Not clear on LED Boost Converter:    Is an output resistor irrelevant ? - Page 1
Forum section: electronics / beginners

author: RJSV
author_rank: Super Contributor
timestamp: May 31, 2023, 05:17:22 am
content: Hi, I've been experimenting with concepts of buck and boost type circuits and not clear about LED brightness adjusting. Also, adding to the intrique, have been viewing the ELECTROBOOM yt videos, concerning voltage multiplying, using the MARX GENERATOR designs.Intuition, it seems, to indicate that putting in a series resistor should modify an LED brightness, but not so sure of that, in a voltage boost arrangement. See diagram where your controller will interrupt the LED briefly, by connecting to ground, but where that action makes for a good pulse of current, through a circuit inductor component (no capacitor), and where that happens at about 230 khz.When that inductor path is then open to flow through LED, the field collapse forces a higher voltage and current through the LED path (to ground).My problem is this: It looks like that would be acting more like a current source (unconditional) vs a supply voltage, and thus putting a series resistor doesn't limit the LED like it would on a voltage ⚡ source, that is, whether 1 kohms or 2 kohms doesn't affect a current source.Most literature out there is also discussing putting an explicit capacitor component in the circuit, getting into resonance issues, (where I'm not so strong on theory there).They mention controlling the LED current by way of which inductor value is incorporated.Haven't tried things yet, on my solar garden light workbench, but ...?It just seems counter-intuitive that an included LED series resistor won't affect brightness in the voltage boost circuit...

author: Manul
author_rank: Super Contributor
timestamp: May 31, 2023, 01:36:26 pm
content: Quote from: RJHayward on May 31, 2023, 05:17:22 amI pressume this circuit to not be actively controlled and working in discontinuous conduction mode. You can view it as decaying current source. It starts with whatever current was flowing through inductor and decays to zero. And it's not only decaying, but it also has finite energy, which means it's decay rate is proportional to voltage and it's not a fixed time constant.With higher value resistor, peak current will stay more or less same (in an ideal world), but inductor voltage will be higher so current will decay faster. Averaged in time, LED power will be less. Same current, but the decay curve is more compressed in time.It easy to see if you consider just one cycle and think about energy. At the begining, some finite energy was stored in inductor. As time goes, every resistive component like a resistor or LED will dissipate energy proportional to their voltage drop times current until all energy is wasted. So every additional dissipative component will steal energy and there will be less left for others.The picture changes if a driver can actively regulate inductor current by changing duty cycle. In that case added resistor will not change LED brightness, because resistor wasted power can be compensated by driver (within it's control limits).

author: jwet
author_rank: Frequent Contributor
timestamp: May 31, 2023, 03:19:18 pm
content: A charged inductor is a current source and LED's are current operated devices. I would include a resistor in that circuit though it will decrease efficiency some- any delay in switching in some fault or startup, will push Vcc/R(inductor) through the LED. If the supply voltage is lower than the LED string voltage, this isn't an issue. Often a resistor below the LED is used as feedback with an op-amp stage if required to explicitly make this a current loop. Maxim makes little boost converter LED string drivers purpose built for this- might be worth looking at those circuits. Creating real DC with a diode and a cap and then voltage biasing with a resistor is less efficient and more expensive. LED's are fine with unipolar AC drive like this. LED's like all diodes have a very non linear IV curve- a constant voltage per decade of current (60 mV/decade of I for silicon sticks in my head). Have fun.

author: RJSV
author_rank: Super Contributor
timestamp: May 31, 2023, 08:33:48 pm
content: Thanks folks, Manual and jwet, for response. Wow, more complexity, but that's why I'm here!My thoughts are along the lines of having that stored energy in the inductor field, that's soon to collapse, but it gives a sense of finite energy, for each cycle. I think the waveform looked to be inductor shorted to ground at about 1/5 or 20 % of each cycle, 230 khz.Someplace I've read had also involved capacitance although that would be parasitic in the simple solar garden light.To be clear, my real question is involving battery charge life, (from one day of sunlight), and whether a series LED resistor could extend that, as the original lamps only last, brightly, until 9 pm or so.It would seem that battery drain is only to create the inductor magnetic field and that (any) LED resistor is significant only during the 'on' time with magnetic field collapsing. However, now that I'm thinking about that time period, the battery voltage IS in series as a voltage source and so would be affected by any (current reducing) resistor...partially.Confusing, really, but I might need to do some actual tests.Purpose of LED resistor being to extend the little light usefulness until later, like midnight.Thanks.

author: Manul
author_rank: Super Contributor
timestamp: May 31, 2023, 08:56:32 pm
content: So why not look for a way to reduce the duty cycle? That's the primary thing which controls output current. Let's say to 10 or 15%. Or if that boost converter is more advanced and has regulation, adjust the feedback. Adding series resistor will not extend battery life.

author: jwet
author_rank: Frequent Contributor
timestamp: May 31, 2023, 10:18:44 pm
content: RJH- that boost converter is not super efficient. The efficiency could be improved by using a boost converter with a FET switch with a low Ron (bipolar switches downs to Vce sat abut 100+ mv). You could also use a lower resistance inductor and bypass the top of the inductor with a low esr cap to keep AC out out of the battery. There is not be gained here- maybe 15% max fpr both. The bigger gain in the system is on the charger side, solar cells are funny non linear sources, you don't need MPPT but its worth looking at the charger. The big hammer might be to replace the Nicad with something decent- I'm sure what's there is bargain basement. Have fun.

author: RJSV
author_rank: Super Contributor
timestamp: June 01, 2023, 01:30:09 am
content: Yes it is bargain basement science, for sure!Was purchasing a few at a time, at the local Dollar Tree Store, buck twenty-five each. Eventually, I bought two flats, each has 64 little garden lights.When the LED is clipped off then you can simply plug the two output leads directly into a small solderless proto board, and do whatever you want with the signal, like try a short little string of decorative lights, as sold in many toy stores or variety stores. Lately I've favored covering the white-blue LED with half of a colored toy egg...scotch taped. These things aren't made for rainy outside conditions.As a circuit analogy, consider a water bucket, filled up and dumped each cycle. So that way I've figured that the 'dump' phase isn't directly related to the 'fill' action, so (by analogy) aside from the fact that the battery is in series, with the coil and collapsing field, the 'dumping' action is separate, so a series resistor shouldn't affect battery drain, if you ignore the battery series contribution (just for a minute here).It might be, maybe 50/50 so that a series LED resistor would affect battery drain, in proportion.Yeah, and reading some other materials that have LC resonant action was difficult, although I grasp the basics of LC resonance.I saw what you wrote, about possibly including a limiting resistor, like maybe 100 ohms, just for the inrush case. Also have considered using a slightly larger solar panel, larger than the 3 cm square. Putting another solar light panel in parallel gives more charging current, in daytime...(I think, lol)...I'm doing what I term to be 'backyard science', using primitive tools, a multimeter etc.But I'm lucky in that a roommate has all the Oscilloscope and other regular bench stuff, when needed.(I do the lawn maintainence, also....as per the movie; Caddyshack)Yes, I'm having a good time, out back, except this year crazy rainy.

author: RJSV
author_rank: Super Contributor
timestamp: June 01, 2023, 01:37:45 am
content: I've done a bunch of other related posts over the year.

author: RJSV
author_rank: Super Contributor
timestamp: June 01, 2023, 01:41:23 am
content: Camera shot while sensor is covered up (simulating night), uses the solar cell for sensing when dark outside.

author: RJSV
author_rank: Super Contributor
timestamp: June 02, 2023, 12:26:06 am
content: So I'm going to try a couple things on the bench, and keep track of results, noting the time after dusk where an output LED gets 'dim', a rough test. My assumptions to test are that the buck inductor voltage and the series battery voltage are roughly equal, and supply a 'white' LED, heavy on the blue side.Might be 1.2 volts battery plus 1.2 inductor contribution (at start of each cycle at 230 khz.)After noting the single red LED time of 'mostly dim', I'll spend the next few nights placing additional LED loads, with perhaps variations in placing an inserted current limiting resistor. That could be 59 ohms, a popular value, or 100 ohms, on each LED separately. A 100 ohm resistor at 3 mA would create a drop of 300 mV, about right in the face of output voltage nominal 3 volts. I think I remember a blue LED might be at that higher characteristic running voltage.The main thing to note right now is I'm assuming that about half of this is going to resemble a simple limiting resistor dropping the supplied output, so that 'about half' the circuit is otherwise considered on the inductor. If that's good logical approach then 'about half' the thing will behave predictably. The resistor portion being the 'dissapative' load mention in a helpful reply above.Give me a week or so, for testing things.thanks