Driving a PNP with a 555
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From: John Fields on 22 Dec 2009 18:46 On Tue, 22 Dec 2009 21:49:59 GMT, nico(a)puntnl.niks (Nico Coesel) wrote: >"pawihte" <pawihte(a)fake.invalid> wrote: > >>Jim Thompson wrote: >>> 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. > >AFAIK the 555 can do less than 50% duty cycle. --- Not without external help. From Signetics' 1979 applications manual: news:5dm2j51e8fcob9e8ha0skll5gjdrgkd7a9(a)4ax.com JF
From: Tim Wescott on 22 Dec 2009 19:09 On Wed, 23 Dec 2009 02:41:24 +0530, pawihte wrote: > Jim Thompson wrote: >> On Wed, 23 Dec 2009 02:07:15 +0530, "pawihte" <pawihte(a)fake.invalid> >> wrote: >> >>> Jim Thompson wrote: >>>> On Wed, 23 Dec 2009 01:31:42 +0530, "pawihte" <pawihte(a)fake.invalid> >>>> wrote: >>>> >>>>> Tim Wescott wrote: >>>>>> On Wed, 23 Dec 2009 01:05:52 +0530, 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. >>>>>> >>>>>> 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? >>>>> >>>>> >>>> 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. >>> >>> >> 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. 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! -- www.wescottdesign.com
From: pawihte on 23 Dec 2009 01:45 John Fields wrote: > On Wed, 23 Dec 2009 02:54:11 +0530, "pawihte" > <pawihte(a)fake.invalid> > wrote: > >> Jim Thompson wrote: > >>> Do you desire variable or fixed duty cycle? >>> >> >> Fixed. There's no precise requirement for the duty cycle but >> it >> should be around 20%. > > 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.
From: pawihte on 23 Dec 2009 01:50 Tim Wescott wrote: > On Wed, 23 Dec 2009 02:41:24 +0530, pawihte wrote: > >> Jim Thompson wrote: >>> On Wed, 23 Dec 2009 02:07:15 +0530, "pawihte" >>> <pawihte(a)fake.invalid> >>> wrote: >>> >>>> Jim Thompson wrote: >>>>> On Wed, 23 Dec 2009 01:31:42 +0530, "pawihte" >>>>> <pawihte(a)fake.invalid> wrote: >>>>> >>>>>> Tim Wescott wrote: >>>>>>> On Wed, 23 Dec 2009 01:05:52 +0530, 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. >>>>>>> >>>>>>> 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? >>>>>> >>>>>> >>>>> 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. >>>> >>>> >>> 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. > > 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.
From: Jamie on 22 Dec 2009 20:19
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.. At least it works for me that way. |