# Need for Temperature Compensation of Current Mirror

I am currently learning about current mirror configurations. I have made two of them so far. Both of them worked as desired but, when heated or cooled, the current through the right side (the side where the output is taken from) decreased or increased significantly with small temperature differences.

simulate this circuit – Schematic created using CircuitLab

$R_{load}$ for both circuits was low or shorted to +10V. Both circuits were set to mirror the current of 500 uA. All transistors were hand-matched (they are all very close to each other as far as beta is concerned).

Without emitter degeneration both circuits were significantly affected by temperature, especially Fig. A, where the current through $R_{load1}$ changed by 100 uA or more (1 second of heating) as I touched either of Q1 or Q2 with a finger tip; but as the transistors Q4 and Q5 were touched with a finger tip, the current through $R_{load2}$ changed by 50 uA (1 second of heating also), which is less then in the first example but still too much.

With emitter degeneration both circuits greatly improved their temperature stability. For example (the $R_e$ added were 1 kOhm) if I refer to Fig. B, the current through $R_{load2}$ changed only by 10 uA (when heated by approx. 1 second), while the result with Fig. A was a bit worse.

Both circuits are improved as emitter degeneration is added to Q1/Q2 or Q3/Q4. In both examples, current through Q1 or Q3 was approximately constant at all times but the current through Q2 or Q5 wasn't even close to that.

• I there any way to compensate either of circuits shown here, due to varying temperature? I thought that Q5 was going to correct the temperature variation error in current but obviously didn't.
• Vbe vs T matching is important not just beta which is an advantage for an IC bandgap Vref. Can you make them thermally coupled but insulated from ambient? – Tony Stewart Sunnyskyguy EE75 Mar 4 '18 at 14:20
• I think you are missing the point. Don't expect to be able to stabilize it if you heat the transistors differentially. All the math falls over drunk-like and vomits. You expect too much. – Andy aka Mar 4 '18 at 14:35
• @TonyStewart.EEsince'75 I understand that other parameters like Vbe, beta, Early Voltage, etc. matters but beta is only parameter that can be easily measured my multimeter. Do you think that thermally coupled mirror would improve the temperature stability? – Keno Mar 4 '18 at 15:16
• yes of course.. but you can test this with simultaneous and differential temp changes – Tony Stewart Sunnyskyguy EE75 Mar 4 '18 at 15:24
• Your problem is mostly differential temperature, but for small differences don't overlook the fact that the current through the set resistor is temperature dependent because of the Vbe drops from the supply. If it was a lower voltage the dependence would be more significant. – Spehro Pefhany Mar 4 '18 at 17:11

The three main steps are

a) Use as much emitter degeneration as you can
b) Match the temperatures of Q1 and Q2
c) Match the dissipation of Q1 and Q2

For (b), at the very least, glue Q1 and Q2 together. Far better is to use a monolithic transistor array like the CA3046, which constains 5 transistors made on the same substrate. For a really hardcore thermally matched pair, the LM394 'SuperMatch' pair uses thousands of transistor die connected like a chess board.

Q5 not only increases the output impedance, but also controls the dissipation in Q4. Play with series drops on Q5 base or emitter to equalise the Q3/4 dissipation match.

A slightly more complicated solution with less bandwidth but much more precision is to do away with Q1, and use an op-amp to drive Q2 to equalise the voltage drops on Re1/2. Replacing Q2 with a FET eliminates any beta variation contribution to the output accuracy well. Then you only need to be concerned about amplifier Vos drift with temperature, and tempco or Re1/2 resistors.

• Match dissipation? Power dissipation? Current should be mostly equal through both Q1 and Q2 but what is happening with the voltage Vce across Q2 is mainly dependent of the load resistance to be applied. If that is what you meant, otherwise I found your very useful. – Keno Mar 4 '18 at 17:57
• @Keno There are significant differences in the VCE for the two BJTs in the Figure A circuit. That can lead to very different heating in the two mirroring BJTs. Figure B, since there is one VBE for Q4's VCE and two VBEs for Q3's VCE, there should be twice the heating in one vs the other, but that's better (at least some mitigation of differences) because of the added Early effect compensating Q5 arrangement. – jonk Mar 4 '18 at 20:11

If you want to keep both transistors at the same temperature, they should have the same dissipation (ie, same current and same voltage). This also smoothes out some of the other error sources (like Early voltage). Your second schematic doesn't exactly achieve this, as the Vce of one transistor is higher than the other. Here we go:

simulate this circuit – Schematic created using CircuitLab

This is a full Wilson mirror and Q3's role is to drop one Vbe to make Q1/Q2's Vce equal.

A cheap source of dual matched BJTs is DMMT3904 and other dual transistors. They are not monolithic, so the matching and temperature tracking isn't as good as the fancy ones, but they're cheap.

If you want ultimate precision though, you would have to use a low-offset opamp.

• I'd written Keno about this, but hadn't yet mentioned the details you added regarding the additional BJT in the full Wilson. Good addition. +1 He's exploring these ideas on protoboards and differentially heating things up to see what happens. (I'm quite impressed by his thorough testing to see behaviors that he needs to then understand better.) None of these circuits, yours or Neil's, discuss methods for beta-compensation. (The emitter resistors are about ISAT/VBE plus temp compensation, not beta.) Since he's doing discrete stuff, must go back 50 years to see how Widlar handled these things. – jonk Mar 4 '18 at 20:56
• Yes, in this day and age, it feels good to see someone who learns electronics and actually experiments and tries to understand the details instead of just slapping an arduino on top of it!... – peufeu Mar 4 '18 at 22:53

To achieve matched current sources, use transistor arrays such as the (original) RCA CA3046. Its now sold by Harris or Intersil. Matching is to 5milliVolts emitter-base, which is about 10%. For better than that, given you have no way to use multiple emitter stripes and inter-digitate them, you'll need emitter degeneration resistors.

• I'd love to see an improved CA3096 where the low lateral PNPs are made to operate comparably with the NPNs in the device. I have need of mixed NPN/PNP on same die. I'll probably have to broker the darned thing if I ever want to get one. – jonk Mar 5 '18 at 0:20
• Motorola used to sell such. I used them to build an active clamp on an ADC's summing node. Was too slow, because I was ignoring Miller Capacitance of the feedback clamp amplifier. Regarding similarly fast NPN and PNP, Harris Corp in Melbourne FLA has dielectrically-isolated opamps, crafted to perform well in radiation-flux-environments, probably so the inertial-guidance systems in warheads will continue to accurately perform during a atomically-busy atmosphere. – analogsystemsrf Mar 5 '18 at 4:49
• @jonk Thank you for the mention of Chabay, months ago. A good read. Regarding transistors on the same die, there will still be transient thermal mismatches at the 114 uS timeframe, assuming the devices are 100micron apart. If FETs with interdigitated stripes (as diffpairs may be done) with Ma to Mb spacing of 10u, the thermal tau will be 100X faster (its inverse squarelaw) at 1.14uS; at 1micron, the thermal tau is 11.4 nanoseconds. – analogsystemsrf Mar 5 '18 at 13:54
• Interesting added information about the time constants. This is outside of my hobbyist experiences, but interesting just the same. – jonk Mar 5 '18 at 16:14
• @ jonk We use these thermal timeconstant effects in the tool Signal Chain Explorer to predict the Thermal Distortion of OpAmp circuits, including heating of the diffpairs due to output current changes (times VDD of opamp, as approximate change in heat). Ditto for resistors. A cubic meter of silicon has thermal Tau of 11,400 seconds, which is the inverse of the physics constant thermal diffusivity. A cubic micron, 1Million X smaller, is one Trillion X faster at 11.4 nanoseconds. – analogsystemsrf Mar 6 '18 at 2:37