Feeding current limited LEDs a higher voltage - Page 1
Forum section: electronics / beginners

author: jirij
author_rank: Contributor
timestamp: September 24, 2023, 12:51:44 am
content: [SOLVED, thanks!]Hello, I'm trying to understand the various ways how to drive LEDs efficiently and studying various LED bulb teardowns gave me some hints.Newer ones seem to use multi-LED-element packages (ie. 18V drop per package) and match their combined voltage to approximate the rectified voltage (ie. SMD2082D or NUD4001) as opposed to switching buck converters.That made me realize an interesting idea -- why do the voltage matching at all? If you employ current limiting, the voltage across the LEDs will theoretically never rise above their forward voltage drop at that current. A more realistic approach would use a capacitor in parallel with the LEDs, ie.Mains power is plugged in, but no current flows through the LEDs (no switching happening)The current limiting logic starts charging up the capacitorAt some point, the voltage across the capacitor reaches Vf of the LEDs and they start conductingThe voltage continues to rise until the current across LEDs reaches ie. a 20mA limitThe voltage across the LEDs is capped at (oscillates around) whatever the Vf is at 20mA, ie. 9VIt doesn't matter if the open loop voltage is 9V, 12V or 320V. Even the capacitor could be low-voltage rated if the switching logic is fast enough.Experimenting with my lab bench PSU, the theory seems to check out in reality ... ?So what's the problem with this? Why nobody seems to do it like this?Is it because the non-zero ESR of that capacitor would cause the LEDs to see voltage spikes caused by the switching?Is it because the capacitor would see high voltage inductive spikes well above its rating?Is it because it would require high-side current switching and that would be harder to do?Why do we instead use inductors to implement buck converters (even in a minimal form, see PT4115 or FL7760) that inherently dissipate unwanted heat?I tried searching the forum, and searching the web, but most LED current limiting is focused on basics with resistors and nobody dives into the interesting stuff.Thanks for the hints!PS: I'm making a lower voltage high-current limiting circuit for a high power LED, not working with mains voltage, but I'd like to understand the above before I proceed anyway.

author: ledtester
author_rank: Super Contributor
timestamp: September 24, 2023, 08:52:03 am
content: With a constant current driver the excess power has to be dissipated by the driver chip itself and that's why you want the supply voltage to be close to the LED voltage.Have a look at how P_D_total is computed on page 4 of the NUD4001 datasheet attached below. That value is limited to 1.13 Watts.

author: golden_labels
author_rank: Super Contributor
timestamp: September 24, 2023, 02:37:43 pm
content: On top of what ledtester said: because it wastes power. If you are running LEDs at 180 mW (based on your 20 mA @9 V example), you do not want to have over 6 W losses in the regulator (20 mA @230 V AC). One strives for maximum efficiency, not creating a heater with a LED indicator.Linear current limiters are sucking in constant current, but not constant power. Switched mode supplies are approximating constant power consumers. Even if the latter is imperfect, it is still better than the linear solution. At least as long as the unwanted loses are low enough, but that becomes true as soon as we leave the domain of small voltage drops.As for using a single switch and an output capacitor, it’s a common trap for young players. Yes, it will lower voltage just fine. The problem is, again, power. In a way not immediately obvious, in particular if one is skipping half of the circuit. Be aware that your power source, no matter its real implementation, is acting as capacitance. So your entire circuit is two capacitors connected by a switch. That’s a specific configuration to be recognized, because capacitor-capacitor or inductor-inductor circuits can’t do conversion efficiently.To see why, observe that to have different voltage on C1 than it was on C2, a part of that voltage must go to somewhere. The only voltage divider you can form is from C1 and the switch (and wires). If the switch (and wires) drops voltage, it means it also consumes power. And since the switch (and wires) can’t store energy, the only option is that they release it — mostly as heat. This is no different from a linear regulator. The situation would be different, if there was some element, which can temporarily store the energy. As it happens, there is such an element: it’s an inductor. And at this point we invented a switched mode power supply.

author: jirij
author_rank: Contributor
timestamp: September 24, 2023, 03:14:31 pm
content: Quote from: ledtester on September 24, 2023, 08:52:03 amThanks, I somehow missed that in the datasheet.I'm still trying to grasp the physics of it, though --Given a high-side current limiter usinga current-sensing amplifier across a sense resistora comparator to compare it against some voltage-divided referencea P-MOSFET to switch the loadit's true that the MOSFET would have basically constant Rds(on) and thus dissipate more power with higher voltage across it.However it would also charge up the output capacitor quicker (second figure in my first post), so at approx. double the voltage it would see half the duty cycle from the comparator.Yes, it would have to deal with higher voltage spikes, but wouldn't the overall dissipated power be the same?---Quote from: golden_labels on September 24, 2023, 02:37:43 pm(And it's my luck that a new reply just came as I wanted to post this.)Thanks, I kind of already realized that that Kirchhoff's voltage law probably applies and the remaining voltage will be across the regulator. But wouldn't the above still apply?The job of the P-MOSFET inside the regulator is to essentially transfer charge between C2 (power source) and C1 (across the LEDs) to a specific voltage across C1 where the LED characteristics cause them to draw the desired current.And it doesn't matter for the dissipated power if that's in long bursts of lower voltage or short bursts of high voltage. .. Does it?Maybe I'm just failing to understand the inductive characteristics of a MOSFET (judging by your hint about an inductor) causing the extra power losses?Thanks!

author: golden_labels
author_rank: Super Contributor
timestamp: September 24, 2023, 06:15:53 pm
content: Quote from: jirij on September 24, 2023, 03:14:31 pmIn the capacitor-switch-capacitor scenario it does not matter. Look at it from the point of view of power equation. The maths are at the end, if you are interested.Quote from: jirij on September 24, 2023, 03:14:31 pmNo, I mean an actual inductor. As seen in the buck converter.To get lower voltage, we need to get rid of some of the energy. If we only used a switch (as in capacitor-switch-capacitor), that energy is lost in the switch and wires. If we use an inductor, the element in which energy is “lost” is that inductor. However, the energy is not really lost: it is used to charge the inductor. Then it is dumped into the capacitor.In the first part of a cycle the inductor is charged, taking in energy from the power source. In the second part of the cycle circuit is switched, so the inductor puts that energy into the “storage” capacitor. Both cycles are balanced, so exactly the right amount of energy is transferred to get just the right voltage.The maths:LED uses some approximately constant power (PLED). After the capacitor is initially charged, it receives and dumps the same amount of energy in each cycle(1), so it can be eliminated from the equation. So the source must deliver the same amount of power (PSRC) as the LED consumes:PLED=PSRC.The average power LED sees is:PLED=VLED·I. Note that I didn’t put a subscript onI, as the average current remains constant from source through the switch and LED. For completeness, power from the source is:PSRC=VSRC·I.Since we know both power values must be equal, we get:VLED·I=VSRC·IDividing both sides byI(current isn’t 0, if we consider energy being transferred), we get:VLED=VSRCTherefore: if there is no losses across the switch+wires, both voltages can only be equal. That is against the goal of having the voltages different. From that we may conclude there must be some losses at the switch+wires. So we get a simple equation, where we sum three values of power, with two of them being fixed. That obviously means the third one is also fixed. So the average power lost in switch+wires must remain fixed, not matter if it’s pulsed or not. Of corse instantenous power will vary, but not the average one.(1)Assuming an ideal capacitor. With a real one the losses in the capacitor are treated as the load (LED).

author: jirij
author_rank: Contributor
timestamp: September 24, 2023, 10:13:27 pm
content: Quote from: golden_labels on September 24, 2023, 06:15:53 pmYeah - I understood that you were describing a buck/boost converter and the energy being stored in the magnetic field of an actual inductor.I was just trying to figure out what I was missing.Quote from: golden_labels on September 24, 2023, 06:15:53 pmThanks for that explanation, but (in my mind) getting rid of the capacitor kind of breaks the idea, because you end up with a single loop which will always have the same current.The idea was to quickly charge up the capacitor using the high-voltage input and low current, according to its time constant, and then discharge it more slowly by the LEDs.I mean - the math is clearly right, and (as you hinted at earlier) I basically re-invented a linear voltage/current regulator which is something I was able to search for and learn the internals of, but I couldn't figure out an intuitive way to understand it for a while. But I think I have it now.The problem was I didn't fully grasp how capacitors behave in a circuit - one can't just charge a capacitor at 230V - it starts at ie. 8V and will behave as if it had a 8V drop across it, so, indeed, the wires and the switch have to take on the remaining 222V.Even if for just a microsecond, even if at a very low current.The higher input voltage and durable enough components will eventually achieve the desired result - providing short bursts of current to charge the capacitor to ~9V.But here is the kicker that solved it for me -- no matter how short the bursts are, we are still spending most of the power on the switch and the wires. Many times more than what the LEDs consume. So that's why the circuit is so inefficient.It took me a while, but thanks for the help!

author: soldar
author_rank: Super Contributor
timestamp: September 24, 2023, 10:47:25 pm
content: Quote from: jirij on September 24, 2023, 12:51:44 amI think you are either confused or mistaken or a bit of both.There are plenty of LED PSUs that are constant current. I have a thread discussing the repair of an LED fixture which is powered with a constant current PSU. The same PSU can power more or less LEDs connected in series as long as the resultant voltage is between 40 and 80 volts.A buck (step-down) PSU can be voltage or current regulated . I think you are mixing things.

author: Zero999
author_rank: Super Contributor
timestamp: September 25, 2023, 09:00:29 pm
content: Quote from: jirij on September 24, 2023, 03:14:31 pmIt's because it's a linear regulator. If you want a wide voltage difference, then use a switched mode.LEDs used to use capacitive dropper and often switched mode, but in the last five or so years, most have settled on a linear regulator. Probably because it's simple, cheap and doesn't need as much attention to be paid to noise. I prefer switched mode myself because it's more efficient and works over a wide supply voltage and isn't affected as much by high temperatures.

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 25, 2023, 09:13:52 pm
content: The only efficient way of driving a led from a considerable higher voltage is using a switching regulator.Power LEDs should work in constant current mode, as they handle high currents and their VF vary with temperature.There're non-isolated buck regulators that keep circuit simple/costs down and provide good efficiency, like the BP2306, BP2336, BP2338...Aliexpress have plenty of them:https://www.aliexpress.com/item/32795185266.htmlThese are limited to 90V by default, but they can handle >300V with few modifications, replacing the output cap and the OVP resistor divider. Current can also be adjusted, max is about 200mA.This is meant for 10-60 LEDs in series, not for a single led!

author: jirij
author_rank: Contributor
timestamp: September 25, 2023, 10:46:10 pm
content: Folks, I don't want to sound rude, but this thread is basically resolved, as I indicated in my last comment. I wanted to avoid switching regulators (as my real use case is a series of 4 LEDs for a total of 12V 6A and 10-20% switching losses would mean a lot of heat) and got an idea that I couldn't disprove as valid, so I asked here.Thanks mainly to golden_labels (who understood my position), I was eventually able to find the error in my thinking and realized why the idea could not work.Quote from: soldar on September 24, 2023, 10:47:25 pmYes, I know, I've done it before. I don't think I implied it cannot be.Quote from: DavidAlfa on September 25, 2023, 09:13:52 pmYeah, that's basically the conclusion I came to in my last comment, after realizing why the proposed linear circuit wouldn't work efficiently....So I think I will either:Modify my existing switching PSU to output the voltage I need (around Vf) and use timer-based duty cycle switching to dim the LEDs - this way, I at least don't add an extra heat-generating converter into the path. I will indeed make sure to account for temperature-related Vf changes and other factors that come up when pulsing this much current.Build something based on a synchronous buck converter, ie. SIC472ED with a big inductor, hoping for >95% efficiency, and use its built-in current-limiting functionality.I really don't want to offend anybody, but I don't think any further contributions here would add any value to anybody - I'm sure there are other topics discussing LED drivers already.This topic was really only about why the idea in the original post wouldn't work.Thanks!

author: Zero999
author_rank: Super Contributor
timestamp: September 26, 2023, 07:13:57 am
content: Quote from: jirij on September 25, 2023, 10:46:10 pmA switching regulator is far more efficient than a linear and it's easy to reduce switching losses to below 10%.If your problem is solved, then you edit the first post and add (SOLVED) to the subject title. That still might not stop you from receiving replies, both useful and completely irrelevant.