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Current mirror ensuring equal current flow of both LEDs

The source is a constant current LED driver. I found that connecting the LEDs in parallel caused thermal runaway with one LED consuming nearly all of the current with its temperature constantly increasing and the other LED would switch off with barely any current flow and would cool down. I know that if allowed to continue, the LED in thermal runaway would eventually be destroyed and this would cause the second LED to consume all of the current, damaging it.

Will connecting the parallel branches to a current mirror solve this issue? It is my understanding that the current in the mirror is set up in Q1 and mirrored in Q2 (and any further branches). Therefore if the current flowing in Q1 were to increase, it would increase in the mirrored branches as well.

However, because it is supplied by a constant current source, does this prevent the current in Q1 from increasing?

If I connect a third branch, will this solution still provide an equal current to all branches?

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  • \$\begingroup\$ As Neil points out, currents mirrors have a 'master-slave' relation. it will prevent the runaway of only one led. if possible, put the LEDs i series. \$\endgroup\$
    – tobalt
    Aug 5, 2021 at 16:00
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    \$\begingroup\$ This exact arrangement is used in some phone displays where there are multiple strings of backlight LEDs. The collector voltage is arranged to be about 1V on Q1. \$\endgroup\$ Aug 5, 2021 at 16:06
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    \$\begingroup\$ constant current LED drivers, however don't use one of the strings as master transistor but a separate one that is biased via a resistor for example. then all the strings use one each slave transistor that mirrors the current of the master (with a fixed gain usually). \$\endgroup\$
    – tobalt
    Aug 5, 2021 at 18:25

4 Answers 4

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A current mirror can only servo the output current if it has the voltage headroom (compliance) to do so.

If D2 is lower voltage drop and tries to hog the current, then Q2 will reduce its current by current mirror action.

If D1 hogs the current, Q2 will be unable to increase current into a higher drop D2, it's out of compliance voltage.

There are several possible solutions. I'm not convinced that the first and second offer enough power efficiency and performance improvement over the very simple third version to be worth their extra complexity. You would have to simulate them carefully, with thermal and LED mismatches, to see which is best for you.

schematic

simulate this circuit – Schematic created using CircuitLab

In the first option, we try to avoid the asymmetry present in the current mirror you've shown. R1 and R2 can have quite low values. R11 and R12 can be zero, in which case we are relying on the VBE matching and present temperature matching of the two transistors. Or R11 and R12 can be finite, which once they are dropping a few hundred millivolts, removes the transistor mismatch issue. This arrangement will have a minimum voltage drop of VBE, and then some for the resistors.

In the next (not well thought-out) option, we are using a low current defined by R3 (if voltage driven) or D13+R23 (if current driven) to control two identical larger currents in the two LEDs. We need finite values for R13 to R15 to effect the current multiplication, dropping at least a few hundred millivolts. We can only set these to zero if we ratio match the transistor areas and match their temperatures, which is not really practical with discrete devices, but is often done in monolithic ICs. Thanks to SimonFitch in comments, the minimum voltage drop with this arrangement is going to be Q4/5 VCEsat + resistors (not as I first suggested VBE+resistors). We also have some 'wasted' current through R3. However, whether that low voltage can be reliably reached in practice depends on matching the drive level to the components.

In the third and simplest option, we say that if options one and two are going to suffer some voltage drop anyway, let's forget about the transistors altogether and just drop several hundred millivolts across dumb resistors, which if the LEDs are from the same batch and in the same thermal environment will result in reasonable current matching.

The fourth, best performance and best efficiency, option is to use one current driver per LED. No wasted voltage headroom, ideal control of both LED currents.

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  • \$\begingroup\$ Will there necessarily be "at least a Vbe drop" from collector to ground in the second case? I'm under the impresssion that there will be a couple of hundred millivolts from collector to emitter when they are nearly saturated, and then you must add the emitter resistor drop too, but that total can be less than Vbe. \$\endgroup\$ Aug 6, 2021 at 16:24
  • \$\begingroup\$ @SimonFitch Well spotted. Q3 is 'least VBE', but Q4/5 are not. I'll update. \$\endgroup\$
    – Neil_UK
    Aug 6, 2021 at 16:41
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Yes, this will work in principle.

In practice with discrete NPN transistors, it might not be very accurate -- the NPNs won't match very well, and won't necessarily run equal currents even with the same VBE.

You can mitigate this by putting small resistances in series with each emitter -- enough to drop about 100 mV at the operating LED current will help significantly.

So, with 500 mA as shown, R = 100mV/500mA = 0.2 Ω (so use 0.22 Ω) would be suitable.

Your circuit will have nearly identical power dissipation in each NPN (about 0.7V*500mA = 350 mW), so you won't have much drift because of thermal differences, but it would still be best to thermally couple the NPNs together.

There is also a small error because of the finite gain (beta) of the NPNs; with a beta of 100, this will cause a 2 % mismatch in currents which is negligible in this type of application.

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    \$\begingroup\$ sorry, but I believe this is wrong. Carefully read Neil's answer. If D1 runs away, this mirror will not stop that. Instead it will ask D2 to draw the same higher current, which it cant because of lack of supply voltage, so D1 will increase further and D2 will shrink. This mirror will not allow D2 to run away though \$\endgroup\$
    – tobalt
    Aug 6, 2021 at 14:41
  • \$\begingroup\$ The more I think about this, the less I understand it. At first I thought this answer was good, and @tobalt's comment surprised me, but I tend to agree with tobalt now. More and more I think that the only way to combat the LED's negative voltage tempco is with a series resistor that over-compensates for it, on the collector side, which renders the whole current mirror solution redundant. Might as well just have two series resistors, one for each LED, and dispense with everything else, or an independent current source for each LED. \$\endgroup\$ Aug 7, 2021 at 2:11
  • \$\begingroup\$ And now I've flipped back. I see that it can work, because the current source will alter its compliance voltage to whatever is needed to maintain 1A, and even a microvolt of change in Vce of either transistor will prompt the source to compensate and provide whatever headroom each transistor needs to balance things. I fully expect to change my opinion again in about an hour. See you all then. \$\endgroup\$ Aug 7, 2021 at 3:13
  • \$\begingroup\$ It's not that an individual LED runs away, but 2 LEDs in parallel might. The simple mirror circuit will maintain D2's current close to that of D1, even if D2 has a few 100 mV additional drop. Given that the currents are about equal, the input constant current will be split evenly between the 2 LEDs. \$\endgroup\$
    – jp314
    Aug 7, 2021 at 6:02
  • \$\begingroup\$ Well @jp314 I guess one can settle the argument with a quick spice run. Even with fully matched transistors. If D1 is a blue LED and D2 is a red LED, the current will split to 0.5 A each. However, with D1 red and D2 blue, the majority of current will be through D1 = runaway. I have separated this argument into an answer of my own below. \$\endgroup\$
    – tobalt
    Aug 7, 2021 at 12:22
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I am posting this answer to settle a lengthy discussion in the comments.

The mirror does not work as e.g. Neil_UK has nicely explained. The reason is that current mirrors operate on a master slave fashion. So they either need a third tree to run as the master or they will protect only one LED from runaway.

This can be simulated nicely with spice:

enter image description here

Even though the transistors are fully matched, the diodes are not, which is simulated by using a RED and a BLUE LED in parallel.

In the left example the current is split well between D1 and D2 (505 and 495 mA respectively). However, in the right example D3 does runaway hogging over 998 mA of the total 1 A current.

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  • \$\begingroup\$ Do you think @Neil_UK's first (leftmost) solution address this imbalance? I believe those base resistors will perfectly manage the master-slave issue, and the emiiter resistors will keep the transistor mismatch problems in check. This is a great topic & question, I haven't though this hard about anything in a while. \$\endgroup\$ Aug 7, 2021 at 13:26
  • \$\begingroup\$ I suspect the difference between red and blue voltage drops in this example are sufficient to exacerbate the imbalance to a point where @tobalt's concerns are valid, but when the LEDs are strings, and the voltage difference across the chains are only a small percentage of the total voltage, there would still be sufficiently effective current division-by-two to avoid runaway. \$\endgroup\$ Aug 7, 2021 at 13:31
  • \$\begingroup\$ My thinking is that if the current regulation in either chain is sufficiently good that any modulation of voltage that it causes by maintaining the current is more significant than the voltage changes occuring due to the temperature of the chain, then runaway won't occur. In other words, current regulation doesn't have to be perfect, just better than the LED's own tendency to increase its forward voltage with temperature. \$\endgroup\$ Aug 7, 2021 at 13:43
  • \$\begingroup\$ @SimonFitch the problem occurs also for small Vf differences. That is the whole point of 'runaway'. Emitter resistors will bound the runaway indeed. But as Neil writes: The resistors work even without a transistor, resulting in his third solution. The current mirror does next to nothing \$\endgroup\$
    – tobalt
    Aug 7, 2021 at 15:13
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I've pored over this for many hours now, and I can't disagree with anything anybody here has said, because the arguments are all solid. So, the biggest question remains, does OP's (@MRB's) circuit eliminate thermal runaway in one of the LEDs?

I had to simulate it, with CircuitLab, which I did for two cases. The original, unbalanced mirror, and a balanced version inspired by the superb suggestions from @Neil_UK.

I use multiple LEDs in series, to simulate a real life chain, and I model the combined drop in forward voltage of them all (due to LED temperature rise) using a voltage source, connected in such a way that as the source's voltage increases it reduces the combined potential difference across the whole chain. Note how the positive end of the source is at the bottom, something I had to do because CircuitLab inconveniently refuses to perfom DC sweeps in the positive-to-negative direction.

schematic

simulate this circuit – Schematic created using CircuitLab

Below are the plots of current and total power dissipation in the left and right chains (left is blue), where I calcluate power as the product of the total potential difference across each chain (including V1, the forward-voltage adjustment) and the current through it. Each plot is a sweep of V1 from 0 upwards, resulting in a progressive reduction of total voltage drop, thereby representing a somewhat naive analogue of increasing temperature.

enter image description here enter image description here

My interpretation of this is that up until about 400mV reduction in combined forward voltage of the left chain, the power in both left and right chains diminishes with temperature. That's enough to prevent thermal runaway in either chain, but it highlights a problem.

If either chain's combined forward voltage is greater than 400mV different from the other's, then this system enters a different regime, where further disparity will result in thermal runaway. Therefore, this circuit is not very effective if the LEDs are not well matched to begin with.

Under 400mV of temperature-induced forward voltage reduction seems to vindicate me and @jp314, who claimed that the circuit will work. However, in the far more likely scenario that the LEDs will not be well matched, @tobalt's opinion seems easily the better judgement.

Now I include resistors to mitigate the "master-slave" aspect of the mirror, as @Neil_UK suggests (though I do not include emitter resistors to combat transistor mismatch). Here's the schematic, with current and power plots:

schematic

simulate this circuit

enter image description here enter image description here

This seems to solve the issue of LED mismatch, because as combined forward voltage continues to decrease in the left chain, well beyond the previous turning point, power in both chains continues to diminish.

In the light of this, it seems to me that the "balanced" current mirror will actually function to prevent thermal runaway in either chain.

I am aware that the way I modelled the LEDs' negative forward voltage temperature coefficient is pretty naive, and it won't surprise me if my approach is heavily criticised, but that's all I have. This thing already had my head spinning anyway, so I welcome anything that will put this to bed, even (and especially) if it completely invalidates all this.

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  • \$\begingroup\$ I encourage you to step V1 also towards negative voltages.. That will reveal the high assymetry of the unbalanced current mirror. \$\endgroup\$
    – tobalt
    Aug 7, 2021 at 16:27
  • \$\begingroup\$ Agreed! I did that, and the power graph does indeed contiue to rise to the left in both cases. However, I thought that this was equivalent to another voltage source in the right chain increasing, and I didn't want to complicate my already long-winded reply. However, I realise that the master-slave relationship invalidates what I just said, due to asymmetry. BTW, thanks, tobalt, your comments have been truly helpful and inspiring. \$\endgroup\$ Aug 7, 2021 at 16:38

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