# Understanding current mirror

Discussion in 'Homework Help' started by umichfan1, Aug 22, 2012.

1. ### umichfan1 Thread Starter Member

Jun 16, 2012
32
0
On pp. 103 and 105 of "The Art of Electronics Student Handbook" (see attached pdf), it talks about the current mirror circuit. I am very confused about how this circuit works. First of all, I cannot see where the input signal is fed into the circuit--it looks like it is just in a steady state. On p. 105, it challenges the reader to explain how the "program side" of the current mirror works. I would guess that the voltage at the base must be 0.6V, because the base is always 0.6V higher than the voltage of the emitter, which in this case is grounded. Then, because base is connected to the collector, the collector must be at 0.6V also, which means that I_C should be about 1 mA. If we use I_C=\beta * I_B (and assume that \beta=100), then that means that I_B should be 0.01 mA. However, this understanding of the circuit includes nothing about negative feedback. What am I missing?

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2. ### panic mode Senior Member

Oct 10, 2011
1,328
305
current mirrors are very simple circuits. in your drawing resistor in Q1 collector is setting current. Vbe is some 0.7V, so current through resistor is
(Vcc-Vbe)/R

to get that to 1mA, Vcc must be 15.7V because R is 15k and programming current is 1mA.

next step is to assume that Q1 and Q2 have same characteristics (matched or both made on same die for example). that means that base current will be same for both transistors because characteristics match and Vbe is equal for both of them. This also means that Ic will be same for both of them because if they are matched, they will have same gain.

now we use Kirchoff's law to examine circuit....

Ic2 is same as Ic1 (so we just call it Ic) but current through programming resistor R is greater than current through load because:
current through R is

Iprogramming = Ic + 2* Ib

notice the difference? It is 2*Ib because base currents of Q1 and Q2 are both going through R, not through load.

granted, Ib<<Ic so one often neglects the difference.

because Ib is determined by transistor characteristics (fixed), R value (fixed) and supply voltage (fixed), it's value does not change.

because Ic=Ib*hfe
and hfe (in theory) is constant, it means that Ic is constant and therefore, Iload does not change... at least as long as load resistance is low enough (lower than R=15k in this case).

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3. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Current mirrors are extremely useful because integrated circuits inherently get nearly perfect matching of the transistors fabricated on a single wafer so the current mirror will track almost perfectly. That's why IC designers locked onto them as the basic design block for IC circuits.

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4. ### crutschow Expert

Mar 14, 2008
13,501
3,375
A current mirror depends upon the matching of the transconductance (Ic/Vbe) between transistors. It's important to note that the base-emitter voltage is not fixed but is a logarithmic function of the base current and thus so is the collector current.

Thus when you pass a given current through the left (diode-connected) transistor it will establish a certain Vbe as required to carry the collector current. This Vbe voltage appears across the base-emitter junction of the right transistor and, if the transistors are matched and at the same temperature, it will carry the same current as the left transistor. The high output impedance at the left transistor collector makes it appear basically as a current source. Thus the right transistor "mirrors" the current flowing through the left transistor relatively independent of the right transistor's collector voltage.

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