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enter image description here

This is the typical application circuit of the MAX4466 "microphone preamplifier" op-amp.

Are these resistors/capacitors purely for generating a bias voltage? If so, aren't there much simpler ways to do this, like a simple voltage divider? Are they for limiting current?

I'm planning to use this with a MEMS microphone. Does this change it a lot? I know the values need to be adjusted to fit the microphone and ADC that would read it.

Update: I found a microphone that fits my application, but I'm not sure about the circuit and the amplifier.
This is the ESP32's ADC input voltage ranges, from this page in the Espressif docs.

ESP32 ADC input voltage table

The microphone's datasheet says its DC Output is 0.69V, in page 2

extract from linked datasheet page 2

Does this mean the voltage only swings from .69V, by a few mV? If so, would this circuit work?
I don't know what values are needed for R1 and R2, and if they are even necessary.

I calculated the values for R3 and R4 according to this formula:
Gain = 1+Rfb/Rin I guessed Rin to be 10k according to the microphone's datasheet, which said minimum output load 10k but this could be completely wrong.
With -44dbV/Pa sensitivity and a likely maximum SPL of 120dbSPL, the maximum peak-to-peak voltage would be 41mV. To amplify this to the ADC's 2.5db attenuation setting, which seems closest centered to the .69V, I calculated a gain of about 13, which then makes Rfb 120k.

proposed circuit

This is possibly complete BS, but as you likely know by reading this, i don't know much about microphones and pre-amps (oops)

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    \$\begingroup\$ what the answers below don't point out is that (1) it is an electret mic, which requires a bias boltage (2) the op amp is running of a single supply which requires the input to be biased at mid rail ("artificial 0V" as it is sometimes called). Once you know this the purpose of the resistors is clear. \$\endgroup\$
    – danmcb
    Mar 21 at 6:31
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    \$\begingroup\$ To be very precise, part (1) is not bias for the electret capsule but power to the JFET sitting inside the capsule. (Electret capsule has the bias burnt-in to the capsule material in contrast to non-electret condensor microphones). This kind of powering is sometimes called plug-in-power. \$\endgroup\$
    – ghellquist
    Mar 21 at 13:52

3 Answers 3

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The first group are bias for the microphone capsule.

the top 2K reduces supply noise into the capacitor, and the bottom 2k provides bias current

The second group set the bias voltage for the op-amp which controls the average voltage on the output terminal.

There are almost certainly simpler ways, eg: you could loose the two 1M resustors and replace the 10nF with wire, but then the DC voltage on input to the op-amp would depend on the microphone and this may not be well behaved. this would result in the DC part on the output terminal also drifting.

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  • \$\begingroup\$ to be clear, the top 2k + 0.1uF form an RC filter which reduces noise input from the power supply. So do the bottom 2k + the same 0.1uF and that one is necessary cause otherwise the bias would short out the audio signal. \$\endgroup\$
    – user253751
    Mar 21 at 4:18
  • \$\begingroup\$ the mic is basically a capacitor so it sits close to +5V. If you didn't have the input bias the circuit just would not work. Those 1M resistors are absolutely needed in a single supply circuit. \$\endgroup\$
    – danmcb
    Mar 21 at 6:33
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    \$\begingroup\$ To be very precise, part (1) is not bias for the electret capsule but power to the JFET sitting inside the capsule. (Electret capsule has the bias burnt-in to the capsule material in contrast to non-electret condensor microphones). This kind of powering is sometimes called plug-in-power. \$\endgroup\$
    – ghellquist
    Mar 21 at 13:54
  • \$\begingroup\$ @danmcb The mic is basically the source-drain path of a JFET to ground, with the gate voltage determining the drain current (assuming the JFET is in saturation) being modulated by the microphone membrane movement. When the voltage is below saturation, the JFET D-S at least acts as a varying resistor. But without any plugin voltage, you'll not be able to detect that variation. \$\endgroup\$
    – user107063
    Mar 21 at 22:02
  • \$\begingroup\$ @ghellquist correct - yes I forgot that electrets are self Biassed. my bad. \$\endgroup\$
    – danmcb
    Mar 22 at 8:15
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I refer to this datasheet. Starting with the microphone and its power/biasing:

schematic

simulate this circuit – Schematic created using CircuitLab

The microphone drops a few tenths of a volt (I assume 0.2V here, but this is a guess, since we have no idea what JFET is in the microphone). The remaining 4.8V is across R1 and R2, leaving about +2.6V at their junction A.

These are average DC levels, and we can create an Thevenin equivalent circuit for the blue boxed region. It becomes easier to see how C1 is stabilising the potential at A, by attentuating any power supply fluctuations over 1.6kHZ.

Resistor R2 is providing bias current \$I=\frac{2.6-0.2}{2k} = 1.2mA\$ for the JFET inside the microphone, which ultimately decides the DC mean potential at B.

That frequency seems rather high, why not cut-off at 20Hz? I assume the reason is two-fold:

  1. Keeping C1 × Rth small allows C1 to charge quickly, establishing the DC operating point quickly, and keeping "power-on thump" (I don't know what the technical name for that is) short.

  2. It's silly not to have an extremely stable supply for this circuit, so I don't expect there to be much noise anyway. However, this device's power supply rejection (PSRR) degrades quickly above 1kHz (page 4): enter image description here
    At low frequencies the IC takes care of noise, but C1 avoids injecting that same noise into the amplifier's input, and compounding the problem. It helps flattens out the overall system's PSRR across the spectrum.

The DC average at B is close to 0V, but sufficiently high that AC components (of only a few tens of millivolts) will never cause B to leave the amplifier's common-mode input range. However, this setup has a gain of 10, so the output amplitude can be hundreds of millivolts. Therefore you must center the output well above that, to keep the output significantly above 0V and below +5V.

This is achieved by the two 1MΩ resistors, which are not there for current limiting, except in the sense that their values are kept high to minimise loading of the microphone signal. They are just a potential divider to provide a stable +2.5V DC bias at the amplifier's input. The 0.01μF capacitor isolates the 0.2V DC on one side from the 2.5V DC on the other, while still permitting AC fluctuations from the microphone to be transmitted to the other side. This use of the capacitor is called "DC blocking" or "AC coupling".

It's worth noting that the 1μF capacitor is responsible for holding an average potential at the inverting input equal to the average potential at the junction of the two 1MΩ resistors (the non-inverting input), so the overall output is also centered around +2.5V. This capacitor, even though it isn't inserted directly into the signal path, as part of the feedback network it performs the same function as "AC coupling". That means that you don't have to use two 1MΩ resistors; their ratio can be chosen to provide whatever mean voltage around which you wish to center the overall output.

That capacitor causes high-pass filter behaviour, but it doesn't affect frequency response in the audible range, since it gives the amplifier a cut-off frequency of about 16Hz.

That's not the only reason the designers chose to shift the signal up in potential, half-way between the supply potentials, with the two 1MΩ resistors. Also this device is only able to sink/source its rated output current when the output is well over +1.5V. On page 6:
enter image description here

As you can see, every element performs a very specific role, and it's not trivial to simplify the biasing without compromising behaviour in some way. That's not to say it's impossible. There are options when using a MEMS.

From what I've read, MEMS microphones do not need biasing, and can be powered directly from the supply rails. Their output is obviously centered around some potential between those extremes. For example, the IM68A130V01 is biased to have a DC output of 1.3V.

This means that the two 2kΩ resistors and the 100nF cap are no longer needed. How this affects power supply noise appearing at the output I don't know. The datasheets are essential to find out this kind of thing.

Lets's say you have an analogue MEMS device with output 2V DC average. If the maximum expected amplitude of its output is 100mV, then it swings between +2.1V and +1.9V. Tthe amplifier's output would be 11 times greater in amplitude, but with an input of 2V average DC, it would be perfectly OK to connect this signal directly to the amplifier:

schematic

simulate this circuit

Just remember that you have to keep OUT well within the power supply potentials. This direct connection is therefore only possible if the MEMS output is centered somewhere near the middle of the supplies. If it's too close to either supply, you'll get clipping, and the MAX4466 has some weird dynamic output resistance when its output is below 1.5V or so.

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    \$\begingroup\$ The PSRR is just for the op-amp itself. Any supply-based ripple coming in via the input will just be a signal which is amplified. \$\endgroup\$
    – Justme
    Mar 21 at 6:22
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    \$\begingroup\$ Indeed, the RC will help filter noise coupling from the supply into the mic signal path, but it doesn't have anything to do with the PSRR of the op-amp itself. \$\endgroup\$
    – ErnestoG
    Mar 21 at 6:26
  • \$\begingroup\$ I wondered why not use a cap of 10µF, and deal with the whole audio spectrum of power supply noise. I figured that the only reason for 100nF is just to deal with the high frequency power noise, that the MAX4466 is rubbish at. I'm not saying it affects the amp's PSSR, I am saying that it avoids compounding that poor performance by injecting the same noise at the input. The entire system's PSSR is improved, not the amp's. \$\endgroup\$
    – Simon Fitch
    Mar 21 at 6:38
  • \$\begingroup\$ @SimonFitch I agree with your conclusion, but I'd still argue that one must add the 100nF decoupling irrespective of the op-amp's PSRR performance. If the op-amp had an excellent high frequency PSRR, it would suck tremendously for it to be ruined by a poorly filtered signal path bias network, wouldn't it? In the end, the system's PSRR would be as good as the component with worst PSRR. \$\endgroup\$
    – ErnestoG
    Mar 24 at 7:03
  • \$\begingroup\$ @ErnestoG why did the designers choose 100nF, and not 10µF? I've assumed until now that they intended to filter above 1.6kHz only, and I can't think of any other reason for that particular cut-off frequency other than what I said. Why 1600Hz if not for the reason I suggested? \$\endgroup\$
    – Simon Fitch
    Mar 24 at 9:23
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The 1MOhm resistive divider is to provide the input of the opamp a path for DC bias currents to flow. Remember that there are transistors in that input which require currents to flow for the circuit to function properly. Without a DC path to ground, when that 0.01uF DC-block capacitor charges up, it will block those bias currents from flowing to or from ground and the op-amp will cease to work properly. Since opamp input bias currents can be very low, it can take several minutes for the cap to fully charge up and thus it can take a few minutes before the circuit stops working.

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