Current Mirror confusion!

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Onigece

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I have seen simple Current Mirror ckt., and I noticed that C-B is short-circuited by a wire. How come that there is current flowing through C-B even if it is short-circuited? I am really confused on how current mirror works.
 

Hi my friend
Is your mean this?:

It is very simple , when you do short circuiting the CB , you will have a simple diode ( be ) . the be will be as a diode . and the current of collector of q2 is about IC2+2ib . if the beta of both transistors become very high the current of both collectors will be same with together.
Best Wishes
Goldsmith
 
And R1 is current programmer , and you can get the constant current that is approximately same with IC2 from Collector of Q1 to the vcc . BTW: your load resistor can not become higher that R1.
Respect
Goldsmith
 
From what I know, in order for a BJT to operate:

The emitter-base should be forward biased (current mirror's Q1 Emitter-Base is forward biased, so its okay) and the collector-base region should be reverse biased (current mirror's Q1 collector-base is not reverse biased but it is short circuited so its not okay).

I can't also just rely on the equations Ic=beta*Ib because for this equation to be true, the B-E should be forward bias while C-B should be reverse bias. And in the case of current mirror, C-B is not reverse biased.

It will be more clear if you can explain or describe in atomic level whats happening on Q1. I understand holes, donors, etc.
 

Again Hi
No , the Emitter- base should be reverse bias , or in the other hand , the base- emitter should be forward bias.
And about BJT : As i know i think the BE junction will break as a zener diode at rivers bias ( at 5.6 volt (approx)).but if you short circuit the BC , you can use the BE as a diode without that break effect.and you can use a diode instead of Q2 but the precision of the circuit with independent diode will be awful . we will try to use a transistor as a diode , because , the specifications of 2 transistor at one type , are alike with together .
So , see below , circuit please ( i wrote equations on that ):

Best Wishes
Goldsmith
 

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