Need a simple low power pre-amp circuit for electret microphone

[scream_er]

Hey guys,

I need a simple low power, high gain pre-amplifier circuit for an electret microphone. I have come across two pre-amplifiers, which are attached. Which one would most idle considering its simple, high gain and low power?

CIRCUIT 1


CIRCUIT 2

 

[cks3976]

Circuit 1 in my opinion. Gain is much better in that circuit compared to Circuit.2

 

[scream_er]

From the first circuit how can I get an output of 6V? What should I change to achieve this?

 

[Pjdd]

Circuit 1 has a gain of 100 and circuit 2 has a gain of 11. But you cannot get an output of 6V with either one. This is because the supply voltage sets a limit on the output voltage level.

Even if you use a 9V supply in Circuit 1, the theroretical maximum voltage swing at the output is ±4.5V peak-to-peak which is about 3.18V rms for a sine wave. This is with an ideal opamp. In practice, the undistorted output will be much less due to imperfections in a real opamp. With an LM1458, we can expect no more than 2.5V rms at most.

---------- Post added at 18:32 ---------- Previous post was at 16:59 ----------

To get a 6V output, one of the following three techniques has to be used -

1) A higher power supply voltage - at least 18V
2) A step-up transformer at the output
3) A bridge circuit

---------- Post added at 18:47 ---------- Previous post was at 18:32 ----------

Scream_er, I've just seen the thread about your "baby crying detection circuit". Is the mic pre-amp intended to be a part of that project? If it is, it would be better if you had started a single thread about the whole project OR at least explained the purpose of the pre-amp even if you felt that it would be best to create separate threads. That way, some of us might be able to offer a better overall solution.

 

[scream_er]

I have few questions which I need answers.
1) Voltage from electret microphone is AC, right?
2) I have a speaker with 8 ohms, 0.5 W and using P=(V^2)/R, I found that the voltage is 2V's. Is it the output voltage from the speaker or the voltage needed for the speaker to function?
3) Op-amp produce DC voltage, right?
4) If the electret microphone produce AC and this is the input to the Op-Amp, how do I calculate the output from Op-Amp (can I use Vout = [1 + [rf/r1]]Vin) considering a non-inverting amplifier)
5) If my microprocessor has an ADC range of 0 - 5V do I need to amplify the voltage from microphone to exactly to 5V to get the best resolution?

Yes, I want to implement this to detect baby cries. But I am yet to understand how to get an output from electret microphone that is heard over a simple speaker. I'm a novice, and I am trying to learn bit by bit.

Thanks

 

[betwixt]

1. The signal voltage (from sound) is AC but it is superimposed on a DC voltage which is usually about 3V for an electret microphone. The exact DC voltage depends on the microphone type and it's load resistor.
2. Loudspeakers don't produce voltage (when used as intended). Your 2V calculation is AC signal to drive the loudspeaker to full volume.
3. It can be DC or AC or a combination of both. In the designs you show it is half the supply voltage as DC with the signal added to it.
4. Correct although you should add a minus before the calculation because it inverts the signal. Use rf/r1 if it is non-inverting.
5. Correct, to get best resolution you have to use the full range of available results so each step is as small as possible. You might be able to use a lower reference (full scale voltage) instead of increasing the input to it but this also degrades the noise performance.

Don't forget the microphone signal is AC and most ADCs only accept positive voltages. You should consider the use of a rectifier and some filtering before the ADC input.

Brian.

 

[scream_er]

So, how exactly will I know how much to amplify the voltage from the electret microphone? Can you provide me with a simple, low power and high gain circuit to achieve 5V. What do mean by use rf/r1 if non-inverting. Here in this link Non-inverting Operational Amplifier - Non-inverting Op-amp it mention 1+(rf/rin)
Thanks

 

[betwixt]

The first schematic you show is an inverting configuration followed by a second inverting amplifier which turns it the right way up again. In an inverting configuration, the signal is fed at the '-' input of the amplifier, in a non-inverting configuration it is connected to the '+' input. Both basically do the same except signals at the '-' input appear at the output upside down.

To know how much gain ( x amplification) you need to reach 5V you first need to know how much signal you are starting with, this is on the microphone manufacturer's data sheet but don't expect it to give a plain simple figure in volts because it depends on how much sound level is being heard by it. Imagine trying to measure the output in absolute silence, if you saw ANY output voltage you would say somethng was wrong! I'm going to make a guess that with 'average' baby crying at say 2m distance you will get around 25mV RMS of signal. You want it to be a peak of 5V to get best resolution from your ADC. If I assume you will use a half wave rectifier and a filter with a time constant of say 0.1 seconds you need to amplify the signal so it's positive peaks are 5V. So the gain you want is 5/0.025 = 200.

You can do this with a single amplifier stage but you might run into stability problems. When you amplify by a large amount there is a risk of some of the output leaking back into the input and this causes howling and screeching noises. It would be better to use two stages of gain with say x20 for the first stage and x10 for the second one. Your output voltage swing has to be at least 10V so look to an amplifier that works on a higher supply voltage than that, preferably 15V or more. It can be done at lower voltages but the circuit becomes more complicated.

Other than those notes, which I have greatly over-simplified, you can use the schematics shown on the amplifier data sheet or the one in the web link given. Just add a diode in series with the output and two components across the ADC input and ground to act as filters, a 100K resistor and 100nF capacitor in parallel,

Brian.

 

[Pjdd]

2) I have a speaker with 8 ohms, 0.5 W and using P=(V^2)/R, I found that the voltage is 2V's. Is it the output voltage from the speaker or the voltage needed for the speaker to function?

Neither. 0.5W is the maximum power coming from an amplifier that the speaker can stand without damage and/or excessive distortion. It will produce sound even if the incoming voltage is only 0.1V (0.00125W) or even less.

Yes, I want to implement this to detect baby cries. But I am yet to understand how to get an output from electret microphone that is heard over a simple speaker. I'm a novice, and I am trying to learn bit by bit.

General-purpose opamps like the LM1458 are meant to be used mainly as voltage amplifiers. This means that they are normally expected to have a high impedance load, which in turn means that they are not expected to deliver high currents to the load. A speaker is a low impedance device. Therefore an opamp (except a special high-power opamp) is not suitable for driving a speaker directly. We need a power amplifier, even a simple one, for that.

So, how exactly will I know how much to amplify the voltage from the electret microphone? Can you provide me with a simple, low power and high gain circuit to achieve 5V.

(To simplify matters, let's stick to voltage amplifiers for the time being and ignore current and power requirements at the output)
An amplifier amplifies an input by a given factor. For example, an amplifier with a gain of 100 produces an output which is 100 times the input. Therefore, to calculate how much amplification is needed to get an output of 5V, we have to know the input voltage which is the output voltage of the microphone. If the microphone's output is 10mV, we have to amplify it by a factor of 500 to get an output of 5V. If the input is 100mV, we need a gain of only 50.

One problem that cannot be avoided in a practical design is that the microphone's output is not constant. Different microphones have different sensitivities. Even for the same microphone, the output will vary with the intensity of the sound it picks up. That intensity depends not only on how loud the sound source is, but also on how close it is to the microphone.

When your baby cries, the strength (voltage) of the mic's output depends on how loudly it cries, how close it is to the mic, whether it is facing towards the mic or away from it, whether the sound is muffled by clothes, room acoustics, and so on. Therefore, the mic's output and the amplification needed to get 5V cannot be calculated to a specific number.

The output from the microphone may be anything from less than 1mV to 100mV, depending on the conditions described.

What can be done, and what is done in designing any audio amplifier, is to estimate a ballpark figure for the output voltage coming from the microphone, design an amplifier with enough amplification to get full power from the softest sound expected, and provide a volume control to manually change the sensitivity. Some audio systems have an automatic volume control (AVC) or automatic gain control (AGC) which compresses the wide volume range to a narrower one.

To sum up, it's impossible to design an amplifier which will produce an output of 5V unless the input is known and constant. What can be done is to estimate the output from the microphone under the circumstances in which it is to be used, and then design an amplifier that will produce an output roughly in the range of 5V.

 

[scream_er]

The voltage from the electret microphone is just few hundred uV. So, I need to amplify the voltage to atleast 1V. I have 3 options, but what do you suggest is the best option. Attached are the chips and circuit considered.

1) Should I use LM358N as or
2) Use dual LM358N for higher gain (since LM358N is dual op-amp) or
3) Use LM386N since it is simpler.

LM 358N

LM 386N

 

[Pjdd]

As I explained in an earlier reply, the single LM358 circuit has a gain of 11. If the mic's output is "a few hundred uV", multiply that with 11 and you get only a few millivolts.

Again as explained earlier, the dual LM358 circuit still gives you a gain of only 100. You need an input of 10mV to get an output of 1V.

The third circuit uses an LM386 power amplifier. It has a gain of 20 in that configuration. You can calculate the output from that using the same principle.
Note: The LM386 can be configured to have a gain of 20, 200 or anything in between.

 

[scream_er]

So, none of the above can amplify the electret microphone to 1V. So, what option do I have? I'm using ATmega328 micro processor which comes with arduino, I found online that people have used both of these chips to amplify voltage from electret microphone. So what can be the solution?

I mentioned 1V because the resolution of the Arduino analog input (using the default 5v analog reference) is around 5mV. So it seems its best if you get 1V but 100mV also seems to be ok.

 

[Pjdd]

It's wrong to say categorically that none of your proposed circuits can amplify an electret mic to 1V. They can. They just need a higher input from the microphone, which means the sound has to be louder or closer to the mic.

Did you read my previous post (#9) thoroughly? It's all explained there.

 

[betwixt]

I think there are several issue to respond to here:

1. you must understand the difference between voltage amplification, power amplification and power output.

Voltage amplification is just that, the number of times by which the input signal is multiplied. A gain of 2 means you get out twice the voltage out than you put in and so on. However, the voltage may not have much 'push' behind it, as soon as you try to use it, the voltage might drop because the load is using up the resource producing it.
Power amplification is a general term meaning the available 'push' is increased so even if the voltage gain is 1 (same out as in) it can bear a heavier load than the input alone could provide.
Power output is a measure of the amount of energy that can be transferred into the load.

To demonstrate with an extreme example, a hearing aid amplifier has high gain but if you tried to hook it up to a stadium speaker system it wouldn't even make a squeak.

2. You alternate between mentioning a loudspeaker load and an ADC load. The two are very different.
A loudspeaker is current driven, the more current you force through the speaker the more the cone moves and the louder the sound becomes. The voltage across it may be very small to make a large current flow.
An ADC is voltage driven, the current it uses is very tiny, maybe one millionth of the power needed to drive a loudspeaker.

Another example, assuming a standard 8 Ohm loudspeaker is used and the voltage available is 5V, the speaker power would be (V^2/R) = 3.125 Watts while 5V at the input of an ADC may only be 0.000025W or even less.

3. The resolution of an ADC is the number of unique steps in result it is capable of resolving. More bits means more reolution because there are more combinations of numbers when you have more bits to count with. You mention 5mV resolution, I'm guessing this is actually a 10-bit ADC which has 1023 steps so each is 5/1023 of a volt = approx 4.89mV. If you put zero volts in the reading out would be 0000000000, if you put 5V in the reading would be 1111111111. If you use less than the 5V range (it isn't always 5V, some devices are different or can use external references) you can't use all the available bits so the range is reduced.

4. An ADC can only measure voltages which are positive with respect to ground and are less than the reference voltage. There is no point in feeding AC directly into an ADC, it certainly wont measure correctly and the negative part of the waveform might even cause damage to the chip. The ADC also 'samples' the input so if you feed it a changing signal the measurement will depend on exactly at which point within the signal you sampled at. Even a steady repetitive waveform may give entirely different measurements depending on which point in the signal cycle you measure. The fix is to use a rectifier and a filter. The rectifier blocks the negative side of the waveform and the filter turns the result into a slower moving positive average which when sampled will give a more consistent reading.

Yet another example: you have no doubt seen audio level meters on consumer equipment or mixing desks. If you just measured the voltage the needle would try to swing back and forth to follow the individual cycles of the sound, it would spend half it's time pegged against the backstop and the rest of the time flying around wildly. What you actually see is a representation of the average sound level, this is what you want your ADC to measure.

Brian.