dual Vs single opamp for stereo mic preamp

another question: If I wanted to pad the output to hook up to a DSLR mic input by using a 100/1 voltage divider, would it be better if the resistor values were large (i.e. 100K/1K) or  could they be smaller (i.e. 10K/100R)?

spaceludwig said:

Hey everyone,

So unit now works. I've been working on the power supply, the schematic of which I'll upload later tonight.

Question: Is there any advantage/disadvantage to using a lower value for the gain pot? i.e. 10K or even 1K rather than 20K? Would it impact the output impedance?

Click to expand...

Assuming the schematic shown in the first post in this thread, the value of the pot sets the maximum gain, Av = 1 + 20k/30.

Minimum gain is basically unity, figure Av = 1 + wiper resistance/20k, and the wiper resistance is much much less than the 20k so consider it zero, and also the 30-ohm resistor is also much much less than 20k so it's swamped out.

-a

Andy Peters said:

Assuming the schematic shown in the first post in this thread, the value of the pot sets the maximum gain, Av = 1 + 20k/30....

Click to expand...

Hi Andy,

I think I was probably not clear. What I meant to ask is if there was some reason not to use resistors of a lower value but maintaining the same ratio, i.e.  10k/15 or 2k/3, etc...  Would choosing these lower values have a negative impact on the functionality of the amplification? 

Below is the 9V power supply. Comments welcome.

You have a typo in that drawing. The +4.5V and Gnd and -4.5V are shorted together.

Speedskater said:

You have a typo in that drawing. The +4.5V and Gnd and -4.5V are shorted together.

Click to expand...

Corrected, thanks.

spaceludwig said:

Andy Peters said:

Assuming the schematic shown in the first post in this thread, the value of the pot sets the maximum gain, Av = 1 + 20k/30....

Click to expand...

Hi Andy,

I think I was probably not clear. What I meant to ask is if there was some reason not to use resistors of a lower value but maintaining the same ratio, i.e.  10k/15 or 2k/3, etc...  Would choosing these lower values have a negative impact on the functionality of the amplification?

Click to expand...

I can answer this. The op amp has to drive the output and the feedback network, so the lower you make the resistor values in the feedback loop, the more the op amp has to work to produce the same voltage output. This means higher distortion, however lowering the feedback resistors lowers noise also, so you have to make a compromise between the drive capability of the op amp and the noise you're willing to tolerate. Most really good op amps can drive 600 ohms all by themselves with really low distortion, so you can use fairly low feedback resistors depending on the load impedance. The load and the feedback network are in parallel, so you have to do the math to figure out the actual load the op amp sees. What is the load impedance you expect for this circuit? You may not need the BD139/140 outputs if it's high enough.

Thanks, that's what I suspected. So ultimately, if I chose low values like 1k/20 I am better off including the the 2 transistors at the output to beef up the signal?

Right now it is working with 1k/75 though I would like to increase the gain ratio.

Question: I've noticed that on all portable recorders when you engage phantom power the battery dies much quicker. However, I've read that a phantom powered microphone typically only consumes in the neighborhood of 2 - 5ma.

I'm just curious as to why the battery would die so quick if the current draw is so small. Does this happen  when the nominal battery voltage (i.e. 9V) is multiplied up to ~48 Volts?

spaceludwig said:

Question: I've noticed that on all portable recorders when you engage phantom power the battery dies much quicker. However, I've read that a phantom powered microphone typically only consumes in the neighborhood of 2 - 5ma.

I'm just curious as to why the battery would die so quick if the current draw is so small. Does this happen  when the nominal battery voltage (i.e. 9V) is multiplied up to ~48 Volts?

Click to expand...

Exactly... 48v/9v means  every mA drawn by phantom supply is 5+ mA from 9V, + losses from conversion.

JR

by 5+ you mean X5, i.e. 9 X 5 = 45V so if the mic's nominal draw is 3ma, after voltage multiplier the draw from the batteryu will be 3 X 5 = 15ma plus losses along the way?

spaceludwig said:

by 5+ you mean X5, i.e. 9 X 5 = 45V so if the mic's nominal draw is 3ma, after voltage multiplier the draw from the batteryu will be 3 X 5 = 15ma plus losses along the way?

Click to expand...

Yes, exactly, Ohm's Law at work.

-a

Ah, ok.

Can the current draw be calculated from the mic's output impedance?

(on second thought a 250 ohm mic would mean close to 200 mA which seems excessive)

spaceludwig said:

Ah, ok.

Can the current draw be calculated from the mic's output impedance?

(on second thought a 250 ohm mic would mean close to 200 mA which seems excessive)

Click to expand...

No...  Phantom supply is common mode, not normal to the mic element, and  has 6.81k series build out resistors in each leg, so you can predict a worst case, both input lines shorted scenario (2x48/6810 mA), but normal use current will vary with mic design. Always less than worst case short circuit.

JR