If we are doing a current-to-voltage converter we can use negative feedback on a common-emitter configuration. Using npn transistors would look like so:

current mirror

We are using a fixed current input and it forces the base-emitter voltage to be the corresponding value according to the Ebers Moll model. So my question is, how can we make the current input exactly what we want so it is mirrored on the other side?

I don't really know how it is possible to feed the exact current we want to the dual-transistors without having to calculate the voltage V1 in equilibrium for \$I_c\$.

If we increase \$I_c\$ then \$V_{BE}\$ would increase, which would make the voltage drop across \$R_1\$ to decrease (in a smaller amount). In order for the control of the current to be exact, when would need to calculate the precise increase we need in \$V_1\$ to reach the desired \$I_c\$ current so the base-emitter voltage is exactly what is needed to drive the second transistor. How can we keep the current exactly how we need it?

I appreciate your help.

  • \$\begingroup\$ A current mirror isn't really an ideal current source. Usually you assume some approximate Vbe and select the drop across R1 to be large enough that deltas in VBE don't affect the current too much. You can add some emitter degeneration resistors too. But you'll have other inaccuracies with this topology like Early voltage effects, mismatch, and temperature effects. \$\endgroup\$
    – John D
    Jul 31, 2015 at 20:40
  • \$\begingroup\$ I don't see negative feedback in your first circuit. \$\endgroup\$
    – Andy aka
    Jul 31, 2015 at 20:49
  • \$\begingroup\$ Andy aka, supposedly the connection from base to collector is negative feedback. \$\endgroup\$
    – OFRBG
    Jul 31, 2015 at 20:52

1 Answer 1


It's only relatively fixed. If the voltage V1 is something like 5V then the current will be around 4.3/R1 under typical conditions and will only change a bit with temperature or for different transistor characteristics. For example it would change by +0.05% per degree C.

The matching of the transistors is another matter- they must be very closely matched - that circuit is pretty much useless unless the transistors are on the same die so that they are both matched and at the same temperature.

  • \$\begingroup\$ So in practice matched transistors reduce the effects of temperature, yet don't make it null? \$\endgroup\$
    – OFRBG
    Jul 31, 2015 at 20:54
  • \$\begingroup\$ If you want a really temperature-stable circuit with desirably high output impedance you'd use a more complex circuit. Nevertheless it is useful- for example the front end of the LM324 op-amp has a 2-transistor current mirror. The Wilson mirror improves output impedance at the expense of another diode drop and more transistors. \$\endgroup\$ Jul 31, 2015 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.