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Let me start off by saying that I'm a complete beginner. I've watched some videos on amplifiers and read about them but it's hard for me to analyze circuits that don't match exactly to the examples I have seen in the past. Also, a lot of the information just goes way over my head. I have no electrical engineering background but I have experimented and tinkered with some Arduino based projects.

For a project I'm working on, I want to make an audio amplifier circuit that takes an electret microphone's output and amplifies it to 300mVpp from a single source power supply. I found a schematic from TI that looks like it could do the job, with some modifications. Here is a link to the place I found it: TIPD181. I built a prototype PCB based on the schematic below and I am very happy with the sound quality. I've built other amplifiers and they all hiss or crackle (completely due to my incompetence). It does sound hollow (not sure if that's the right word for it) or maybe lacking bass? That could be due to the cheap speakers that I was using but regardless, the final use of this will not be to drive a speaker. I am hoping to create a visual spectrum analyzer. I will be taking the output of this circuit and putting it into a MSGEQ7(Datasheet). I know that there are prebuilt microphone breakout boards that have AGC and adjustable gains like the MAX9814 but I'm enjoying the learning process of building microphone circuit. I'm hoping to eventually cut out the MSGEQ7 and create my own switched capacitor filters but I'm saving that for another time.

TI Schematic

Here is what I'm thinking about when trying to adapt this schematic to my needs and this is where I'm just not getting it but maybe you can help. Again, I'm a total beginner. I can take criticism but know that I'm not trying to waste anyone's time so if you think I'm a lost cause then please ignore me.

  1. From the overview on the TI website, it says that this circuit will amplify the mic output to line levels voltages. A quick google search tells me that's .448V to 1.228V. This is higher than the 300mVpp that I require so I think I will need to reduce the gain.
  2. That leads me to having to find out the Vpp output of the microphone without the amplifier to know how much gain I need. Right? Microphone's Datasheet How would I figure that out from the datasheet? I think it has to do with the sensitivity value, right?
  3. Then I need to find the voltage gain. To me, this is in the inverting configuration so the gain would be -R2/R1 which currently is 5.47. Am I correct in this? If I am, do I need to add the microphone's impedance (2.2k Ohms) to R1 for an accurate gain value? After that I can see exactly what this circuit is outputting and adjust R1 / R2 to my needs, right?
  4. What is the purpose of R4 and R6 is this schematic? Is R4 to adjust the output impedance? If so, why? Is R6 for a high pass filter? If so, why?
  5. When I measure the output of this circuit on my oscilloscope, the center of the signal starts high (I think Vcc/2 or what I put into the non inverting input. I can check this later if it's crucial for this analysis) and then slowly drops to 0V over time. Is that normal operation?

Any other suggestions or criticisms are welcome. If I'm missing something, please let me know.

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  • \$\begingroup\$ What level and range of sounds, at the microphone transducer, do you expect to experience? \$\endgroup\$ Commented Jun 22 at 4:09
  • \$\begingroup\$ (When you let "the system" number your lists as per post editor help, you don't end up with two questions labelled 3.). Downside: limited choice of styling.) \$\endgroup\$
    – greybeard
    Commented Jun 22 at 5:49
  • \$\begingroup\$ Thanks for the replies. I can see why auto number would be much better. Thanks for editing it to correct that issue. I'm not very knowledgeable about what you are asking @periblepsis so please don't take this the wrong way if I'm not answering your question correctly. I would want the microphone to be able to pick up the frequency range of 63Hz to 16kHz. As to the level, I'm slightly confused. Does that mean the voltage level? If that's the case, I am unsure. I would like to be able to calculate what to expect so I can adjust the amplifier accordingly. Let me know if I'm way off. \$\endgroup\$
    – Ivory
    Commented Jun 22 at 14:27
  • \$\begingroup\$ @Ivory I'm just asking about the sound pressure levels you expect to handle. These are density variation waves in air. These strike the microphone. It converts them to electronic signals. You should read up on tables that exist discussing the various kinds of loud sounds and what kind of sound pressure levels they generate. Knowing this range, and knowing your microphone details, the electric signals you need to handle can be worked out. But it is a process. \$\endgroup\$ Commented Jun 23 at 12:12
  • \$\begingroup\$ ohhhhh I see what you mean now. I think I seen some videos about this. Would this be dB SPL? I am planning to use this microphone inside my home to measure peaks of music playing. I'm thinking it would be in the 50-70 dB range \$\endgroup\$
    – Ivory
    Commented Jun 23 at 12:56

1 Answer 1

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[...]gain would be -R2/R1 which currently is 5.47. Am I correct in this? If I am, do I need to add the microphone's impedance (2.2k Ohms) to R1 for an accurate gain value?

If microphone impedance is 2.2k, it gets combined with R1 like this: \${{2.2k \times 13.7k}\over{2.2k + 13.7k}}=1.896k\$
Then voltage gain is \$-{{R_2}\over{1.896k}}\$
If you need more gain, increase value of R2.

What is the purpose of R4 and R6 is this schematic?

R4 keeps the amplifier stable, and protects it from short-circuit loads.
R6 is a "pop" protector. You may notice that on power-up it will still "pop" as C5 discharges. But if you power up, then somewhat later connect to your spectrum analyzer, the connection should be pop-free, because R6 has discharged C5.

4.) When I measure the output of this circuit on my oscilloscope, the center of the signal starts high (I think Vcc/2 or what I put into the non inverting input. I can check this later if it's crucial for this analysis) and then slowly drops to 0V over time. Is that normal operation?

You're seeing C5 discharge - that's normal. It is standard to remove DC offsets from amplifier outputs. That's basically what C5, R6 does. In this case, the amplifier is biased at one-half Vcc so that audio can swing more positive and more negative. We call this a DC-offset. DC-offset is not wanted by the MSGEQ7.


Your experiment with a speaker load sounded "tinny". Yes, this can be caused by an unboxed speaker. But an audiophile (boxed) speaker would still sound tinny because C5 is too small to properly drive an 8-ohm loudspeaker. A 100uf capacitor would help restore bass.
But C5=2.2uf should be just fine for driving the high-impedance MSGEQ7 without losing bass. The impedance of MSGEQ7 is 1M ohm, much higher than 8 ohms.

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    \$\begingroup\$ R1 does not contribute to the signal gain. \$\endgroup\$
    – John Doty
    Commented Jun 21 at 22:33
  • \$\begingroup\$ Thanks for the answer! @glen_geek It clears up a lot. I do have a follow up on the connection from the amplifier to the MSGEQ7. This is probably a very basic and simple question but I'm using 9V for the amplifier circuit but the MSGEQ7 operates at 5.5V max. Can the output of the amplifier go directly into the MSGEQ7 without damaging the MSGEQ7? I would think it can because the amplifier should only be outputting 1.2V max? Now that I know the gain calculation, how do I find the microhpone's base output voltage so I can multiply it by the gain to get to 300mVpp? \$\endgroup\$
    – Ivory
    Commented Jun 22 at 14:36
  • \$\begingroup\$ @JohnDoty Does that mean I should only use the microphone's impedance in the gain calculation? \$\endgroup\$
    – Ivory
    Commented Jun 22 at 14:39
  • \$\begingroup\$ @Ivory Yes. In the inverting configuration, impedances to other sources (VCC in this case) do not affect the gain. \$\endgroup\$
    – John Doty
    Commented Jun 22 at 16:36
  • \$\begingroup\$ @JohnDoty Not even the microphone impedance? what would I use as R1 in the gain calculation? \$\endgroup\$
    – Ivory
    Commented Jun 22 at 17:46

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