Blue LED at 3.3V
in [Design]
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From: John Larkin on 30 May 2010 20:18 On Sun, 30 May 2010 18:49:16 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Fri, 28 May 2010 09:13:40 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Fri, 28 May 2010 09:54:50 -0500, John Fields >><jfields(a)austininstruments.com> wrote: >>> >>I should hope it works. It's a simple charge pump. Your latest example >>puts about 5.5 mA average forward current into the LED. You have >>proved that my charge pump works, even with the unrealistic schmitt >>model. > >--- >Well, I must admit that I dropped the ball on this one, but I think >you have too, in that you haven't considered how much current an HC14 >(which is what I think the OP said he wanted to use) can source or >sink using a 3.3V supply. Wrong again! I don't know why you keep doing this. My sketch didn't specify HC logic. All I did was post a scribble that had a schmitt oscillator in it, followed by a cap+diode charge pump. That was the idea I offered. You offered zero ideas. I fact, an HC14 would easily charge-pump 5 mA average current into a blue LED. But the HC series is ancient, and most more modern cmos logic has even more drive capability at 3.3 volts. > >I'd supply the numbers _I_ found, but since you seem to think it's >everyone's job but yours to flesh out the skeletons you dangle in >front of them, I'll leave that up to you, if you choose to pursue it. I offered an idea. For free. I didn't volunteer to do a detailed design. > >What I'm suggesting is that if the duty cycle is low enough, even if >the LED is 100% efficient in turning current into light, it'll be >invisible. Sure, less current makes less light, and you can break any circuit if you use sufficiently stupid values. But all of your sims pushed reasonable currents into the LED, so your "doesn't work" and "can't work" and "cockamamie circuit" comments were doubly bizarre. Your own sims are working examples of my circuit; you just didn't realize it. >It's not me that's holding it back, if _you_ want to prove it works, >flesh it out. Why bother? I *know* that it works, and you have already proved that it works. And you proved that the schmitt+resistor thing *does* oscillate, as another geezer swore it wouldn't. John
From: John Larkin on 30 May 2010 20:43 On Sun, 30 May 2010 19:14:34 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Fri, 28 May 2010 22:06:53 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Fri, 28 May 2010 19:56:36 -0400, Bitrex >><bitrex(a)de.lete.earthlink.net> wrote: >> >>> >>>> Bitrex's circuit likes an asymmetric duty cycle, which is maybe why he >>>> has the thresholds where they are. The asymmetry struck me >>>> immediately, because an HC gate oscillator would be nearly symmetric. >>>> >>> >>>I'm still new to LTSpice, I was a little confused about how the model >>>sets the thresholds and hysteresis. In datasheets the hysteresis is >>>defined as the difference between the high transition point and the low >>>transition point, in the model Vh seems to be the difference between the >>>high and low transition points and the midpoint, which is defined by the >>>Vt parameter. I think I've rectified most of my errors now...:) >> >>If the Spice definitions aren't clear, you can just look at the gate >>input triangle to see where it actually switches. >> >>Most cmos schmitts will have their switch limits straddling Vcc/2, > >--- >None that I've seen... Name a few whose input switch limits don't straddle Vcc/2. Then I'll name a few that do. >--- > >>maybe a bit less. > >--- >How much is "a bit less"? Couple tenths of a volt maybe, at 3.3 supply. Most cmos inputs transition a little below Vcc/2. It's not guaranteed, and there are lots of cmos parts around, but that's the pattern. Hadn't you ever noticed? > >Take a look at Philips' HC14. > >it looks more like ECL than CMOS. Insane. ECL input transitions are at about 1.35 volts below Vcc, independent of Vcc-Vee. And their outputs swing about two junction drops. The NXP HC14, with 4.5 volts Vcc, has datasheet typical switch levels of 2.38 and 1.40. The mean of those two is 1.89. Vcc/2 is 2.25. So the stuff I said is right. >--- > >>Take a look at A simple RC (or R with no obvious C!) should make a >>pretty symmetric square wave. > >--- >Now there's the John Larkin I love... > >Vague, himself, while demanding quantitative data from his critics... What don't you understand about "pretty symmetric"? Hint: the pulses you simulated weren't very symmetric. >--- > >>I wonder how big R can be. 1 Gohm easily. Maybe 1 Tohm if everything >>is clean. An open cmos gate input will hang high or low, whichever way >>you leave it, for seconds at least. I had one floating cmos level last >>week that took about 15 seconds after powerup to change state. That >>might roughly correspond to a pA of leakage. > >--- >See what I mean? A ballpark estimate of gate leakage current is useful. More than a ballpark is useless for design purposes. Knowing that the leakage is in the low pA range would allow one to use a 10M or 100M or 1G oscillator resistor with perfect safety. You can't do that with a classic 555. So, how big a timing resistor is it safe to use with a CMOS 555? John
From: Don Klipstein on 30 May 2010 21:17 In article <jno5061i4ghd8r6uo2hvvjrtgqg46jb280(a)4ax.com>, John Fields wrote in part: >On Fri, 28 May 2010 09:13:40 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>With that in mind it's clear that even with a perfect driver, your >>>>>circuit, which pushes 150mA spikes which decay to essentially 0mA in >>>>>4ns every 10 microseconds or so through the LED, won't work. >>>> >>>>It actually does work with a perfect driver, as your sim shows. >>> >>>--- >>>If the criterion determining "works" is whether the LED will emit >>>light which can be detected by the human eye in a casual manner, then >>>I submit that 4ns wide pulses occurring at a rep rate of 100kHz won't >>>quite fill the bill. >> >>Average current makes light. Eyeballs respond to average light. Are >>you suggesting that X microwatts of light are visible if DC, but >>invisible if the same average amount of photons arrive in bunches? > >--- >Not at all. > >What I'm suggesting is that if the duty cycle is low enough, even if >the LED is 100% efficient in turning current into light, it'll be >invisible. 150 mA decaying to near zero over a 4 nanosecond pulse repeated every 10 microseconds: (Did I get that right?) Suppose the average current during the pulse is 50 mA. Doing that for 4 every 10,000 nanoseconds would make the average current 20 microamps. Since a usual blue LED is likely to be more efficient at 50 mA than at 20 uA, I would expect it to be brighter with this than with 20 uA steady DC. (Ratio of photometric output to current usually peaks around 1.5-4 mA with these LEDs.) So, I would expect brightness typical of 30-40 microamps steady DC. That sounds to me on the dim side for an indicator LED, but I do expect this to be visible. If this blue LED is switched to 3.3V through a MOSFET, such as in a CMOS IC, then the full 3.3V is available. A fair number of blue LEDs nowadays on average need no more than 3.3V to push 20 mA through them, and that would make them so bright at 20 mA that some of these come with warnings to not stare into them. So, it sounds easy to me to get half a milliamp or a milliamp through them from 3.3V without a boosting circuit. And at this much current, many blue LEDs get plenty bright. -- - Don Klipstein (don(a)misty.com)
From: John Fields on 31 May 2010 04:58 On Sun, 30 May 2010 17:18:11 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sun, 30 May 2010 18:49:16 -0500, John Fields ><jfields(a)austininstruments.com> wrote: > >>On Fri, 28 May 2010 09:13:40 -0700, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>On Fri, 28 May 2010 09:54:50 -0500, John Fields >>><jfields(a)austininstruments.com> wrote: >>>> >>>I should hope it works. It's a simple charge pump. Your latest example >>>puts about 5.5 mA average forward current into the LED. You have >>>proved that my charge pump works, even with the unrealistic schmitt >>>model. >> >>--- >>Well, I must admit that I dropped the ball on this one, but I think >>you have too, in that you haven't considered how much current an HC14 >>(which is what I think the OP said he wanted to use) can source or >>sink using a 3.3V supply. > >Wrong again! I don't know why you keep doing this. --- Doing what? Didn't he say he needed to use a 3.3V supply? Didn't he say was thinking about using an HC14? --- >My sketch didn't specify HC logic. All I did was post a scribble that >had a schmitt oscillator in it, --- Ok, have it your way, but, no matter what you _think_ you posted, that scribble didn't include anything that would oscillate. --- >followed by a cap+diode charge pump. >That was the idea I offered. You offered zero ideas. --- Right. I offered critique. --- >I fact, an HC14 would easily charge-pump 5 mA average current into a >blue LED. But the HC series is ancient, and most more modern cmos >logic has even more drive capability at 3.3 volts. > >> >>I'd supply the numbers _I_ found, but since you seem to think it's >>everyone's job but yours to flesh out the skeletons you dangle in >>front of them, I'll leave that up to you, if you choose to pursue it. > >I offered an idea. For free. I didn't volunteer to do a detailed >design. --- Bret Cahill rides again. --- >>What I'm suggesting is that if the duty cycle is low enough, even if >>the LED is 100% efficient in turning current into light, it'll be >>invisible. > >Sure, less current makes less light, and you can break any circuit if >you use sufficiently stupid values. But all of your sims pushed >reasonable currents into the LED, so your "doesn't work" and "can't >work" and "cockamamie circuit" comments were doubly bizarre. Your own >sims are working examples of my circuit; you just didn't realize it. > > >>It's not me that's holding it back, if _you_ want to prove it works, >>flesh it out. > >Why bother? I *know* that it works, and you have already proved that >it works. > >And you proved that the schmitt+resistor thing *does* oscillate, as >another geezer swore it wouldn't. --- Again, my original critique had to do with the lack of accuracy of your scribble, which is what both of us geezers agreed upon. End of story, as far as I'm concerned; rave on if you like...
From: John Fields on 31 May 2010 08:17
On Sun, 30 May 2010 17:43:00 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sun, 30 May 2010 19:14:34 -0500, John Fields ><jfields(a)austininstruments.com> wrote: > >>On Fri, 28 May 2010 22:06:53 -0700, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>On Fri, 28 May 2010 19:56:36 -0400, Bitrex >>><bitrex(a)de.lete.earthlink.net> wrote: >>> >>>> >>>>> Bitrex's circuit likes an asymmetric duty cycle, which is maybe why he >>>>> has the thresholds where they are. The asymmetry struck me >>>>> immediately, because an HC gate oscillator would be nearly symmetric. >>>>> >>>> >>>>I'm still new to LTSpice, I was a little confused about how the model >>>>sets the thresholds and hysteresis. In datasheets the hysteresis is >>>>defined as the difference between the high transition point and the low >>>>transition point, in the model Vh seems to be the difference between the >>>>high and low transition points and the midpoint, which is defined by the >>>>Vt parameter. I think I've rectified most of my errors now...:) >>> >>>If the Spice definitions aren't clear, you can just look at the gate >>>input triangle to see where it actually switches. >>> >>>Most cmos schmitts will have their switch limits straddling Vcc/2, >> >>--- >>None that I've seen... > >Name a few whose input switch limits don't straddle Vcc/2. Then I'll >name a few that do. --- Well, I'm talking "straddle" in the strict sense, which would mean that the typical switch points would be equally displaced about Vcc/2. But, OK. For the Philips case we have: Vt-typ Vt+max 1.4V 2.38V | | 0V--------------+--------+--+---------------------4.5V | 2.25V Vcc/2 Straddle? That's almost sidesaddle! Even more intersting, the minimum and maximum thresholds for HC: Vcc/2 2.25V Vt+min | Vt+max 1.7V | 3.15 | | | 0V---------+--------+--+--+--------+--------------4.5V | | 0.9V 2.0V Vt-min Vt-max Which says that for a device with a Vt+min of 1.7V, Vt-min will, presumably, lie between 1.7V and 0.9V. Your turn now; let's see even _one_ that straddles Vcc/2 symmetrically. --- >>--- >>How much is "a bit less"? > >Couple tenths of a volt maybe, at 3.3 supply. Most cmos inputs >transition a little below Vcc/2. It's not guaranteed, and there are >lots of cmos parts around, but that's the pattern. Hadn't you ever >noticed? --- Huh??? If both transition points are below Vcc/2 how can they straddle it? --- >>Take a look at Philips' HC14. >> >>it looks more like ECL than CMOS. > >Insane. ECL input transitions are at about 1.35 volts below Vcc, >independent of Vcc-Vee. And their outputs swing about two junction >drops. --- Oops... I was looking on the HCT page. --- >>>Take a look at A simple RC (or R with no obvious C!) should make a >>>pretty symmetric square wave. >> >>--- >>Now there's the John Larkin I love... >> >>Vague, himself, while demanding quantitative data from his critics... > >What don't you understand about "pretty symmetric"? Hint: the pulses >you simulated weren't very symmetric. > --- Geez, John, you'll probably have to send that Katana in for sharpening what with all the straw men you keep chopping down! --- >>>I wonder how big R can be. 1 Gohm easily. Maybe 1 Tohm if everything >>>is clean. An open cmos gate input will hang high or low, whichever way >>>you leave it, for seconds at least. I had one floating cmos level last >>>week that took about 15 seconds after powerup to change state. That >>>might roughly correspond to a pA of leakage. >> >>--- >>See what I mean? > >A ballpark estimate of gate leakage current is useful. More than a >ballpark is useless for design purposes. --- For a sloppy designer or a sloppy design, that would probably be true. --- >Knowing that the leakage is >in the low pA range would allow one to use a 10M or 100M or 1G >oscillator resistor with perfect safety. You can't do that with a >classic 555. --- Apples and oranges, since in normal use the leakage of the cap is so high that it swamps the 10pA typical trigger and threshold leakage currents. --- >So, how big a timing resistor is it safe to use with a CMOS 555? --- If the cap had absolutely no leakage and you used this circuit: VCC | +-----------|------+ | +-------+8 | [Rt] | +----+----+ | | | 4|_ Vcc | | | +-O|R OUT|-+-->OUT | | |3 +----6-|TH | | |__ | +-----O|TR | | 2| GND | [Ct] +----+----+ | U1 7555 |1 +-----------+ | GND Then it would depend on how much the currents into and out of TH and TR- (10pA at Vcc = 5V) affected the accuracy of the timing. You figure it out... JF |