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Parallel power transistors

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Advanced Member level 1
Nov 29, 2003
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Hi all,

I need to parallel four 2N3771 in an output power stage of a power supply but the hFE are very different like:

First 2N3771 hFE=49, Second 2N3771 hFE=29, Third 2N3771 hFE=19 and fourth 2N3771 hFE=10.

How do I equalize the gain to have a current splitter or equalizer to get the same current flowing thru the 2N3771 equally?

Can you use a 30 mOhm Mosfet instead instead?

If this is for repair of an old design, I understand.

The Vce(sat) is 2V/15A max or Rce=133 mOhm. Four of them will be 33 mOhm max. This is based on Ib= 10% of Ic or Ic/ib=10
The same is true for 4V/30A max for Rce but using more bias with Ic/Ib=30A/6A or 5:1

Thus depending on your supply current being shared and min max voltage drop I would suggest you insert custom Rbe values for each transistor instead of shared driver with common bases and insert values ranging from 0 to 500 mOhm. But see the full design would help.

Emitter resistors are the usual way to achieve equal current sharing in parallel circuit. Extensively used in BJT high power audio amps and linear power supplies.


yes, use emitter resistors if possible.

In case they don´t have individual base resistors, then the hFE is the minor problem.
The bjt that carries the most current will heat more than the others. This leads to decreased V_BE. This leads to increased current, this to increased see what i mean? It amplifies itself, until one bjt is killed...


Generally a bad idea to try to match currents when the hFE is so mismatched and depends greatly on how you measured hFE vs the actual operating range from no load to full load. Since worst case is full load, I would suggest the approach I suggested, if you have no other choice and choose the Rb instead of Re where the power Watt rating resistors will be smaller.

Naturally all devices must be at the same temperature on the same heatsink but variances due to in Rjc thermal resistance are unavoidable but help mitigate thermal runaway. I find that the optimal passive solution is to choose series equalizer R's that are in the same range as the ESR of the devices, in this case the Vbe diode ESR when saturated.

Below I chose Rb * hfe =6 approximately for each case. The equivalent output impedance would be seen at each emitter as an increase of Rb/hfe.

This is a rough cut showing the voltage drop across the series pass transistors as 8V which affects heat loss in the transistor. Since Ib is < 10% of Ic, most of the temperature rise will be the Vce*Ic which is also the effective ESR or Rce of each device. I expect less variation in Rce than hFe, so this should balance the transistors just as effectively as using separate Re's which will get hotter.


In the above simulation, I chose a DC of 4V and AC of +/-1V to drive the output current of 40A into the load of 0.1. This is purely for current sharing and the actual output voltage could be anything higher, as I assume the bypass is an emitter follower arrangement. It could be 40V and 1 Ohm load for 40A and give the same results. The supply could be 8V above the 40V with a reasonable variation, but in this case is just showing the drop above the regulated output voltage.

Unless you give more details, nothing more I can say.
Very thank you for the replies, all of you help me a lot. SunnySkyGuy the complete circuit I'm using it is here, of course the 2N3055 will be replaced with 2N3771 with base resistors you suggest. I'll add the image below.



My comment to the circuit:

good old linear power supplies...

While your circuit seems to work and the technical problems with the transistors are solved...

Are you aware of the huge ammount of dissipated power?
Output voltage range is from 12.9V ... 16.8V (Pb battery charger?) and output current is rated up to 20A, so the dissipated power may be around 500W.
This is in the range of a cooking plate or hair dryer. A huge aluminum heat sink with additional fan is needed.
I recommend to add a thermal fuse at the heatsink to switch down the power supply at overtemperature.

A modern switch mode supply could output the same but about 1/10 of dissipated power = heating.


of course the 2N3055 will be replaced with 2N3771 with base resistors you suggest

The circuit depends on industry standard emitter resistor feedback to achieve equal current share. Additional base resistors are just useless.

The base resistor variant can give slightly lower voltage drop, but you have to select individual resistors depending on transistor current gain. Never saw this option implemented in a professional design.

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