What is the reason of that 1 kilo ohm connected to the microphone? Why always too high resistors are used? I couldn't find any explanation. How do we calculate these values?
Asked
Active
Viewed 1.5k times
1
-
On what grounds do you consider 1k *too high*? Too high for what? Like most components, a microphone is designed to be used in some practicular way, hence the only good reason for using a particular value is that the designers of the microphone intended you to do so. So why use another value? – Wouter van Ooijen Apr 06 '12 at 08:35
-
let me be more clear. if i have a microphone, and wanna make an amplifier, how am i gonna decide of that pull up resistor!s value? nothing is written on microphone. depending on what i will decide the value of the resistor? – user16307 Apr 06 '12 at 08:40
-
For lack of a datasheet for your microphone you could use datasheets for similar microphones. http://www.primo.com.sg/ourproducts-jap-microphone.html shows mostly 2 k, with a few exceptions (1k, 5k6, 15k). – Wouter van Ooijen Apr 10 '12 at 06:36
5 Answers
4
The reason there's a pull-up resistor is to supply power to the built-in FET impedance converter:
The mic diaphragm itself doesn't need any bias, since it's an electret mic. "Electret" means "permanently electrically charged". So the diaphragm is always already biased, even without any power applied. So the resistor and bias voltage has no effect on the frequency response or other properties of the diaphragm.
The resistor value does affect the gain of the FET, however, since this is a common source amplifier. Increasing the resistor (within reason) will increase the output voltage level.
(and 1 kilohm is not a high resistor value. If the diaphragm was not electret, you'd need to supply your own bias to it, through a resistor, and for minimal self-noise, this is typically in the gigaohm range, which is 1 million times as large.)
endolith
- 28,494
- 23
- 117
- 181
2
The microphone is an electret type, which needs a bias for the capacitor, which it basically is. Some electret microphones also have an amplifier built-in, which also needs power. The larger the resistor from \$V_{CC}\$ the larger the voltage drop caused by the supply current, though for an unamplified electret the current is low and the resistor may be bigger.
A too small resistor will cause the signal to be attenuated. \$V_{CC}\$ is ground for AC signals and a small resistor will cause too much current to go that way.
In practice you'll often see higher resistor values, like 10k\$\Omega\$.
stevenvh
- 145,145
- 21
- 455
- 667
-
1Couple of things wrong with your answer: 1) The Electret element inside the microphone does NOT need bias. The bias is part of the material that makes up the microphone dielectric. 2) Every Electret mic that I have seen has a built-in source-follower FET. This FET is what needs the external bias that you supply to the microphone capsule. Typical output impedance for most of the Electret mics that I have worked with measures to be close to 1k but that can vary widely. – Dwayne Reid Aug 30 '15 at 01:03
1
Examples of Condenser Mic specs with their corresponding data sheets can be found on DigiKing (see Electret Condenser Microphones).
Most devices specify both a typical/standard operating voltage along with a max voltage.
They also provide in the data sheet the typical current consumption (0.5mA appears a lot).
Armed with these facts, it's straight forward to determine the resistor pullup value.
$$Pullup\; resistor = \frac{supply\;voltage - operating\;voltage}{current\;consumption}$$
Example using a 9V supply and a 0.5mA current consumption with a 2V standard operating voltage
$$\dfrac{9V-2V}{0.5mA} = 14K\Omega$$
Ricardo
- 6,134
- 19
- 52
- 85
Paul Carew
- 11
- 4
1
Microphones can generate voltages by themselves sometimes, but often they need to be powered by an external source: in this case, the resistor is used to provide a supply to the microphone without forcing it's output voltage.
The signal generated from the microphone will be a variable current which will cause a varying drop over the resistor. Note that there is a series capacitor to AC-couple the signal, which means removing the common mode voltage (DC component).
And by the way, 1 kOhm is not a too big value: consider that 1 Ampère in electronics is often too big, and milliAmpères are far more common.
clabacchio
- 13,481
- 4
- 40
- 80
-
do you mean that the current should be very small before entering the amplifier. is that the reason for that 1 kilo ohm instead of 100 ohm. is 1 k resistor operates as a voltage divider? – user16307 Apr 06 '12 at 08:27
-
@cmd1024 1kOhm is not too high, please read my answer and study some theory: in electronics, 1kOhm is a quite common value, and even small for some applications – clabacchio Apr 06 '12 at 08:29
-
thanks i think i got the picture. sometimes it is too hard to grasp from formulas the main idea. many books even dont mention these reasons. do you know any good reference for beginners like me? – user16307 Apr 06 '12 at 08:33
-
There is a lot of posts over here for books, look at the top voted questions; Art of Electronics is often recommended, even tough I've never read it – clabacchio Apr 06 '12 at 08:46
-
I agree with clabacchio. Most textbooks that explain resistors and the like use component values that are easy to write down (maybe 1 - 50 ohms), but are of little practical use in electronics. In electronics resistors *easily* range from 1 - 1.000.000 ohms. – jippie Apr 06 '12 at 09:01
-
is that because not to burn the equipment by lowering the current with high resistors? – user16307 Apr 06 '12 at 09:21
-
@cmd1024 try to understand this: it's not because there s a 'k' before that the resistor is big: a kilogram is not so heavy after all, right? – clabacchio Apr 06 '12 at 09:25
-
guys i know, but my the question is the differnce between using 0.00000000000000001 ohm and 1 k ohm. ok "not high" that is a relative term. forget that. the question is what is the reason to use 1 k ohm instead of 0.5 k ohm. offff.. – user16307 Apr 06 '12 at 09:33
-
maximum rated current, energy efficiency, frequency response, amplitude range. Things you would normally get from a datasheet. The circuit will quite probably work equally well with 470E or 2k2. Consumer grade electronics is not an exact science. – jippie Apr 06 '12 at 13:35
-
@jippie well, I would rather say that it's made of trade-offs and specific requirements, but every choice should have a reason behind – clabacchio Apr 06 '12 at 13:43
1
The microphone as drawn in the circuit diagram is an electret microphone. It basically is a capacitor of which the plates can slightly move further and closer apart, by sound. A capacitor without an electric charge does nothing, so a slight electric charge is applied to the capacitor through the 1k resistor. Now when the plates of the capacitor start moving closer/further apart (caused by slightly changing air pressure from sounds), the capacity of the device changes with the distance between the plates and while the charge on its plates takes a relatively long time to change, the result is a varying voltage across the microphone.
These devices have a very high impedance for audio frequencies and therefore you often see an small buffer amp built close to it, so it can drive a long cable.
This type of microphone is very cheap and audio quality is high.
jippie
- 33,033
- 16
- 93
- 160
-
-
11k is not very high, it is rather low. The idea is that the capacity of the mic does not discharge while its plates are moving. If you manage to keep the charge on its plates constant, then the output voltage follows the recorded sound in detail. The 1k resistor allows the capacitor to discharge rather fast, which will influence frequency response and output signal swing. The lower the series resistor, the lower the response for low frequencies. – jippie Apr 06 '12 at 08:32
-
so we can also use 1 ohm? same principle isnt it? no reason why 1 kilo ohm rather than 0.001 ohm? – user16307 Apr 06 '12 at 08:36
-
theoretically ... yes, but the signal you get from it will be astronomically small. In practice I'd go for 10k or 100k maybe even more. – jippie Apr 06 '12 at 08:43
-
this is the point i dont get. if the resistor is 0.0001 ohm why signal is astronomically small? shouldnt it be opposite according to ohms law I=V/R. so i was thinking less resistance will cause higher current. where am i wrong? – user16307 Apr 06 '12 at 08:46
-
The signal is basically a current because dQ=I, so to have big voltage given that V=RI you need big R – clabacchio Apr 06 '12 at 08:52
-
In this application you want current to be very small. This type of mic requires a charge on its plates and it gives the best results when this charge does not change over time. In that case the voltage across the microphone follows the movement of the plate. If you manage to keep the charge 'number of electrons in the plates' equal, then the voltage across the device varies with distance between the plates. – jippie Apr 06 '12 at 08:56
-
1Note: An **electret** microphone almost universally has a FET inside to amplify the change in voltage created by the movement of the plates. Electrets mics have a static charge applied to the plates, it's not a function of the bias voltage either. See the [wikipedia page on electret microphones](http://en.wikipedia.org/wiki/Electret_microphone) – Connor Wolf Apr 06 '12 at 09:28
-
This is incorrect. The bias voltage is not applied to the microphone diaphragm at all ("[electret](https://en.wikipedia.org/wiki/Electret)" means the diaphragm has a permanent electric charge). The bias voltage is used to power a built-in FET impedance converter, so this whole discussion about resistor values is irrelevant. – endolith Jul 09 '18 at 21:32