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Microphone - voltage or current source?

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alphacat

New Member
Hey,

When using a microphone as a small signal input to an amplifier, is it considered to be a voltage source or a current source?

If its a voltage source, then when speaking into it at a certain level, it would always output the same voltage no matter what is the load (as long as it doesnt exceed its power rating).

If its a current source, then when speaking into it at a certain level, it would always output the same current disregarding the load (again, taking power rating into account).

So, what is that?
Thanks.
 

alphacat

New Member
When you want to amplify your voice, you use a mic as an input to an amplifier, right?

Then I considered that mic as a small signal source.

What is wrong with that assumption?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
When you want to amplify your voice, you use a mic as an input to an amplifier, right?

Then I considered that mic as a small signal source.

What is wrong with that assumption?
Nothing - your assumption about current source or voltage source is where you are confused.

It's essentially an alternator - a coil moving in a magnetic field.
 

Hero999

Banned
I suppose it could be considered as an AC voltage source.

The internal impedance will depend on the type of mic. A dynamic will have a low impedance, 600Ω typical, an electret will be equal to the ballast resistor and a piezo will be qutes high about 50k.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
I suppose it could be considered as an AC voltage source.
But not as the OP suggested it.

The internal impedance will depend on the type of mic. A dynamic will have a low impedance, 600Ω typical, an electret will be equal to the ballast resistor and a piezo will be qutes high about 50k.
None quite right really.

Dynamics often are nominally 600 ohm, but the actual impedance is often much lower. Makes no real difference, as all connections are for voltage transfer, not power.

Electrets are the 'ballast' resistor and the output impedance of the FET in parallel, not just the resistor.

Piezo are MUCH higher than 50K, and the higher impedance you can feed it in the better.
 

alphacat

New Member
It's essentially an alternator - a coil moving in a magnetic field.
Thanks for that information Nigel.

A coil moving in a magnetic field generates an electromotive force, which is measured in Volts.
So if the mic turns audio signal into Volts, why isnt it true that it is a voltage source?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thanks for that information Nigel.

A coil moving in a magnetic field generates an electromotive force, which is measured in Volts.
So if the mic turns audio signal into Volts, why isnt it true that it is a voltage source?
You specified a voltage source as providing a constant voltage, and a current source as providing a constant current - neither are true. For that you need a constant voltage source, or a constant current source.

Both current and voltage will vary (together) as you speak - but current shouldn't be a concern anyway, as you shouldn't be drawing current from a microphone.
 

alphacat

New Member
First, I'm sorry for not having here a mic so i could check it out myself.

Please tell me where am i wrong.
Assume that i'm speaking into the mic at the same rate and volume, constantly.
I measure the mic's output with a multimeter and it constantly reads 10mVac.
Now, as I continue to speak into the mic (in the same way as before), I connect a 10MΩ resistor in parallel to the mic.
I measure again the mic's voltage.
Wont it continue to read 10mVac?

And if now I connect a 5MΩ resistor in parallel to the mic, wouldnt the multimeter continue to read 10mVac?
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
First, I'm sorry for not having here a mic so i could check it out myself.

Please tell me where am i wrong.
Assume that i'm speaking into the mic at the same rate and volume, constantly.
'Whistling' is the term you're looking for.

I measure the mic's output with a multimeter and it constantly reads 10mVac.
For a start, don't try and measure it with a multimeter.

Now, as I continue to speak into the mic (in the same way as before), I connect a 10MΩ resistor in parallel to the mic.
I measure again the mic's voltage.
Wont it continue to read 10mVac?

And if now I connect a 5MΩ resistor in parallel to the mic, wouldnt the multimeter continue to read 10mVac?
No to both, it will read slightly less - in these EXTREME cases you're mentioning VERY slightly less - but less all the same.

As you're trying to do voltage transfer, the load impedance needs to be a minimum of five times the source impedance - in your examples with a 600 ohm mike and a 10Meg or 5Meg load, apply ohms law and do the maths - but don't round the result (even at that it may be too far down in the zeros anyway).

But putting essentially no load at all on it, doesn't make it a constant voltage.
 

alphacat

New Member
Thanks a lot Nigel.

I wasnt trying to do a voltage transfer, since I was talking about measuring the voltage on the mic itself, and not on the load as you probably thought i intended.

That is the point i was trying to make.
We apply a constant audio signal on the mic's input.
If the mic is to generate voltage, then no matter what load we connect to it, 10MΩ or 5MΩ, the mic's voltage would remain the same, correct?

But if the mic was to generate current, then connecting different loads to it, would have created different voltage drops on the mic.

Moreover, I see that you also analyzed the mic as a "voltage source" with a source resistance in series.

Its not the normal voltage source that we know, but what i was trying to say is, that i think we should treat the mic as a voltage (not current) input, and therefore connect it to a an amplifier with large Rin, and not to an amplifier with a small Rin, which we would have done if we treated the mic as a current input.
I dont understand whats wrong with such thinking.

Perhaps i should have used the term voltage input instead of voltage source?
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
That is the point i was trying to make.
We apply a constant audio signal on the mic's input.
If the mic is to generate voltage, then no matter what load we connect to it, 10MΩ or 5MΩ, the mic's voltage would remain the same, correct?
But those aren't really loads - they are MUCH too high to be considered loads. But even such high loads will have an effect, consult ohms law, just that it's far too little to be significant.

Look at it this way - you have a 500 gallon tank of water, you remove one teaspoon full of water, does the level in the tank change? - yes it does, but by so little you can't tell.

Now remove 200 gallons of water, does the level change now? - obviously yes, because it's NOT a constant level, just as a mike isn't a constant voltage.

But if the mic was to generate current, then connecting different loads to it, would have created different voltage drops on the mic.
You can't have one without the other, and a mike is just as much a current generator as a voltage one.

Moreover, I see that you also analyzed the mic as a "voltage source" with a source resistance in series.

Its not the normal voltage source that we know, but what i was trying to say is, that i think we should treat the mic as a voltage (not current) input, and therefore connect it to a an amplifier with large Rin, and not to an amplifier with a small Rin, which we would have done if we treated the mic as a current input.
I dont understand whats wrong with such thinking.
You're looking at it in the wrong way, and making it FAR more complicated than it is - it's VERY, VERY simple:

Connect a mike to an input impedance of between five and ten times it's nominal impedance - so for a 600 ohm mike, an impedance of between 3000 and 6000 ohms. Nothing else to be concerned about.
 

alphacat

New Member
Connect a mike to an input impedance of between five and ten times it's nominal impedance - so for a 600 ohm mike, an impedance of between 3000 and 6000 ohms. Nothing else to be concerned about.
Hey,
This is exactly my point.

What you said in this section i quoted is that the source which drives your amplifier (no matter whats the source is, regulator, mic, etc) is one who provides voltage.
Yes i know that voltage goes hand to hand with current, but if this source was to provide current, then wouldn't you say the opposite of what you just said - meaning wouldn't you say to connect that source to an input impedance of between 1/10 to 1/5 times its nominal impedance?

In my studying of amplifiers and their inputs, i divided voltage inputs and current inputs according to these:
Large input impedance needed (related to source impedance) => voltage input (and the other way).
Small input impednace needed (related to source impedance) => current input (and the other way).

Thats why i placed the mic in the voltage input's group.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
In my studying of amplifiers and their inputs, i divided voltage inputs and current inputs according to these:
Large input impedance needed (related to source impedance) => voltage input (and the other way).
Small input impednace needed (related to source impedance) => current input (and the other way).
Not really still - you're talking about voltage transfer OK, but the other is POWER transfer (not current). For power transfer the maximum possible efficiency is only 50%, and to achieve this the load impedance must exactly equal the source impedance. If the load is higher or lower, efficiency drops.
 

alphacat

New Member
I see.
Thanks for sharing this Power issue, i havent learned about it yet, but its always good to know about new matters. :)

Thanks for all of your (huge) help Nigel. :)
 

The Electrician

Active Member
When using a microphone as a small signal input to an amplifier, is it considered to be a voltage source or a current source?
Any source of electrical energy, and a microphone is such a source, although of only a small amount of energy, will have an impedance.

If its impedance is very high, infinite in the extreme ideal case, then it is a current source.

If its impedance is very low, then it's a voltage source.

If its impedance is other than infinite or zero, then it's neither a (perfect) current source nor a voltage source. It's a source with finite impedance.

A dynamic microphone, and a small speaker used as a microphone, will have a fairly low impedance as microphones go.

Ribbon microphones (before the transformer) have a very low impedance.

A condenser (without a preamp) microphone will have a high impedance.
 

alphacat

New Member
Hey,

Isnt the source's type (current or voltage) also depnded on how the impedance is connected to the source?
Whether in parallel or in series?
 

The Electrician

Active Member
Hey,

Isnt the source's type (current or voltage) also depnded on how the impedance is connected to the source?
Whether in parallel or in series?
The impedance I'm talking about is not something you can connect to or disconnect from the source. It is an inherent property of the source; it's built-in to the source.

For example, if I take a new AA alkaline cell and try to measure its output current capability with a 100 Amp meter, the meter will look like a short on the output of the cell, and the meter won't limit the current. When I do this I get a current of 12 amps. But if I do the same thing to a NiMH rechargeable cell, I get a current of 25 amps. Why the difference? Why isn't the current 100 amps, anyway?

It's because of the internal impedance of the cell itself. You can't disconnect this impedance because it's a property of the materials of which the cell is made.

A microphone will have an internal impedance (the source impedance) which is an inherent property of the microphone. Different microphones will have different impedances.

This impedance isn't something you can get rid of; you just have to accept and work with it.

If you connect the microphone (or a battery, or a generator, or any source of electrical energy) to a load, and the load might be a resistor which will heat up, a motor, or the input to an amplifier. The load will have its own inherent input impedance, and this impedance will interact with the impedance of the source which drives it.

To get maximum power transfer from a source to a load, it is necessary to match the impedance of the load to the impedance of the source.

Search for "impedance concept" on the web:
Understanding Impedance

The impedance concept has applicability in many places in the physical world. For example, there is such a thing as "mechanical impedance". The transmission in a car is there to match the impedance of the engine (at various RPMs) to the impedance of the drive train at various speeds.
 
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mneary

New Member
alphacat, this gets back to trying to map practice into theory. As you know, in theory, a typical signal source can be modeled in either of two ways. You choose a current source with a resistor parallel, or a voltage source with a resistor in series.

Now you ask which model is a 'real' component? A device is not a model. It's a real device. We choose models that work best for our analysis.

As you know from theoretical analysis, the current source in parallel with a resistor is equal to a voltage source with a series resistance. The physical device isn't different just because we choose a particular model.
 
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