EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Logan on March 10, 2022, 04:12:44 pm
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Hi friends.
Sometimes I need a simple LED indicator, I usually just put a resistor in serial. But as the voltage goes higher, the efficiency becomes worse. At 12V, the resistor already begin to warm up.
Are there any way to do this more efficiently, in a way that is not too complex?
Thanks!
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I use ultra-bright LED diodes. They might be bright enough with only about 0.1 mA.
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Yeah, use more efficient leds, which need less current, and if you want, parallel two resistors to spread the voltage drop across both.
There are also led drivers that are very simple 2-3 pin devices which are like a resistor... and you can buy them with default current of 5-10mA ... but for a single indicator led it's not worth it.
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For small diodes (5-20ma), the best is to first lower the voltage using a buck converter (To provide best conversion efficiency) to something slightly higher than the max led voltage drop, so it'll require very small resistors and barely waste any power.
But if not a battery-powered application and they're just few leds, it's not worth it.
Driving the led with 10mA at 12V would waste ~100mW in the limiting resistor, will get slightly warm but nothing to worry about unless you use very small resistor sizes:
(https://i.stack.imgur.com/0GqgI.jpg)
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If it's AC, a capacitor might be usable (in part). At low voltages, you'd need kind of a big cap though, several uF. Film capacitors not so practical there; ceramics plenty available in such ratings, albeit relatively expensive just for lighting an LED. If this is DC then obviously this is irrelevant.
Note that there are two kinds of green LED: traditional old GaP greens have poor to terrible efficiency (particularly bad at low currents), and a ~2.2V Vf. The new kind is GaInN based, Vf ~ 3V just like blue LEDs, but green emission (which is achieved with, something about nanodots or a quantum well, I think?). Despite the higher voltage drop, the efficiency is so much better, plus efficacy (around green is the "brightest" color to human vision), that they are in all respects better.
And needless to say (from all the blue spots we burned in our eyes, mid-2000s onward), blues are usually quite high efficiency. Reds are reasonable too, in the classic AlGaAsP alloy; yellow and IR are in the same family too, with efficiency worsening as you go up in color. So, yellow is still alright, but green kinda sucks (note GaP is just AlGaAsP with Al and As turned to zero).
Speaking of efficacy: the eye curve is dropping off steeply in the red range. Deep red is quite a dim color to the eye. There's some advantage in skewing it even just a little orange. Such "high efficiency reds" are available. The color is very subtly orangish, enough that you'll probably never notice it short of a side-by-side comparison.
Interestingly, GaP and GaInN greens are noticeably different, GaP being just slightly yellowish, GaInN slightly cyanish. It's quite apparent, side by side; but no one's going to mistake either one for anything other than "green". :-+
Anyways, for any kind of DC arrangement, the only thing you can do is, use multiple in series, or add a ground-return terminal and power converter.
The former might actually be a little feasible, but it depends. For sure it's the way to go for discrete LEDs on board, that you just need to stack up a bunch say for a background light, or to trace out a shape, or whatever. If they all need to be independent, that's harder to pull off in the same way, however. As for the converter, obviously that's not going to fly, not just for individual indicators.
Apparently there are some high-voltage LEDs, but I don't think anything is widely available outside of illumination? Saw that on bigclive the other day, some multichip solution, or maybe it's monolithic after all? A bunch of diodes wired in series, all clustered together acting as one big chip rated 24V. Also the faux filament stuff, lots of teeny chips wired in series, almost use it like glowing wire. Not sure you'd use either for indication though, and certainly not in colors (filtered LEDs give terrible washed out colors!).
If they're on most of the time, it might do to run them in series anyway, from a current source, and short out a few using a suitable switch (maybe a BJT across each respective diode, or set of diodes, that needs to be blanked?). The load is always the same, so the more LEDs turned out, the CCS drops the remainder, and the hotter it gets (instead of the same heat distributing over all the LEDs).
Or just add the 3.3-5V supply and run them all on that. Preferably with current-limiting drivers, very easy to do.
But these are getting very much towards design options. Whereas, if this is like replacing incandescents in an existing product, yeah, don't even worry about it really, it's still more efficient even with the resistor in there -- just try to stick the resistor somewhere it can dissipate its heat.
Tim
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Thank you everyone!
I think I may just live with it... Other methods are just not worth it as you mentioned.
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simple,cheap，fyi.
[attachimg=1]
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simple,cheap，fyi.
(Attachment Link)
That will regulate the voltage, but it's slightly less efficient than an plain old LED & resistor.
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Do you mean two resistors in SERIES to spread the voltage drop (and power)??
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simple,cheap，fyi.
(Attachment Link)
that's actually burning more power... now you have the base current and the current through the zener to contend with.
you can't destroy energy. you still need to drop x amount of voltage and deliver y amount of current so voltage x current is still power. you are just burning it off in different parts.
the onyl way to make this lower current is to go the switching route.
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yes,that can`t improve efficiency(To be precise, at the same LED current, compared with direct series resistance, the efficiency is reduced.)
The purpose of this circuit is to keep the BRIGHTNESS of the LED from changing with the supply voltage .