This is a basic 3-transistor current mirror:

enter image description here

I see some circuits add an extra resistor from emitter of Q3 to \$V^{-}\$. Just like this:

enter image description here

What's the purpose of this resistor?

  • \$\begingroup\$ What's the purpose of this resistor? Just think about what would happen if 1) that resistor was an open and 2) the resistor was a short. Would that make the current mirror work ? And actually this is not a current mirror as a current mirror has an input current while this circuit uses an input voltage. And that also gives a hint why that resistor is needed. \$\endgroup\$ Nov 19, 2016 at 15:21
  • \$\begingroup\$ Isn't @diverger referring to a resistor (not present in his schematic) that would connect Q1 and Q2's base junction (and Q3's emitter) to the negative supply? \$\endgroup\$
    – DavideM
    Nov 19, 2016 at 15:26
  • \$\begingroup\$ Yes, you are right. And that resistor is not shown in the circuit. It's just placed between the '+' and '-', and connect the emitter of Q3 and negative supply. \$\endgroup\$
    – diverger
    Nov 19, 2016 at 15:46

1 Answer 1


I'll draw up the schematic you are talking about:


simulate this circuit – Schematic created using CircuitLab

Rather than just tell you the answer (I will, but allow me a moment), let's start by just laying out the nodal equation for \$V_x\$ shown in the schematic:


Adding \$R_2\$ makes only one difference here. There are two directions to head, in considering "why" it's added.

One direction is to focus on the impact on \$Q_3\$: (1) Increased emitter current; and, (2) reduced \$r_e=\tfrac{k T}{q I_C}\$; and, (3) The ability to set \$Q_3\$'s emitter current so that it is relatively independent of the base currents for \$Q_1\$ and \$Q_2\$.

The other direction is to focus on the impact of "current in" and "current out" of the node. I think it is this latter option that matters more here.

Imagine that there is some unmanaged capacitance sitting at \$V_x\$ (to \$V_-\$.) \$Q_3\$'s emitter can charge up this capacitance up quite actively, having access to the current through \$R_1\$ multiplied by its \$\beta_{Q3}\$. But it can only be discharged through the relatively much smaller base currents (the same current through \$R_1\$ but this time divided by \$\beta_{Q_1}\$ and \$\beta_{Q_2}\$.) This asymmetry leads to an undesirable response to higher frequency changes.

Adding \$R_2\$ provides a separate way to sink current out of the node and to help balance the ability to sink and source current at \$V_x\$.

This is also achieved another way that you may also encounter:


simulate this circuit

None of these are high precision circuits for discrete designs. They lack careful consideration of operating temperature differences, beta mismatch, and \$V_{BE}\$ mismatch, just to name a few reasons why. A lot of good work went into designing around those issues back in the 1960's. So if you are considering building these on a protoboard, you might want to research beta compensation resistors and .. well, probably Wyatt's pretty nifty design, too. (It's interesting entirely on its own.) You can also consider getting pairs of BJTs built on a common substrate for better thermal matching (BCV61 and BCV62, for example) and, in some cases, much better beta matching as well (BCM61 and BCM62, for example.)

  • 1
    \$\begingroup\$ Any link to "Wyatt's pretty nifty design" ? \$\endgroup\$
    – G36
    Nov 20, 2016 at 10:38
  • 1
    \$\begingroup\$ @G36 Wyatt tried to get people's attention to it on EDN: edn.com/design/other/4351571/… \$\endgroup\$
    – jonk
    Nov 20, 2016 at 18:04
  • 1
    \$\begingroup\$ @G36 He originally named it "Cascode Peaking Current Source" and developed it as part of an RF IC project, working as a Senior Engineering Fellow at Honeywell, where he then designed ICs. He may be still working as the Chief Engineer at Exelis in Clearwater, Florida. But I'm not sure. \$\endgroup\$
    – jonk
    Nov 20, 2016 at 18:07
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    \$\begingroup\$ @G36 It's performance is astounding. The output is perfecctly flat with an input voltage between 2 V and 10 V and an input current range from about \$50~\mu\textrm{A}\$ to perhaps \$200~\mu\textrm{A}\$. And it actually uses the positive tempco of its resistors (aluminum in ICs, but metal film are close enough) to compensate the BJTs. The result is spectacular. You have to add the resistor tempco to get the right results in Spice. But once you do that? Wow! \$\endgroup\$
    – jonk
    Nov 20, 2016 at 18:14
  • \$\begingroup\$ @jonk: The figures links for that paper are all bad. Can you provide other links. \$\endgroup\$
    – diverger
    Nov 21, 2016 at 7:12

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