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PNP as switch with different emitter voltage

J

Jamie

Jan 1, 1970
0
Spehro said:
On 17/01/2011 2:41 AM, John Fields wrote:

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


On 17/01/2011 12:20 AM, Sylvia Else wrote:
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...?

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


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.

---
Indeed; that's why I'll stick with my original solution of an N MOSFET
(with, conceivably, no gate resistor) being driven by the 4013's Q
output or an NPN with a gate resistor being driven from the same
point.

That is, if Q isn't being loaded so heavily by what's already there
that my scheme won't work.
_
If it won't, and he's forced to use Q, then I'd bite the real-estate
bullet and use two NPN's, like this:

+18
+18 |
| [BUZZER]
[10k] |
| C
+--[10k]--B 2N3904
| E
_ C |
Q>--[10k]--B 2N3904 GND
E
|
GND

No Zeners, no tricks, no marginal operation, just a rock-solid switch.


You don't need the 10K between C and B of the 2N3904s.

He could also use one of these, plus one resistor:

http://www.onsemi.com/pub_link/Collateral/MUN5211DW1T1-D.PDF
Hey, thats a good idea, dual package logic transistor..
 
J

Jon Danniken

Jan 1, 1970
0
Jon said:
I don't think so.

Assume the piezo requires Ip=15mA and Vp=? volts, for now.
When the piezo is sounding at 15mA, your R2 must be <=
(18-Vp)/Ip or [1200-(200/3)*Vp]. Of course, if Vp is 18 then
R2=0 and you are obviously in trouble. So you need to get by
on less than Vp=18V with the buzzer. I suspect you can. But
let's continue on a different path.

Now assume the piezo is off and Q1 is turned on. At this
point you've got to cause Q1's collector to drop below some
voltage required for the buzzer to work. The arrangement you
have above doesn't really allow you to do anything but treat
the BJT as a switch, so you need to figure that there will be
18V across R2 in order to turn off the piezo, so Ic*R2=18.

I haven't looked up your output capability when high, but I
recall reading Sylvia saying you are guaranteed 440uA for HI.
Assume it is for now. You can plan on say beta=30 for the
BJT.

Then Ic can be as much as 30*440uA or 13.2mA. But also
recall that Ic*R2=18. So from this we compute R2=18/13.2mA
or 1364 ohms.

Which is a problem because 1364 ohms * the piezo's 15mA is
more than 18V drop across R2. Which means the piezo will be
off either way. It just won't work.

Even if you assumed that beta=100, R2 would be about 410 ohms
and 410 * 15mA = 6.15V. Which means your piezo would have to
work okay with less than 12V. Assuming it still drew 15mA,
then. Likely, it draws maybe a little less meaning the
voltage across it is a little higher and it finds the balance
point at something less than 15mA and something more than 12V
to work with.

You could try it, I suppose. Maybe try R2=390 or 450?

Aye, thanks Jon. I think even if I did get it all dialed in it probably
wouldn't be very stable, if the temperature change.

Jon
 
J

Jon Danniken

Jan 1, 1970
0
Pimpom said:
There can be no hard and fast rule because the measured Hfe
depends on the levels of current and Vce at which it was measured
(even if we disregard the effects of temperature). Hfe decreases
with Vce, rapidly as saturation is approached. Ic/Ib = 10 is
usually used for specifying Vce(sat) in datasheets. My own rule
of thumb is to use that one-tenth rule (Ib = Ic/10) for low-beta
devices and Ib = Ic/20 or Ic/30 for higher betas. These are for
absolute peace of mind. I sometimes use Ic/40 or Ic/50 in a pinch
for types like the BC547B I mentioned earlier (PNP version =
BC557B).

The KN2907 is obviously a clone of the popular PN2907 which is
itself a plastic version of the venerable 2N2907. 0.8mA of base
current should be quite adequate. You probably won't notice any
difference with 0.5mA. Even if the transistor is not driven into
hard saturation, that kind of load will not burn it up at any
level of base drive.

Aye, indeed you are correct. I got the circuit breadboarded earlier today,
and although I was seeing 0.7mA with a 22k base resistor, with a 16v zener
on the base, it only drew 0.1mA. I subbed in a 5.6k resistor, and I'm just
a hair below 0.5mA.

The peizo beeper is getting it's full voltage though, so I'll stick with
this combination. Since the transistor has an Hfe of 170, running it at a
fifth of that should give me adequate saturation.

Thanks again for the suggestion to use the zener, it's working perfectly,
and I only had to bugger piggyback the existing circuitry at one point (to
pull off the 18V before the regulator).

Jon
 
J

Jon Danniken

Jan 1, 1970
0
Spehro said:
Where it is specified is more than adequate, usually Ic/Ib = 10. I
often use 20, unless operation at temperature extremes is required
(and even then, self heating may help the cause, but it requires
analysis).

Thanks Spephro. I ended up going with 0.5mA Ib to control my 15mA load, so
I'm going with 30. The transistor is measured at 170, so I think that
should be good.

Jon
 
J

Jon Danniken

Jan 1, 1970
0
Jon said:
[snip]
Any ideas?

Thanks to everyone who contributed to this thread. I ended up going with
Pimpom's suggestion of a zener on the base, which is working well for me.

I ended up learned a lot more than I had set out to do, which, of course, is
half the fun of doing a project like this, and I printed a few notes to
stick in my Engineer's Notebook for the next time I have a funny project.

Thanks again for all of the suggestions, they are all very much appreciated.

Jon
 
F

Fred Bartoli

Jan 1, 1970
0
Pimpom a écrit :
No. The OP wants to turn the beeper on when the FF output is low.

So use the FET common gate tied to Vcc, and hook the FET source to the
logic output.


Just one transistor...
 
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