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Driving a PNP with a 555

S

Spehro Pefhany

Jan 1, 1970
0
I need to project the beam a considerable distance. Hence the
high peak current. It's also why I want to have the option of
increasing the peak current above 250mA. Most 5mm IR LEDs I've
seen are rated for 1A peak, 100mA continuous.

Well, possibly the most elegant circuit would be a CMOS 555 driving a
P-channel MOSFET gate directly, assuming a reasonable supply voltage
so as not to be too close to the maximum Vgs.

Eg. http://www.diodes.com/datasheets/ds31125.pdf
http://www.nxp.com/documents/data_sheet/ICM7555.pdf

Total cost about 30 cents in 1K, even from Digikey, and it could
switch as much as a couple of amperes with 20% duty cycle, and the 555
won't need any extra bits to get 20% on-time.

Of course if it's a school project you may have to use the parts in
the parts crib, eh?
 
P

pawihte

Jan 1, 1970
0
Spehro said:
Well, possibly the most elegant circuit would be a CMOS 555
driving a
P-channel MOSFET gate directly, assuming a reasonable supply
voltage
so as not to be too close to the maximum Vgs.

Technically, yes, it's an elegant solution. But P-channel MOSFETs
are not easily available here and the object was to minimize
component count. Otherwise the 12V power supply is high enough to
allow inserting a zener diode or an LED in series with the base
resistor to raise the turn-on threshold.
Eg. http://www.diodes.com/datasheets/ds31125.pdf
http://www.nxp.com/documents/data_sheet/ICM7555.pdf

Total cost about 30 cents in 1K, even from Digikey, and it
could
switch as much as a couple of amperes with 20% duty cycle, and
the 555
won't need any extra bits to get 20% on-time.

Of course if it's a school project you may have to use the
parts in
the parts crib, eh?

Cost is not a primary concern here. I think I'll just reduce the
b-e shunt resistor from 1k to 220 ohms. That will steal 3-3.5mA
from the base drive, but still leave about 20mA which should be
enough.
 
P

pawihte

Jan 1, 1970
0
Tim said:
I think it was saturating at 2-3 Vbe's. I could go check.

The important part is getting the B-E resistors small enough so
the
transistor is certainly on or off. Which actually, with 1k and
1k, it
should only be turning off with less than 1.2V (if it's 1.8V,
the PNP
might never fully turn off!). Hmm, I should probably change
those
resistor values then.

I'm also thinking along the same lines. I'd originally intended
to use 470 ohms for drive and 1k as the b-e shunt. Reducing the
shunt to 220 ohms will give me better peace of mind. It will
divert 3-3.5mA from the base drive, but the remaining 20mA should
still be enough to saturate the transistor. If not, I could halve
each resistor and still stay well within the 555's output
capability.
 
P

pawihte

Jan 1, 1970
0
Jamie said:
No.
The output of a 555 is not low due to a low side pulling on it
there
for, you should not see biasing effects being generated from
some low
side source of the 555.
THe output of a 555 on the high side is a emitter, so what
you have
there, using that pull up R, will actually bring the base to
the VCC
when the 555 is in the high state..
That's what I thought, but I thought I'd better check with you
guys in case there was something I missed.
 
S

Spehro Pefhany

Jan 1, 1970
0
So has most of the crap you sell on Ebay.

Don doesn't have any 555s at the moment. ;-)


Best regards,
Spehro Pefhany
 
T

Tim Williams

Jan 1, 1970
0
Don Lancaster said:
Use a PIC instead.
Then throw away the 555.
It has been obsolete for decades now.

Wait, who is this? Jan Lancaster? Don Panteltje?

Tim
 
P

pawihte

Jan 1, 1970
0
Don said:
Use a PIC instead.
Then throw away the 555.
It has been obsolete for decades now.

Would you use an Allen nut-bolt where a nail does the job just as
well?

A 555 is cheap (10 US cents each retail), available everywhere,
requires no programming tool or knowledge. Try to beat that with
a PIC. In any case, the versatility of a PIC is no advantage
here, and a PIC will still need to interface with the power
driver stage, which is what this thread is about.
 
P

pimpom

Jan 1, 1970
0
Michael said:
It think you would be way better off switching an npn on the
low side. I imagine you won't need a b/e resistor or any of
that
headache. Others have shown how to get a low duty cycle using
diodes, but you
can do it with just two resistors if you think out of the box:

,----R1-----,
| |
| ,--R2-+
| | |
,--+-----+-----+----+--,
| 8 7 6 5 |
| / |
| / |
| / |
| 1 2 3 4 |
'--+-----+-----+----+--'
| |
= out
|
gnd

Obviously the value of R1 has to be at least twice R2 for the
circuit to work at all, because the cap voltage has to fall
below 1/3
Vcc for the astable multivibrator function.
I don't have the time to calculate resistor values for a 20%
duty
cycle right now. You can do it by adjusting a pot until you
get
exactly 20%, then measure the resistance and replace the pot
with
fixed resistors.

Not very practicable. The required relative values of R1 and R2
are too critical to have oscillation *and* a low duty cycle.

The data sheet gives t2 (output low) = [R1*R2/(R1 +
R2)]C*ln[(R2 - 2*R1)/(2*R2 - R1)]. This requires making R2 very
close to 0.5R1 for a duty cycle much lower than 50%. Even at R2 =
0.49R1, the duty cycle is still about 32.76%. Keeping R2 >0.49R1
and <0.5R1 while having to adjust either of them for frequency is
just not practicable.

So it still seems the simplest solution is to keep Rbe low
relative to the base drive resistor. Next comes inserting a zener
diode or an LED in series with the base drive.
 
J

JosephKK

Jan 1, 1970
0
I hate to be a spoil sport, but with total component cost coming in at
under a buck, why don't you just breadboard it up and see what happens?

Hey i vote for that.
 
P

pawihte

Jan 1, 1970
0
Bob said:
I hate to be a spoil sport, but with total component cost
coming in at
under a buck, why don't you just breadboard it up and see what
happens?

There's always that of course, and the cost factor is certainly
trivial. But working things out on paper and thinking through the
whys, hows and ifs have their own reward. I'll bet this thread
has prompted more than one reader to channel their thoughts in a
direction they never took before.

I'm not against empirical techniques and I employ them at times,
but they do have limitations. For instance, a circuit may work
with a specific set of parts under one particular environment,
but may become unsatisfactory with a slight change of one or more
parameters. Thanks for your interest.
 
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