How does the current mirror work?

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lucky6969b said:
https://en.wikipedia.org/wiki/Current_mirror

I don't understand if the positive rail is +9V and negative rail is -9V
Why the current is 18V - 0.65V / Rc?
Anyone explain this?
Thanks
Jack
Click to expand...

don't know where you are getting this equation from in your reference link, but it would seem that --

18 = +9 - (-9) is the total voltage differential
0.65 = PN junction drop
Rc = resistance in the path

then I = V/R
where V across the R will be (18 - 0.65)
and R = Rc.

hence your equation is most probably I = (18 - 0.65) / Rc
which is easy enough to understand ?
 

I don't know if we are referring to nodal analysis here. Where is the reference point?
Thanks
 

Is the reference point right at the middle of T1 and Rc, I mean how do you get 18V - 0.65V
Counting from the collector of T1?
Thank
 


No voltage drop across T1? Sorry, I am quite new to current mirror circuit
Why not IcRc + Ic(Resistance of transistor 1) = 18V
18V / Rc + Resistance of transistor 1) = Ic
 

lucky6969b said:
No voltage drop across T1? Sorry, I am quite new to current mirror circuit
Why not IcRc + Ic(Resistance of transistor 1) = 18V
18V / Rc + Resistance of transistor 1) = Ic
Click to expand...

(sigh)

the 0.65v for PN junction IS the voltage drop across T1
Transistors and PN junction's and such active devices are usually not considered as resistances, except in special cases.

With the Collector & Base shorted, the effective part of this transistor is simply the Base - Emitter pn junction. almost exactly a diode.
the standard voltage drop across a silicon pn junction is usually taken in the range of 0.6v to 0.7v, and will not vary beyond these limits during forward conducting and reasonable currents.
 


Thanks, I understand some of the parts now. The transistor is acting like a diode, but 18V - 0.65V, if the 0.65V is he drop across the transistor, but why not considered Rc which might have a resistance of several Kilo ohms or Mega on its own. if so, the voltage drop across it might be ~17V
please point out where I am wrong here
 

Rc controls or sets the current, as shown in the equation. If you increase Rc, then the current will reduce, so that drop across Rc stays exactly at 17.35volts. Ohms law : I = V/R

Once this current is set, then any value of load resistance R can be connected on T2, and it will have the exact same current flowing through it also. Thats why its called a current mirror.

HOWEVER - you are right in that if RL (the one on T2) is made too large, then the system will not work as a proper current source. But it does work for ANY RL which can range from 0 upto Rc. And thats how it has to be used.
 

Okay, I remember now. The current of a transistor circuit doesn't depend on the DC voltage, but the base current/voltage. So the Ic doesn't depend on Rc.
Right?

- - - Updated - - -

Ok, get it, the potential difference all over is 18V with a 0.65V transistor difference. divided by Rc...
 

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