# PNP as switch with different emitter voltage

Discussion in 'Electronic Design' started by Jon Danniken, Jan 16, 2011.

1. ### Jon DannikenGuest

Hi, I am trying to take advantage of an extra flip/flop output in an
existing circuit. The driving signal is either 0V or +5V, but the emitter
voltage (to drive the switched load) is +18V.

This is the circuit I am planning to use:

+18V
|
Rb |
___ |/
On=0V ---|___|-----| PNP
Off=5V |<
\
|
.-.
| |
'-'
|
|
===
GND

I am going to try getting the base resistor to be such at the transistor is
off when the IC is at 5 VDC, and on when the IC is at 0 volts.

Besides fiddling with the base resistor, is there a different topology, or
modification I can make to the existing one to make this work better?

I do realize that putting an NPN transistor off of the IC's output would
solve this nicely, but space is at a premium, and I would like to do this
with just the one transistor.

Any ideas?

Thanks,

Jon

2. ### PimpomGuest

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.

3. ### Sjouke BurryGuest

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?

4. ### Guest

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
333ohms.
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.

5. ### Sylvia ElseGuest

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

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.

6. ### Sylvia ElseGuest

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:

0.6 = (18 - 2.5) * R2/(R1 + R2).
0.6R1 + 0.6R2 = 15.5R2

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.

7. ### Sylvia ElseGuest

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

But as a one off...?

Sylvia

8. ### JamieGuest

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

jamie

9. ### PimpomGuest

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

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.

10. ### Jon DannikenGuest

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

Jon

11. ### Jon DannikenGuest

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

12. ### Jon DannikenGuest

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

13. ### PimpomGuest

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.

14. ### PimpomGuest

You may not even have to do that. See one of my other replies
heavier loads reliably at much higher frequencies with a single
CMOS logic output.

15. ### Jon KirwanGuest

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

16. ### Jon DannikenGuest

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

17. ### Jon DannikenGuest

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

18. ### Andrew HolmeGuest

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.

19. ### Andrew HolmeGuest

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

20. ### Jon KirwanGuest

meant 'of 300uA'

Jon