Low-voltage rail for transistor bases?

Question

Hello.

When power transistors are used as switches, the base current can be significant, as the saturation beta may be as low as 10. On the other hand, the base _voltage_ need only be at Vbe(sat), and any additional voltage needs to be dropped by e.g. a resistor. Given the high currents, this voltage drop may be a significant loss.

This suggests to me that ideally the base of a low-side NPN switch would be powered by a low-voltage rail, only slightly higher than Vbe(sat). This way the conduction loss would be minimized.

Has such a rail been used with bipolar switches? If not, what's wrong with my thinking?

(Please don't tell me to use MOSFETs because they are better. I know that already. I'm looking for understanding, not solutions.)

Lauri

John Fields · Accepted Answer

--- If you had a low voltage rail just to use as a base-to-emitter supply, you'd still have to switch it into the base, which would require another transistor: (View using a fixed-pitch font.) Vcc>-----------------------+                            | Vbb>-------------+         |                  |        [RL]                  E         |      | O--[R]--B PNP       C   -->|           C--[R]--B NPN      | O-+                 E           |                 |  GND>----+-----------------+ Which would work, and the dissipation would be lower than driving the NPN's base from Vcc. --  JF

John Larkin · Answer

Couple of issues:

Vbe_sat isn't really well defined. If you jam a fixed voltage onto the base, the base current could vary a lot with temperature, or between different transistors.

Transistors have equivalent emitter resistance on-chip, and more in the leads and circuit board traces. At high currents, the drop in the total emitter resistance reduces the voltage across the actual, internal b-e junction, so reduces the base current, which is what you don't want to do. And if you tune it for the loaded condition and then there were no load current for some reason, the base current might get huge.

I practice one rarely sees (ie, I've never seen) a circuit that tries to drive a switching transistor base at a fixed voltage. There's usually enough resistance in the driver to force the current to be pretty constant against changes in Vbe.

Mosfets are better!

Phil Allison · Answer

"Lauri Alanko" < ** No doubt -  but it would still have been a 5 or 12 V rail.   But the DC voltage being switched might be several hundred. ...  Phil

whit3rd · Answer

Yes, of course. Old NIM (nuclear instrumentation module) power for switching transistors included +24, +12, +6, -12, -24V power supplies, and it was common to make use of whatever was most convenient and efficient from the plethora.

Vacuum tube radios typically had A battery for filaments (1 to 6 V), B battery for tube bias (20 to 30V) and C battery for plate supplies (as high as you can get).

Sometimes, you see Vbb for base drive voltage supplies, Vcc for collector.

John Fields · Answer

--- The "B" battery was for the plate supply and the "C" for bias. --- --  JF

Low-voltage rail for transistor bases?

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