Driving a PNP with a 555
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From: John Fields on 23 Dec 2009 16:07 On Wed, 23 Dec 2009 12:31:18 +0530, "pawihte" <pawihte(a)fake.invalid> wrote: >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. --- 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] +----+----+ [0.1�F] | .. | |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 = I�R D = 0.025A� * 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 = I�RD = 0.26A� * 51R * .2 ~ 0.69 watts, so a standard 51 ohm +/- 5%, 1 watt resistor would be fine. JF
From: Don Lancaster on 23 Dec 2009 16:36 pawihte wrote: > Tim Williams wrote: >> "pawihte" <pawihte(a)fake.invalid> wrote in message >> news:hgrc5p$1ld$1(a)news.eternal-september.org... >>> 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. >> > > 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. > > Use a PIC instead. Then throw away the 555. It has been obsolete for decades now. -- Many thanks, Don Lancaster voice phone: (928)428-4073 Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552 rss: http://www.tinaja.com/whtnu.xml email: don(a)tinaja.com Please visit my GURU's LAIR web site at http://www.tinaja.com
From: John Fields on 23 Dec 2009 20:37 On Wed, 23 Dec 2009 14:36:08 -0700, Don Lancaster <don(a)tinaja.com> wrote: >pawihte wrote: >> Tim Williams wrote: >>> "pawihte" <pawihte(a)fake.invalid> wrote in message >>> news:hgrc5p$1ld$1(a)news.eternal-september.org... >>>> 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. >>> >> >> 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. >> >> > > >Use a PIC instead. >Then throw away the 555. >It has been obsolete for decades now. --- And yet... ;) JF
From: krw on 23 Dec 2009 21:02 On Wed, 23 Dec 2009 19:37:54 -0600, John Fields <jfields(a)austininstruments.com> wrote: >On Wed, 23 Dec 2009 14:36:08 -0700, Don Lancaster <don(a)tinaja.com> >wrote: > >>pawihte wrote: >>> Tim Williams wrote: >>>> "pawihte" <pawihte(a)fake.invalid> wrote in message >>>> news:hgrc5p$1ld$1(a)news.eternal-september.org... >>>>> 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. >>>> >>> >>> 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. >>> >>> >> >> >>Use a PIC instead. >>Then throw away the 555. >>It has been obsolete for decades now. >--- >And yet... ;) Are there two Slowman's here?
From: Tim Williams on 24 Dec 2009 00:28
"Don Lancaster" <don(a)tinaja.com> wrote in message news:7pfgq4Fdc1U1(a)mid.individual.net... > 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 -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |