-di/dt base drive of bipolar power transistor. - Page 1
Forum section: electronics / projects

author: Circlotron
author_rank: Super Contributor
timestamp: November 05, 2019, 10:32:30 am
content: 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.

author: bd139
author_rank: Super Contributor
timestamp: November 05, 2019, 12:03:25 pm
content: 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.

author: Circlotron
author_rank: Super Contributor
timestamp: November 05, 2019, 01:22:59 pm
content: 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.

author: bd139
author_rank: Super Contributor
timestamp: November 05, 2019, 01:56:16 pm
content: 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.

author: Benta
author_rank: Super Contributor
timestamp: November 05, 2019, 05:46:37 pm
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 05, 2019, 11:12:29 pm
content: I think there's going to be something about drift of stored charge, namely: even if you sweep out B-E charges by reverse biasing it hard, there's still free charge in the rest of the base, crossing over to the collector, over which you have no immediate control. Hence the limit on fT and tr.The B-C junction also takes time to reach full rating. Consider the picture as the charge begins to clear: the B-C depletion region widens (as Vce rises), but charges continue streaming out of the base (accelerated rapidly across the depletion region by the electric field), and collector current remains high. (This would be relevant for switching a resistive or inductive load, less so for a capacitive load.) The depletion region width is limited by charge density, so it stays narrow, and the breakdown voltage stays low -- and the ample supply of charge carriers ensures that impact ionization will occur (avalanche breakdown) if voltage rises too quickly. Basically, it looks like dV/dt limited switching. Some time later, charges settle down and full Vceo or Vcbo (as the case may be) can be applied. This is diagrammed in the turn-off RBSOA, not often seen in BJT datasheets but the switching types (MJE13009, etc.) often have it.Evidently the storage effect is much stronger than the dynamic breakdown effect, so t_stg is a few microseconds on these types, while t_f is a few hundred nanoseconds.The dynamic breakdown effect can be relevant to PN junction diodes in inductive circuits, where recovery current charges the inductor, and when it lets go (depending on the diode's softness factor), the excess energy causes voltage to ping back and forth, alternately (momentarily) forward-biasing and avalanching the diode. (Which, I suppose, lengthens recovery time even further, as some of that energy is turned back into free charge carriers, preventing the diode from turning off in one simple action.)On a more practical note, I haven't experimented much with extreme base drive. Stronger drive only seems to make smaller types go faster (e.g., a 2N4401 can switch in 10ns when driven with hFE(off) in the 5-10 range I'd guess), but I've not formally tested this on power switching transistors.Probably the closest I've tested, was this self-driving circuit,https://www.seventransistorlabs.com/Images/LED_Light2.pnga similar thing is used in classic ATX power supplies, but half-bridge.The characteristic of this circuit is, turn-on is at constant force hFE(sat) (given by turns ratio), and turn-off is given by drive transformer inductanceplus saturation. So it can potentially be quite brutal in terms of base turn-off; I don't have a direct measurement of Ib2 however.I've plugged in a number of transistors to see how they perform in this type of circuit; MJE15028 works better than TIP31C, for example. 2SD1273 is quite sluggish (but it's always quite sluggish, perhaps more due to Rbb' than base thickness -- which is likely quite thin given its high hFE?). 2N3055 varies with manufacturer (unsurprisingly, old ones are terrible, while planar Motorolas perform as well as MJE15028 -- same process?). 2SC4821 (RF/video transistor) is fast, but maybe not as fast as expected (perhaps because of too low hFE(sat)?). MJE13009 works a little slower than MJE15028, not surprising given its higher voltage rating.One could set up a circuit like this, with some current sense resistors (or current transformer?) and with emitter at common ground, to get a clearer idea of what's going on.Tim