Are there two Slowman's here?
Driving a PNP with a 555
Dec 22, 2009
50 Replies
So has most of the crap you sell on Ebay.
Offworld checks no longer accepted!
Don doesn't have any 555s at the moment. ;-)
Best regards, Spehro Pefhany
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
Wait, who is this? Jan Lancaster? Don Panteltje?
Tim
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Obviously, or he would have tried to sell them here.
Offworld checks no longer accepted!
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.
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
Thinking outside the box won't work in this case because the lowest duty
cycle you can get is about 25%.
Using this circuit: (View in Courier)
.+V>---+-----------------------------------+
. | |
. | +---------+ |
. +--[40k]-+-[R2]--7-O|D Vcc|-8---+
. | | _| |
. +--------6-|TH R|O-4--+
. | |__ | |
. +-------2-O|TR OUT|-3---|-----+
. | | GND | | |
. [1nF] +----+----+ [0.1µF] [1K]
. | |1 | |
.GND>-----------+---------------+----------+-----+
and varying R2 in 1% steps, I got:
||
-->| |= 20k, only a single output pulse appears at
power-up:
JF
Hey i vote for that.
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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