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I'm confused by the reasoning used to solve part a) in this question and when solving current mirror problems in general. From what I've seen the common method is to say that \$I_{B1}=I_{B2}\$ since \$V_{B1}=V_{B2}\$ and the transistors are identical. However, in the question it also states that \$V_{BE ON}=0.7V\$ which implies a vertical line at \$V_{BE}=0.7V\$ on a \$I_{B}\$ vs \$V_{BE}\$ graph i.e. a diode I V curve. However, if this were the case then we can't be sure that Q1 and Q2 share the same base current since each could have any base current along the vertical ranging from 0 to infinity. Can someone explain why we can say that \$I_{B1}=I_{B2}\$?

  • \$\begingroup\$ Simply assume that you have perfectly matched BJT's. Also notice that if Ib1 = Ib2 then Ib2 = Io/β thus Ie3 = Vbe/R2 + 2*(Io/β) and Iref = (Vcc - 2Vbe)/R1 \$\endgroup\$
    – G36
    Commented Jun 18, 2023 at 12:33

2 Answers 2


Strict math and the approximative nature of "ON-state Vbe=0,7V" clash here. The maker of the problem tries to find those persons who intuitively can reject the strict math and think Vbe is not exactly 0,7V, it depends on Ib and that happens in the same way in Q1 and Q2. But it's still so close 0,7V that the calculation by assuming Ib1=Ib2 and Vbe=0,7V is useful. The found people are seen useful in future practical works.

  • \$\begingroup\$ I had a feeling this would be the reason. Thank you for clarifying. Just a quick follow up, would it be correct to say that \$I_{B3}\$ is not equal to \$I_{B2}\$ nor \$I_{B1}\$ since Q3's base is not shorted to Q1 and Q2 base though we would approximate all their voltages to all be 0.7V? \$\endgroup\$ Commented Jun 18, 2023 at 12:24
  • \$\begingroup\$ The base current of Q3 is different than the base currents of Q1 and Q2. It depends on how much emitter current from Q3 to qive to R2 current 0,7V/R2 plus the base currents to Q1 and Q2. \$\endgroup\$ Commented Jun 18, 2023 at 12:43
  • \$\begingroup\$ Yep ok thanks for the help. \$\endgroup\$ Commented Jun 18, 2023 at 13:10

Depending on the particular application a PN junction (diode or base-emitter junction here) can be thought (desired) to have:

  • zero forward voltage (rectifier diodes)

  • 0.7 V forward voltage (voltage stabilizers acting as low-voltage Zener diodes)

  • nonlinearly changing forward voltage (log and antilog converters, current mirrors)

When two non-linear but mutually inversed converters (I-to-V log and V-to-I antilog here) are cascaded, the whole converter I-to-I is linear. This is the idea of the current mirror.

So the third case is valid here and not the second as advised.


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