to the FET in the electret mic.
 
audioguru said:
The resistor that powers the mic is also its load. A 2.2k resistor is lower than the output resistance of trhe FET in the mic so it reduces its output level. A 10k resistor feeding it results in a higher output level if it has a high enough source DC voltage so it can provide about 0.5mA to the FET in the electret mic.

/me futzes with a calculator....

Aha! And the math works out better to boot, from a 9V supply. Thanks for the explanation!


Torben
 
Torben said:
OK. I don't think my guitar gets much going on up that high. It might but I can't hear it with my bat detector. :)

So is this for miking a guitar?, probably OK for recording, but NOT for live performance - they are EXTREMELY sensitive to feedback!.
 
Nigel Goodwin said:
So is this for miking a guitar?, probably OK for recording, but NOT for live performance - they are EXTREMELY sensitive to feedback!.

Yep. Kinda like a piezo on a banjo (you'd be surprised how often you see that...)


Torben
 
The 10uF capacitor does not pass DC so the inputs of the opamp can have the same DC voltage. Then the DC gain of the opamp is 1 and its output DC voltage is the same as its input DC voltage which is half the supply voltage.
Then the output can have a wide voltage swing.

The capacitor passes AC to set the correct amount of negative feedback. The AC gain is (22k/1k) + 1= 23.
 
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