Do you have a comprehension problem? It's clear to me what he is doing.
He's rotating the leads on the DVM. He'll get 6V DC either way.
Did you notice he stated "In Circuit"? That should have rang a bell.
Lets try this.
How about if the 6VDC(+) volts on the cathode side of the diode..
In message , Eeyore writes
Totally unnecessary as you say but if you have one it can become like a third hand if you do a lot of analogue repair work.
Extremely useful device, similar to the component test on Hameg scopes if it's what I think it is. With a little experience you can spot faulty components very easily in circuit, leaky caps and analogue ICs included.
-- Clint Sharp
Ok, that helps a bit. The D44H8 is used to switch the lights on/off. It's an NPN, meaning the base must be about .6v more positive than the emitter for the transistor to turn the lights on. The
2N4403 is a PNP, meaning the base needs to be about .6v more negative than the emitter to turn that transistor on. You can look at the base circuit for the D44H8 to see where the + comes from that is needed to turn the transistor on.They are probably using the 2N4403 to control the D44H8, with some sort of input for checking whether the power supply is providing DC for the unit. The idea would be to drive the D44H8 hard so that it saturates, when mains power drops. Likely, the zener would also be in the circuit to cause the D44H8 to turn off when battery voltage drops to roughly
5.7 volts.Is the zener the diode that you are shorting out? Can you read the part number on the zener? (From that you can find its voltage rating.) It is probably something like 1N4733
Also, these lights usually (always?) have a test button or switch. If it has one, the switch circuit can provide helpful clues about the circuit. Tracing it will lead you to the circuit that turns the D44H8 on. Also you can use your ohmmeter to find out if the switch is normally open or normally closed. Do that test only with the battery removed and the AC power off.
Ed
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