Current through the chain would be limited by the weakest link?
Given the OPs question he might well be better off with 3x D NiMH cells in a piece of plastic drainpipe and avoiding the electronics altogether.
-- Regards, Martin Brown
Martin Brown 's
YEs tHem oLD GaS LIGhtS wOrk REaLlY WelL
However, a switch mode current source can save power and get useful energy from the battery below where a resistor will work..
Or less, at the top end. It depends on the requirements.
True. But it can also take rechargeable batteries into an abusive regime where it fails to recharge or suffers severely shortened life.
-- Regards, Martin Brown
Or not. It's all about requirements.
a zener would do or maybe two diodes drop
Not making your life any easier are they. The easiest answer is one resistor feeding each lamp, so with 3 lights you use 3 resistors.
8 leds @ 20mA ea = 160mA. So you can use a 12 ohm resistor there, power rating 0.5w or more. It will get hot. Or small Rs of 5.6 & 6.8R in series 240mA so 8.6 ohms at least 0.5 watt - or small 3.3 ohm and 4.7ohm resistors in series 2.2 ohms, at least 1 watt. Or 4x 8.6 ohm small Rs in parallel.If you pick resistors of say double those power ratings they wont get so hot.
If anyone wonders I've calculated these for 6.5v.
NT
a "6v" AGM is 7.2V when fully charged.
6V is likely to destroy the LEDs.a series resistor is a better simple solution.
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14.4v is charging voltage for flooded 12v lead acids, and they'll charge at 13.6v too. I dont believe youre going to get 7.2v from the battery itself.
NT
The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
On Sunday, March 30, 2014 10:28:13 PM UTC+1, Maynard A. Philbrook Jr. wrote :
rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
No :) That would reduce life due to convertor losses.
NT
Maybe I missed something but for what ever reason, I don't think the math is working in your favor?
Jamie
te:
in rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-u p the 6V voltage to 24-30 V as required.
eProve it.
** Impossible to reduce the battery drain and have the same power. ** Also impossible, cos they all have different current drains. ** Insanity.
Just adds conversion loses to the battery drain.
.... Phil
-- Instead of just blurting out opinion, why not do the work that shows what numbers are working against him and post that work? John Fields
rote:
rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
I kinda expected you'd see the mistake you'd made. The transforming action gets you back up to the battery current you started with, plus more due to convertor losses. You dont gain a thing.
NT
It's so f'ing obvious where the screw up is. Go back and read it again, thoroughly this time.
I can understand you sliding down the other side of the hill, but really, I didn't think you were anywhere near the valley, yet!
P.S. If it makes you feel better, at least you still have one friend, Bloggs. One lies, the other swears to it.
Jamie..
-- You can't support your opinion with fact, so you play the game as if it's you who's in charge and then introduce all sorts of delaying tactics hoping for a miracle to come along and save your miserable ass from the roach spray.
Amps = .65 amps
rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
Ye gods did you ever miss the point.
?-((
Spare us all the nuisance deletion exercises and stop feeding the troll/village-idiot. ...Jim Thompson
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