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Old 28th April 2001, 01:31 PM   #1
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Can anyone explain what a current mirror does and where I can find examples and models of this circuit?

I had the idea of using 2 CSS's, but I was told to try a CM.
I'm not sure how it works.

Thanks!

Vince
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Old 28th April 2001, 01:53 PM   #2
Geoff is offline Geoff  United Kingdom
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Vince

See the article 'Current sources, sinks & mirrors' at the ESP Audio pages.

http://www.sound.au.com/articles.htm

Geoff
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Old 27th May 2001, 12:23 PM   #3
ppl is offline ppl  United States
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See the Burr Brown Application Note AB-165. This is a great Resource on this. as well as Current Sources.
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Old 7th July 2012, 12:53 PM   #4
tsiros is offline tsiros  Greece
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I'd like a detailed explanation on how a current mirror works.

so far every "explanation" i've seen is the same, merely stating what it does, not why it works. I don't have a problem if it goes technical to painful detail - in fact i would much prefer it.
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Old 7th July 2012, 01:58 PM   #5
sreten is offline sreten  United Kingdom
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Quote:
Originally Posted by tsiros View Post
I'd like a detailed explanation on how a current mirror works.

so far every "explanation" i've seen is the same, merely stating what it does, not why it works.
I don't have a problem if it goes technical to painful detail - in fact i would much prefer it.

Current mirror - Wikipedia, the free encyclopedia

Hi,

What it does and why it works are essentially the same thing.
Painful detail really doesn't help understanding the basic operation.

The current in one device (fixed or variable) defines a terminal
voltage that is used to control the current through a similar device.

e.g. the current through a diode determines a voltage drop, apply
this the the base-emitter diode of a transistor and that voltage
determines the current through the transistor. (In practice small
resistors may be added to improve the matching of devices.)

The diode can be a matching transistor wired as a diode.

More advanced current mirrors reduce the errors due to hfe
between the two currents and address high frequency issues.

rgds, sreten.
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Last edited by sreten; 7th July 2012 at 02:05 PM.
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Old 7th July 2012, 02:03 PM   #6
jcx is offline jcx  United States
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the input side converts the current to a voltage - the voltage controls the output device to pass the same (or ~ linearly porportional) current as the input

matched BJT junctions' do this through their log/exponetial Vbe, current relations

Resistors are of course much more linear converters of I to V, so resistor "degeneration" is often used in both sides of the current mirror to limit semiconductor process mismatch errors

Last edited by jcx; 7th July 2012 at 02:26 PM.
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Old 7th July 2012, 02:09 PM   #7
DF96 is offline DF96  England
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Here is a simple explanation. The current through a BJT depends on the voltage between base and emitter (it follows an exponential law - see Ebers-Moll). It is almost independent of the collector-emitter voltage.

Take two identical BJTs. Join their emitters together and ground them, and join their bases together. On one of them, the input BJT, join the collector to the base. Put a current in here. Most of the current will flow through the collector connection. A tiny fraction (roughly 1/beta) will flow through the base connection and set up a voltage at the base. This voltage Vbe will be exactly the right voltage to enable that collector current to flow. This same voltage is across the base and emitter of the output BJT, because they are joined together. Hence that will have the same collector current as the input BJT.

I have described the simple two-transistor current-mirror, which has limited performance. You can add emitter resistors and extra transistors to improve performance but the basic idea is the same. If you think that a BJT is controlled by base current you will never understand current-mirrors.
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Old 7th July 2012, 04:59 PM   #8
tsiros is offline tsiros  Greece
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i appreciate your replies!

Quote:
Originally Posted by sreten View Post
Painful detail really doesn't help understanding the basic operation.
as a physicist, i like painful detail...

Quote:
Originally Posted by sreten View Post
e.g. the current through a diode determines a voltage drop, apply
this the the base-emitter diode of a transistor and that voltage
determines the current through the transistor. (In practice small
resistors may be added to improve the matching of devices.)
i like something like this, only more detailed
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Old 7th July 2012, 05:01 PM   #9
tsiros is offline tsiros  Greece
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Quote:
Originally Posted by DF96 View Post
A tiny fraction (roughly 1/beta) will flow through the base connection and set up a voltage at the base.
why?
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Old 7th July 2012, 05:08 PM   #10
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Because of the way a transistor works. You need to have a reasonable grasp of transistor physics to be able to see through these things.
If, in the above explanation by DF96, you insert 1mA into the input collector, that current can only flow when the base current is about 1/beta times the collector current.
And that small current into the base can only flow when the base-emitter potential is around 0.6V, so you will see that.
And since that 0.6V is also on the Vbe of the output transistor, that one will try to conduct the same collector current as the one in the input collector (assuming matched devices).

Actually, it's even easier to visualize (you DID draw the circuit, did't you ;-)
The collector current imposed caused the Vbe to rise until it gets to around 0.6V. At that point, base current starts to flow, and that in turn causes collector current to flow. Because the base current is only 1/beta the collector current, the bulk of the inserted current flows into the collector.

jan didden
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Last edited by jan.didden; 7th July 2012 at 05:12 PM.
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