LED flashlight regulated vs unregulated? - Page 1
Forum section: electronics / projects

author: drummerdimitri
author_rank: Frequent Contributor
timestamp: August 12, 2019, 12:31:11 am
content: I have some flashlight's I'd like to test in terms of having their power supplies.My theory is that a regulated flashlight will try to maintain a constant power draw so as the battery voltage drops with use, it will increase the current draw by a proportional amount correct?Unregulated flashlights would just follow ohms law and act like a resistor right?I could then characterize the LED driver's behavior through my Rigol DP832A if my methodology is correct.

author: BrianHG
author_rank: Super Contributor
timestamp: August 12, 2019, 01:14:49 am
content: Quote from: drummerdimitri on August 12, 2019, 12:31:11 amClose, but not quite. Remember, the LED is a diode with a greater than exponential curve in current draw as voltage increases. So, as the battery voltage drops, the current drops far quicker than if it were a resistive incandescent bulb (whose resistance actually decreases as its glow decreases making the battery drain a little faster as well) or a fixed resistor value.In fact as the battery voltage comes close to the LED's turn on voltage, you may be drawing 1-2ma from the batteries while only 0.5v higher, and the LED will draw 20ma or more.

author: Audioguru again
author_rank: Regular Contributor
timestamp: August 12, 2019, 01:35:26 am
content: If you use a Lithium battery (throw away or rechargeable type) then its voltage does not drop much as it runs down then you might not need a voltage regulator.An LED needs a current-limiting resistor or circuit.

author: edavid
author_rank: Super Contributor
timestamp: August 12, 2019, 01:56:01 am
content: Quote from: drummerdimitri on August 12, 2019, 12:31:11 amNot necessarily.A 1 cell alkaline/NiMH flashlight will use a boost circuit, and will draw constant power from the battery, and supply constant current to the LED.However, a 1 cell Li-ion flashlight willusuallysometimes use a linear current regulator, and draw constant current from the battery.

author: james_s
author_rank: Super Contributor
timestamp: August 12, 2019, 03:46:07 am
content: Quite a few Li-ion powered flashlights use a buck regulator, at least the decent ones do.

author: digsys
author_rank: Unknown
timestamp: August 12, 2019, 04:08:06 am
content: All of the above PLUS a yes / no ..A dc-dc converter will have loss, usually 12-18% (ie 82-88% efficiency), unless you are lucky and find one with a 92% efficient converter, which is still 8% loss.Can you detect a (slight?) light output difference if you were ~12-18% lower in voltage?What I am alluding to is - IF the battery voltage is close to the "ideal" LED light output voltage for most it's battery life, there is no point adding extra complexity.If you're after longer run time, use a "fatter" battery, or put them in parallel. IF you can't match your "ideal" LED type / configuration to an "ideal" battery type / voltage, then YES, you need to think about dc-dc converters of the constant current type.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 12, 2019, 09:35:34 am
content: Basically at least three options:#1: Cheap-ass: li-ion cell runs between about 4.1-2.8V (under load); LED Vf at rated current is about 3.2V and exponential; very small-value series resistor is used to limit current. Light is very bright and the LED runs very close to or beyond max ratings with full battery, and dies out almost completely well before the battery is fully exhausted (somewhere around 10-30% state-of-charge). You can still accidentally overdischarge the cell by leaving it on. Typical in cheap products.#2: DC/DC converter regulates the LED current, and as you say, the input current rises when the battery voltage drops. The DC/DC has an under-voltage cutoff to protect the cell from overdischarge; the light goes out suddenly,#3: like #2 but with some intelligence which gradually lowers the LED current to "show" the user that the battery is running out, extending runtime near the end, at least producing some light output for a longer time, instead of dying suddenly.#1 and #2 have similar efficiency. The resistor drops about 0.4-0.5V, or about 10-15% on average; a good DC/DC converter is also about 85-90% efficient. #1 is best when you can charge the cells to optimum voltages, and only use part of the capacity; think about only charging to 4.0V, and only discharging to 3.7V, to avoid running either too bright (and too hot; inefficient as well), or too dim. But the capacity sucks if you do this.

author: drummerdimitri
author_rank: Frequent Contributor
timestamp: August 12, 2019, 12:02:38 pm
content: I guess the only way to know is to characterize each flashlight individually to see how it behaves since it seems there are many ways the manufacturer could have designed the LED driver.I tried digitizing the current (attached image) on one LED I suspected was driven by PWM. Can I safely assume that the PWM signal is 7518 KHz in this case or should I be using a photoresistor instead?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 12, 2019, 01:41:51 pm
content: You can get a relative measurement with a photodiode or photoresistor, yes. You need to calibrate at a few points (photodiode) or most of the curve (photoresistor), and preferably it should be done in a dark enclosure with a consistent optical setup (everything clamped, etc.).If you have access to the internals, you can replace the LED with a schottky diode (if its supply is AC, often the case for boost type circuits) and filter cap, and connect an electronic load to emulate an LED while measuring the DC power output. If it's switching, it shouldn't be far off to measure the input power just as well, and assume a modest efficiency (80-95%?).Should be pretty obvious in any case, once you start varying Vin.Incidentally, about switchers -- everything should stop when the cell hits 3.0V or so, as further discharge is hazardous. The control may simply shut off, in which case, the voltage then slowly recovers, and then it flashes again, and so on. Or it may throttle down, effectively giving you a flashlight that dims like an unregulated one does, but all the dimming is squished towards the fully-discharged end, not gradual throughout the charge curve.Tim

author: Zero999
author_rank: Super Contributor
timestamp: August 12, 2019, 08:21:23 pm
content: Quote from: edavid on August 12, 2019, 01:56:01 amIt depends, I've seen both linear and switched power supply circuits running LEDs off a lithium battery, although the former is more common.

author: tunk
author_rank: Super Contributor
timestamp: August 12, 2019, 08:51:49 pm
content: You can find tests of some LED drivers here:https://lygte-info.dk/info/indexLedDrivers%20UK.html

author: Datman
author_rank: Regular Contributor
timestamp: August 13, 2019, 12:19:21 pm
content: Some 3W flashlights with lithium rechargeable battery doesn't have a regulator or a series resistor. They only rely on the wires resistance and the LED internal resistance. So, a similar flashlight draws something like 1.4A at full charge, while maximum LED current is about 1A, and current decreases during discharge of the battery.Many years ago I bought in a shop a flashlight very similar to this:https://www.amazon.it/Levoluzione-della-Torcia-Militare-autorit%C3%A0/dp/B00DEXFLWCI opened it and I saw that there wasn't a regulator inside, and current was much higher than maximum admissible current for the LED, then I modified the flashlight adding three AMC7135PKFA paralleled to obtain a total current of about 1A. Their dropout voltage is only 120mV, then nominal current is guaranteed until battery voltage is only 0,12V more than LED conduction voltage.