# Current Mirror Explanation Needed.

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#### vdi_nenna

Paid Member
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

#### ppl

See the Burr Brown Application Note AB-165. This is a great Resource on this. as well as Current Sources.

#### tsiros

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.

#### sreten

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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#### jcx

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

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#### DF96

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.

#### tsiros

i appreciate your replies!

http://en.wikipedia.org/wiki/Current_mirror
Painful detail really doesn't help understanding the basic operation.

as a physicist, i like painful detail...

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

#### tsiros

A tiny fraction (roughly 1/beta) will flow through the base connection and set up a voltage at the base.

why?

#### jan.didden

Paid Member
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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#### tsiros

well, the current flowing into the p doped semi between the n doped semi, controls the flow of current, i'd rather not get into the QM behind it, but i never thought it can work in reverse?

how can i impose a current?

this can't work with only one transistor, right? not the current mirror, the appearance of a voltage at the base when forcing a current through C/E

#### DF96

The voltage appears at the base because it is connected to the collector. A BJT is not a superconductor, so if a current flows it will cause a voltage drop. The voltage drop is across both base-emitter and collector-emitter as they are connected together.

You appear to be thinking about a BJT in current terms. As I said, you will never understand current mirrors (or many other BJT circuits) if you think in this way. Think in terms of applying a voltage to base-emitter.

Imposing a current is easy (approximately). You just need a battery and a resistor. If the resistor is the dominant resistance in the circuit then you have (approximate) control of the current. Remember, Ohm's Law says nothing about causality: does voltage cause current or current cause voltage?

#### jan.didden

Paid Member
well, the current flowing into the p doped semi between the n doped semi, controls the flow of current, i'd rather not get into the QM behind it, but i never thought it can work in reverse?[snip]

If you have a transistor, B connected to C, the E grounded and the C/B through 1k to +15V, what do you think will happen?

jan

#### tsiros

i will experiment and see how it goes

thank you all for your patience, effort and time!

#### jan.didden

Paid Member
What will be the voltage across the 1k you think?

jan

#### tsiros

15 V minus the Vbc drop?

#### jan.didden

Paid Member
No, B is connected to C, remember? So Vbc = 0...
But you're close. You really must draw it, then it's clear. Remember, pencil and paper ;-)

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#### tsiros

npn, right?

15 V minus the Vbe drop, usually 0.6 V

edit: i really should stop relying on mental images for things with which i haven't got much experience :/

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#### jan.didden

Paid Member
You're doing fine. So we have +15V on one side and Vbe (0.6V) on the other side. Ohms law says there will be 14.4mA through that 1k.
Where does that go?

#### tsiros

doesn't have anywhere else to go, to ground.

is the transistor operating in saturation right now?

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