Driving a PNP with a 555
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From: pawihte on 24 Dec 2009 07:23 Don Lancaster wrote: > pawihte wrote: >> Tim Williams wrote: >>> "pawihte" <pawihte(a)fake.invalid> wrote in message >>> news:hgrc5p$1ld$1(a)news.eternal-september.org... >>>> Your 555 output also sources the base drive for the 2N4401 >>>> on >>>> the >>>> high state. If it still goes high enough to turn the 2N4403 >>>> off, my >>>> application should have a better margin. But just to be >>>> sure, >>>> did >>>> you ever observe the swings with a scope? >>> I think it was saturating at 2-3 Vbe's. I could go check. >>> >>> The important part is getting the B-E resistors small enough >>> so >>> the >>> transistor is certainly on or off. Which actually, with 1k >>> and >>> 1k, it >>> should only be turning off with less than 1.2V (if it's 1.8V, >>> the PNP >>> might never fully turn off!). Hmm, I should probably change >>> those >>> resistor values then. >>> >> >> I'm also thinking along the same lines. I'd originally >> intended >> to use 470 ohms for drive and 1k as the b-e shunt. Reducing >> the >> shunt to 220 ohms will give me better peace of mind. It will >> divert 3-3.5mA from the base drive, but the remaining 20mA >> should >> still be enough to saturate the transistor. If not, I could >> halve >> each resistor and still stay well within the 555's output >> capability. >> >> > > > Use a PIC instead. > Then throw away the 555. > It has been obsolete for decades now. Would you use an Allen nut-bolt where a nail does the job just as well? A 555 is cheap (10 US cents each retail), available everywhere, requires no programming tool or knowledge. Try to beat that with a PIC. In any case, the versatility of a PIC is no advantage here, and a PIC will still need to interface with the power driver stage, which is what this thread is about.
From: Michael Robinson on 26 Dec 2009 17:08 "pawihte" <pawihte(a)fake.invalid> wrote in message news:hgr72q$uko$1(a)news.eternal-september.org... >I want to use a classic 555 timer IC to drive the base of a PNP transistor >through a resistor, the emitter of the transistor being tied to the 555's >Vcc. The 555 datasheet gives a graph for the high-state output voltage vs. >sourcing current, but not when the load is tied to Vcc. > > Vcc > -------------------------- > | | > .|. | > | | | > | | | > '-' | > | | > 555 out ___ | |< > ------------|___|------| > |\ > | > | > > What I'm concerned about is: Is there a possibility that the high-state > output of the 555 drops low enough below Vcc to partially turn on the PNP > transistor? I could increase the turn-on threshold of the transistor with > diodes, an LED or a resistive voltage divider, but I'd like to avoid that > if it's not needed. > It think you would be way better off switching an npn on the low side. I imagine you won't need a b/e resistor or any of that headache. Others have shown how to get a low duty cycle using diodes, but you can do it with just two resistors if you think out of the box: ,----R1-----, | | | ,--R2-+ | | | ,--+-----+-----+----+--, | 8 7 6 5 | | / | | / | | / | | 1 2 3 4 | '--+-----+-----+----+--' | | = out | gnd Obviously the value of R1 has to be at least twice R2 for the circuit to work at all, because the cap voltage has to fall below 1/3 Vcc for the astable multivibrator function. I don't have the time to calculate resistor values for a 20% duty cycle right now. You can do it by adjusting a pot until you get exactly 20%, then measure the resistance and replace the pot with fixed resistors.
From: pimpom on 27 Dec 2009 13:42 Michael Robinson wrote: > "pawihte" <pawihte(a)fake.invalid> wrote in message > news:hgr72q$uko$1(a)news.eternal-september.org... >> 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. >> > 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.
From: John Fields on 27 Dec 2009 13:51 On Sat, 26 Dec 2009 17:08:47 -0500, "Michael Robinson" <nospam(a)billburg.com> wrote: > >"pawihte" <pawihte(a)fake.invalid> wrote in message >news:hgr72q$uko$1(a)news.eternal-september.org... >>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. >> >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. --- 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] +----+----+ [0.1�F] [1K] .. | |1 | | ..GND>-----------+---------------+----------+-----+ and varying R2 in 1% steps, I got: |<-----------T2--------->| -->| |<--T1 __ __ ____| |_____________________| |_____ ohms ms ms % R2 T1 T2 D 19.3k 2.74 7.68 35.7 19.6k 2.74 8.53 32.1 19.8k 2.74 10.8 25.3 19.9k 2.84 9.58 29.6 So not only is a duty cycle of 20% impossible to attain, note the extreme sensitivity to duty cycle changes with changes of R2, where for a change in resistance of 200 ohms from 19.6k to 19.8k (about 1%) the change in duty cycle is about 20%. This suggests that if a duty cycle of 20% was attainable it would be very difficult to achieve and maintain stable, where with either of the diode schemes 20% is easily achievable and easily tuned. Version 4 SHEET 1 880 748 WIRE 224 192 -128 192 WIRE 544 192 448 192 WIRE -128 256 -128 192 WIRE -80 256 -128 256 WIRE 32 256 0 256 WIRE 64 256 32 256 WIRE 224 256 144 256 WIRE 512 256 448 256 WIRE 32 320 32 256 WIRE 176 320 32 320 WIRE 224 320 176 320 WIRE 608 320 448 320 WIRE 480 384 448 384 WIRE 608 416 608 320 WIRE 176 480 176 320 WIRE 512 480 512 256 WIRE 512 480 176 480 WIRE -128 512 -128 256 WIRE 480 512 480 384 WIRE 480 512 -128 512 WIRE -128 528 -128 512 WIRE 176 544 176 480 WIRE -128 624 -128 608 WIRE 176 624 176 608 WIRE 176 624 -128 624 WIRE 544 624 544 192 WIRE 544 624 176 624 WIRE 608 624 608 496 WIRE 608 624 544 624 WIRE -128 688 -128 624 FLAG -128 688 0 SYMBOL Misc\\NE555 336 288 M0 SYMATTR InstName U1 SYMBOL voltage -128 512 M0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 5 SYMBOL res -96 240 M90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 58 VTop 0 SYMATTR InstName R1 SYMATTR Value 40k SYMBOL res 48 240 M90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R2 SYMATTR Value 19.8k SYMBOL res 624 400 M0 WINDOW 0 -45 40 Left 0 WINDOW 3 -59 71 Left 0 SYMATTR InstName R3 SYMATTR Value 1000 SYMBOL cap 192 544 M0 WINDOW 0 -33 32 Left 0 WINDOW 3 -39 58 Left 0 SYMATTR InstName C1 SYMATTR Value 1E-7 TEXT 0 656 Right 0 !.tran .1 uic TEXT 456 176 Left 0 ;1 TEXT 456 240 Left 0 ;2 TEXT 456 304 Left 0 ;3 TEXT 456 360 Left 0 ;4 TEXT 200 368 Left 0 ;5 TEXT 200 304 Left 0 ;6 TEXT 200 240 Left 0 ;7 TEXT 200 176 Left 0 ;8 Note that with R2 >= 20k, only a single output pulse appears at power-up: JF
From: JosephKK on 28 Dec 2009 05:20
On Wed, 23 Dec 2009 12:43:23 -0800 (PST), Bob <stephensdigital(a)gmail.com> wrote: >On Dec 23, 9:50 am, "pawihte" <pawi...(a)fake.invalid> wrote: >> Jamie wrote: >> > pawihte wrote: >> >> >> 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.. >> >> That's what I thought, but I thought I'd better check with you >> guys in case there was something I missed. >> >> > At least it works for me that way. > >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. |