Driving BJT's

The terminal voltages of a BJT are drastically different. Is there an issue when driving BJT's as with MOSFETS gate source voltage? I've done some testing and it seems one doesn't have to worry too much about driving the base of a bjt because of the diode characteristics.

One NPN I plan on using for a capacitor multiplier is rated at V_CB=1100V, V_CE=800V, V_BE=7V. Since this will be used in a sort of "high side" configuration I'm unsure of how to deal with keeping V_BE within spec. From what I understand it is not a big deal unless I'm running huge currents into the base?

If this is the case then do I need to worry about extreme voltage spikes such as putting a zener across the base and emitter or is this even a waste?

My question is mainly do I need to worry about the base-emitter voltage spec in a similar fashion as mofsets gate-source or for most applications is it not a big deal?

Thanks

Reply to
James Rollins
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The Vbe rating of the transistor is it's reverse breakdown voltage, and should be avoided, just as exceeding the rated Vce should be avoided.

Bipolar transistors require drive current, rather than drive voltage.

Driving high voltage bipolar transistors as switches will require care to avoid saturation, if higher switching speed is intended. This can be achieved using Baker clamps to divert turn-on overdrive currents.

Turn-off time requires reverse base current to remove stored charge from the quasi-saturated junctions. This must be voltage limited and is most effective if applied with a controlled di/dt.

Because of the SOA limitations of bipolar switching transistors, the driving of capacitive loads is NOT recommended, without load-line tailoring through the use of inductive absorbers, or possible use of cascode emitter switching methods, which can extend the SOA.

ST Micro (and possibly NXP and ON Semi) may have some good app notes that originate from products developed by Thompson-CSF, Philips and Motorola.

motorola AN873 AN875 AN951 AR119 AR120 AR131 AR317

philips SC06

RL

Reply to
legg

A baker clamp at Vce(sat) =3D 5V? I'd like to see that. (You'll need a feedback circuit to handle that, I think, or at least a zener and an ugly current shunt.)

Curiously, I've entertained some possibilities, for the purposes of protecting beam tetrodes. Now, switching speed isn't a problem for tubes (they're like MOSFETs with a hell of a lot less C_miller!), but the screen pulls a hell of a lot of current if you let the plate voltage drop too low. So something baker-clampey would be good.

In the end, I just went with a series screen resistor and a bypass capacitor. It's dumb and it works. %-) Vpk(sat) ~ 30V at 500mA isn't too bad for something that's essentially a 5kV, 1A MOSFET with 1/5th the gain.

Switching speed may not be a problem; HOTs are usually spec'd for

2-3us t_stg (and that's with Ib2 =3D -2*Ib1, IIRC). Which is pretty damn long, but typical for HOTs. Just means you have to turn off sooner, really; t_f is usually pretty sharp (~200ns), so switching loss doesn't have to be high.

Tim

Reply to
Tim Williams

On a sunny day (Fri, 15 May 2009 19:14:08 -0700 (PDT)) it happened James Rollins wrote in :

For Vbe watch out mainly for reverse voltage, keep the reverse voltage low, for example by connection a reverse biased diode across base and emitter. For the rest a transistor in common base configuration is a current amplifier, the current you drive into the BE junction in forward direction, is appearing beta times bigger in the collector. It would take a huge current to get more then about a diode drop for Vbe in forward direction... so that likely will not be your problem. There is some charge storage effect, you need to be able do sink (say short the base to the emitter) fast enough to get rid of the charge there, if you want to switch fast. This may require a special RC network in the base drive.

Reply to
Jan Panteltje

The man is talking about driving bipolar transistors. Baker clamps can be designed for whatever saturation voltage is anticipated as acceptible. Vcesat of even KV bipolars is generally specified below

2V.

All very interesting, but not of much immediate use to the OP.

RL

Reply to
legg

Easy to do, if you _really_ want to ;-)

...Jim Thompson

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Reply to
Jim Thompson

From your other post, it seems that your 'capacitor multiplier' is actually a linear ripple filter applied to a HVDC souce.This has nothing to do with switching transistor drive requirements.

Main problems will be behaviour under overload and fault conditions. A reverse EB diode and base current limiting resistor can protect the EB jn against overstress, but this does not address output short stress on the SOA of the regulating transistor.

RL

Reply to
legg

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