If you want 10x you would set the 100K pot at 19.8K. Voltage gain = pot setting divided by R5 (2.2K) +1

 

No I mean 10x max, I’d change the 2.2k to 10k, right?

 

The 2.2k resistor and the 22uF capacitor cut frequencies below 3.3Hz (-3dB) so that 20Hz is not reduced and the capacitor charging time is fairly short.
If you change the 2.2k resistor to 10k without reducing the 22uF capacitance then the capacitor will take a noticeable 1/4 of a second to charge while playing distortion.

 

No I mean 10x max, I’d change the 2.2k to 10k, right?

Actually appx 11.11K for a gain of 10. You have to add 1 to the division of the two resistance values.

 

Ok. So is it better to reduce the pot to say 20k to limit gain to 10x? Or with 100k pot, 11.1k resistor and a different cap? How do I calculate that? Which is better?

 

Actually, I think that comes from 1/(2piRC). So then with a 10k, I’d want to reduce that cap to 4.8mfd. Right? Or change pot, which I better?

 

The TI data sheet suggests a 100 pf in parallel with RL instead of in series. Im
Not sure what to do with that

 

I think that a mic preamp needs a gain at least 50 or 100 so that the volume control on the power amplifier is not at maximum all the time and can be turned up when the mic sounds are low levels.

 

Is there any concern over changing R3 from 100k to 1M?

 

The TI data sheet suggests a 100 pf in parallel with RL instead of in series. Im
Not sure what to do with that

They do not suggest the capacitor. Instead they are showing that the circuit works and does not oscillate with the capacitor which might be a coaxial cable.

 

I just came across this thread for the first time. I am rather familiar with phantom power.
Those two wire microphone circuits are TOTALLY DIFFERENT from the three-wire "phantom Power" circuit, which is intended to be balanced with respect to common to reduce noise pickup. Balanced mic circuits are always lower impedance.
For the power supply in post #1, the 470 ohm resistors between sections should be larger, 4700 ohms or even 47,000 ohms, That will help quite a bit with the filtering.

 

If you increase R3 to 1M in the opamp biasing then the + input of the opamp will be out-of-range of the input common-mode voltage range needed, causing the opamp to not work.

 

For the response to be flat (not -3dB) at 80Hz then you calculate a cutoff frequency of (80/5=)16 Hz.
What is the input resistance of the mic input on your amplifier?

hey this divided by 5 stuff. How does that math work? Is that an approximation?

 

Calculation? Simply look at a graph of the frequency response of an RC highpass filter.
At its -3dB cutoff frequency the level is 0.707 times.
It begins to noticeably drop off a little at cutoff x 5.

 

I think the math I’m looking for is this, based on -20 dB/decade:

The cut-off frequency point is 70.7% or -3dB (dB = -20log VOUT/VIN)

 

I'm trying to understand what the 22uF (C5) does in this mic preamp circuit. C2 and C4 are coupling caps, and C1 and C3 are bleeding off power supply AC ripple. But I rarely see capacitors to ground in op amp designs. What does C5 do and how does one calculate it's value? Maybe it's related to C1 and C3?

 

C5 allows the opamp to amplify AC audio at various levels up to 46 times but it prevents the opamp from amplifying the 4.5V DC bias voltage. Then the opamp has a voltage gain of 1 for DC.

 

The 46x you refer to is the 100k pot / 2k2. So the 22uF is a DC blocker or at least limits it to a gain of 1 which is less than the gain circuit? Whereas C1 abd C3 are smoothing caps for the AC ripple. They seem similar in value, but diffferent in function.

 

The terminology needs to be worked-out first.
C1 and C3 are Power-Supply "Bypass-Capacitors".
R1 is a "Bias-Voltage" Capacitor that provides power for the Internal-Amplifier in the Microphone.

The Microphone has a very High-Impedance-Output,
meaning that it can only drive a very high-value Resistance.
For this reason,
the Volume-Control components are incorporated into the "Feed-Back" Circuit,
( instead of the Input ),
and therefore the Load on the Output of the Microphone will have a consistent,
and very light Load on it at all times.

The Volume-Control-Pot, and R5 make up an adjustable Voltage-Divider,
which provides more, or less, Feedback to the Inverting-Input of the Op-Amp to control its Gain.

C5 creates a "High-Pass-Filter",
which will not reduce the Feedback at very low Frequencies,
but will allow a reduction in Feedback for normal Audio-Frequencies.
Less Feedback = more Amplifier-Gain.
At very low-Frequencies, all the way down to DC,
the Feedback will always be 100%, which results in a Gain of ( 1 X Gain ).

The maximum-Gain possible is determined by the value of R5.
The Formula is 100K / 2.2K = 45.5 + 1 = 46.5 X Gain.

The minimum Gain possible is ( 1 X Gain ),
because the Slider on the Pot will be directly connected to the Op-Amp-Output,
which will then provide 100% Feedback to the Op-Amp's Inverting-Input.

One useful change that could be implemented in this Circuit is by
changing R2 and R3 from 100K to 1M.

The only drawback to this Circuit is possibly an objectionable Noise level,
( even though the selected Op-Amp is a good choice ),
If the Gain was split between 2 Op-Amps,
the "Noise-Floor" could possibly be lowered to virtually zero.

Further reductions in Noise can be had by using 1/2-Watt rated, 1%, Metal-Film-Resistors.
.
.
.

 

The terminology needs to be worked-out first...

Very thorough explanation. I appreciate the time you took. I'm still trying to understand some of the concepts and terminology. I'm a mechanical engineer so I understand the basics, but not a lot of the practical stuff.

Why would 1/2W resistors be better than 1/4W... assuming the same 1% metal film? They are so big, I'm trying to conserve space in my 2x2" 1590L. But if there was one or two or three specifically that would benefit from 1/2W, I could try it out.