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If we have a linear power supply, which uses parallel power transistors (NPN) as series element, with emitter current sharing resistors, then which is the maximum acceptable difference of current between 2 parallel transistors?

For example, if the total current is 3A, and we use 3 NPN transistors in parallel each with 0.1Ω emitter resistors, then which is the maximum acceptable difference of current between 2 parallel transistors?

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  • \$\begingroup\$ Do you mean "what is the maximum expected difference?" The reason I ask is that only you can say what is the "maximum acceptable" difference. That's a decision you need to make based upon many things you haven't disclosed. \$\endgroup\$ – jonk May 24 '19 at 21:25
  • \$\begingroup\$ Yes, I mean the "maximum expected difference". \$\endgroup\$ – mike_mike May 24 '19 at 21:35
  • \$\begingroup\$ What's the non-ideality factor, \$\eta\$, for the BJTs under consideration? You can expect \$\eta\cdot V_T\cdot\operatorname{ln}\left(N\right)\$ change in \$V_\text{BE}\$ for each \$N\times\$ the collector current, ignoring the Early Effect. For small signal BJTs (which yours are not), \$\eta=1\$ so this means about \$60\:\text{mV}\$ for each factor of 10 in collector current. The emitter resistors provide a linear drop vs collector current. So it's pretty easy to compute the current sharing. Do you need help with the mathematics? Do you need to take into account the Early Effect (what's VCE?) \$\endgroup\$ – jonk May 24 '19 at 21:49
  • \$\begingroup\$ You specify "with 0.1 Ohm emitter resistors" thereby placing constraints on the result. Better (probably) is to ask "If I wish to ahieve xxx sharing results and cannot afford more than yyy wasted energy / loss of voltage headroom, what do I need to do?" For example, with 1A per channel you obviously "waste 0.1V drop across the resistors. If Vout is 5V that's 0.1/5 = 2% loss of Vin as "headroom" across the resistors. Vce min is probably 0.5V minimum (depending on transistors) and probably several times that. ... \$\endgroup\$ – Russell McMahon May 24 '19 at 21:57
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    \$\begingroup\$ Why then was 0.1 Ohms chosen? Just because that's what people do, or ...? That may be a fine value, but, working back from result specification will tell you. \$\endgroup\$ – Russell McMahon May 24 '19 at 21:57
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Estimate the variability of \$V_\text{BE}\$

If you pulled and tested 100 adjacent BJTs from the same reel, at low enough currents that the Ohmic base and emitter resistances weren't much of an issue, you might see a spread of \$20\:\text{mV}\$ (or \$\pm 10\:\text{mV}\$) for the \$V_\text{BE}\$ across the group. And that's an exceptional situation. If you look at parts across different part manufacturers and at different times, it would be twice that much or more.

You are also specifying a situation for BJTs carrying \$I_\text{C}\approx 1\:\text{A}\$. Such devices will often include substantial (and varying) Ohmic base resistance and some Ohmic emitter resistance, which can account for almost half of the \$V_\text{BE}\$! For example, I wouldn't be shocked to find \$V_\text{BE}\ge 1\:\text{V}\$, with almost half that appearing across the internal parasitic Ohmic resistances between base and emitter.

So, considering other factors as well, I think it would be reasonable to plan for a spread of \$100\:\text{mV}\$ (or \$\pm 50\:\text{mV}\$) across your devices.

Implications on \$I_\text{C}\$

In general, if you raise \$V_\text{BE}\$ by about \$60\:\text{mV}\$ then the collector current for will be \$10\times\$ as much as before. So you can already see that the above \$\pm 50\:\text{mV}\$ implies anything from 10% to 1000% for the collector current between devices. (As much as 100:1 between them.) If realized in practice, you can well imagine the current sharing would be terrible.

That doesn't even take into account the fact that if only one of the BJTs just happens to take on most of the total current, it will heat up dramatically and in the process take even more current and heat up still more. If you didn't do anything to manage this situation, you would probably be better off just using one BJT able to handle all the current requirements and forget about "current sharing."

Emitter resistors for each BJT

To make current sharing work well (and it works better if you used a compound BJT like a Darlington or Sziklai), an emitter resistor is often added to each BJT.

You should plan a voltage drop across the emitter resistor that is perhaps 4-5 times as much as the expected spread. In your case, with a planned \$I_\text{C}\approx 1\:\text{A}\$, this means \$R\approx 470\:\text{m}\Omega\$. You could get by with less (or more, if you can afford to waste the power.) But not a lot less.

Let's consider the worst possible case: two BJTs which are \$100\:\text{mV}\$ apart from each other in \$V_\text{BE}\$ if their collector currents were equal. Without the emitter resistors, one of the BJTs would hog 98% of the planned \$2\:\text{A}\$ current, leaving only 2% for the other BJT. The hogging BJT would then overheat and would probably wind up taking 99% or more of the total current. Not a good situation.

But with \$R\approx 470\:\text{m}\Omega\$ in their emitters, that wouldn't happen. Instead, in very simplified terms, one of the emitter resistors would have \$100\:\text{mV}\$ more voltage across it than the other one. But this means only a difference of current in the emitter resistors of \$\frac{100\:\text{mV}}{470\:\text{m}\Omega}\approx 200\:\text{mA}\$ difference. So one of them might have \$I_\text{C}\approx 1.1\:\text{A}\$ and the other would be \$I_\text{C}\approx 900\:\text{mA}\$. A 20% spread. This means that you should expect to see about \$\pm 10\%\$ across the collector currents. A much better situation.

If you can tolerate more variation between collector currents, you can lower the value of \$R\$. If you need to tighten it up more, you can increase \$R\$. (Of course, I've used \$\pm 50\:\text{mV}\$ variation between BJTs as a rough but useful guide when you don't have better information. If you do have better information, then by all means use it.)

As a rough rule, \$R=\frac{\Delta V_\text{BE}=V_{\text{BE}_\text{MAX}}-V_{\text{BE}_\text{MIN}}}{\Delta I_\text{C}=I_{\text{C}_\text{MAX}}-I_{\text{C}_\text{MIN}}}\$. \$\Delta V_\text{BE}\$ is just the worst case \$V_\text{BE}\$ spread you feel you need to deal with and \$\Delta I_\text{C}\$ is the worst case collector current spread you want to permit. Or \$R=\frac{\Delta V_\text{BE}}{2\cdot I_\text{C}\cdot \%}\$, where \$I_\text{C}\$ is your target collector current and \$\%\$ is the percent variation in collector current you can tolerate (where 0.1 represents 10%.) Put still more simply, and keeping the \$\Delta V_\text{BE}=100\:\text{mV}\$ estimated spread for BJTs, this works out to reserving \$500\:\text{mV}\$ for \$R\$ if you are willing to tolerate a \$\pm 10\%\$ collector current variation.

Reality is more complicated, of course. The above is admittedly highly simplified. But it provides a first level approximation that you can reasonably use.

Your value of \$R=100\:\text{m}\Omega\$ would imply (from the above) that you are willing to accept \$\pm 50\%\$ variation in collector currents. That may be okay with you. Or not.

Other approaches

You might also consider using added BJTs to "sense" the emitter resistor current and amplify that difference in controlling each power BJT (or compound BJT arrangement.) This can permit a smaller voltage drop across the emitter resistor and tightens up the design a bit. And if you want to go that far, you might be better off with using those current-sense BJTs to control power MOSFETs instead of power BJTs.

But I'll leave such ideas for another question.

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