How to make a clean amplified microphone analog to digital conversion?

Question

I've asked a couple of questions relating to this project the past couple of days, but I cant seem to put it all together.

I hooked up an electret mic into an opamp and gave output to my arduino microcontroller. The ADC on the microcontroller converts a range of 0 to 5 vV to a 10-bit number (0 to 1023).

I tried 3 different amp chips:

• LM386 - I got feedback this chip was no good for this purpose, since its not opamp, and it didn't work correctly as expected.
• LM358 - works
• UA741 - works, amplifies more than LM358

I followed this schematic exactly (except for I messed around with resistor values to get good gain): I used 50k ohm for R5 and 10 ohm for R2.

The problem is that the output from the latter two chips is not "clean". The analogRead() on the Arduino is always reading a non-zero value even when I make no noise in the mic. The reading reacts properly when I make noise, but the "zero" value is non-zero. Sometimes the "zero" value even flickers throwing off the reading all the time. Hopefully that made sense.

Can you help me sort this out?

Just as unimportant, extra information: I'm trying to make something like this, eventually.

Answer 1

Get rid of the output capacitor. That circuit was probably meant to produce a signal around zero, so the capacitor is there to block the 1/2 Vdd offset. However, the microcontroller wants to see the signal centered around 1/2 Vdd, so just get rid of the capacitor.

Microcphones do need a lot of gain. Electrets can be sensitive, but you still might need a voltage gain of 1000. The gain in your circuit is the ratio of R5 to R2, but this only works within limits of what the opamp can do.

The values you mentioned above would give you a gain of 5000. That's a lot more than you should try to get from a single opamp stage. Not only will the offset voltage be multiplied by this gain, but the opamp won't be able to provide that over the full frequency range. At 1 MHz gain-bandwidth, you'll only get that gain somewhat below 200 Hz. Even a 1 mV input offset becomes 5 V after amplification by 5000.

R2 is also the impedance seen by the microphone after the input capacitor. You need this to be somewhat larger than the impedance of the microphone with its pullup and the input capacitor at the lowest frequency of interest. 10 Ω is way too small for that. 10 kΩ would be a better value.

Try two stages with a gain of 30 or so for starters and see where that gets you. That's a gain it can handle over reasonable frequencies with enough headroom left for the feedback to work. You also need to capacitively couple the two stages so that the input offset voltage does not accumulate thru all the stages.

Edit: Added circuit

I didn't have time to draw a circuit last night when I wrote the reply above. Here is a circuit that should do it:

This has a voltage gain of roughly 1000, which should be sufficient for a reasonable electret microphone. I may be a little too much, but its easy to add some attenuation.

The topology is rather different from your circuit. The most important single thing to note is that it doesn't try to produce the whole gain in one stage. Each stage has a gain of about 31. That leaves plenty of gain headroom at the maximum audio frequency of 20 kHz for the feedback, so the gain will be nicely predictable and flat accross the audio frequency range since the MCP6022 has a typical gain-bandwidth product of 10 MHz. The limiting factor will most likely be the microphone.

Unlike what I said before, the two stages do not need to be capacitively coupled to prevent the offset voltage accumulating along with the gain. That is because in this circuit, each stage has only a DC gain of 1, so the final offset is only twice the opamp offset. These opamps have only 500 µV offset, so the final offset is only 1 mV due to the opamps. There will be more due to the mismatch of R3 and R4. In any case, the output DC will be plenty close enough to 1/2 the supply to not eat into the A/D range in a meaningful way.

The DC gain of 1 per stage is achieved by capacitively coupling the feedback divider path to ground. The capacitor blocks DC, so each stage is just a unity follower for DC. The full AC gain is realized as the capacitor (C3 in the first stage) impedance gets small compared to the lower divider resistor (R7 in the first stage). This starts to happen at about 16 Hz. One drawback to this approach is that the time constant to settle is C3 times R7+R5, not just R7. This circuit will take a couple of seconds or so to stabalize after being turned on.

Answer 2

As you say, the digital value will be from 0 to 1023. The middle of this range is not 0, it is 512 (which corresponds to a voltage of around 2.5). For silence, you should see something around the middle of the range like this. Doesn't have to be 512 exactly, but it should be close. This is called "DC offset". The signal is shifted upward and centered around 2.5 V.

If you're measuring 2 V and seeing ADC values around 400, then it's basically working fine.

The sound waves go from negative to positive pressure. If the center point were 0, and the signal could only be measured between 0 and 1023, then negative pressure values (-1023) would be cut off.

Also, it will always fluctuate a little due to the noise floor of the ADC. (And there will always be some audio noise in the room no matter how quiet you are.)

Answer 3

Can you post the spec sheet on that microphone? There is no reason you should need a gain of 5000 with a electret mic unless you have a bare unit with no internal FET. If that is the case the preamp needs to look much differently.

Additionally the circuit you used isn't terribly conducive to being used as a pre-amp for an electret mic.

I'd recommend:

R5/R4 sets the gain and can be adjusted without screwing with the input impedance of the circuit. R3 can be from 2k -> 10k ish. 10k will tend to improve distortion performance, if you adjust this too low you should rethink the values for R1 and R2 to fix the input impedance.

Its also very important that the power supply is adequately decoupled as any noise will feed into the microphone.

As the other answers mentioned your "zero" point will be ~512 when you read the ADC and will fluctuate a bit no matter what you do.

If your goal is blinking lights in response to level you shouldn't be taking instantaneous readings with an arduino anyway since i doubt your going to be able to sample fast enough to make it respond well. Instead do peak or average level detection in the analog domain and set the averaging period proportionally to whatever your sampling rate will be.

EDIT: More on doing this with a peak detector

The issue you will have here is that the arduino has a relatively limited sampling rate, I think your maximum is going to be about 10khz which means you can only resolve a 5khz audio signal max. That is with the arduino doing very little except running the ADC, if you need to do any real work (and you do some to get level) the sampling rate will be lower.

Remember your taking discrete samples of the raw signal, just because you have a full range sine wave feeding into the ADC doesn't mean you won't get readings of 0 from the ADC, you'll get samples at various points of the wave. With real music the resulting signal will be quite complex and you'll have samples all over the place.

Now, if all you are trying to measure is the level of the input signal, and don't care about actually getting a digital representation of the signal, then you can use a simple peak detector after this pre-amp to do so.

What this does it turn your audio signal into a voltage that represents its peak level. When you measure this voltage with the ADC you'll have an immediate value representing the level of the signal at the time the reading was taken. You'll still have a bit of wobbling as sound is a complex, always varying waveform, but this should be easy to deal with in software.

A peak detector without hold is really just a rectifier with a filter on the output. In this case we need to deal with low level signals and maintain accuracy so we need to do a little bit more than what would be done for your average rectifier circuit. This family of circuits is called "precision rectifiers".

There are about a billion different ways to do this but I'd go with this circuit, it seems to work best when using a single supply. This would go after the pre-amp circuit already discussed and the input could be AC coupled or not, despite it running from a single supply it will actually work just fine with negative input voltages as long as you don't exceed the available peak-to-peak voltage from the op amps.

OP1 acts as a (nearly) ideal diode which gets around the usual issue of voltage drop across the diode when rectifying. Almost any small signal diode will work for D1, something with a lower forward voltage drop would increase accuracy but I doubt it will matter for your use.

C1 and R4 act as a low pass filter to smooth the output, you can play with their values to match up the performance to what your trying to do (and your sampling rate).

You can probably use the same op amp model your using in the pre-amp but Rail-to-Rail and high slew rate are ideal for this circuit. If you have a stability issue increase R1,R2 and R3 to 100k ohm.