1
\$\begingroup\$

I'm trying to amplify an electret microphone signal with an inverting op amp (LM358). I don't have access to an oscilloscope at the moment but with a voltmeter I read around 12mV at V1 and around 12mV as well at V2. For testing I am just playing a 1kHz test tone by the mic. Does this mean my op amp is not actually amplifying the signal. If my gain is -100 currently, then I should be getting 1.2V at the output of the op amp correct?

On a side note, if the microphone is outputting a varying voltage centered at 0, shouldn't the voltmeter give me 0V? What is measured by the voltmeter, is it peak to peak? Average? Also I am a first year EE student so please take into account my limited technical knowledge.

Yes the ground rails on the left and right are connected (not shown in pic)

Thank you

enter image description here

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ See breadboardtronics.wordpress.com/2012/08/28/… on biasing. \$\endgroup\$ Nov 23, 2023 at 21:08
  • \$\begingroup\$ You need +/- supplies to have the output go below (about) zero volts. Also make sure you have the polarity of the microphone capsule correct. \$\endgroup\$ Nov 23, 2023 at 21:19
  • \$\begingroup\$ Yes microphone polarity is correct. \$\endgroup\$
    – HiThere
    Nov 23, 2023 at 21:25
  • 1
    \$\begingroup\$ Bias the non-inverting input up to mid-supply using two 10k resistors in series between the voltage rail and 0 v and then add a 10uF capacitor from the mid-supply bias voltage to 0 V. \$\endgroup\$
    – user350400
    Nov 23, 2023 at 21:29

2 Answers 2

1
\$\begingroup\$

Where is the voltage supposed to come from that ever could get the "-" input below the "+" input? You either need a -9V supply for the opamp, or you need to put its "+" pin on a higher operating point, like using a voltage divider (2×4.7k, for example) and a smoothing capacitor (22µF, say) to arrive at 4.5V. Of course, if you do that, you will have this as DC on the output and will want an output capacitor for removing it, too.

\$\endgroup\$
8
  • 2
    \$\begingroup\$ Your first sentence asks a valid question. I suggest that rewording it as a statement of what is required would be more helpful when addressing a beginner. \$\endgroup\$
    – Russell McMahon
    Nov 23, 2023 at 22:05
  • \$\begingroup\$ Thanks for the response. I figured to do something like that. I added a voltage divider (2 10K resistors) and I have half of supply going into the + input. I have a 10uF cap going from the + input to ground. Lastly I added a 0.1uF capacitor between the output of the op amp to V2. The issue is now I am getting 0V at V2 and half of supply voltage at V1. \$\endgroup\$
    – HiThere
    Nov 23, 2023 at 23:49
  • \$\begingroup\$ another thing I notice is that when I test the microphone circuit alone (ie removing everything to the right of V1 and probing V1), I get a voltage that drops (I am assuming something to do with the cap) slowly until it gets to about 2mV where it hovers around there regardless of whether I play a sound or not. Can somebody explain what is going on here I am so lost. \$\endgroup\$
    – HiThere
    Nov 24, 2023 at 0:00
  • \$\begingroup\$ @HiThere "The issue is now I am getting 0V at V2 and half of supply voltage at V1." How is that an issue? That's the correct DC bias at V1 and an output idle voltage of 0V (admittedly, you actually need some load to get the output capacitor the proper DC charge, so you could add a 22k resistor to ground there if you have no other load). That's exactly what you should be seeing. \$\endgroup\$
    – user107063
    Nov 24, 2023 at 0:36
  • \$\begingroup\$ @user107063 Could you see if I am understanding this correct: is V2 zero (I tried with a load resistor it was 0V as expected) because the signal is centered at 0V and the voltmeter is simply reading the average voltage which is 0V? And at V1 it is reading 4.5V but really it is a very small signal from the microphone centered at 4.5V? \$\endgroup\$
    – HiThere
    Nov 24, 2023 at 0:48
0
\$\begingroup\$

The signal generated by the mic and 2.2k bias resistor will have a DC offset that you decouple with the 1uF capacitor. Because of this decoupling, you have an (ideally) pure AC signal, meaning the voltage can go above and below 0V (negative and positive voltage).

You are powering the op amp with a single supply (9V battery), meaning you are only amplifying half your signal, the positive half.

To solve this without using a dual (+9V and -9V) supply, you need to add a DC bias to your signal (setting a reference voltage, Vref) at the non-inverting (+) pin of your op amp, and then use the circuit below. Note: Vref is usually half the supply voltage, so 4.5V if your supply is 9V. This gives you the full amplitude range of the op amp in single supply operation.

In this case, you want to calculate b to be 4.5V while also getting the amplification you want of m = 100. The math is simple and the document linked below shows you how to do the calculations.

enter image description here

enter image description here

The source of the screenshots and more on the topic of single supply operation of op amps can be found here.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.