I am studying the current mirror circuit.

I already went through similar questions and answers on this forum. But I was unable to get clarity from the answers. I am just a beginner. I request if you can answer in simple terms in steps for my benefit. I myself had asked similar question but I was not able to understand the answers. I went back and tried to understand and still was not able to get clarity. So, please help.

  1. What is the purpose of using a diode connected BJT? Won't a normal Diode be sufficient in place of BJT?

Current Mirror Circuit

I understand it is a very simple circuit but I am not able to understand how does this circuit establish current in Q1 and Q2 to be same?

Can someone explain me the basic concepts to help me understand this?

  1. What is the necessity to establish a constant current using this method? Why can't we use a single transistor and control the Vbe and Ib of it to produce a constant current? Why do we need two transistors to complicate this? Please help me understand this.

  2. While considering Q2, proper biasing is giving by Q1. Vbe of Q1 and Q2 are same. But should it mean that collector current of both the transistors are same? I understand that both are identical transistors in their geometry. But how does the current be same?

  3. I also understand that Q1 must be in saturation and for Q2 to establish a constant current, Q2 should be in active region,right? So that, if the voltage Vout is increased,Q2 will vary its Vce to provide constant Iout. Is this correct?

  • \$\begingroup\$ Quote: "Vbe of Q1 and Q2 are same. But should it mean that collector current of both the transistors are same?" Yes, of course. I am afraid, your problems in understanding result from the fact that you still believe that the base current would physically determine/control the collector current. \$\endgroup\$
    – LvW
    Commented May 8, 2020 at 7:56
  • \$\begingroup\$ Be careful with discrete mirrors, to have them perform well the BJT needs to be matched i.e. coming from the same piece of silicon. They sell them but they are somewhat expensive \$\endgroup\$ Commented Feb 25, 2021 at 7:10

2 Answers 2

  1. The V-I characteristics of the diode-connected transistor and the controlled transistor need to be the same. You wouldn't get that with a diode and a transistor.

  2. Current mirrors are used in integrated circuits, where transistors are cheap and most other components are expensive. How would you control Vbe to get the collector current you want? How about using the cheapest component available to do the job? That cheapest component is a transistor.

  3. The collector currents are the same by virtue of the fact that the transistors are matched. Because they are matched, the Vbe generated in Q1 matches the Vbe needed to generate the same collector current in Q2

  4. Q1 is not in saturation. Saturation happens when the base-collector diode is forward biased enough that current would flow from base collector in the absence of emitter current. Vbc of Q1 is zero, so the junction there is not forward biased at all.

  • 1
    \$\begingroup\$ High performance audio amplifiers also use current mirrors using dual transistors in a 5 wire sip package. Common base and common emitter can be had as well. \$\endgroup\$
    – user105652
    Commented Jan 26, 2019 at 4:45
  • \$\begingroup\$ Thank you for the answer. But I still have some doubts. 1. I googled for the VI characteristics of a BJT and a diode connected transistor. Both seem to be the same. What is the difference? 3. So, the amount of current through Q2 will be same independent of the resistor value(if connected)? Can you provide an analogy? 4.So, Q1 and Q2 are in active region? In what regions are both the transistors operating? Thanks \$\endgroup\$
    – Freshman
    Commented Jan 27, 2019 at 11:56
  • \$\begingroup\$ Diode-connected transistors hold to the theoretical V-I characteristics over a wider range of currents than diodes do. And nothing is going to match a given transisor's V-I characteristic better than that transistor's twin brother. \$\endgroup\$
    – TimWescott
    Commented Jan 27, 2019 at 22:27
  • \$\begingroup\$ What resistor are you talking about? Q2 will have nearly the same current as Q1. There is no analogy, it just is. Have you worked through the math? \$\endgroup\$
    – TimWescott
    Commented Jan 27, 2019 at 22:28
  • \$\begingroup\$ Yes, Q1 and Q2 are both in the active region. They're clearly flowing current, and neither is in saturation, there's only one thing left. \$\endgroup\$
    – TimWescott
    Commented Jan 27, 2019 at 22:29

The transistor appears to be a diode but still has some current gain when near saturation.

Even though the hFE may only be 20% ~ 30% of max hFE it still controls the current as a function of Vbe. Vce is not Vce(sat) @ rated current but = Vbe.

For low currents, the bulk resistance differences and massive hFE differences with a line Vce on the mirror vs the saturated reference.

So it works much better than a diode as a base voltage controlled current source. This makes them match better with mirror loads and with changing reference loads to select the mirrored current.

Although hFE is constant In the linked example, I choose the reference side left as hFE=10 and the mirror BJT with hFE=1000. You can swap the BJT with the diode and see the difference when the left switch draws more current.


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