PNP as switch with different emitter voltage

You may not even have to do that. See one of my other replies about driving two loads with the other FF output. I've driven heavier loads reliably at much higher frequencies with a single CMOS logic output.

The OP says this is on the order of seconds (manual shut-off.)

Assume Q2 nearing saturation sufficiently well with beta=50, then the Q2 base current should be about 15mA/50 or 300uA. The on/off control needs to be able to sink 300uA, plus a little more for R2 (say 10% of 200uA) and Q1 base (likely only 2-3uA.) R1 needs to supply that when the on/off control is at or near 0V. Assume about 1V for compliances needed by Q1's Vbe and the sinking output near 0V... so:

R1 = (5V-1V)/(300uA+30uA+3uA) = 12k R2 = 0.7V / 30uA = near 22k

Close enough for student work. Could add a speed-up cap across R1.

Jon

Yes, I had thought about piggybacking onto the output already in use, but I thought it would be less intrusive to use the other output (there is already a trace with an unused solder pad out of it).

Jon

The problem is that an NPN on this output will give me the opposite from what I am after. I am trying to switch on the load from a 0V signal, and turn it off with a 5V signal.

NPN would work if I was trying to switch the load with a 5V signal, but that is not what I am doing.

Jon

Good student work, Jon ;-) ...Jim Thompson

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You have been given two different ways of connecting the NPN. The configuration where the NPN base is at 5V and the emitter is driven by the logic output does what you want.

It needs some resistors in there too, to limit the current. Try it in LTSpice.

meant 'of 300uA'

Jon

Sadly, I've never taken or audited a single course in elecronics. But let it never be said I cease being a student. :)

Now, what about a bit of a snubber resistor on the base of Q1, too?

Jon

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You must have missed my original post in this thread.

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You must have missed my original reply in this thread.

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And, as a consequence, you're now being forced to use two transistors,
a Zener, and a few resistors.

My original solution allowed you to use a single transistor and
resistor.

Why did you find that objectionable or did you even read my post?

That is what you need.. Its a 2A Gate driven output via an optical input, which you can drive with your circuit.

This optical has either low or high side, or use both of you wish :)

1 R to drive it from your logic circuit. If you want high side, connect the Vcc to your 18 volts. If low side is what you want? connect the Vee to the common and cycle the 18 volts through your buzzer to the output (6)

P.S. I wouldn't push that 2A's continuos but I think you can drive that buzzer of yours which should be far under that load. And you may want to put a snubber on it.

Jamie

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How's he going to get 10mA to drive the LED from a 4013?

Interesting, thanks Jon, I hadn't seen a switched emitter before.

I'd still like to do this with only one transistor, though. I'm wondering if something like this would work:

When the IC output was 0, the piezo would draw current through R2. When the IC output was high, the transistor would conduct, pulling down the voltage at the collector.

Dunno if that might work, though. It'd depend on if R2 would be small enough to still allow enough current through the piezo, while not burning out the transistor, but still big enough to pull down the collector enough to shut down the piezo (~3 volts).

Jon

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Yeah, why bother when it's not necessary?

Something wrong with using a transistor to drive the LED ? THe idea was to keep what he had and put that PNP on the logic supply line to drive the LED via a R instead of the buzzer. The output of the coupler then could be connected to what ever he pleases with in it's range..

you did notice I said "with your circuit"

Jamie

Do you need to ground one side of the load? If you can run the piezo between V+ and a switch to ground, all you need is an NPN and a resistor. Even better, just an N channel mosfet, 2N7002 maybe.

John

He could get more leeway by putting a diode between the emitter and 18v, and the cost of reducing the voltage across his load, which is now revealed to be a piezo beeper.

However, that pushes up the component count again, and space was stated to be an issue.

Sylvia.