Calling All 'Lectronicians !
in [Harley-Davidson]
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From: 73Sportster on 11 Jun 2010 08:33 "spacecriter \(Bill C\)" <spacecriter(a)sysmatrix.net.yercrank> wrote in news:Gq6dneXdDJZz3o3RnZ2dnUVZ_sWdnZ2d(a)sysmatrix.net: > > 73Sportster wrote: >> "spacecriter \(Bill C\)" <spacecriter(a)sysmatrix.net.yercrank> wrote >> in news:NLWdndQ5BqFERJPRnZ2dnUVZ_uCdnZ2d(a)sysmatrix.net: >> >>> >>> 73Sportster wrote: >>>> "spacecriter \(Bill C\)" <spacecriter(a)sysmatrix.net.yercrank> wrote >>>> in news:zp6dnR-z-7RbvpPRnZ2dnUVZ_sKdnZ2d(a)sysmatrix.net: >>>> >>>> <a buncha stuff that I snipped> >>>> >>>> Wanted to let you know that ya'll were missed this weekend. Talked >>>> about you too! That's what you get for not showing up. >>>> >>>> Rosie, tell the missus I said "Hi!" >>> >>> >>> heh, shoulda known...Was wondering why my ears were ringin' >>> >>> Missed all you folks too. I bet nobody had any fun either <g> >> >> We tried, but it was *this* close... > > That's what I figured. I've never had any fun with y'all either :-P > > 'tho, there was that time when we lost Kenny...may have been more than > once...came close to having fun then. > Yeah. He's still got this thing about electric fences. <G> At Roger's, I offered to have someone escort him to the road, but he wasn't going for it. >> >>> >>> Don't know if you saw my post across the street, but one of our >>> cabana boys was graduating Saturday. We now have a Dentist in the >>> family woo hoo! >>> >> >> I heard. Congratulations. > > Oh good, that means Karen got Lisa's phone message. She was > wondering. I was wondering too, I'd been asking her to call all week. > Yep, message was received. And Karen let everyone know why you weren't there. > We're really proud of the kid. Gonna be proud parents again in a > couple weeks. He's getting married too. He really didn't think about > all the pressure he'd put on himself by scheduling the wedding so > close to his graduation. He found out that planning a wedding, > honeymoon, getting incorporated, prof licensing, searching or a job, > and working though his final grad requirements was challenging when > done all at once. I've only said 'told ya so' a couple times. He just > rolls his eyes. Makes me feel kinda warm inside to know I can still > get a response. > I know what you mean. It's fun to embarrass them in front of their friends, too. > The other Cabana boy now lives in Charlotte. I think you met him at > the last MISFIT south of Asheville (Chimney rock?). He's been making > noises about us going to the next MISFIT. With everything else going > on I'm not planning on it, but who knows. Hope to see you guys there > if we do. > Wow, we haven't been to a MISFIT in forever... Not so sure we'll make this one either. We'll see. Rosie
From: spacecriter (Bill C) on 11 Jun 2010 09:59 Snag wrote: > spacecriter (Bill C) wrote: >> Snag wrote: >>> spacecriter (Bill C) wrote: >>>> So snag, which way are you going for this? >>>> >>>> BTW, if you do use a voltage reg, you still need to use a current >>>> limiting resistor, or you could blow the LEDs. >>> >>> Yeah , I just tonight found that out . I might be in over my head >>> here ... so I guess I better learn to swim ! >>> BTW , I will be going with a voltage reg , set at 10v . 5 LED's in >>> series that want a nominal 2v . So how does one size a current >>> limiter ? Got a big bag of 1/4w carbon film resistors in various >>> sizes ... bet I'm gonna need the specs on my LED's . >> >> 1/4 W ought to be OK >> >> Use the calculator I sent you. It will give you the nearest common >> resistor value. >> >> Ignoring the voltage regulator, plug in 12, 10, and 20. >> >> It says you should use 120 ohms and it will burn .04 watts. Well >> within .125 W. >> >> You could dial it up a litle to derate the LEDs. >> >> Using 13.5, 10, and 18 yields a 220 ohm resistor. 1/4 W still works. >> >> -- >> Bill C. "I am NOT lost... I'm *exploring*" > > Your email (??) must be lost in cyberspace ... I found a calculator , > using 5 of the 2v/25ma LED's in series and 10v it calls for a 1ohm > 1/4 watt resistor for each string . Shorter strings/higher voltages > result in much higher resistor values , but 1/4 w is still > recommended . My main concern if I use an unregulated power supply > is the widely varying voltages . I wonder if they'll be bright enough > at the lower voltages if I allow for the higher ones ... this circuit > may see anywhere between 11 or so up to 14 volts . Not an e-mail. It was a post in this thread. Here's part of my previous response: you don't have to use a voltage regulator in this > application. Another solution is to put in a current limiting > resistor. That is, use the resistor in series with the LEDs to drop > the voltage down to 10 volts. Pick the highest current you want > through the LED and use this calculator to pick a value. > > http://www.quickar.com/noqbestledcalc.htm > > I don't believe you need to use one R per string, but I would use > more than one for the whole circuit. Maybe one R per 3 strings? Just a > guess on my part. Remember, the more strings conneted through > one R drives more current through the resistor, and it will have to > be rated for more power dissipation. > > Also, you may want to put in a ceramic capacitor around the whole > thing (paralle to all of it) to cut down voltage spikes (ignition > noise). OK so bear with me because I know the theory, but I don't do design every day. The diodes basically have the same voltage drop no matter how much current is going through them. As you've seen, too much current releases the magic smoke that makes all eletronics work. See, I told you I knew the theory ;-) assUme-ing the diodes you have produce a nominal 2 V drop (NOTE if this is value is different, then EVERYTHING below changes a little) we can do some cipherin' 5 diodes in series will drop 10 volts. A series resistor will have to drop the difference between the supply voltage and 10V. Now say that the supply is 12V. This means there will be 2V across the R. If you want to limit the curent to 20mA, we can calculate the size of the R to be 2/.02=100 Ohms. That would be great *IF* the supply voltage was always 12V. But like you point out, it changes. The charging system on the bike usually keeps it under 14.5V right? (Really, that was a serious question. I can't remember what the max voltage is on a typical vehicle, so this may be off a little. I'm just going to assUme its correct at this point). Anyway, knowing the max voltage the circuit needs to handle will allow us to pick a resistor which will limit the max current and keep all that magic smoke bottled up. So let's put the max curent at 20mA. Max Suply voltage: 14.5V 5 diode voltage drop: 10V This sets the resistor at 225 Ohms. And the max power this resistor will dissipate is ~ 90mW But here's another rub (ain't there always?). Common resistor sizes usually don't match what we calculate. So we can either mix several standard sizes to get the calculated value. Usually only do that if this is a precision application with a narrow range of variation. So normally you will just pick the nearest, bigger, common size. Bigger because the current we used to get the size was the max we ever want to see, so a bigger R value will keep the current under that cap. The calculator I used picked 270 Ohms. That was all well and good for fixing the upper end current. But, like you pointed out, what happens when the supply voltage dips? We can figger (that's a real word, trust me) the current using the other prameters above. Supply: 11V 5 diode voltage drop: 10V Resistor value: 270 Ohms Yields current of 3.7mA. Are the diodes bright enough with that little current? Heck, I don't know. I suspect it is enought to turn them on, but they are probably pretty dim. So, now the "fun" begins. Gotta start thinking about how we can get this low supply current up and still stay under the high supply cap. One way is to increase the voltage difference between the supply and the diode string. How? By reducing the number of diodes in the string from 5 to 4. Now the original calculations turn into: max curent: 20mA. Suply voltage: 14.5V 4 diode voltage drop: 8V The calculated R is: 325 Ohms, and the next higher one is 330 Ohms. When the supply drops to 11V, the current is: 9mA. This is better, but I still don't know if you will think they are bright enough. And according to the principle of TANSTAFL* there is a consequence. In this case, it makes the power dissipated in the R higher. It is now around 1/8 watt at the high end of the suply voltage. This means that you have to use one of these resistors per 4 diode string. That may be OK, but let's say you want to put 3 strings in parallel for each R. Now we have: max curent: 60mA (3 strings in parallel @ 20 mA each) Suply voltage: 14.5V 4 diode voltage drop: 8V Calculated R: 108 Nearest R: 120 Current @ 11V: 9 mA Resistor Power dissipation @ 60 mA: 320 mW More TANSTAFL: This means you need a 1/2W resistor but you need 2/3 fewer of them. Also, if 1 diode in a group of 12 fails, it will obviously take out the string of 4 it is in, but the current in the ramaining 8 may increase enough to allow the smoke to escape from them as well. Also, you probably still want a cap around all of this to kill spikes from the ignition. I'm too rusty to figure out what you want for this, but any big ceramic or tantalum cap should help. Could prolly use an electrolytic here too, but I wouldn't use one cuz they have the potential to leak. Hope this helps. Send me the data sheet to the diodes or Ping me offline if you want to twiddle some of the numbers. Or just post it here. I'm sure some of the lurkers want to see this too, and I appreciate others looking over my math. -- Bill C. "I am NOT lost... I'm *exploring*" *TANSTAFL -> there's no such thing as a free lunch
From: Snag on 11 Jun 2010 11:34 spacecriter (Bill C) wrote: about the LED calculator : > > Not an e-mail. It was a post in this thread. > OK , I missed it the first time I read that post . Since I already ordered the regs , I might as well use them . Also got a buncha resistors , but will need to get some .1 and 1 Mfd caps for filtering . I might be opening a market here for exotic bits for some of the black guys in the neighborhood that ride . Lengthened a kickstand for a 2cam CB750 chopper yesterday , these guys have got money and like to have one-off stuff .... I've been asked to consider a front downtube lengthen for another guy . Tested the alternator rotor for the kickstand guy ... might get some work from these folks ! -- Snag "90 FLHTCU "Strider" '39 WLDD "PopCycle" BS 132/SENS/DOF
From: danl on 11 Jun 2010 11:49 On Fri, 11 Jun 2010 09:59:03 -0400, "spacecriter \(Bill C\)" <spacecriter(a)sysmatrix.net.yercrank> wrote: > >Snag wrote: >> spacecriter (Bill C) wrote: >>> Snag wrote: >>>> spacecriter (Bill C) wrote: >>>>> So snag, which way are you going for this? >>>>> >>>>> BTW, if you do use a voltage reg, you still need to use a current >>>>> limiting resistor, or you could blow the LEDs. >>>> >>>> Yeah , I just tonight found that out . I might be in over my head >>>> here ... so I guess I better learn to swim ! >>>> BTW , I will be going with a voltage reg , set at 10v . 5 LED's in >>>> series that want a nominal 2v . So how does one size a current >>>> limiter ? Got a big bag of 1/4w carbon film resistors in various >>>> sizes ... bet I'm gonna need the specs on my LED's . >>> >>> 1/4 W ought to be OK >>> >>> Use the calculator I sent you. It will give you the nearest common >>> resistor value. >>> >>> Ignoring the voltage regulator, plug in 12, 10, and 20. >>> >>> It says you should use 120 ohms and it will burn .04 watts. Well >>> within .125 W. >>> >>> You could dial it up a litle to derate the LEDs. >>> >>> Using 13.5, 10, and 18 yields a 220 ohm resistor. 1/4 W still works. >>> >>> -- >>> Bill C. "I am NOT lost... I'm *exploring*" >> >> Your email (??) must be lost in cyberspace ... I found a calculator , >> using 5 of the 2v/25ma LED's in series and 10v it calls for a 1ohm >> 1/4 watt resistor for each string . Shorter strings/higher voltages >> result in much higher resistor values , but 1/4 w is still >> recommended . My main concern if I use an unregulated power supply >> is the widely varying voltages . I wonder if they'll be bright enough >> at the lower voltages if I allow for the higher ones ... this circuit >> may see anywhere between 11 or so up to 14 volts . > >Not an e-mail. It was a post in this thread. > >Here's part of my previous response: > >you don't have to use a voltage regulator in this >> application. Another solution is to put in a current limiting >> resistor. That is, use the resistor in series with the LEDs to drop >> the voltage down to 10 volts. Pick the highest current you want >> through the LED and use this calculator to pick a value. >> >> http://www.quickar.com/noqbestledcalc.htm >> >> I don't believe you need to use one R per string, but I would use >> more than one for the whole circuit. Maybe one R per 3 strings? Just a >> guess on my part. Remember, the more strings conneted through >> one R drives more current through the resistor, and it will have to >> be rated for more power dissipation. >> >> Also, you may want to put in a ceramic capacitor around the whole >> thing (paralle to all of it) to cut down voltage spikes (ignition >> noise). > >OK so bear with me because I know the theory, but I don't do design every >day. > >The diodes basically have the same voltage drop no matter how much current >is going through them. As you've seen, too much current releases the magic >smoke that makes all eletronics work. See, I told you I knew the theory ;-) > >assUme-ing the diodes you have produce a nominal 2 V drop (NOTE if this is >value is different, then EVERYTHING below changes a little) we can do some >cipherin' > >5 diodes in series will drop 10 volts. A series resistor will have to drop >the difference between the supply voltage and 10V. Now say that the supply >is 12V. This means there will be 2V across the R. If you want to limit the >curent to 20mA, we can calculate the size of the R to be 2/.02=100 Ohms. > >That would be great *IF* the supply voltage was always 12V. But like you >point out, it changes. The charging system on the bike usually keeps it >under 14.5V right? (Really, that was a serious question. I can't remember >what the max voltage is on a typical vehicle, so this may be off a little. >I'm just going to assUme its correct at this point). Anyway, knowing the >max voltage the circuit needs to handle will allow us to pick a resistor >which will limit the max current and keep all that magic smoke bottled up. > >So let's put the max curent at 20mA. >Max Suply voltage: 14.5V >5 diode voltage drop: 10V > >This sets the resistor at 225 Ohms. >And the max power this resistor will dissipate is ~ 90mW > >But here's another rub (ain't there always?). Common resistor sizes usually >don't match what we calculate. So we can either mix several standard sizes >to get the calculated value. Usually only do that if this is a precision >application with a narrow range of variation. So normally you will just >pick the nearest, bigger, common size. Bigger because the current we used >to get the size was the max we ever want to see, so a bigger R value will >keep the current under that cap. The calculator I used picked 270 Ohms. > >That was all well and good for fixing the upper end current. But, like you >pointed out, what happens when the supply voltage dips? We can figger >(that's a real word, trust me) the current using the other prameters above. > >Supply: 11V >5 diode voltage drop: 10V >Resistor value: 270 Ohms > >Yields current of 3.7mA. Are the diodes bright enough with that little >current? Heck, I don't know. I suspect it is enought to turn them on, but >they are probably pretty dim. So, now the "fun" begins. Gotta start >thinking about how we can get this low supply current up and still stay >under the high supply cap. One way is to increase the voltage difference >between the supply and the diode string. How? By reducing the number of >diodes in the string from 5 to 4. Now the original calculations turn into: > >max curent: 20mA. >Suply voltage: 14.5V >4 diode voltage drop: 8V > >The calculated R is: 325 Ohms, and the next higher one is 330 Ohms. > >When the supply drops to 11V, the current is: 9mA. This is better, but I >still don't know if you will think they are bright enough. And according to >the principle of TANSTAFL* there is a consequence. In this case, it makes >the power dissipated in the R higher. It is now around 1/8 watt at the high >end of the suply voltage. This means that you have to use one of these >resistors per 4 diode string. That may be OK, but let's say you want to put >3 strings in parallel for each R. > >Now we have: > >max curent: 60mA (3 strings in parallel @ 20 mA each) >Suply voltage: 14.5V >4 diode voltage drop: 8V > >Calculated R: 108 >Nearest R: 120 > >Current @ 11V: 9 mA > >Resistor Power dissipation @ 60 mA: 320 mW > >More TANSTAFL: This means you need a 1/2W resistor but you need 2/3 fewer >of them. Also, if 1 diode in a group of 12 fails, it will obviously take >out the string of 4 it is in, but the current in the ramaining 8 may >increase enough to allow the smoke to escape from them as well. > >Also, you probably still want a cap around all of this to kill spikes from >the ignition. I'm too rusty to figure out what you want for this, but any >big ceramic or tantalum cap should help. Could prolly use an electrolytic >here too, but I wouldn't use one cuz they have the potential to leak. > >Hope this helps. Send me the data sheet to the diodes or Ping me offline if >you want to twiddle some of the numbers. Or just post it here. I'm sure >some of the lurkers want to see this too, and I appreciate others looking >over my math. Vewey, vewey intewesting. danl
From: Eiron on 12 Jun 2010 02:20
On 11/06/2010 01:24, spacecriter (Bill C) wrote: > So snag, which way are you going for this? > > BTW, if you do use a voltage reg, you still need to use a current limiting > resistor, or you could blow the LEDs. Use an LED driver chip. It costs almost nothing and does the job properly. -- Eiron. Alazzurra (1985) GSXR1100L (1990) XL883L (2008) |