Howdy Wouter,

At 11.27 2015.03.03, Wouter van Ooijen wrote:
>Electron schreef op 03-Mar-15 om 10:15 AM:
>> Hi Wouter,
>>
>> At 16.27 2015.02.28, Wouter van Ooijen wrote:
>>> 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) transistor.
>>>> 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 resistor (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)=20
>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 any=
way).
>>>> Now the question: do you think that the Rb resistor is really necessar=
y?
>>>> If so, why?
>>>>
>>> Static case: Without that resistor the off-current (leakage) of the NPN
>>> will be amplified by the worst case (in this case: highest) Hfe of the
>>> PNP. The resulting current will flow through the PNP, even when it is
>>> supposed to be off. The Rb is there to 'take' this current without
>>> causing the BE junction of the PNP to conduct. So this resistor must be
>>> less than 0.6V/(NPN leakage).
>>>
>>> Dynamic case: When the NPN switches off, the charge on the base of the
>>> PNP will keep it conduction for some time. The Rb resistor will amek it
>>> switch off faster.
>>>
>>> Suggestion: add a 5V zener from NPN base to ground, and put the 2k in
>>> the emitter lead. That makes the NPN stage a constant current sink, so
>>> you will have more predictable properties.
>
>If 3.3V, make the zender 3.3V too.
>
>Choose the emitter resistor as 2.7V / (PNP base current).
>> But will it still be possible to turn it ON and OFF via the (3.3V) PIC?
>
>Yes, of course.

Today I ran some simulations.. I understood that if I force a certain volta=
ge
at the base of the transistor, of course I will have that minus the Vbe dro=
p
at the emitter. But here comes the interesting part (that I never thought o=
f),
as I can "force" a constant voltage on the emitter, per Ohm's law I can als=
o
force a current through it. So if e.g. I force 3.3V on the base, I will hav=
e
2.7V (3.3V - 0.6V drop) on the emitter, now if I place a 1kohm resistor
between the emitter and ground, the current through the resistor will be a
constant 2.7V/1k =3D 2.7mA
That's how and why the constant current sink works.

Thanks again, your suggestion was enlightning beyond its practical value.

With kind regards,
Mario



>Wouter
>
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