From: John Woodgate
Subject: Re: Best way to power array of LEDs?
Date: Thu, 7 Nov 2002 05:44:24 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Thu, 7 Nov 2002 07:44:32 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Nucharin W. Jansen
wrote (in <3DC99AD4.2AB0FE52@fanwap.com>) about 'Best
way to power array of LEDs?', on Thu, 7 Nov 2002:
>Yes, I mean in series.
>But two strings can not use the same R values.
>So sad it very hards to calculate the current and R.
>the forward voltage drop does not identical in each LED.
>and it changes when current change too.
>strings 1 : ( 12 volt - ( 3.453 + 3.6324 + 3.833 ) ) / 40 ohm = 27 mA
>Now I want 20 mA , How to calculate the resistor ?
>Since forward voltage change when the current change too.
A very slight change in temperature would change your 4-figure voltages
significantly. You are looking for a degree of precision which is
meaningless. Consider the voltages to be 3.45, 3.63 and 3.83 V. Between
27 and 20 mA, they will hardly change. So your resistor has to drop 12 -
10.91 V at 20 mA, and is thus 54.5 ohms. A standard 56 ohm would be OK.
While the current won't be *exactly* 20 mA, it will be close enough not
to make a significant difference to the light output.
However, this resistor value is much too low to keep the current
substantially constant if the (12 V - just 2-figure accuracy) supply
voltage varies. Furthermore, 1.09 V is not really enough voltage to
operate a simple electronic constant-current circuit. With just two LEDs
in series, the simple series-resistor technique is much more
satisfactory. 12 - (3.63 + 3.83) = 4.54 V. 4.54/20 mA = 227 ohms - use
220 ohms, giving 20.6 mA. Then at, say, 12.2 V supply, we get 4.74 V
across 220 ohms, which means that the current is 21.5 mA. If that is not
'constant' enough, 4.54 V is quite enough to operate a simple electronic
constant-current circuit. Just a single transistor, two resistors and
two diodes (or a red LED) would do.
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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