Howdy,
thank you for your replies.

I reckon that I have to give more details, sorry:

Input voltage may go up to 400V, but it rarely does, however the switch
is not to charge the output capacitor to such high voltages, never.

I want to charge output capacitor to about 5V, and of course protect it
(by switching OFF the transistors) when high tension spikes (up to 400V)
are present in the input.

Nevertheless, it may still be that the switch is ON even for relatively
high input voltages (i.e. considerably higher than the 5V output target
voltage) if the latter needs to charge up (e.g. turn OFF if >=3D5.1V, turn
ON if <=3D4.9V).
But the input impedance is relatively high, and anyway series inductance
(input comes from a magneto) prevents the current to rise too quickly.

If shorted, the input may source up to 1A peak, but it would take time to
build up, due to the high inductance, and anyway I turn it ON only when
the output capacitor goes below a certain voltage (usually 4.9V) and turn
it OFF when the capacitor exceeds 5.1V

There's also 400V active input voltage protection (a tranzorb) to filter
eventual spikes that go beyond 400V.

It's in short a circuit that I use to get power for a MPU (and other stuff)
from an AC generator that is very noisy (i.e. switched ON and OFF for other
purposes (high tension generation on another circuit, controlled by the MPU
too)).
The 200K "leak" resistors allows the MPU to powerup at startup, when the
generator turns slowly. Then the MPU takes over the process and eventually
starts generating high tension too (via another circuit it controls too),
other than handling the depicted circuit to maintain a ~stable 5V output.
However this value (5V) can be changed if needed, e.g. to power more voltag=
e
hungry stuff (the MPU of course takes it through a Vreg, it's a 3.3V PIC).
Sometimes I needed it @ 8.5V to power a sensor that required higher voltage=
..
It's nice to control all of this in software.

Now that I think of it though, a considerable amount of energy may be waste=
d
in the base junction of the PNP transistor. But then again how could I prev=
ent
it?

With the new info I given in mind, does the 2K resistor appear still too lo=
w?
I have chosen it empirically, i.e. it allowed saturation of the PNP transis=
tor
at the target voltages when I was wanting it to be ON. We talk about 10-20m=
A
max really, as the voltage inputs drops significantly when the switch is ON=
..

Thank you.

Cheers,
Mario


At 13.33 2015.02.28, RussellMc wrote:
>Power in 2k is V^2/R =3D 80 Watts.
>This seems "a little high" [tm]
>
>If  Vcc / Vdd is available at reset time then a base pullup to there will
>save you the HV drive resistor.
>
>As Vasile says re Rb - **YES**
>(Even in low voltage circuits it's a really good idea to have some high
>value R there)
>
>I_2K max =3D 400V/2k =3D 200 mA.
>What order of output current are you switching?
>That allows 2A at Beta of 10 (800 Watt) and 20A at 400V at Beta of 100 (8
>kW).
>
>
>          Russell
>--=20
>http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive
>View/change your membership options at
>http://mailman.mit.edu/mailman/listinfo/piclist

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