PNP for soft switch and reverse battery protection

Oh, duh.. my mistake... please ignore my post. George h.

Hah, Thanks, that would work.

George H.

This PNP transistor will have no problem with Vbe reverse breakdown

Ok, I breadboarded the PNP circuit in both the normal orientation of the transistor and with the transistor placed backward. Since the base is the middle pin, I just turned the transistor around 180 degrees.

The two circuits are shown here:

In the normal orientation, the 2N3906 functioned normally. Opening the base via J1 turned off the transistor completely, at least to the limit of my meters, at any V+ up to 12V.

In the reverse orientation, at low V+ levels, it also functioned normally, but with a bit less output current. This presumably results from the lower beta.

However, as I ramped up the V+ power supply, C-E current began to flow regardless of what was happening at the base - even if the base was completely open. Using the best meter I have for measuring very low currents, which is my old analog meter on the 50 uA scale, I found that the needle just began to move when V+ reached 8.03V, and current continued to increase as I increased V+ further, reaching 75 uA with V+ at about 11.9V.

This current was minimized with the pullup resistor (your R2, my R1) connected to the collector. "Off" current increased if I connected it to the emitter where you had it.

So this is either a characteristic of the 2N3906, or perhaps I have two of them that just happened to fail in the same way, or there's something about the resistor values I used that's causing this result.

But if none of those apply, and notwithstanding what Spice may have to say, it's possible that this is a characteristic of all PNP transistors, or possibly all bipolars. And if that's the case, then what I wanted to do, which is use one device for both switching and polarity protection, will work fine at, say, 5V or less, but not if you're using a 9V battery.

I guess the good news is that for 3.3V or 5V, this actually works, and you don't even have to turn the transistor around because the reverse voltage is less than the Vebo maximum. You can switch the power on and off with the PNP, and you get free polarity protection to boot. But it appears this doesn't work at higher voltages - unless of course I screwed something up that can be corrected.

If anyone wants to try confirming my test, I would be interested in the results. Meanwhile, unless there's a fix I haven't thought of, it looks like the backwards P-channel MOSFET will be needed for polarity protection for the 9V battery.

An interesting fact is that many (most?) PNP BJTs actually have quite high Veb breakdown voltages. They may not say so on the specs, but when you measure them, you discover this is true. Consider, as an example, the LM339 comparator's PNP input transistors, Q2 and Q3. They can handle 36 volts of reverse biasing.

Those are laterals, though, which is a bit of a special case--aiui they don't have super high emitter doping.

Cheers

Phil Hobbsh

Use a Pmos as polarity protection and you can soft start the gate. Jamie

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soft starting the gate doesn't help much when the body diode is conducting

you need two pfets

Ok, I miss understood the problem...

First off, you still should use a PChannel for protection because of the body diode in the reverse state and haveing the gate pulled low will force it fully on in the correct state. Otherwise without the gate pulled low, it'll pass voltage with one diode drop worth of loses

Put another Pchannel after that one orientated the other way as the circuit switch. The first one can be tied on at all times, the second you simply pull the gate to common when you want it.

Basically you get a switch and protection that has very little drop.

Yes, I know, I miss understood the question, I thought he was looking for good protection and didn't see where he was also looking for switching.

Anyway, I followed that up with another one after the fact.

Thinking some more about this, somewhere in my history I remember using an protective switch IC, ready made.. But I guess he maybe looking for heavy loads too ?

Jamie

Correct! Discretes, on the other hand, DO have heavily doped emitters. ...Jim Thompson

You have R1 the wrong way around in the emitter-toward-load mode. And it should be about the same value as the bias resistor.

And lose the diode. ...Jim Thompson

With the transistor in the backwards orientation, and V+ at

I made all three resistors 47K, and removed the diode.

I still get C-E current with the base open. If I move R1 to the emitter side, it INCREASES the OFF current from 75 uA to

I tested three NPN transistors the same way - connecting C or E to 12V through a 47K resistor with the other connected to ground, but with the base open, and got similar results. If I connect C and E in normal polarity, no current flows (because there is no base current), but when I reverse the polarity of the C and E connections, I get current - even though there is no base current.

So this effect appears to hold for all my bipolar transistors so far. What I don't know is whether the voltage at which this current flow begins to happen is related to any datasheet parameter.

What is you load current? I think you're doing something wrong... don't open the friggin' base! ...Jim Thompson

As I said, I changed all the resistors to 47K. So the base pullup resistor is 47K, the base resistor is 47K, and the load resistor is 47K. I removed the diode. And the power supply is 12V.

When I insert the transistor backwards, the ON performance is essentially the same as when it's in normal orientation. The base current is about .25mA, and load current is also .25mA, both of which are right for 47K resistors.

The problem comes when I try to turn the transistor OFF by disconnecting the base resistor from ground, leaving only the base pullup resistor connected. If the base pullup is connected to the collector (i.e. - directly to the 12V power supply), I still get 75 uA of current through the load. If instead I connect the pullup to the emitter as you suggest, the load current increases to 190 uA. The problem of course is that with the base pullup tied high, the transistor should be completely off, and there shouldn't be any load current at all.

The OFF load current only happens if the transistor is in the opposite orientation from normal, and even then only if the power supply is above about 8V (for my 2N3906's at least).

Even though this effect doesn't appear to involve the E-B voltage, I do wonder if the Vebo rating of a transistor affects the 8V threshhold. But I don't have a high-Vebo part to test.

If anyone actually breadboards this (not Spice) and gets a different result, please let me know.

What is "motor"? ...Jim Thompson

Oooops! I misunderstood your original post. I only tried reverse voltage protection, not turn-off via the transistor.

Do you really need to turn off via the transistor rather than just a switch at the battery? ...Jim Thompson

...Jim Thompson

I believe that is "meter", the thing in series with the voltage source on the left.