Jean T. wrote...
Normal operating condition:
- I have a 18V supply
- emitter at +18V
- base pulled-down to 0V (ground) through a 10K resitor
- I added a diode between base and emitter, anode to the base
- then the collector "outputs" +18V (less 100mV may be) to some load
Good, you have properly protected the emitter-base junction.
Now if we reverse the 18V supply:
- the diode protects the base-emitter
- I end up with a full reverse 18V accross collector-emitter
(there is a minimal load at collector)
If I read your schematic correctly, the 10k will cause 1.7mA of
current through your base-emitter protection diode, and the base
will be at about -17.3 volts. The PNP's collector-to-base diode
will be forward biased and the collector will be pulled down to
about -16.6 volts. It'll easily carry all the current your load
provides. The 2n3906 won't be damaged, but your load might be!
If you add a second diode in series with the PNP's collector, as
Fred suggested, then under the normal supply connection you'll
get about +17.2 volts out, and under reverse supply you'll have
an open circuit. If you use a Schottky diode, you'll normally
get about 17.5 volts.
If your goal is simply to provide a low-voltage drop protection
against reverse input supply, that can be accomplished with a
single P-channel MOSFET, wired as shown,
| +18V in
| o---------D S--- +18V out, with
| G the load to gnd
| |
| gnd
It may seem amazing, but you get out Vin less a small drop for the
current through the FET's low Rds(on) resistance, and for reverse-
input (negative-voltage) fault conditions, you'll get nothing!
The drawing above doesn't show the FET's intrinsic substrate diode,
which conducts in the desired direction, creating a ~ +17V output.
| ,--|>|--,
| o-------+-D S-+-- out
| G
| |
| gnd
The substrate diode provides enough gate voltage to turn on the FET,
which is then strongly on, in parallel with the diode. For example,
if you use an IRF9Z34 FET, it'll be about 0.07 ohms* (0.14 ohms max),
so you can pass 8 amps with less than 0.5V drop. If you pass more
than 10A, the FET's diode substrate conducts in parallel with the
channel, but sadly the diodes in most p-channel FETs are real wimps,
so it won't help out much over the already good-enough Rds(on).
The IRF9Z34 will hold off 60V, but watch out, if the input voltage
can ever exceed +/-20V, even for an instant, you'll need to protect
the FET's fragile gate with a zener diode, like this,
| +V in Low-voltage-drop high-current
| o---------D S-----+-- +V out reverse-protection "diode"
| G |
| | 15V | FET R(on) +/-Vmax
| +--|>]--' ------- ----- -------
| | 100k irf9z34 0.14 60V
| '---/\/\--- gnd irf9540 0.2 100V
| irf9640 0.5 200V
The FET's rating must be higher than the maximum reverse voltage
that could be presented, but you can P-type FETs rated up to 500V.
* from the irf9z34 datasheet figure 1, Characteristic Curves.
Thanks,
- Win
(email: use hill_at_rowland-dot-org for now)
