Driving a PNP with a 555

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.

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

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www.wescottdesign.com

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.

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

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"I'd rather have a bottle in front of me than a frontal lobotomy"

Its 200 mA

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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.

Thanks for the tip.

I won't.

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

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.

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?

String a trimpot between 2 diodes (facing opposite ways) wiper to pins 2 & 6 (IIRC) and the common of the 2 diodes to pin 7.

With the pot at one extreme the cap will charge faster than it discharges, swing the pot the other way and the cap will discharge faster than it charges.

That will give you a duty cycle of not far short of 0 to 100%.

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.

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.

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.

That's what I thought, but I thought I'd better check with you guys in case there was something I missed.

For people who cant get binaries you can also get it here.

Full pdf is available at the top of the page right side.

Other 555 stuff on the main page as well.

So what's the history of the LM122? Was it marketed as an improvement to the

555?

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?

--- Here's the circuit I posted earlier in that LTspice circuit list with a few refinements: .+15>--+--------------------------------+-----------+ . | | | . [6k2]R1 +---------+ | [51R] . | |_ | | | . +-------------7-O|D Vcc|-8---+ |A . | R2 | _| | [LED] . +---[30k]---+--6-|TH R|O-4--+ | . | | |__ | | C . +-[1N4148>]-+-2-O|TR OUT|-3---|--[240]--B 2N4401 . | | GND | | Q1 E . [1nF] +----+----+ [0.1µF] | . | |1 | | .GND>--------------+---------+----------+-----------+

Assuming 250mA through the LED and forcing Q1's beta to 10 means that we'll have to force 25mA through its base-to-emitter junction in order to light up the LED.

From Signetics' 555 data sheet we get a maximum drop of Vcc - 1.5V at the 555's output when it's sourcing 25mA, and from the 2N4401's data sheet we get a Vbe(sat) of about 0.9V with a collector current of 250mA, so the base resistor's resistance needs to be:

Vcc - (Vbe(sat) +(Vcc - Vdrop(555))) Rb = -------------------------------------- Ib 15V - (0.9V + 1.5V = -------------------- 0.025A

= 250 ohms

270 ohms is a standard 5% part and would dissipate, where D is the duty cycle:

Pd = I²R D = 0.025A² * 270R * .2 ~ 0.04 watts,

so a standard 270 ohm +/- 5%, 1/4 watt resistor would be fine.

When the LED is off the output of the 555 will be low, and since it won't be sinking current, its output voltage will be very close to 0V, forcing the base of Q1 close enough to GND that only a miniscule charge, leakage current, will flow through the LED and the collector-to-emitter junction of Q1.

For the LED side of the circuit, Vce(sat) of Q1 will be about 0.175V with 250mA of collector current, And Vf of the LED will be about 1.5V (That's for a typical IRLED with 100mA If through it. At 250 mA it'll be somewhat higher, so plug that number into the following to get real values) for the same current, so the current limiting resistor should be:

Vcc - ((Vce(sat) + Vf) Rs = ------------------------ If

15V - (0.175V + 1.5V) = ----------------------- 0.25A

= 53.3 ohms

51 ohms is a standard 5% value, and since you're looking to throw the beam a far distance it might be a good choice.

Looking for how much current it'll allow through the LED we can use the numbers we got earlier and say:

E 13.33V I = --- = -------- ~ 0.260A = 260mA R 51R

Not bad, and if it turns out to be excessive you can always decrease the duty cycle to compensate.

The resistor, working at 20% duty cycle will dissipate:

P = I²RD = 0.26A² * 51R * .2 ~ 0.69 watts,

so a standard 51 ohm +/- 5%, 1 watt resistor would be fine.

JF

Use a PIC instead. Then throw away the 555. It has been obsolete for decades now.

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Many thanks,

Don Lancaster                          voice phone: (928)428-4073
Synergetics   3860 West First Street   Box 809 Thatcher, AZ 85552
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And yet... ;)

JF