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Topic: Reverse voltage protection using PNP transistor (Read 2 times) previous topic - next topic

PetriH

Hi! I'm working on reverse voltage protection for a battery powered device. I have a bunch of 2N5401 PNP transistors so I was planning to do it like this:



The load is 50mA (max), hfe of the transistor is 50, so a base current of 1mA should turn the transistor fully on.

But now I'm confused... the datasheet of the transistor says that max emitter base voltage is only 5V and I was planning to use a 6V battery pack. Is this a problem?

My theory is that the voltage drop over the emitter-base junction is 1V (= base-emitter saturation voltage in the datasheet), so that the base resistor needs to drop the remaining 5V. So the resistor value should be... R = U/I = 5kohm? 

Did I get this all backwards? If so, how do I calculate the correct resistor value, or is 6V too much for the circuit to handle?

(I know I could use a P-channel MOSFET for better results but I don't have the required parts here right now so I'd like to try the PNP transistor circuit first)

jremington

You should not try to exceed the Vebo rating of 5 V. Why not use a diode for reverse voltage protection?
"It seems to run on some form of electricity"

Nick Gammon

http://www.youtube.com/watch?v=IrB-FPcv1Dc
http://www.gammon.com.au/electronics

jackrae

I agree with the simple diode concept - been used since diodes were invented.  By the way what protects your transistor from "reveresed-reverse" voltage

AmbiLobe

Your circuit is a good one. The pnp transistor is used correctly. The resistor value 5k is good and it is safe to use 2.2k ohms instead.
I am going to get going.

Chris1448

A guy could just use a bridge rectifier. they can be had as small as a transistor these day's.
Depends on your current and voltage..
Its down right idiot proof

PetriH

Thanks for the replies guys! The reason I don't want to use a diode is the forward voltage drop (around 0.8V). I'm using a step-up/step-down voltage regulator which has a cut-off voltage of 2.7V. With a diode the cut-off voltage would rise to 3.5V. So when the battery pack drops below this the device will no longer function.

I just tried the PNP transistor circuit with 6V and it seems to work. I measured a voltage drop of 0.8V volts in the emitter-base junction and 5.2V on the base resistor. So the transistor should be safe in this case.

However, in the reverse voltage case, the base of the transistor will be high shutting down current flow through the transistor. Emitter will be at 0V and base at 6V (since there is no current flow through the base). WIll this break the transistor? Any ideas how to fix this before I start blowing up transistors :)

bombasticbob

I have to wonder what kind of regulator might be damaged by reverse voltage, and how could the reverse voltage get there anyway?  You aren't paralleling power supplies, are you?  if you have a battery, use a standard off-the-shelf boost/buck regulator to produce a nice clean "whatever voltage" (typically 3.3v I'd imagine).  Linear Technologies makes a lot of them, as one example.

I would expect the regulator to simply stop charge-pumping if voltage went above its threshold.

But you might damage OTHER components.  For that, maybe a series fuse and a zener crowbar to prevent overvolts.  Dunno what else you might try.

PetriH


I have to wonder what kind of regulator might be damaged by reverse voltage, and how could the reverse voltage get there anyway?

I'm using Polulu's S7V8F5 step-up/step-down converter and it does not have reverse voltage protection. The reverse voltage can get there by connecting the battery the wrong way around :)

jremington

Can't you use a polarized connector, so that it is impossible to connect the battery backwards?

The PNP transistor solution is not a good one, for the reasons that you have already stated. Most junction transistors won't survive a reverse base-emitter bias of more than a few volts. Even if they do temporarily withstand a backwards battery, down the road they will likely fail and perhaps, take everything else with them.
"It seems to run on some form of electricity"

Chris1448


If left unchecked and connected wrong, something is going to get hot and eventually burn out.





Nick Gammon

http://www.gammon.com.au/electronics

cjdelphi

Um...

Simply place a diode in series with the base pin?

Nick Gammon

If you watch the video I linked there are a number of problems with that. Two are:


  • The voltage drop across the diode

  • The heat dissipated by the diode caused by the voltage drop

http://www.gammon.com.au/electronics

PetriH

#14
Oct 16, 2013, 12:50 pm Last Edit: Oct 16, 2013, 12:56 pm by PetriH Reason: 1

What was wrong with the MOSFET solution?

Nothing, it's just that I don't have a MOSFET with low enough gate threshold voltage  (I'm going to order some soon though).

I also found this document by Maxim, where they have exactly the same reverse voltage protection circuit with a PNP transistor (see fig 2 in the pdf):
http://pdfserv.maximintegrated.com/en/an/AN636.pdf

So I'm still puzzled by this. Why would they show the circuit if it doesn't work? Shouldn't they know this stuff better?

EDIT: i can't find it mentioned in the 2N5401 datasheet that reverse voltage on the base-emitter will break the transistor... is this common knowledge and for this reason skipped in the datasheets, or what's going on?

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