Repairing/understanding PNP in charging circuit

Looks like they're using the transistor as a diode, silly but it works.

If you don't have a schematic, trace it out and draw one. Often you have to re arrange what you get into a more recognizable form. Research the devices. This is called "reverse engineering". The cost of this for a small item you describe might be $200-$500 worth of labor depending on research obstacles. This can be a fun thing to do if you are into challenges. Whole lot clearer when we are looking at the same page.

I have heard of transistors being used as Zeners, Varactors or other Diodes. Some devices use the same package as transistors.

What makes you think that the device in question is a PNP transistor ? Does it actually *have* a third leg that isn't connected anywhere ? The reason I ask this is that there is a range of wire-ended fuses which are in a TO92 package, just like a transistor, but they physically have only two legs at the two 'corners' or the package. These devices are typically marked "Nx" such as "N10" for instance. What is the descriptor silk screened on the PCB for this device ? If it really is a transistor, it is likely to be something like "Qx" or "Trx", but if it is a fuse of this type, it will likely be "ICPx"

Arfa

Welcome to "Learn By Destroying(tm)". You're off to a great start.

Because it's probably a power FET and not a PNP transistor. Are there any identifying marks on the device? A photograph perhaps?

Electrons flow from negative to positive. Holes flow from positive to negative. PNP transistors have the collector more negative than the emitter. Therefore, electrons will flow from collector to emitter in a PNP. It's the same if you think of "electric charge flow" as the flow of electrons. If you're confused, you have plenty of company. This might help:

(or make things worse).

I don't think the transistor went anywhere. It's still in front of you. Where did you expect it to go?

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Jeff Liebermann     jeffl@cruzio.com
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I tracked the part down to

Silk screen on the part is FZT 789A

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The transistor is only connected on the E and the C (tab) the other two are not connected to anything.

I tracked the part down to

Silk screen on the part is FZT 789A

The device having only collector and emitter connected, makes no sense at all. A transistor connected in this configuration, would represent essentially an open circuit (not a diode as someone else suggested) and no current would (or even *could*) flow between these terminals.

I see from the data sheet that it is in fact a surface mount device, so presumably all three terminals and the tab are at least soldered down to pads ? Could it be that the base connection is actually underneath the device - maybe even via a thru' plated hole ?

Arfa

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I started to trace this out, and found what you had said. There was a trace under the transistor that was painted over. And was very hard to see.

I made this trace diagram of the circuit.

By jumping the C and E on the transistor it works as expected when not plugged into the dc power(12v) which is expected as then it is connected just as the DC 12v power is. The only thing it appears stopping the battery power from reaching the device load is the transistor and the base resistor.

Aha! It's a temperature-compensated Zener diode, made by using the reverse breakdown of the BE junction and the forward CB diode. It should have about 7 V at breakdown for lowest temperature coefficient; datasheet suggests you will get 9V or so from this part..

Either that, or there IS some kind of connection to the base, and it evaporated or is hiding from you...

Purpose-build compensated diodes (like 1N821) are expensive by

I started to trace this out, and found what you had said. There was a trace under the transistor that was painted over. And was very hard to see.

I made this trace diagram of the circuit.

By jumping the C and E on the transistor it works as expected when not plugged into the dc power(12v) which is expected as then it is connected just as the DC 12v power is. The only thing it appears stopping the battery power from reaching the device load is the transistor and the base resistor.

Looking at your schematic, which maybe doesn't look *quite* right, then if R3 is good, the transistor should be on. You could try measuring it in-circuit with the battery disconnected, and you should see sensible readings. Failing that, remove it and read it. I don't know how much experience you have of getting devices like this off a board, but if you only have access to 'conventional' soldering equipment rather than hot air rework equipment, you need to be careful that you don't 'lose' any traces or pads.

Use a good quality desolder braid with a good sized iron - preferably a temperature controlled one of perhaps 50 watts max - to remove as much solder as you can with the braid, from both the pins and the tab. Then heat the tab alone with a small scalpel under its edge, until the solder flows enough to be able to slightly twist the blade to lift the tab no more than

1mm off the board. Then heat the pins, all at once if possible, and repeat the blade twisting under the device body, to lift them up from the board a little. This can be difficult to achieve, if the manufacturer has kindly glued the device to the board, before flowing it ...

You should now be able to use a fresh piece of desolder braid to get the remaining solder under the pins and tab, and the device should come off the board cleanly, and with no damage. It's not easy, but if you can solder ok, and understand about not overheating solder pads and causing them to delaminate from the board, then it's not overly difficult, either.

Arfa

It is entirely possible that I missed something, the traces are hard to see and hide under a few components. I spent quite a bit of time double checking with a multi meter for no resistance to help find beginning and end of traces. only if they made sense and would not have crossed another trace.

Testing the resistance on the PNP in circuit no battery it seems to change every time I test it.

To get a proper reading I will have to pull it out.

With battery in circuit, there is no resistance reading between C and E I get 10.5 volts to B

11.9 volts to E 2.4 volts on C (not sure where the 2.4 volts is coming from)

This should be letting the 11.9 volts through to C but its not

When this was working "properly" there was a parasitic load on the battery that would drain it within a a few days. which was the original reason for opening this up. Was to put a switch on the battery connection to disable the battery when not in use.

So there would be two switches. Original to turn device on. New switch to disconnect battery.

If the transistor problem eludes me too long, just removing it and using the switches to control it manually would suffice. Although if the device is on while charging the battery will not be disconnected from the circuit via the pnp.

OK. You should not read any resistance between collector and emitter i.e. it should be open circuit. For measuring junction resistances on bipolar semis, I have always preferred an 'old fashioned' 20k ohms per volt analogue multimeter - in my case an AVO 8 MkIV. Digital meters can give very misleading resistance readings across transistor junctions. With an analogue ohm meter, with the red probe to the base pin, you should read around 700 or

800 ohms to the collector and emitter, with the black probe, for a silicon PNP transistor. All other combinations of test probe polarity and pins, should read open circuit.

As far as the voltage readings that you are getting, they would seem to be wrong. With 11.9v to the collector, the base voltage should be about 0.6 to

0.7v below that - i.e. around 11.3v, not 10.5v. Assuming that R3 is in good order, this voltage difference is a fixed function of a silicon PN junction, so any deviation from that figure would tend to indicate a faulty device. The 2.4v 'output' that you are seeing on the collector, is probably just a voltage that is 'battering its way through' the device from the emitter input voltage. My next move would be to get some proper resistance readings for the device junctions, removing it if necessary. From the information you have given, I think that you are going to find that the transistor is indeed faulty.

Arfa

I should have clarified my post better. By no resistance reading between collector and emitter I meant it was closed circuit. I was expecting open circuit. But then everything would be working.

Thanks for the assistance on this. And I can relate to the advantages of using a analog meter. I once trouble shot a wiring harness in a car. And was getting a rogue voltage reading on a DMM when there should have been none. An anolog meter drained the residual charge it was getting from surrounding wires and then it would read as expected(no voltage). But until this was pointed out to me I was confounded. As it was just enough to light a dash led when the harness was unplugged.

Ill track down an analog and try to pull this out this week.

That's why even the chepest digital multimeters have a "diode test" mode that measures the forward voltage at 1 to 4 mA (depending on the multimeter, check the manual). Low-power-transistors should read about 600mV B-E and B-C, and open circuit C-E.

Regards, Michael Karcher

Yes, Michael, I've been in the service business long enough to understand the differences between a digital DMM and its diode test facility. However, it's not at all uncommon to get unstable readings, particularly if you are measuring 'in-circuit', and significant (in terms of actual digits) differences between two equally good devices. Even the temperature of your fingers can cause the digits to shift, as can the long cooling down period if the device has just been unsoldered from a board. And of course, differences between different categories of device.

When training others, I have found that this can give rise to confusion in students, who expect to see 'absolutes' on meters with digital readouts. An analogue meter used for this purpose gives a simple go / no go on a junction that in my experience, is a better than 99% indicator of a good or faulty device. The OP said at the start that he was a self-trainer in electronics service. He actually stated in one post that measuring the junction resistances in-circuit, resulted in a different reading every time. This is typically what you find when using a digital meter. Often, in these cases, a good old analogue meter will give a stable and consistent reading, which will allow you to make a sensible evaluation. This was the point that I was trying to make to the OP.

Arfa

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I started to trace this out, and found what you had said. There was a trace under the transistor that was painted over. And was very hard to see.

I made this trace diagram of the circuit.

By jumping the C and E on the transistor it works as expected when not plugged into the dc power(12v) which is expected as then it is connected just as the DC 12v power is. The only thing it appears stopping the battery power from reaching the device load is the transistor and the base resistor.

It would appear that with no DC1, the battery would forward bias U1 through R3 and power the device through SW1 unless the EB juction was wacked. With DC1 active, D1 keeps the base Hi and U1 is cut off. Battery is charged through D3 and device is powered by D2. D4 protects the device from spikes.

I pulled the transistor from the pcb and tested it. Using a diode test there was no V reading between the B and E using the leads either way. It was open both ways. A replacement PNP that I had provided a .6 v reading on the diode test(only one way as expected). Once soldered into the circuit everything is working now as it once was.

Thanks for the help everyone. esp Arfa I learn so much more working on something then reading about it.

- Eric