how to save the wasted current from a resistor? - Page 1
Forum section: electronics / beginners

author: crowzing
author_rank: Newbie
timestamp: May 31, 2016, 05:42:05 pm
content: if i have this led circuit working just fine but i want to save the energy wasted by the resistor what should i do then?i mean i don't want to waste anything. give the led what it needs and save the rest to a battery or something and that's it.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 31, 2016, 05:57:37 pm
content: What you probably want is a switch mode constant current source, eg:http://www.onsemi.com/pub_link/Collateral/NCP3065.PDF

author: alsetalokin4017
author_rank: Super Contributor
timestamp: May 31, 2016, 07:30:22 pm
content: Or even simpler, for single or just a few normal LEDs, the AP2502 and (optional) a PWM signal for variable dimming.

author: Dave
author_rank: Super Contributor
timestamp: May 31, 2016, 07:34:38 pm
content: Quote from: alsetalokin4017 on May 31, 2016, 07:30:22 pmNope. You're still wasting power, just in a MOSFET inside the chip instead of a resistor.

author: Zero999
author_rank: Super Contributor
timestamp: May 31, 2016, 07:56:08 pm
content: Quote from: crowzing on May 31, 2016, 05:42:05 pmWhat's the: battery voltage, LED current, forward voltage?If it's unregulated, as simple squarewave oscillator driving a MOSFET, inductor and flyback diode, forming a buck circuit will do.The attached example drives the LED at 270mA, when run from a 6V supply. For lower currents, reduce the duty cycle, use a CMOS 555 and you may be able to omit the MOSFET and use the 555's internal MOSFET.

author: retrolefty
author_rank: Super Contributor
timestamp: May 31, 2016, 09:07:17 pm
content: Quote from: Dave on May 31, 2016, 07:34:38 pmWasting far less with a PWM/MOSFET switch then a simple resistor.

author: KL27x
author_rank: Super Contributor
timestamp: June 01, 2016, 02:05:14 am
content: QuoteAre you sure about that?

author: rsjsouza
author_rank: Super Contributor
timestamp: June 01, 2016, 02:24:59 am
content: Quote from: crowzing on May 31, 2016, 05:42:05 pmIn the circuit you showed, the resistor will dissipate energy as heat according to the Joule equation:\$P=V_{Resistor}I_{LED}\$As you suspect, this energy can be used to heat something you need (water, your body, etc.) or even help aThermal battery, but I am pretty sure this is not very practical. In other words, the energy is mostly wasted to the environment.With that in mind, the solutions others proposed are tailored to increase the efficiency of your circuit, thus decreasing the wasted heat generated.

author: EEVblog
author_rank: Unknown
timestamp: June 01, 2016, 03:04:42 am
content: What you want is the most efficient constant current source you can get.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 01, 2016, 03:18:07 am
content: Quote from: Hero999 on May 31, 2016, 07:56:08 pmUhhh....Might make a smoke generator though.This must have a current feedback loop to work. Otherwise it's no different than connecting the battery (now an arbitrary fixed voltage, rather than 6.0V or whatever) directly to the LED.Tim

author: digsys
author_rank: Unknown
timestamp: June 01, 2016, 03:19:02 am
content: The ol' law of diminishing returns !! If you have a high power LED, requiring 1A++, there's heaps of ways to save a LOT of wasted energy.IF you're talking about a 10-20mA LED, or even a high efficiency 1-2mA, it gets tougher, and often quiescent currents-switching losses etcoutweigh the savings. In some of these cases, it may be more efficient to use capacitor switching.

author: batteksystem
author_rank: Regular Contributor
timestamp: June 01, 2016, 03:20:51 am
content: Quote from: KL27x on June 01, 2016, 02:05:14 amI would say whether a resistor will burn more energy than a constant current control circuit depends on the LED driving current. For large driving current (>700mA) I do think there is significant saving.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 01, 2016, 03:28:30 am
content: Quote from: batteksystem on June 01, 2016, 03:20:51 amWhat's the power lost in the resistor? P = (Vsupply - Vled) * IfWhat's the power lost in the current sink? P = (Vsupply - Vled) * If + Vsupply * IbiasThe current sink can only ever belessefficient, because (except for some boutique parts that don't need it) some additional bias current is needed.The only reason we use CCS for LEDs is where the wide range of constant brightness is needed.Tim

author: KL27x
author_rank: Super Contributor
timestamp: June 01, 2016, 05:18:40 am
content: ^Beat me to it.I had to look up the datasheet to figure out what I assumed Dave already knew."linear current sink."Ok, next question.So how does PWM help?Is it possible that 1 amp at 10% duty cycle can be visibly brighter to a human being than 100% duty cycle at 100mA? I actually think this can be the case, given the properties of the LED, but perhaps only with very minimal current that is not close to the LED's max rating. I am thinking that at some point near the knee that doubling the current might possibly more than double the light output? (Of course near the max rating this is for sure the reverse... and PWM would be less efficient in terms of light output). Or maybe that strobed light is more efficiently used by the human eye? I dunno, but I was open to hearing all about it. I was wondering if this is what batteksystem was going to get at. Or maybe he just didn't realize his LED driver of choice is a linear current sink or that PWM in general wastes just as much energy for the given amount of current. W/e the voltage drop you are getting rid of linearly, it obviously doesn't matter if it's 100% duty cycle at X amps or 10% duty cycle at 10X amps... it's the same.As far as I know, the most direct and efficient way to reduce that waste is to select the power supply to produce as little overhead as possible. Just enough to provide a fairly stable/constant current over operating temp range. By arranging the LEDs, selecting the appropriate power supply, and/or using DC converter in order to convert some of that excess overhead voltage into more current at a lower voltage. You will still want a series resistor, but a smaller one that is dropping less voltage.Using a DC converter WITH constant current output may or may not be as efficient at any given power supply voltage, but it will be more stable. The active current limiting/adjusting part is a red herring that doesn't actually increase efficiency. It's the buck converter (and boost converter, if you're talking about extracting more energy from sagging battery output; so yeah, from a battery, this sort of switchmode IC with current regulation will certainly be king) where you reclaim that wasted energy. Essentially, an LED uses only X volts. If you don't want to waste power, feed it with as close to X as possible to give the desired results.OP, to keep it simple: If FVD is 3.0V, and you can provide 3.1V with a small series resistor to drop the excess 0.1V, AND IF the brightness/current draw over the temperature range and battery life is acceptable... that will waste only 1/10th of what happens when you use a 4.1V power source. The tradeoff is your brightness/current draw curve is going to not be as flat, considering battery output curve.

author: Zero999
author_rank: Super Contributor
timestamp: June 01, 2016, 08:52:44 am
content: Quote from: T3sl4co1l on June 01, 2016, 03:18:07 amNope, there's no need for current regulation. If the duty cycle of the oscillator is set correctly then the current will stabilise to a safe level.

author: danadak
author_rank: Super Contributor
timestamp: June 01, 2016, 10:24:19 am
content: This might be useful -https://ledshield.wordpress.com/2012/11/13/led-brightness-to-your-eye-gamma-correction-no/https://en.wikipedia.org/wiki/Persistence_of_visionRegards, Dana.

author: batteksystem
author_rank: Regular Contributor
timestamp: June 01, 2016, 11:20:51 am
content: Quote from: KL27x on June 01, 2016, 05:18:40 amWhat I said is "constant current control circuit", which is something like NCP3065. Of course for Linear CCS, the efficiency will not improve.

author: Dave
author_rank: Super Contributor
timestamp: June 01, 2016, 12:54:12 pm
content: Quote from: KL27x on June 01, 2016, 05:18:40 amIf the PWM frequency is sufficiently high for the brightness to appear constant (150-200Hz and up), it all just comes down to how many photons hit your retina. The eye's response is a low-pass thing so it makes zero difference if the light is pulsed or constant.But here's the kicker: If you drive LEDs at a significantly higher peak current, not only is the power dissipated on the LED going to be higher (higher voltage drop), it is going to have a slightlylowerbrightness. The efficiency of the LEDs decreases as you increase the current (google 'Auger effect in LEDs'), so you will always get slightly less light output out of an LED with PWM than you would with the same mean DC current.If you want the best efficiency, you need an SMPS that feeds the LEDs with a constant current. Simple as that.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 01, 2016, 02:11:51 pm
content: Quote from: Hero999 on June 01, 2016, 08:52:44 amThere's no control for LED voltage, as a ratio of supply. Or for LED current in general. See:SPICE models are fairly typical, what I had on hand. Note the output stage is severely overbuilt; an NDT3055 would be more than enough here. The FDD8647 isn't even as big as the IRF3717. And a B130 diode would be more than enough, too. (The RC damper was added to improve transient response, but dissipates little power.) The LED model is LXML-PWC1, from the manufacturer, but mind it only models forward DC parameters. But that's fine here. And the inductor is lossless, which is quite unreasonable, but we'll just take that as understood.H sources and X blocks are just for measuring currents. Their results aren't shown below, and they have no effect on the circuit.Here's what I_led looks like for 6, 9 and 12V input:It's just barely BCM at the lowest input voltage, and current risesvery sharplyfor higher input voltage. The operating frequency was about 62kHz.On the upside, efficiency (at 6V) was about 90%. In reality, that would be 70-80%, dominant losses being the inductor, and switching losses in the transistor.Tim