GEOFEX - Cheap and Good Polarity Protector

[waltk]

There's a description (and circuit) for a "cheap and good polarity protector" on the GEOFEX site here: http://www.geofex.com/Article_Folders/cheapgoodprot.htm

I drew my own version of the schematic, and created a layout from it.  The finshed build doesn't work right, however.  It's supposed to completely prevent current flow in the wrong direction, right?  My build passes current with a very small voltage drop when the input polarity is correct, but it also passes about -1V when the input polarity is reversed.  I first tried it with 2n3904/2n3906 transistors, and when that didn't work right, I tried a 2n4401/2n4403 pair - same result. 

So I guess the NPN transistor is turning off when the polarity is reversed, but it doesn't seem to shut off the PNP transistor.

Anyone else have experience with this circuit?  Here's the schematic (the BT2 battery symbol is not really a battery, just the output):

[R.G.]

So I guess the NPN transistor is turning off when the polarity is reversed, but it doesn't seem to shut off the PNP transistor.

Lemme go back and dig out my original work on that. I may have decided that -1V isn't harmful or something similar.

Or I could have made a mistake. 

I've built them, and they do seem to protect the pedals, but ...

Lemme go look.

[R.G.]

Didn't get a chance to breadboard it, but I did some more sim work.

There is a small amount of reverse voltage that gets through. It depends on the impedance attached to the output. With 1M, it's about 0.5-0.6V. With 1K attached as a load, it drops to a much smaller voltage.

What happens in your circuit if you put a 10K resistor on the output and reverse the battery?

[trjones1]

I've used this polarity protection a few times and I've gotten results similar to what RG describes.  In real world use I've hooked up an 18v adapter with the wrong polarity to these pedals and have had no bad results.  I like this circuit a lot.

[waltk]

What happens in your circuit if you put a 10K resistor on the output and reverse the battery?

Thanks for looking into this R.G.

I'm using a regulated 9V supply (9.00 V on my Fluke 87).  With normal input I get 8.98V output.  With reverse input and a 10K resistor on output I get -.155V.  ...not so bad.  Maybe wouldn't hurt anything.

If I use a 16.51V input, I get 16.49V with correct polarity, but -6.1V with reverse polarity (all with 10K across the output).

I'm having fun building and experimenting with this, but in the back of my mind I'm thinking the <200mv drop of a single series shottkey 1n5819 isn't so bad for simple polarity protection.

-Walt

[R.G.]

The thing is, any reverse voltage less than 0.6V can't turn on the substrate diode of silicon devices, so it's no practical danger to ICs. That is in fact the fundamental operation of a reverse diode protector. What's different about this circuit and a reverse diode is that the reverse current is not unlimited like it is with a reverse parallel diode.

This circuit is designed as it sits especially for 9V circuits, and I might do it differently for bigger voltages. There is a weak point in this circuit, and that is the reverse breakdown voltage of the base/emitter of the NPN. That's generally 6-8V for modern NPNs. And I believe that the voltages you see on the output actually come from current leaking through the reverse-broken base-emitter. I can imagine that as the reverse voltage increases, yeah, you do get more reverse voltage on the output.

I think that a diode in series with the NPN base or emitter would fix that, although I have not done the work. It's just an idea.

The MOSFET polarity protector works well, as does a Schottky diode. The MOSFET version gets you down to tens of millivolts lost, the Schottky to 200-300mV. But when I posted those, people complained that they couldn't find MOSFETs, they're expensive, complex, etc. and was 1N4004 as good as a Schottky, etc.

[waltk]

Thanks R.G.

The thing is, any reverse voltage less than 0.6V can't turn on the substrate diode of silicon devices, so it's no practical danger to ICs.

Didn't know that.  So the -.155 I get from reversing the polarity with a 9V input isn't going to hurt anything.

[jonnyeye]

The reverse voltage is due to the reverse leakage current through the NPN's B-C junction (and the current through the 4.7M resistor and the PNP's B-C junction, but this is small in comparison).  In order to keep the reverse voltage small, the easiest way is to add another load in parallel with the pedal, whose resistance is high when the correct polarity of voltage is applied (so as not to increase current consumption) but low when the polarity is reversed (to create as small a voltage differential as possible)...

Perhaps perversely, it's the old reverse-biased diode trick.  The main advantage to the rest of the circuitry then becomes current-limiting when reverse voltage is applied, although if R1 were made slightly smaller, D1 could be replaced by an LED to give a visual cue that the polarity is reversed.

[PRR]

> reverse breakdown voltage of the base/emitter of the NPN.

That's why you put the diode in. Clamped to -0.6V.

> reverse leakage current through the NPN's B-C junction (and the current through the 4.7M resistor

Not just reverse leakage.

My analysis says there's 8.3V across the 4.7Meg, thus 2uA toward Q1 Base and Q2 collector resistor. Q2 is shut-off hard, so 2uA flows into Q1 Base.

Remember that _BOTH_ B-C and B-E junctions are diodes. Put an arrowhead on Q1 collector. With "wrong" battery voltage, the "B-C" junction is FORWARD biased. The B-E junction is reverse biased. If this were two un-coupled diodes, B-E would block. But they are coupled, and the forward current in "B-C" injects carriers just like we normally do with B-E junction. The blocked B-E junction passes current just like we expect a collector to do.

This current is Beta times base current, except we use the REVERSE Beta. All modern BJTs are asymmetrical and optimized for large forward Beta and maybe small reverse Beta. So the 2uA base current is not multiplied by 100-400 as we normally expect, but by some smaller number.

The 2N3906 model I have here uses nominal BR=4.977. At the current we have arrived at, it interpolates as BR=4.82. This fits the excess precision Ib=1.769uA Ie=8.529uA numbers from analysis.

But do we "know" BR? Another jellybean BJT model shows BR=0.737. I have not seen a spec-cite for reverse Beta in 40 years; we get NO promises from the makers. I'd have to re-read A.S. Grove to recall how BF and BR are skewed 100:1, to guess if it is reliable or if we could find parts with "high" BR.

Perhaps the 4.7Meg could be higher, to reduce base current and multiplied collector current. At some point we hit Iceo, which is very small for cool Silicon but rises rapidly with temperature. However with the 4.7Meg omitted and temp run up to 50 C, I get 13pA output leakage, versus 11uA with the 4.7Meg.

> it's no practical danger to ICs.

YES. No danger at all.

> What's different  ... the reverse current is not unlimited

Correct.

Reverse voltage does NOT blow-up semiconductor junctions. Neither does reverse current. (*)

They blow-up because in many cases, when you reach a certain voltage, the current rises to "infinity" and then the POWER dissipated is "infinity". A B-E junction with no current limiting will be high impedance to 7V and maybe 10 ohms for higher voltages. With 9V backward that gives maybe (9-7) or 2V across 10 ohms which is 0.2 Amps, times the 9V is 1.8 Watts, which will melt a typical 0.3W junction quickly. However if there is even 1K of added resistance in there, 1,010 ohms total, we have 2mA times 9V is 0.018 Watts which is quite safe.

(*) While junctions work after reverse breakdown, there can be small shifts of parameters. If you match a pair to 1mV, then zap one, they may become a 2mV match. Also very low-noise parts may increase their hiss, though you may need a sharp tool to prove it. However sub-mA currents are not usually considered damaging; the problem is short several-mA spikes.

R.G.'s circuit WORKs; while you may measure a part-volt out of it, the voltage, current, and POWER available is far too small to hurt any part. It works over -24V to +24V, giving ~~0.3V the wrong way and 23.985V (0.02V loss) the right way.

If you are really scared of 0.3V at 10uA, omit the 4.7Meg AND test (while hot) for a leaky PNP. The cleanest Silicon will work this way; if you get a bad-day PNP it may work better with the 4.7Meg. Since "with 4.7Meg" is utterly safe, I suggest you go ahead with the plan as posted.

[PRR]

A similar plan could be developed around a P-type FET. However P-FETs (JFET or MOSFET) are not common parts like PNPs.

AMZ Jack has an essay about using transistors in the Reverse Beta mode. He's using Ge parts, and many older Ge "alloy" transistors were very nearly symmetrical. You cut a thin N slab and put a dot of P on each side, heated until the two P-dots nearly touched inside the N. That's quite different from multiple diffusions on on surface, "planar", how all Silicon is made now.



While this drawing is not to-scale, it shows the idea. Carriers injected into the B space are far more likely to hit the C pin than the E pin. This gives large Beta in normal connection, small Beta in reverse connection.

[merlinb]

Is Q2 even required? It seems to me that it could be replaced with a single diode. (In fact, the emitter-base resistor looks suspiciously superfluous too...)


(In SIM I also get about 8V reverse bias across the BE junction of Q1. How does this not damage Q1 when it is rated for Vbe ~5V max?)

[R.G.]

Is Q2 even required? It seems to me that it could be replaced with a single diode. (In fact, the emitter-base resistor looks suspiciously superfluous too...)


(In SIM I also get about 8V reverse bias across the BE junction of Q1. How does this not damage Q1 when it is rated for Vbe ~5V max?)

Yeah, I kinda went through these issues in the original design. For such a simple circuit, there are a deceptively large number of issues hiding in there.

The NPN is not needed if all you want is to let current through and are not worried about how low you drive the voltage across the PNP. Of course, if you can't get the voltage drop across the PNP down under 0.5V, there's no reason not to use a silicon diode, and if you can't get it under 200mV there's no reason not to just use a 1N5817 Schottky.

A sillicon PNP can saturate to well under 50mV if driven to a low enough saturation gain.  That's one reason you need the NPN - it bangs the PNP into saturation as hard as you want to set it, limited only by that 3.3K base drive resistor.

The base-emitter resistor on the PNP is perhaps superfluous, but I learned to set up switching transistors back when germanium leakage was a hot topic, and a base-emitter resistor to pull base leakage under the base-emitter turn-on voltage was de rigueur. It forces even somewhat leaky transistors to turn off when you want them to, and I have a hard time just trusting silicon to not leak at all.

The reverse voltage across the NPN base-emitter is held to 0.7V by the diode. The reverse current through this diode is limited to the reverse voltage minus one diode drop by the 10K resistor which sets the forward base current to the NPN. In reverse mode, the NPN's base-collector, which is acting like a "base-emitter" for reverse mode, is reverse biased by the applied voltage, so it tries not to conduct as well as it can. The current through the NPN's "collector" in the reverse mode is limited by the fact that current leaking through it is pulled to the reverse voltage by the 10K on the base, which torpedoes the reverse-mode gain. There is current through the PNP base-emitter, and that's where the majority of the reverse leakage current which lets some reverse voltage appear on the output happens. This reverse voltage is significant to an open circuit, under about a diode drop or two for 1M, and under a diode drop for 10K and under.

As to damage, I checked and none of the currents in the base-emitters is damaging, even if they do break, because they're limited by series resistors. This does damage their noise performance, but then they are not being used as low noise preamplifiers. Silicon junctions are not damaged in the "don't work any more" sense by being reverse broken (think of zeners and avalanche diodes).

The circuit could probably get by without the two base-emitter resistors. The other parts are either needed to work, or needed to work well, at least by the goals I set for myself.

[R.G.]

A similar plan could be developed around a P-type FET. However P-FETs (JFET or MOSFET) are not common parts like PNPs.

It was. http://www.geofex.com/Article_Folders/mosswitch/mosswitch.htm from 11-14-99, copied as an application of Bob Pease's article in EDN.

This was met in 99 and later with cries of MOSFETs, especially P-channels being hard to find, expensive, easy to damage, etc. The "cheap and good" design was an attempt to make it both easier and cheaper to build with available parts.

[merlinb]

A sillicon PNP can saturate to well under 50mV if driven to a low enough saturation gain.  That's one reason you need the NPN - it bangs the PNP into saturation as hard as you want to set it, limited only by that 3.3K base drive resistor.

But the best the NPN can do is to itself saturate, allowing max base current for the PNP! The NPN doesn't appear to be used as anything more than a switch. Exactly the same result obtains if you use a diode instead (and possibly a smaller resistor, say 2.7k, although even with 10k it SIMs perfectly well up to 100mA load current). If anything, the NPN hinders the base current of the PNP, since the NPN might not be fully on, whereas a diode allows max current regardless.

The reverse voltage across the NPN base-emitter is held to 0.7V by the diode.

I was referring to the BE voltage of the PNP, which reaches about 8V during 'blocking mode', as Walt and PRR noted.

[R.G.]

A sillicon PNP can saturate to well under 50mV if driven to a low enough saturation gain.  That's one reason you need the NPN - it bangs the PNP into saturation as hard as you want to set it, limited only by that 3.3K base drive resistor.

But the best the NPN can do is to itself saturate, allowing max base current for the PNP! The NPN doesn't appear to be used as anything more than a switch. Exactly the same result obtains if you use a diode instead (and possibly a smaller resistor, say 2.7k, although even with 10k it SIMs perfectly well up to 100mA load current). If anything, the NPN hinders the base current of the PNP, since the NPN might not be fully on, whereas a diode allows max current regardless.

It's a matter of degree and intent.

Sure, a diode can do that. Using an NPN sharpens up the turn on, and can turn on the PNP at lower voltages where the PNP with just a resistor and diode might be current starved. The NPN can be fully saturated on the PNP's base current with very small base current of its own, 1/beta of the PNPs current. I thought it was a fair trade for being able to ensure the PNP was banged on really hard and being able to better control when and how that happens.

Depends on what performance you want. My circuit is certainly not the only way to do this.

I was referring to the BE voltage of the PNP, which reaches about 8V during 'blocking mode', as Walt and PRR noted.

OK.
I did at one time have a diode in series with the PNP base for just such an emergency. For 9V, it didn't improve the actual performance much, so I took it back out. The leakage through the NPN backwards and the PNP base account for the original poster's note. In practice, it seems to work OK without it. It's OK to put it back in, too. That means the circuit turns on at a higher point in the ramp up of the input voltage, which may or may not matter. It cuts the reverse leakage, which is good, but without it the circuit doesn't seem to let components die, so I decided it was OK to leave out.

As I'm always harping on, there probably isn't a perfect anything unless one defines what "perfect" means very closely. Both simpler and more complex circuits do this. This is just one compromise.

[PRR]

> the {PNP} emitter-base resistor looks suspiciously superfluous too...

I mentioned the theory of this: "Perhaps the 4.7Meg could be higher.... However with the 4.7Meg omitted and temp run up to 50 C, I get 13pA output leakage, versus 11uA with the 4.7Meg."

In simulation, omitting the 4M7 gives a million times less output leakage.

If you believe the simulation, it IS totally superfluous, actually "hurting" the (still excellent) performance.

I accept that 999 of a thousand small PNPs will not need the 4M7; but like R.G. I mistrust loose bases and feel that "If you get a bad-day PNP it may work better with the 4.7Meg."

> Is Q2 even required?

If you _know_ your load is "light", then that plan works too.

If hFE is as low as 100, and your load gets near 100 ohms (90mA at 9V) then the PNP comes out of deep saturation and has voltage drop similar-to or greater-than a simple diode blocker.

You or I may know by inspection if a particular pedal may suck 90mA; such learned insight is not universal. (Witness the proposal to run a 50 Watt amp with a 1044 voltage booster.) R.G. squandered the extra dime for for "robust": it will stay saturated for about any load which will fit in a pedal.

> The leakage through the NPN backwards and the PNP base

My untrustworthy sim says the main leakage is the 4.7Meg times the reverse Beta of the PNP. That the NPN is dang-near dead (not even leaking significantly), but the output leakage is greater than the 4.7Meg's current alone.

Walt and R.G.'s observations suggests that reverse Beta of their parts is unity or less, which appears to be a reasonable value (and certainly more trustworthy than SPICE models worked far outside "normal" conditions).

[alparent]

OK I'm going a bit off topic.....but I need to recap.

1 - If I only use a diode in series I'm draining to much power from my battery.
2 - If I use the diode in parallel. I don't drain power, but I can I can blow the thing up if I leave it for to long.
3 - If I use the MOSFET, It's an almost perfect solution (can get the proposed replacement MOSFET for 50 cents at Futurlec) but there was something about adding a Schottky in there?
4 - Now this new cheaper solution (but alot of parts for the little power board I'm working on)
5 - And I reed also that using a 1n5819 in series isn't so bad for simple polarity protection.

Is that about right?

[merlinb]

> Is Q2 even required?
R.G. squandered the extra dime for for "robust": it will stay saturated for about any load which will fit in a pedal.

But RG's design isn't more robust by using the NPN. Replacing the base resistor with 3.3k makes the diode version equally as robust as RG's version, less three parts. (A 2N3906 is rated for 200mA max)

[PRR]

> Is that about right?

No.

A series diode turns a 9.0V supply into a ~~8.5V supply. This is often very acceptable.

A shunt-diode SHORTS OUT a reversed supply. A battery will go flat in minutes. A small cheap wall-wart may start to smoke. A beefy wall-wart will smoke the diode and THEN reverse-volt your precious pedal. (You could add a fuse and let that be the burn-out point.)

The MOSFET does not need protection up to 20V per spec (probably 40V in real life, if you like to push the ratings; "20V" is an arbitrary spec). Once in-circuit it is fine; the danger is handling the naked MOSFET. Most "power" MOSFET (even dinky ones) will survive handling. (We old guys remember when an RF MOSFET would fail at a touch.)

Does your pedal really need reverse protection? Many don't.

[alparent]

Thanks for that PRR.

So if I want to keep the series diode approach. Is a 1N5819 better? Or will all diodes turn my 9v into 8.5v

Or is there a better diode I should use?