0
\$\begingroup\$

I just wonder how precise of 2-bjt current mirror circuit then I found this circuit from link form circuitlab.com enter image description here My question is

  1. How much we can rely on 2-bjt how to determine how precision there are?
  2. How 4-bjt improve precission of 2-bjt circuit?

For my inpectation, 2-bjt will added base current to one side which make offset depend on bjt DC gain (beta). So the left side current will offset by 2betamirror_current. The 4-bjt approach just balance base current on 2 side of current path. BTW, these not included imperfection from semicondutor manufacuturing. The mismatch of VI curve between base and emitter and difference beta value can effect the precision as well.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Note that the collectors that drive the bases could be buffered with emitter followers, further improving static performance. Imagine replacing the B-C short with an emitter follower with collector connected to supply and thus bypassing base currents outside of the mirror. Q3+Q4 not only compensate to an extent for the base current asymmetry, but also stabilize the collector voltages of Q1+Q2, further improving accuracy. \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Mar 11, 2022 at 16:22
  • 1
    \$\begingroup\$ I'd like to know why you care. Are you trying to create a stable current sink/source this way? Or is there some other reason why, in particular, you are looking at a mirror? Your question also implies a desire for understanding sensitivity of the mirror to various parameter variations. And that is a whole other and long topic. Finally, accuracy is one thing, stability vs variables another, precision still another, etc. See terms. \$\endgroup\$
    – jonk
    Commented Mar 11, 2022 at 16:58
  • \$\begingroup\$ @jonk Because I use this in my circuit sometime and i just assume that both current is equal but now I wan't to know how it can be worse (1% is acceptable) \$\endgroup\$
    – M lab
    Commented Mar 11, 2022 at 22:09
  • 1
    \$\begingroup\$ @Mlab 1% what exactly is acceptable? Accuracy is hard to come by. Precision is easier. And over what operating temperature range? Even copper itself exhibits 3900 ppm/K. Some circuits, such as the Wyatt Cascode Peaking Current Source, actually use this fact to gain stability over temperature. You may imagine that 1% is just a hand-wave towards "easy to do." But if this is accuracy you want? No so much. Even for stable precision over a non-trivial temperature range can make this less than easy. It's no hand-wave to me. \$\endgroup\$
    – jonk
    Commented Mar 12, 2022 at 2:59
  • \$\begingroup\$ @jonk If I use current mirror as current source, then I use that current pass through unknown resistant. I want to find that resistant by reading current and multiply with known current from current source. so I want to make sure that mirror current is +- 1% from design value. that how I define my problem. \$\endgroup\$
    – M lab
    Commented Mar 12, 2022 at 16:21

2 Answers 2

Reset to default
2
\$\begingroup\$

The two transistor mirror offsets the cúrrents by 2Ib. Adding the third transistor (Q3) creates a Wilson mirror which cancels out the error due to the base currents. Adding the fourth transistor (Q4) doesn't add any error caused by base currents but adding the fourth transistor creates an improved Wilson mirror which removes Early Effect error in Q1 & Q2 by making their collector - Emitter voltages equal at 1Vbe drop.

\$\endgroup\$
2
\$\begingroup\$

In practice if you used discrete transistors such as 2N5401 the matching and temperature differences would limit the accuracy. The simulation does not reflect reality in this regard.

Even if the transistors were monolithic the dissipation in Q3 is different from that in Q4 by more than 70mW which will cause a thermal drift in output current.

This is not a configuration used in modern "test and measurement equipment" as claimed. One would use an op-amp (plus one or more transistors). Some of the biasing in the op-amp would use current mirrors most likely, but they do not have a first-order effect on accuracy.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ This is not a configuration used in modern "test and measurement equipment" as claimed. << That why I'm confusing I thought that this circuit canbe use in measurement \$\endgroup\$
    – M lab
    Commented Mar 11, 2022 at 22:12
  • 1
    \$\begingroup\$ Useful in measuring the student grade maybe. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 11, 2022 at 22:18

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.