Electron schreef op 28-Feb-15 om 11:10 AM:
> Hello,
> please take a look at the following circuit:
> http://oi57.tinypic.com/256h0fa.jpg
>
> I use a high voltage (500V) PNP transistor as a high side switch.
> With a PIC pin I turn it ON or OFF, through the NPN high voltage (500V) t=
ransistor.
> The 200K resistor is to partially turn ON the transistor when the PIC is =
in tri-state (e.g. during brief reset time). Then the PIC will be either ON=
 or OFF, so it will be able to withstand also the leakage from that 200K re=
sistor (2 mA worst case, the protection diodes can withstand it). 200K and =
2K resistors are of course rated for 400V (actually I use a couple of 200V =
equal resistors (half resistance) in serie).
>
> Basicly the PIC knows the voltage of both sides, and turns ON the switch =
when the left side voltage is higher than the right side voltage, with some=
 exceptions.
> It's like a diode, but being in software it allows me to do things that a=
 diode cannot do.
> Switching speed can be in the KHz region. The impedance is not really low=
 (currents are 100 mA peak max, and limited by the source anyway).
>
> Now the question: do you think that the Rb resistor is really necessary?
> If so, why?
>

Static case: Without that resistor the off-current (leakage) of the NPN=20
will be amplified by the worst case (in this case: highest) Hfe of the=20
PNP. The resulting current will flow through the PNP, even when it is=20
supposed to be off. The Rb is there to 'take' this current without=20
causing the BE junction of the PNP to conduct. So this resistor must be=20
less than 0.6V/(NPN leakage).

Dynamic case: When the NPN switches off, the charge on the base of the=20
PNP will keep it conduction for some time. The Rb resistor will amek it=20
switch off faster.

Suggestion: add a 5V zener from NPN base to ground, and put the 2k in=20
the emitter lead. That makes the NPN stage a constant current sink, so=20
you will have more predictable properties.

Wouter van Ooijen

--=20
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