Amplifier for electret microphone

[Bassalisk]

So I posted a question regarding amplifying signals from electret microphone.

I had a problem, in which I had to design such amplifier which uses unipolar power supply.

I was suggested LM386, and I got somewhere. Its very neat little amplifier, but I still need more.

I tried a lot of things. My goal is: low noises, and 4-5 V at the output. So yes very large amplitude.

Currently I am using 9V battery for my power source, so I don't know how much power I can get out of it.

I tried a lot of things.

http://pokit.org/get/b446414024786bd4ba37c3ce805f9784.jpg

I tried this with a little bjt pre-amplifier. It was better but not as good as I wanted it.

I tried cascading these
http://pokit.org/get/1e2ed15bad7f97f328fd12f3fcdff81b.jpg

I get very weird sound at the output. Generally I need help with these amplifiers, because I am shooting things at random, because my knowledge of amplifiers is only theoretical, and as I see its a whole different story when you try to put it together on breadboard. I got 3 LM386, and few LM741 op amps, I got over a 70 BJTs both npn and pnp, help me design this amplifier, because I am currently nowhere.

I may also add that I want this so high because I will be using capacitors at the output, to integrate this signal. So I need certain voltage, that can charge capacitors, ergo 4-5[max amplitude of the signal] V.

 

[AlephZero]

You seem to be missing the fact that an "electret mic" contains not just the microphone but also the first stage of an amplifier, to convert the impedance to a reasonably low level.

Your circut needs a resistor and capacitor to complete the "internal amplifier" circuit, to get a useful signal from the mic. See

http://en.wikipedia.org/wiki/Electret_microphone
http://www.zen22142.zen.co.uk/Prac/ecm.htm

If you connect the output from the capacitor in those circuits to Vin of your amp circuit, things should work better.

There's a complete circuit here. It's not quite the same as yours (which doesn't matter), but you should be able to see what's going on.
http://www.zen22142.zen.co.uk/Circuits/Audio/lf071_mic.htm

 

[jim hardy]

386 datasheet shows enough output swing for your application provided cap isn't just huge
16 0hms it'll drive to about 7Vp-p wth 9 volt supply. See rightmost chart in first row page 5of11 in PDF, page 4 of National's datasheet

http://www.national.com/mpf/LM/LM386.html#Overview
click datasheet PDF icon

I hope Aleph's post helps you bias the microphone

as best i recall the input to 386 is deceptively simple to bias - just don't feed ANY dc to either input, but provide a path for dc FROM each input to common.. it pumps out a small bias current from each input. See how last application circuit in datasheet, AM radio amp, uses a capacitor to block DC from detector. Your electret will need same treatment.

If you need some easy gain look at LM324 single supply opamp. Not very fast but cheap and easy to use.

old jim

 

[Bassalisk]

You seem to be missing the fact that an "electret mic" contains not just the microphone but also the first stage of an amplifier, to convert the impedance to a reasonably low level.

Your circut needs a resistor and capacitor to complete the "internal amplifier" circuit, to get a useful signal from the mic. See

http://en.wikipedia.org/wiki/Electret_microphone
http://www.zen22142.zen.co.uk/Prac/ecm.htm

If you connect the output from the capacitor in those circuits to Vin of your amp circuit, things should work better.

There's a complete circuit here. It's not quite the same as yours (which doesn't matter), but you should be able to see what's going on.
http://www.zen22142.zen.co.uk/Circuits/Audio/lf071_mic.htm

I did power electret mic like that. I am using 10 nF as the output from the microphone, and I am powering it through 10k resistor.

I heard that shunt resistors are needed or something, I really don't know what shunt resistors are for...

 

[Bassalisk]

386 datasheet shows enough output swing for your application provided cap isn't just huge
16 0hms it'll drive to about 7Vp-p wth 9 volt supply. See rightmost chart in first row page 5of11 in PDF, page 4 of National's datasheet

http://www.national.com/mpf/LM/LM386.html#Overview
click datasheet PDF icon

I hope Aleph's post helps you bias the microphone

as best i recall the input to 386 is deceptively simple to bias - just don't feed ANY dc to either input, but provide a path for dc FROM each input to common.. it pumps out a small bias current from each input. See how last application circuit in datasheet, AM radio amp, uses a capacitor to block DC from detector. Your electret will need same treatment.

If you need some easy gain look at LM324 single supply opamp. Not very fast but cheap and easy to use.

old jim

I will try that.

I don't know why I am getting this. It worked fine when I used +9 and -9 V supply with the LM741, two stage amplifiers. I don't know why I am getting so little gain...I should add that I am currently getting, with this setup, some 0.5 V max.

 

[jim hardy]

Wow I'm sorry you're having that trouble.

How much gain are you getting?

Looks like there's nothing magic about the electret
here's a hobbyist preamp,, see schematic
http://www.sparkfun.com/products/9964

The TI datasheet they link is interesting see fig 7 page 11.

I am using 10 nF as the output from the microphone,...
..
I don't know why I am getting so little gain...

hmm note schematic of LM386 internals.
Input resistance is ~ 50K Ω

10nf is 50K at 318 hz
try a little bigger DC blocking cap?

 

[Bassalisk]

Wow I'm sorry you're having that trouble.

How much gain are you getting?

Looks like there's nothing magic about the electret
here's a hobbyist preamp,, see schematic
http://www.sparkfun.com/products/9964

The TI datasheet they link is interesting see fig 7 page 11.




hmm note schematic of LM386 internals.
Input resistance is ~ 50K Ω

10nf is 50K at 318 hz
try a little bigger DC blocking cap?

Yes about that, what determines the size of cap? What am I getting by using bigger or smaller capacitors?

 

[jim hardy]

I should add that I am currently getting, with this setup, some 0.5 V max

Is that unloaded?

.5/200 = 2.5 mv in...

 

[Bassalisk]

Is that unloaded?

.5/200 = 2.5 mv in...

My AC voltmeter, when I am blowing into microphone is giving me this, briefly.

 

[jim hardy]

Yes about that, what determines the size of cap? What am I getting by using bigger or smaller capacitors?

When cap impedance equals amplifier's input impedance , it drops the input voltage.

So that makes your amp a high-pass at 1/ωC = Rin which is ~ 50kΩ at 318 hz
Bigger cap lowers corner frequency

 

[Bassalisk]

When cap impedance equals amplifier's input impedance , it drops the input voltage.

So that makes your amp a high-pass at 1/ωC = Rin which is ~ 50kΩ at 318 hz
Bigger cap lowers corner frequency

I will try that tonight. Currently I am trying to clean the mess I made over last 2 weeks with all electronic projects I had. I found that underneath all this I had a desk :D

 

[jim hardy]

I have a garage full of projects, am no stranger to clutter..

I'm feeling a little embarassed - maybe 200 just isn't enough gain for your electret mike.

 

[Bassalisk]

I have a garage full of projects, am no stranger to clutter..

I'm feeling a little embarassed - maybe 200 just isn't enough gain for your electret mike.

When I used dual op amp, the gain that was ok was about 290.

First one was 4,9 and the second was 59,8.

But the thing is, I think in that setup I amplified all frequencies equally, because of dual supply.

But with single supply I have less "window" of usable frequencies.

Again, this was the setup that worked:

http://pokit.org/get/641fcaca23c811ed20ebb0c16c0866c1.jpg

 

[jim hardy]

Ahh i see..

You want to capture the peaks?

LM386 has output centered at mid supply, so in quiet would try to hold 4.5 volts at pin 5..

I think you'll have to capacitively couple into your holding caps.

draw this into that "gain of 200" LM386 circuit:

1. Replace speaker with a diode, preferably Schottky for low voltage drop, cathode up to 250uf's - side. Call it D1
2. Add a second diode(also Schottky if you can) to junction of 250uf- and D1. Call it D2.
D2's anode goes to junction of 250uf-/D1cathode, and D2's cathode hangs out to right. We'll connect it in a minute...
3. Connect your holding capacitors to D2's cathode.

Now is it a charge pump transferring peaks into your holding caps? or did i type before thinking again?.

Something doesn't seem quite right yet, but an experiment would say if on right track.
Probably adjust value on 250 uf for better setling time.

This one is interesting !

old jim

 

[Bassalisk]

Ahh i see..

You want to capture the peaks?

LM386 has output centered at mid supply, so in quiet would try to hold 4.5 volts at pin 5..

I think you'll have to capacitively couple into your holding caps.

draw this into that "gain of 200" LM386 circuit:

1. Replace speaker with a diode, preferably Schottky for low voltage drop, cathode up to 250uf's - side. Call it D1
2. Add a second diode(also Schottky if you can) to junction of 250uf- and D1. Call it D2.
D2's anode goes to junction of 250uf-/D1cathode, and D2's cathode hangs out to right. We'll connect it in a minute...
3. Connect your holding capacitors to D2's cathode.

Now is it a charge pump transferring peaks into your holding caps? or did i type before thinking again?.

Something doesn't seem quite right yet, but an experiment would say if on right track.
Probably adjust value on 250 uf for better setling time.

This one is interesting !

old jim

I am aiming for this:

When people are talking in background, I want that noise to charge up my capacitors, so that that voltage across those capacitor can trigger something, or in other words, I want to integrate that noise.

So I don't want my trigger to go off on an amplitude, like a delta(somebody screams), but rather continuous noise.

There is a Schmitt's trigger after this circuit, that is comparing this voltage across capacitors, with some reference voltage.


I don't have Schottky diodes, but I do have small signal 1N4148 ones, I think they will have to do. I will try this and come back with results.

Thank you mr. Hardy!

 

[jim hardy]

Great !

1N4148's should work okay i'd think...

Another thing i don't know about is effect of that big capacitive load on amplifier.
But without a 'scope I'm not sure how to check for it.
A quick experiment would be to place a small speaker in series with the 250uf, say between amp's output pin and + side of 250uf cap.. see if circuit behaves more like the 741's did. Also that little extra impedance would slow the charge time giving you an averaging effect that clips short duration peaks.
If capacitive load turns out to be problematic, might you get away with 1uf integrating cap and 1 meg potentiometer, and 1 or 2uf coupling cap instead of 250??

old jim

 

[Bassalisk]

Great !

1N4148's should work okay i'd think...

Another thing i don't know about is effect of that big capacitive load on amplifier.
But without a 'scope I'm not sure how to check for it.
A quick experiment would be to place a small speaker in series with the 250uf, say between amp's output pin and + side of 250uf cap.. see if circuit behaves more like the 741's did. Also that little extra impedance would slow the charge time giving you an averaging effect that clips short duration peaks.
If capacitive load turns out to be problematic, might you get away with 1uf integrating cap and 1 meg potentiometer, and 1 or 2uf coupling cap instead of 250??

old jim

Ok I am trying those things right now, will come back.

 

[Bassalisk]

Ahh i see..

You want to capture the peaks?

LM386 has output centered at mid supply, so in quiet would try to hold 4.5 volts at pin 5..

I think you'll have to capacitively couple into your holding caps.

draw this into that "gain of 200" LM386 circuit:

1. Replace speaker with a diode, preferably Schottky for low voltage drop, cathode up to 250uf's - side. Call it D1
2. Add a second diode(also Schottky if you can) to junction of 250uf- and D1. Call it D2.
D2's anode goes to junction of 250uf-/D1cathode, and D2's cathode hangs out to right. We'll connect it in a minute...
3. Connect your holding capacitors to D2's cathode.

Now is it a charge pump transferring peaks into your holding caps? or did i type before thinking again?.

Something doesn't seem quite right yet, but an experiment would say if on right track.
Probably adjust value on 250 uf for better setling time.

This one is interesting !

old jim

Btw. what do I do with D1 anode?

 

[jim hardy]

It goes to common - sorry..

 

[Bassalisk]

It goes to common - sorry..

You are a wizard. That trick actually worked.

I am detecting the level of sound by LED driver bar. I just need to work out the numbers with regarding resistors, to achieve desired trigger level!

I will keep you posted.

Can you explain what you did?
:D :D

 

[Bassalisk]

When I understand what you did with diodes, I have to tweak this a bit get slower charge/discharge rate at capacitor.

When I put large capacitor, it goes over 12 V and that is not desriable, because my voltage has to be in 0<V<12[0<V<10 desirable]

 

[jim hardy]

Sorry, Bassalisk, been away all day. Went to a gunshow with some old retired friends.

Glad it worked ! or appears to anyway..

I suggest print this and sit down with your schematic. Read one line at a time and work in your head.

For simplicity let's take the diodes and capacitors one at a time.
Further allow me to use, for explaining this, ideal diodes with zero forward voltage drop.

On left side of your 250uf is an AC voltage centered at Vbatt/2.
Let me call Vbatt/2 five volts for simplicity, probably you're closer to 4.5

And let's start out assuming two volt p-p signal out of LM386, so voltage on left side of 250uf swings between +4 and +6 volts.

What's happening on right side of 250uf?
D1 prevents that node from going negative but allows it to go positive.
So when left side drops to +4 volts, right side is held at zero by D1.
The 250uf now has 4 volts across himself , because he acquired some charge when D1 conducted.
So, were there no D2 and no integrating capacitor, as signal swings from +4 volts to +6, the right side of 250uf and D1's cathode would swing from 0 to +2 volts.
That's the same 2 volt swing as on other side of capacitor(4 to 6). Note that's 4 volts DC across Mr 250uf..

However -
Since there IS a D2 and integrating capacitor, something more happens:
When node formed by {right side of 250uf and D1cathode and D2anode} tries to go positive, current flows through D2 into integrating capacitor.
That same current must also flow through Mr 250uf because it can't get out via D1, he's reverse biased..
So our 2 volts of signal change divides between the 250 and the integrating capacitor. If the capacitors are equal the voltage will divide equally. Let us pretend for a moment they ARE equal - it simplifies our mental gymnastic.
Two volts divided between integrating cap and Mr 250uf is one volt each.

That'd place one volt across integrating capacitor, and one volt more across 250 uf.
250's left side is at 6 volts, right side at 1 volt instead of 2; 5 volts across Mr250uf.

If signal again returns to +5, right side of Mr 250uf goes to zero
Integrating capacitor is left at +1 volt and D2 traps it there

On next downswing of AC cycle, 250uf is again forced to +4V on left and zero on right.
D2 has trapped the 1 volt on integrating capacitor.
When AC cycle swings up again,
left side of 250 goes to 6 volts, as before
right side of 250 starts toward +2 volts as before but this time no current flows through D2 until voltage exceeds the +1 across integrating capacitor.
So this time we transfer only half as much charge, adding only 1/2 volt to integrating capacitor.
So we have across integrating capacitor 1 volt from last cycle and 1/2 volt more from this cycle...
D2 traps that 1.5 volts there.
And Mr 250uf was left with 4.5 volts across himself, +6 on left and +1.5 on right.

Next cycle we get to 1.75 volts
and asymptotically approach 2 volts.
That's because as integrating capacitor accumulates charge it becomes increasingly difficult to pump in more charge. From Mr 250's point of view it looks like we're approaching open circuit behind D2..

Voltage on integrating cap won't get quite to 2 volts because of resistor across integrating cap- when it bleeds charge at same rate we add charge, voltage will stabilize there.

Now complicate this by adjusting for the forward voltage of diodes, and set capacitors unequal, and have varying input signal...

This might not be a true "charge pump" circuit but it's the basic principle.
It's analogous to filling a leaky bucket(integrating cap & bleed R) with a ladle(Mr250).

I hope that puts it across.
We could figure out the relevant equations I'm sure but they'd have to be tweaked for diode drops and nature of sounds in your room.

Relative sizes of your ladle and bucket will adjust your response to transients.
Myself I'd model it in basic until had a good "feel" for it, then experiment with a real circuit and a handful of capacitors in a real room.Keep us posted - Sounds like an interesting project !
And accept my apology for this low-tech approach in explaining.
Please throw back at me anything that's unclear, or plain wrong.
Exaggeration and arm-waving is part of good communication.

old jim

 

[jim hardy]

When I put large capacitor, it goes over 12 V and that is not desriable, because my voltage has to be in 0<V<12[0<V<10 desirable]

From a 9 volt battery? I hope not, if so I've missed something (and that's not unusual for me) ..

that might be a symptom of LM386 not happy with large capacitive load - try a few ohms, just five or ten, between pin 5 and Mr250.

Yungman - have you ever used 386 this way?

 

[Bassalisk]

From a 9 volt battery? I hope not, if so I've missed something (and that's not unusual for me) ..

that might be a symptom of LM386 not happy with large capacitive load - try a few ohms, just five or ten, between pin 5 and Mr250.

Yungman - have you ever used 386 this way?

To add, this was an AC signal, was strange to me too...
reading the big part, be right back.

I think this term is called an overvoltage. We learned something like this at class..

 

[Bassalisk]

Ok I read it.

And I actually have to thank you for the type of language you use, because it is much more easy to understand.
I understood everything, at least I think I did. I do have problems with AC vs DC analysis. Those are still intersected in my head, so analysis like the one you did helps a lot regarding developing an engineering way of thinking. For that, I am very thankful.

First of all, I have to tell you that you are absolutely right about the pumping charge.

http://pokit.org/get/9c5f7eaab733085e4d49432526d25ac9.jpg

This is the circuit

http://pokit.org/get/a3543599eff05adec2b11808527511be.jpg

And this is what I am talking about.[Sorry for the mess]

See those 2 LEDs that are on?

Those were on even though my room was quiet, which means, there is some residual charge left in that capacitor, which means circuit works, that diode is locking that charge, and that 47k is discharging, but only until some point so something is left over.

http://pokit.org/get/f603a2bc64828e778f20a85121857d94.jpg
This is actual circuit.

So I will have to modify this to get following:

I want that capacitor to charge slowly, and then discharge slowly. So continious talking in the room (loud talking) would slowly charge our capacitor, and that would be visible with that LED driver. Discharging slowly is optional, better said, it doesn't matter.

After it reaches a certain level, this signal will trigger a monostable, but this working nicely, I need to figure out how to charge that capacitor slowly.

Putting a resistor in series with integrating capacitor?

This is working with 10 nF.

If put like 47 uF, I get voltages up to 13 V at that integrating capacitor [AC voltmeter]

Yes he is charging slowly, but he doesn't know when to stop :D
10 nF is currently best regarding maximum voltage. But not as good regarding the rate of charging.

one more thing though, I don't want that residual charge, how can we get rid of it?Funny thing happened actually :D

When I first assembled the circuit you suggested, I noticed that all LEDs were on. I was very confused (didn't know how circuit works), so I used a voltmeter to measure the drop. Then as I put the voltmeter probes in, I noticed that LEDs slowly went off. Then I figured ahaaaaa! something is holding this charge on the integrating capacitor[across which I measured the voltage] and the capacitor discharged through voltmeter! So that is how 47k pot came into the picture :D

 

[jim hardy]

I want that capacitor to charge slowly,

I'd think that were your C1 smaller than your C2 that would slow down the charging of C2 - it would requirte many cycles to move the charge into C2.
As it is, 1 uf C1 can fill 0.01 uf C2 to overflowing on first cycle.

If put like 47 uF, I get voltages up to 13 V at that integrating capacitor [AC voltmeter]

Yes he is charging slowly, but he doesn't know when to stop :D
10 nF is currently best regarding maximum voltage. But not as good regarding the rate of charging.

AC voltmeter ?? What does that meter report when you connect it across the battery?
Some meters will report DC as higher AC, some report it as zero...
Let's be sure we know our test equipment's individual quirks.

If it's truly AC then i'd say the amplifier breaks into oscillation with that large capacitive load.
That's why i suggested LM386, it's made for driving a speaker so has fairly robust current output capability.
Two possible fixes for oscillation::
1: Smaller capacitors - charge rate will depend on ratio of C1::C2, a smaller C1 charges C2 slower(less coulombs transferred per cycle), and helps reduce capacitive load on LM386.
Try C1 = 1/10th C2, then 1/100th
2: Some impedance between amplifier output pin and C1. Since 386 can drive a speaker, connect a small one between pin 5 and C1. Or a resistor in 5 to 10 ohm range.

A speaker might be good for experimenting, if it whistles you know it's oscillating. (Just don't point it at the microphone)

Bassalisk I'm enjoying your experiment. You're in waters I've never sailed before !
If i learn something every day, might know a little someday.

Hang in there !

 

[Bassalisk]

AC voltmeter ?? What does that meter report when you connect it across the battery?
Some meters will report DC as higher AC, some report it as zero...
Let's be sure we know our test equipment's individual quirks.

It gives me 20 V. [on a 9V battery] Now there is something you don't see every day.

I tested it on mains and I get 230 V like it should give.


Ok I will try with different capacitors. I will get back with results.

And if you never sailed these waters before, imagine how I feel :D

 

[Bassalisk]

hmmm.

I tried smaller capacitor. It behaved as I expected, it charges til one point but won't discharge through that pot. Simply pot is too big. [Only when I lower the pot I can discharge it]

Tried the speaker part. No it doesn't oscillate, I can hear myself barely, with a lot of noise in background, but the noise doesn't bother me.

I am not outputting it on a speaker anyway.

I will try different setup with pot.

 

[Bassalisk]

Well somehow, I made the oscillation go away. Or at least I am not getting 12 V AC any more. I fiddled with capacitor values, but I think my voltmeter providing a shunt resistance for that capacitor. I am going to put a megaohm resistor in parallel with integrating capacitor, to see what happens.

And my AC voltmeter is back to normal with his feelings regarding the voltages across capacitors.

 

[jim hardy]

It gives me 20 V. [on a 9V battery] Now there is something you don't see every day.

Okay so the 13V was probably more like 6.
It's the type meter that half-wave rectifies AC, then reports average multiplied by a scale factor. That's okay so long as we are aware of it.
DC multipled by that scale factor is more than the DC was. If you reverse the leads it'll probably read zero not peg downscale.
One must know his test equipment - just like people what it says isn't always what it meant..

Only when I lower the pot I can discharge it

That sounds like either an open spot in potentiometer near the end, or there's a small signal making us charge the capacitor.
Check pot with ohm-meter over full range of adjustment.

You are getting close i believe.

What are values of C1 and C2 now?

I may have not made this clear - opamps generally dislike large capacitive load because they can't deilver enough current to change the voltage across capacitor quickly (i = dv/dt)
so they can't make their feedback and burst into oscillation.
So we picked a speaker driver opamp with fairly hgh current drive.
That's why i am nudging for a smaller C1 & C2, with maybe a very few ohms in series with amp outout pin. 1uf C2, 1 meg pot, and 0.1uf C1 is what my intiution says would keep LM386happy in audio range. Give pin 5 wire a few turns around a ferrite donut (from junk PC power supply) perhaps.

old jim

 

[Bassalisk]

Okay so the 13V was probably more like 6.
It's the type meter that half-wave rectifies AC, then reports average multiplied by a scale factor. That's okay so long as we are aware of it.
DC multipled by that scale factor is more than the DC was. If you reverse the leads it'll probably read zero not peg downscale.
One must know his test equipment - just like people what it says isn't always what it meant..
That sounds like either an open spot in potentiometer near the end, or there's a small signal making us charge the capacitor.
Check pot with ohm-meter over full range of adjustment.

You are getting close i believe.

What are values of C1 and C2 now?

I may have not made this clear - opamps generally dislike large capacitive load because they can't deilver enough current to change the voltage across capacitor quickly (i = dv/dt)
so they can't make their feedback and burst into oscillation.
So we picked a speaker driver opamp with fairly hgh current drive.
That's why i am nudging for a smaller C1 & C2, with maybe a very few ohms in series with amp outout pin. 1uf C2, 1 meg pot, and 0.1uf C1 is what my intiution says would keep LM386happy in audio range. Give pin 5 wire a few turns around a ferrite donut (from junk PC power supply) perhaps.

old jim

ferrite donut

you mean, this?

http://pokit.org/get/9bb10dafa57c717930fa3f66b378c293.jpg

This is from an old motherboard.

This in an inductor right? If it is, this is my first use of inductors :D:DI will try the capacitors and get back to you, I got some trimmers laying around, so I think they will do.

 

[jim hardy]

That's a ferrite donut okay !
You could use a smaller one,
but objective is just to provide some impedance for high frequency between amplifier and those caps... that one would do and it's handy (if that's a junk supply - don't wreck a good one).

 

[jim hardy]

Getting on to bedtime here - see you tomorrow

Have fun - I'm learning with you on this one.
I hope Yungman is watching, and will point out any errors of thinking.

old jim

 

[jim hardy]

Dont give up !

 

[Bassalisk]

Dont give up !

Im not giving up :D I just had tons of math on my back. This a project for my soul, but my faculty still wants my soul too :D

I've been deriving some equations for entropy and all, I will get back on the project tomorrow probably.

Don't worry, I will keep you posted! I am probably 2h from finishing it, thanks to you of course.

[for some strange reason, didn't get the notice for your last post sorry :( ]