8
\$\begingroup\$

I'm trying to understand the unusual current mirror circuit below. The current controlled by R2 is mirrored to R1. The strange feature of this circuit is that Q1's collector and base aren't connected directly as in a normal current mirror, but through two diode-connected transistors (Q3 and Q4).

What is the purpose of Q3 and Q4? I haven't seen this on current mirrors before. I looked at other current mirror circuits (Wilson current mirror, cascode current mirror) and this doesn't match. In LTspice, this mirror circuit functions, but I don't see any advantages to it. It seems to work just as well if I replace Q3 and Q4 with a wire.

This circuit is inside a chip I'm reverse engineering. The two-diode current mirror appears multiple times, so there must be some reason for it, but I can't figure it out. Any ideas? Does this circuit have a name?

schematic

simulate this circuit – Schematic created using CircuitLab

(Ignore the part numbers in the schematic.)

\$\endgroup\$
2
  • 3
    \$\begingroup\$ I just call it a "buffered current mirror." You can go to Analog Devices, though, and drop down to 11.7 at this page: wiki.analog.com/university/courses/electronics/text/… They use a slightly longer name for it. \$\endgroup\$
    – jonk
    Mar 15, 2017 at 22:39
  • 1
    \$\begingroup\$ I've put that circuit together and the output impedance goes up with the addition of each of the emitter followers (Q3, Q4). The improvement with discretes on a breadboard isn't the same as it could potentially be inside a chip with much better match between Q1 and Q2. It's perhaps worth investigating what effect the mismatch of Q3/Q4 has on the operation, say heating them up, or even better, cooling them down a lot relative to Q1/Q2 pair. But I can get output impedance to change a bit due to thermal effects just by swapping either Q1 or Q2 with a different part from the box, if R2 is lowered. \$\endgroup\$ Feb 24, 2022 at 15:27

2 Answers 2

14
\$\begingroup\$

Q3 and Q4 are not "Diode connected".

They are arranged as emitter followers to reduce the error that would result from the base current for Q1 and Q2 being derived from the current reference flowing through R2.

The gain of Q3,Q4 will reduce the current required by a factor of Hfe**2 (maybe 10,000 times) so that virtually all the current from R2 flows through Q1 rather than some being diverted.

They will also mean that Q1 is operating at a higher Vce (two Vbe drops) this will affect the current through R2 but may put Q1 into a better operating point.

A further improvement would be to put resistors in the emitters of Q1 and Q2. The resistor only needs to drop a couple of tenths of a volt to significantly improve the accuracy by reducing the effects of mismatch between Q1 and Q2. A resistor such as 470 ohms would be suitable.

\$\endgroup\$
3
  • \$\begingroup\$ In what way are Q3 and Q4 emitter followers? \$\endgroup\$
    – user253751
    Mar 15, 2017 at 23:37
  • 3
    \$\begingroup\$ It helps to look at the picture the other way up - being PNPs it si drawn with the supply at the bottom. The signal goes in the base of Q4 that then goes into the base of Q3. The output of the pair feeds the other two transistors. \$\endgroup\$ Mar 16, 2017 at 0:29
  • \$\begingroup\$ Oops, I didn't realise they were PNP. \$\endgroup\$
    – user253751
    Mar 16, 2017 at 0:32
2
\$\begingroup\$

This BJT current mirror configuration is also called "beta helper". In case you need to search for more information.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ You could make your answer a lot better by including information about beta helpers. \$\endgroup\$
    – CHendrix
    Mar 20, 2017 at 12:55

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.