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| -di/dt base drive of bipolar power transistor. |
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| Circlotron:
Totally obsolete tech but interesting nonetheless. I remember reading in an old ST publication “The Power Transistor In It’s Environment” that when you want to turn off a BJT fast (well, fast for those days) you had to drive the base negative and pull reverse current out of it until the emitter-base junction becomes reverse blocking, similar to reverse recovery of a diode. If you pull too much reverse current the transistor turns partially off very quickly but proceeds the rest of the way very slowly. The accepted solution was to have a small inductor in series with the base so that the reverse base drive would ramp up at a controlled rate. This would lead to optimum switch off speed. But instead of going with too much reverse base drive, why couldn’t you just find an optimum constant reverse current until it switches off. I would have thought that sweeping out that stored charge would be more effective with a constant but reasonable current. |
| bd139:
It was a very very long time ago I did anything related to this so please excuse me if I'm talking rubbish. AFAIK that doesn't actually help the saturation time decrease if you pull the current out of the base. Only way to do that is keep the transistor out of saturation using a Baker Clamp or something. |
| Circlotron:
Yes, a Baker clamp will keep it out of saturation and consequently reduce the storage time, but what I am thinking of is the reduction of fall time. |
| bd139:
Think it is the base-emitter diode capacitance and whatever remnants of storage charge you have to wait for on that. Pulling current out of the base will only fix the former. |
| Benta:
On a saturated BJT, the problem is the stored charge in the junction, not the capacitance. Pulling negative current from the base to switch the BJT off is an effective way of evacuating the stored charge and speeds up turn-off significantly. |
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