PNP as switch with different emitter voltage

Insert a 15-volt zener diode in series with the base resistor. This way, the zener will break down and drive the base only when the FF output is in the low state. Depending on the gain of the transistor, I suggest 1.5-3.3k base resistor. It's also a good idea to place a resistor across the b-e junction to bleed off any leakage current through the zener diode. Something like 10k should be OK.

Very unstable, and the slightest error will blow the circuit attached to the output. Connect to a save voltage instead, like 5 or 3.3 V depending on the type of destination. If you insist on this config. at least put a bleeder resistor between base and emitor. That makes it simple to reach an off-state,13 v on rb and r-bleed produce less then .7 volt on the base-emitor, and when switching, 18 v on them and Vbe gets a bit higher. Example:13 volt:Rb 12k,rbe 560 ohm, ~.5 volt, so off. 18 volt: , ~.75 volt so on(sort of). But it is a lousy solution anyway.

The way you did it, the transistor will be on all the time, never giving 0 volt out. So what output do you need? And does it HAVE to be a pnp transistor?

Not going to work. The base-emitter is always reverse biased and the collector-base is always conducting (actually the transistor will operate with reverse beta but won't switch). If you reverse the emitter and collector with it's emitter hooked to +5V and the load in the collector, it should work, but you're only going to get a ~5V swing. You will get ~15mA if the load is

Not going to happen. There will always be current flowing in the collector-base diode. Reverse beta is really crappy so it's not going to switch.

Reverse the emitter and collector and tie the emitter to +5V.

Other than the fact that it won't work, it looks great.

What are you trying to achieve? 18V across the load when the logic turns on?

Cannot turn off either sex of bipolar transistor that way; the circuit is linear. Try fiddling with a FET and a zener..

I assume you've put the transistor the wrong way up by mistake.

If you attach the emitter to +18V and the collector to the load, and then stick a resistor between the base and +18V, then by a suitable choice of resistor values, you can make the transistor switch.

Except that I suspect that your inputs are not really 0 and +5. In particular, your input may not be willing to sink current at 5 volts. Depends on the IC. A diode to 5 volts could cure that (but watch out that you don't ask the PS to sink current in the process), but it will still be pulling the IC output a bit above 5, and you'd have to check that that's OK.

However with such an arrangement you also have the problem of whether you can actually pull enough current from the base to switch the transistor with a 15ma load, which is going to depend on the actual properties of your IC and the beta of your transistor.

You'd be driving the transitor into saturation, which means there'll be a turn-off delay. What switching rate are you dealing with?

Sylvia.

What's on the left of Vin is never asked to source current, but presumably there's no doubt it can sink current when it's driven low. The doubt would be as to its behaviour when driven high - usually it would expect to source current in this state, so will it still sink it? But you're assuming that it will, so...

Consider Veb when the transistor is on to be 0.6V (Veb because it's PNP).

Suppose we want to switch when Vin traverses 2.5V.

Then:

0.6R1 = 14.9R2

R1 = approx 24 * R2

Vb would only rise to 17.5 (Veb = 0.5) volts when Vin is 5v, but that should be enough to turn the transistor off.

The main issue I see with this arrangement is whether it is possible to sink enough base current to saturate the transistor given the value of load, which depends on the thing IC driving Vin, and the transitor beta, but I said that.

If it's an ideal IC, and can sink an infinite amount of current in both states, then it's just a matter of choosing the resistor values with knowledge of the worst case beta.

It would be more difficult if the OP wanted the opposite switching behaviour.

Sylvia.

I would add that I wouldn't want to use this approach in a circuit intended for mass production. A combination of adverse IC output characteristic, resistor value and transistor variation would mean it was asking for trouble.

But as a one off...?

Sylvia

I am sure by now, most have seen the problem.

Your transistor is backwards.

jamie

On Mon, 17 Jan 2011 01:51:47 +1100, Sylvia Else wrote:

--- Maybe, but still kinda scary unless you could pin everything down.

In this sim the one on the left is using standard 5% resistors and the other four are the permutations with the resitors at the tolerance limits...

Version 4 SHEET 1 1524 932 WIRE -752 -64 -1024 -64 WIRE -640 -64 -752 -64 WIRE -240 -64 -512 -64 WIRE -128 -64 -240 -64 WIRE 272 -64 0 -64 WIRE 384 -64 272 -64 WIRE 784 -64 512 -64 WIRE 896 -64 784 -64 WIRE 1296 -64 1024 -64 WIRE 1408 -64 1296 -64 WIRE -752 -32 -752 -64 WIRE -240 -32 -240 -64 WIRE 272 -32 272 -64 WIRE 784 -32 784 -64 WIRE 1296 -32 1296 -64 WIRE -640 32 -640 -64 WIRE -128 32 -128 -64 WIRE 384 32 384 -64 WIRE 896 32 896 -64 WIRE 1408 32 1408 -64 WIRE -880 80 -912 80 WIRE -752 80 -752 48 WIRE -752 80 -800 80 WIRE -704 80 -752 80 WIRE -368 80 -400 80 WIRE -240 80 -240 48 WIRE -240 80 -288 80 WIRE -192 80 -240 80 WIRE 144 80 112 80 WIRE 272 80 272 48 WIRE 272 80 224 80 WIRE 320 80 272 80 WIRE 656 80 624 80 WIRE 784 80 784 48 WIRE 784 80 736 80 WIRE 832 80 784 80 WIRE 1168 80 1136 80 WIRE 1296 80 1296 48 WIRE 1296 80 1248 80 WIRE 1344 80 1296 80 WIRE -1024 176 -1024 -64 WIRE -912 176 -912 80 WIRE -640 176 -640 128 WIRE -512 176 -512 -64 WIRE -400 176 -400 80 WIRE -128 176 -128 128 WIRE 0 176 0 -64 WIRE 112 176 112 80 WIRE 384 176 384 128 WIRE 512 176 512 -64 WIRE 624 176 624 80 WIRE 896 176 896 128 WIRE 1024 176 1024 -64 WIRE 1136 176 1136 80 WIRE 1408 176 1408 128 WIRE -1024 320 -1024 256 WIRE -912 320 -912 256 WIRE -912 320 -1024 320 WIRE -640 320 -640 256 WIRE -640 320 -912 320 WIRE -512 320 -512 256 WIRE -400 320 -400 256 WIRE -400 320 -512 320 WIRE -128 320 -128 256 WIRE -128 320 -400 320 WIRE 0 320 0 256 WIRE 112 320 112 256 WIRE 112 320 0 320 WIRE 384 320 384 256 WIRE 384 320 112 320 WIRE 512 320 512 256 WIRE 624 320 624 256 WIRE 624 320 512 320 WIRE 896 320 896 256 WIRE 896 320 624 320 WIRE 1024 320 1024 256 WIRE 1136 320 1136 256 WIRE 1136 320 1024 320 WIRE 1408 320 1408 256 WIRE 1408 320 1136 320 WIRE -1024 384 -1024 320 WIRE -512 384 -512 320 WIRE 0 384 0 320 WIRE 512 384 512 320 WIRE 1024 384 1024 320 FLAG -1024 384 0 FLAG -512 384 0 FLAG 0 384 0 FLAG 512 384 0 FLAG 1024 384 0 SYMBOL pnp -704 128 M180 SYMATTR InstName Q2 SYMATTR Value 2N3906 SYMBOL res -656 160 R0 SYMATTR InstName R4 SYMATTR Value 1200 SYMBOL voltage -1024 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V3 SYMATTR Value 18 SYMBOL voltage -912 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V4 SYMATTR Value PULSE(5 0 0 1e-6 1e-6 1 2) SYMBOL res -768 -48 R0 SYMATTR InstName R5 SYMATTR Value 240 SYMBOL res -784 64 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R6 SYMATTR Value 5100 SYMBOL pnp -192 128 M180 SYMATTR InstName Q4 SYMATTR Value 2N3906 SYMBOL res -144 160 R0 SYMATTR InstName R10 SYMATTR Value 1200 SYMBOL voltage -512 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V7 SYMATTR Value 18 SYMBOL voltage -400 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V8 SYMATTR Value PULSE(5 0 0 1e-6 1e-6 1 2) SYMBOL res -256 -48 R0 SYMATTR InstName R11 SYMATTR Value 228 SYMBOL res -272 64 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R12 SYMATTR Value 4845 SYMBOL pnp 320 128 M180 SYMATTR InstName Q5 SYMATTR Value 2N3906 SYMBOL res 368 160 R0 SYMATTR InstName R13 SYMATTR Value 1200 SYMBOL voltage 0 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V9 SYMATTR Value 18 SYMBOL voltage 112 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V10 SYMATTR Value PULSE(5 0 0 1e-6 1e-6 1 2) SYMBOL res 256 -48 R0 SYMATTR InstName R14 SYMATTR Value 228 SYMBOL res 240 64 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R15 SYMATTR Value 5355 SYMBOL pnp 832 128 M180 SYMATTR InstName Q6 SYMATTR Value 2N3906 SYMBOL res 880 160 R0 SYMATTR InstName R16 SYMATTR Value 1200 SYMBOL voltage 512 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V11 SYMATTR Value 18 SYMBOL voltage 624 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V12 SYMATTR Value PULSE(5 0 0 1e-6 1e-6 1 2) SYMBOL res 768 -48 R0 SYMATTR InstName R17 SYMATTR Value 252 SYMBOL res 752 64 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R18 SYMATTR Value 4845 SYMBOL pnp 1344 128 M180 SYMATTR InstName Q7 SYMATTR Value 2N3906 SYMBOL res 1392 160 R0 SYMATTR InstName R19 SYMATTR Value 1200 SYMBOL voltage 1024 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V13 SYMATTR Value 18 SYMBOL voltage 1136 160 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V14 SYMATTR Value PULSE(5 0 0 1e-6 1e-6 1 2) SYMBOL res 1280 -48 R0 SYMATTR InstName R20 SYMATTR Value 252 SYMBOL res 1264 64 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R21 SYMATTR Value 5355 TEXT -1010 352 Left 0 !.tran 0 10 .1 uic

--- JF

If you have room for a zener, and resistors, why not room for an NPN?

Tie NPN base to +5V

Resistor from emitter of NPN to logic signal

Collector of NPN to base of PNP

Resistor from base of PNP to +18V

Emitter of PNP to +18V

Load from PNP collector to ground

Resistor value calculations left as an exercise for the student ;-) ...Jim Thompson

Yeah. I actually started out meaning to add "if there's room", but forgot about it.

I used the zener technique in a design last year. The difference is that the digital chip and the PNP transistor used the same 12V supply. It would have worked without the zener if I'd kept to my initial design, but I added a load between the chip output and ground. This pulled the output below Vdd in the HI state, low enough to turn the PNP tr on. Inserting a 5.6V zener and changing the base resistor value solved the problem.

Thanks Pimpom, I hadn't thought of that. I'll pick up a zener tomorrow and play around with that.

Jon

Indeed I did, I went with what the ascii conversion software gave me instead of flipping and mirroring it. My apologies for not noticing that.

It's coming off of a 4013 flip flop, and I did actually measure the outputs to be +5 or 0. The other output from the IC is inverted, and in use (driving an NPN), hence my desire to use the PNP on the other output.

Measured in seconds. The load is a piezo beeper that goes off until I get annoyed enough to go push the reset button.

Thanks,

Jon

Hi Sjouke, it doesn't have to be a PNP, but the logic of the IC output (5V when I want this off, 0V when I want this on) suggested that to me. The output is driving a 15mA piezo beeper.

I'm going to try the suggestion of the zener on the base (and a BE resistor) and see where that gets me.

Thanks,

Jon

What's the load on the NPN? Is it similar to or lighter than the beeper? If so, the 4013 can drive the two loads from one FF output if the transistors have reasonably high gain.

For example, a BC547B or a 2N3906 can switch two 15 mA loads with ~1 mA base current. A 2.2k-3.3k resistor plus the 4013's internal output resistance will supply more than enough base current to saturate the transistor with a 30mA load. If it's not convenient to parallel the two loads and drive them both with the existing NPN transistor, you can add the extra NPN and drive it with the same FF output via a 3.3-4.7k base resistor. There's no need to go to the trouble of using a PNP on the other FF output.

ter

You can do this with a NPN either with base at +5V, emitter driven, or with emitter GND, base driven, but not with a PNP transistor (unless you want to use some fussy level-shift tricks). A PNP in conduction has to be controlled by voltages (base vs. emitter) at the positive rail.