On Fri, 22 Oct 2004 16:33:20 -0400, Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote:>On Fri, 22 Oct 2004 19:39:45 GMT, the renowned Jonathan Kirwan ><jkirwan@easystreet.com> wrote: > >>On Thu, 21 Oct 2004 09:18:45 -0400, Spehro Pefhany >><speffSNIP@interlogDOTyou.knowwhat> wrote: >> >>>It's caused by parasitic inductance in the base (for example from long >>>wires) and parasitic capacitance across the emitter resistor. The >>>inductance causes a negative real impedance looking into the emitter, >>>which can cause oscillation over some range of conditions. >> >>I know the exact equations are a little different here, but let me try to put >>this into grossly simplified dynamic considerations that ignore the precise >>quantities: >> >>Looking at the 0V to 1.5V rising edge case, the inductance of long wires to the >>base (I've found figures like 10 nH/cm) would limit the rate of current change >>to (V2-v1)/L (V2 being the pin drive voltage and V1 at the base initially held >>to zero by the BJT capacitance.) This rising current is needed charge/discharge >>the base-emitter and base-collector capacitances, before the transistor can >>respond accurately. (For simplification, I'll just say that the transistor >>remains "OFF" until these capacitors are charged, rather than worry about the >>exact details.) At this point, the BJT goes into the normal region and >>operates, driving the collector rapidly down in voltage. This is coupled via >>newly recharged base-collector capacitor back to the base, which now drives >>downwards, shutting the BJT off again until the new (V/L) can again recharge the >>BJT capacitor pair. Etc. This will be a damped oscillation on this rising >>edge, an RLC thing, with the R being R1||R2 and the C being CJC+CJE? > >Not necessarily damped. Gory details: > >http://home.tiscali.be/kpmoerman/electronics/notes/efollow/efollow.htmFirst thing I notice without bothering to read further is that this doesn't use a collector resistor. This would eliminate the R of the RLC, since I believe that the equivalent is R1||R2 and with one of those zero, it's just LC, which doesn't damp. In other words, I'm still thinking my very gross simplification may still hold here -- without having tried to read the "gory details" on the page (which I will do.)>>Roughly speaking, is this about it? And if so, I'd expect a different behavior >>on the falling edge case. > >The bias is different when the output is low, so it's a bit harder for >it to oscillate. > >Here's a simulation I did using a 2N4401: >www.speff.com/emitter_follower.pdfTry adding a collector resistor and what's that 180pF cap doing there??? I don't wire up things *that* bad!!>With a lot more capacitance across the 180R it will oscillate >continously, once started, even if the input goes low. > >In practical terms this may be an okay configuration for the purpose >intended if layout etc. is good enough, but with a lash-up, maybe not.Jon
PIC18 transistor LED drivers
Started by ●October 20, 2004
Reply by ●October 22, 20042004-10-22
Reply by ●October 22, 20042004-10-22
On Fri, 22 Oct 2004 21:25:26 GMT, the renowned Jonathan Kirwan <jkirwan@easystreet.com> wrote:>On Fri, 22 Oct 2004 16:33:20 -0400, Spehro Pefhany > >>Here's a simulation I did using a 2N4401: >>www.speff.com/emitter_follower.pdf > >Try adding a collector resistorWith 100 ohms (drops 2.4V at 24mA) and 10pF get continuous oscillation.> and what's that 180pF cap doing there??? I don't wire up things *that* bad!!(we're talking about the OP, not *you*, of course) ;-) Anyway, it's a 180 ohm resistor (for 24mA nominal) and 8pF shunt capacitance. Not that out of line. This kind of setup is insensitive to loads on the collector, but very fussy about the base and emitter, so it's not that badly suited for this application if the layout is reasonable. Personally I'd rather slap a 1K resistor in the base, which kills the ringing or oscillation *and* protects the micro against faults in the driver. www.speff.com/with_base_resistor.pdf
Reply by ●October 22, 20042004-10-22
On Fri, 22 Oct 2004 17:53:55 -0400, Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote:>This kind of setup is insensitive to loads on the collector, but very >fussy about the base and emitter, so it's not that badly suited for >this application if the layout is reasonable. Personally I'd rather >slap a 1K resistor in the base, which kills the ringing or oscillation >*and* protects the micro against faults in the driver.Sounds good to me, sacrificing only a tiny bit of I(C) predictability for those benefits. I see your point! Jon
Reply by ●October 24, 20042004-10-24
Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote in message news:<avgdn05bp26eo94gnuigmrsl3qhq0dn4i7@4ax.com>...> On 20 Oct 2004 12:18:52 -0700, the renowned csembro@yahoo.com (Chuck) > wrote:snip> >Last year I was driving the LED's from a simple 12V DC power supply. > >Their cathodes are connected together and to ground. The anodes > >connected to appropriate resistors.> x------[Rx]----|<|---- +12 (12V) > | \\\ > | LED > | > B |/C > -----[2K7]-------| > |\E 2N2222A, 2N4401 etc. > | > | > 0V (5V) -----------x----------------------- 0V (12V) > > > Only tie the 12V supply ground to the 5V supply ground at one point. > (Do not allow the LED current to flow through the 5V ground circuit). > > If you're using 30mA maximum, then a forced beta of 20 would imply a > base current of 1.5mA, so 2.7K is about right. > > To calculate Rx, you need to know the LED forward voltage Vf. If > you're not sure, use 2.5 for yellow or green LEDs, 3V for white or > blue, and 2.0 for red.snip> Spehro PefhanyThanks Spehro, Your circuit looks exactly like several others I've found on the web. But it inspires me to ask two more questions. How do I know what amount of base current will drive the transistor to saturation? Is this what Hfe is used for? One of my problems in design is that all the LED collectors are tied to ground. In order to use your design, I'll have to unsolder all my wires and tie the anodes together instead. Is there a way to avoid all this rework? Chuck
Reply by ●October 24, 20042004-10-24
On 24 Oct 2004 14:41:41 -0700, the renowned csembro@yahoo.com (Chuck) wrote:>Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote in message news:<avgdn05bp26eo94gnuigmrsl3qhq0dn4i7@4ax.com>... >> On 20 Oct 2004 12:18:52 -0700, the renowned csembro@yahoo.com (Chuck) >> wrote: > >snip > >> >Last year I was driving the LED's from a simple 12V DC power supply. >> >Their cathodes are connected together and to ground. The anodes >> >connected to appropriate resistors. > >> x------[Rx]----|<|---- +12 (12V) >> | \\\ >> | LED >> | >> B |/C >> -----[2K7]-------| >> |\E 2N2222A, 2N4401 etc. >> | >> | >> 0V (5V) -----------x----------------------- 0V (12V) >> >> >> Only tie the 12V supply ground to the 5V supply ground at one point. >> (Do not allow the LED current to flow through the 5V ground circuit). >> >> If you're using 30mA maximum, then a forced beta of 20 would imply a >> base current of 1.5mA, so 2.7K is about right. >> >> To calculate Rx, you need to know the LED forward voltage Vf. If >> you're not sure, use 2.5 for yellow or green LEDs, 3V for white or >> blue, and 2.0 for red. > >snip > >> Spehro Pefhany > >Thanks Spehro, > >Your circuit looks exactly like several others I've found on the web. >But it inspires me to ask two more questions. How do I know what >amount of base current will drive the transistor to saturation? Is >this what Hfe is used for?Yes, sorta. I know the hFE of those parts is fairly high (at 20-30mA Ic and at normal temperature, and a bit of extra drop won't fry the transistor), so to drive them well into saturation I use Ic/Ib = 20 (that's what I mean above by "forced beta" = forced hFE). You might pick 10 or 25 as well- it's a fairly conservative number, but often parts are actually spec'd at 10. See, for example, the graph on page 3- top right for data http://www.fairchildsemi.com/ds/2N/2N3904.pdf>One of my problems in design is that all the LED collectors are tied >to ground. In order to use your design, I'll have to unsolder all my >wires and tie the anodes together instead. Is there a way to avoid >all this rework? > >ChuckIf there's no common between the supplies, you can tie the + sides together (if there *is* already a common this step could damage a lot of stuff), then flip the above circuit, using a PNP transistor. You'll make the input "low" to turn the LED on: +5, +12V (12V supply) 2K7 | ___ |< Microcontroller -|___|- -| 2N3906/2N4403 etc. high = off |\ | | .-. | | Rx | | '-' | | / V / LED - / | ----------------0V (12V supply) (-7V relative to microcontroller ground) 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
