Driving a PNP with a 555

[pawihte]

I want to use a classic 555 timer IC to drive the base of a PNP
transistor through a resistor, the emitter of the transistor
being tied to the 555's Vcc. The 555 datasheet gives a graph for
the high-state output voltage vs. sourcing current, but not when
the load is tied to Vcc.

Vcc
--------------------------
| |
.|. |
| | |
| | |
'-' |
| |
555 out ___ | |<
------------|___|------|
|\
|
|

What I'm concerned about is: Is there a possibility that the
high-state output of the 555 drops low enough below Vcc to
partially turn on the PNP transistor? I could increase the
turn-on threshold of the transistor with diodes, an LED or a
resistive voltage divider, but I'd like to avoid that if it's not
needed.

 

[pawihte]

If it's a CMOS 555 then the output will drive to the rail, near
enough.

If it's a bipolar 555 then chances are the output is a
totem-pole,
which (if I remember correctly) won't drive _to_ the +V rail at
all
vigorously, but get in the way of a pullup at all. In fact (if
I
remember correctly) this was one way of interfacing bipolar
parts to
CMOS, if you didn't mind a bit of a speed hit. So check.

Since your bias network provides that pull-up, you're probably
fine.
To really drive things fast you may want a resistor from the
pin to
+V, before the base current-limit resistor.

Thanks. It's bipolar and, according to the datasheet, the output
is a totem-pole NPN-NPN push-pull. I want to drive an IR LED at
38kHz at about 250mA peak with a 20% duty factor. Do you think
it's OK as is?

 

[Tim Williams]

This works fine for me:
http://webpages.charter.net/dawill/Images/High Voltage Supply.gif

The MJE18008 switches in under a microsecond. I forget if I boosted the
base resistors at all -- 470 or even 220 ohms would be better on the 2N4403.

Tim

 

[pawihte]

Bipolar 555 or CMOS? CMOS, no problem. Bipolar (assuming 5V
supply),
use 750 Ohms base-to-emitter, 1.5K base-to-555-output.

Thanks. It's a bipolar 555, quasi-complementary NPN-NPN output
with a 12V supply. I want the transistor to drive some IR LEDs in
series at about 250mA peak, 0.2 duty. I thought I'd use 470 ohms
as the base drive resistor and perhaps 1K base-emitter. Do you
think that's not enough to ensure that the transistor will be off
when it should be?

 

[pawihte]

Refresh my memory, what's the sink current of a bipolar 555?

200mA source or sink.

You can't connect one end of LED to plus rail?

I considered that and 200mA might be good enough. But I'd rather
have the option of using a higher LED current or additional
LED-resistor combos in parallel.

 

[pawihte]

This works fine for me:
http://webpages.charter.net/dawill/Images/High Voltage Supply.gif

The MJE18008 switches in under a microsecond. I forget if I
boosted
the base resistors at all -- 470 or even 220 ohms would be
better on
the 2N4403.
Tim

Thanks for your interest. As mentioned in my opening post, the
object is to avoid unnecessary component clutter.

 

[Tim Williams]

Thanks for your interest. As mentioned in my opening post, the object is
to avoid unnecessary component clutter.

The whole thing fits on a couple square inches of PCB, no big deal, but I
was more illustrating the NPN and PNP drivers both work fine with the values
shown.

Tim

 

[Jon Kirwan]

Refresh my memory, what's the sink current of a bipolar 555?

Datasheet says sink and source is up to 200mA. But the curves only go
up to 100mA on this sheet and the spec doesn't provide a guarantee
output voltage when sinking 200mA -- only at 100mA and only at Vcc of
15V, which is 2.5V. The curves show a nice 2V output at 100mA at Vcc
of 5V, but that's probably typical and not a guarantee and it's almost
twice that shown for 100mA at Vcc of 15V (which is closer to 1V on the
graph.)

You can't connect one end of LED to plus rail?

Not at 250mA. The darned thing has no guarantee low-out even at 200mA
and the guarantee at 100mA is 2.5V! Hauling it to 250mA might work
but even if Vcc minus that drop is adequate, I'd start to worry about
package dissipation. The packages vary from 100C/W to 200C/W,
roughly, and the junction temp should not exceed 150C. So 1/2 watt
would be about the max I'd want to mess around considering. At up to
2.5V drop and 250mA... that's already exceeded.

I think the OP is definitely right to use a discrete.

Jon

 

[pawihte]

The whole thing fits on a couple square inches of PCB, no big
deal,
but I was more illustrating the NPN and PNP drivers both work
fine
with the values shown.

Oops. Sorry, I missed the point. Got it now.

Your 555 output also sources the base drive for the 2N4401 on the
high state. If it still goes high enough to turn the 2N4403 off,
my application should have a better margin. But just to be sure,
did you ever observe the swings with a scope?

 

[pawihte]

Swap phase and use an NPN booster? Then you have the advantage
of a
true "OFF" state.

The 555 wouldn't go below 50% duty in the opposite phase. At
least not with the basic astable circuit. I haven't investigated
to see if it's possible to change that with some manipulation.

 

[pawihte]

Do you desire variable or fixed duty cycle?

Fixed. There's no precise requirement for the duty cycle but it
should be around 20%.

 

[Nico Coesel]

The 555 wouldn't go below 50% duty in the opposite phase. At
least not with the basic astable circuit. I haven't investigated
to see if it's possible to change that with some manipulation.

AFAIK the 555 can do less than 50% duty cycle.

 

[Tim Williams]

Your 555 output also sources the base drive for the 2N4401 on the high
state. If it still goes high enough to turn the 2N4403 off, my application
should have a better margin. But just to be sure, did you ever observe the
swings with a scope?

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.

Tim

 

[Jamie]

I want to use a classic 555 timer IC to drive the base of a PNP
transistor through a resistor, the emitter of the transistor
being tied to the 555's Vcc. The 555 datasheet gives a graph for
the high-state output voltage vs. sourcing current, but not when
the load is tied to Vcc.

Vcc
--------------------------
| |
.|. |
| | |
| | |
'-' |
| |
555 out ___ | |<
------------|___|------|
|\
|
|

What I'm concerned about is: Is there a possibility that the
high-state output of the 555 drops low enough below Vcc to
partially turn on the PNP transistor? I could increase the
turn-on threshold of the transistor with diodes, an LED or a
resistive voltage divider, but I'd like to avoid that if it's not
needed.

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..


At least it works for me that way.

 

[Jamie]

Thanks. It's bipolar and, according to the datasheet, the output
is a totem-pole NPN-NPN push-pull. I want to drive an IR LED at
38kHz at about 250mA peak with a 20% duty factor. Do you think
it's OK as is?

Hmm. that's a large IR LED? are you sure about the current demand?

The 555 will do 200 mA on its on.. That in it self can drive 4..5
average LED's

 

[Jamie]

Refresh my memory, what's the sink current of a bipolar 555?

You can't connect one end of LED to plus rail?

...Jim Thompson

Its 200 mA

 

[pawihte]

If you don't already have it, download LTspice IV, free, from:

http://www.linear.com/designtools/software/

and run this:

---<snip>-------

It just so happens I downloaded the current version a few days
ago, but I have practically no experience with simulations. This
is probably a good time to start learning.

 

[pawihte]

You shunt one of the resistors with a diode -- IIRC the one
from the
discharge pin to the cap, but you'll want to check that one.

Basically the charge current for the cap flows through the
series
combination of two resistors, but the discharge current just
flows
through one when the discharge pin goes low. This makes the
charge
always slower than discharge. So to get any old duty cycle you
shunt
that one resistor with a diode from discharge pin to cap, and
then the
cap only 'sees' the resistor from V+ to discharge pin when it's
charging.

Thanks for the tip.

If I remember. Don't go committing your design to a circuit
board
without checking!

I won't.

 

[pawihte]

Datasheet says sink and source is up to 200mA. But the curves
only go
up to 100mA on this sheet and the spec doesn't provide a
guarantee
output voltage when sinking 200mA -- only at 100mA and only at
Vcc of
15V, which is 2.5V. The curves show a nice 2V output at 100mA
at Vcc
of 5V, but that's probably typical and not a guarantee and it's
almost
twice that shown for 100mA at Vcc of 15V (which is closer to 1V
on the
graph.)


Not at 250mA. The darned thing has no guarantee low-out even
at 200mA
and the guarantee at 100mA is 2.5V! Hauling it to 250mA might
work
but even if Vcc minus that drop is adequate, I'd start to worry
about
package dissipation. The packages vary from 100C/W to 200C/W,
roughly, and the junction temp should not exceed 150C. So 1/2
watt
would be about the max I'd want to mess around considering. At
up to
2.5V drop and 250mA... that's already exceeded.

I think the OP is definitely right to use a discrete.

I considered all those factors before but didn't include the
details to keep my post from becoming too lengthy. Thanks for the
confirmation.

 

[pawihte]

Hmm. that's a large IR LED? are you sure about the current
demand?

The 555 will do 200 mA on its on.. That in it self can drive
4..5
average LED's

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.