Calling All 'Lectronicians !
in [Harley-Davidson]
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From: Old Crow on 12 Jun 2010 07:09 "spacecriter (Bill C)" <spacecriter(a)sysmatrix.net.yercrank> wrote in message news:JO6dnYPBYty33I_RnZ2dnUVZ_oudnZ2d(a)sysmatrix.net... > > 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.> -- <Whack!> > Bill C. "I am NOT lost... I'm *exploring*" > *TANSTAFL -> there's no such thing as a free lunch > Re-read your Heinlein, Bill. I just read Starship Troopers for like the 500th time. TANSTAAFL "There's Absolutely No Such Thing As A Free Lunch" That man wrote so much like my father talked it was scary. -- Old Crow '82 FLTC(P) 92" '87 FLTC '61 F-100 302/C-6 BS#133, SENS, TOMKAT, SLOB#13, MAMBM
From: spacecriter (Bill C) on 12 Jun 2010 07:48 Old Crow wrote: > "spacecriter (Bill C)" <spacecriter(a)sysmatrix.net.yercrank> wrote in > message news:JO6dnYPBYty33I_RnZ2dnUVZ_oudnZ2d(a)sysmatrix.net... >> Bill C. "I am NOT lost... I'm *exploring*" >> *TANSTAFL -> there's no such thing as a free lunch >> > > Re-read your Heinlein, Bill. I just read Starship Troopers for like > the 500th time. > TANSTAAFL > "There's Absolutely No Such Thing As A Free Lunch" > That man wrote so much like my father talked it was scary. I actually remember the phrase from an Economics course I took a lifetime ago. Freedman wrote a book with that theme. Was so long ago, I got it wrong <sigh> Seems like somebody told me the memory was the second thing to go. Though, I can't remember who. Your pop sounds interesting. -- Bill C. "There *ain't* no such thing as a free lunch"
From: Snag on 12 Jun 2010 08:46 Eiron wrote: > 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) Did some looking around , and they cost a bit more than "next to nothing" for what I want to do . The cost per power unit doing it the way I'm pl^^^ considering is less than a couple of bucks apiece . Got all the parts bought to build five units except the filter caps , and I don't expect them to be all that pricey . -- Snag "90 FLHTCU "Strider" '39 WLDD "PopCycle" BS 132/SENS/DOF
From: Eiron on 13 Jun 2010 03:39 On 12/06/2010 13:46, Snag wrote: > Eiron wrote: >> 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. > > Did some looking around , and they cost a bit more than "next to nothing" > for what I want to do . The cost per power unit doing it the way I'm pl^^^ > considering is less than a couple of bucks apiece . Got all the parts bought > to build five units except the filter caps , and I don't expect them to be > all that pricey . The first LED driver chip I looked at, for $2, needs one resistor and will convert 12v to 85mA at 40 volts (constant current output) So if you want 20mA per LED, you could use one chip to drive 4 strings of 20 LEDS with 4 resistors to balance the current. That's $3 to drive 80 LEDS. But as you only have 93 LEDs left after trying to drive them from a voltage source, you would use one chip to drive 3 strings of 15 LEDs (with 3 resistors). There are higher current drivers available. I was just considering the first one I saw in a catalog. But it shows that it's often cheaper to do it properly. -- Eiron. Alazzurra (1985) GSXR1100L (1990) XL883L (2008)
From: Snag on 13 Jun 2010 07:10
Eiron wrote: > > The first LED driver chip I looked at, for $2, needs one resistor and > will convert 12v to 85mA at 40 volts (constant current output) > So if you want 20mA per LED, you could use one chip to drive 4 strings > of 20 LEDS > with 4 resistors to balance the current. That's $3 to drive 80 LEDS. > > > -- > Eiron. Care to share the source and part number ? -- Snag "90 FLHTCU "Strider" '39 WLDD "PopCycle" BS 132/SENS/DOF |