Need Help Calculating Size of a Capacitor to Light LEDS - Page 1
Forum section: electronics / projects

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 03:56:54 am
content: If I want to find the value of a capacitor to deliver an average approx. 5-mA over a 5.85V(fully charged) and 5.6V(low limit of charge) to two in-series super-efficient blue-white LED's, over a period of .5 Sec, how big a capacitor do I need to do this?No series resistor is to be used, just no power wasted, using only the internal resistance of the LED's in series.I measured 5.72V to give the required brightness and the voltage across the LEDs was measured at 5mA current.The LEDS each glow fairly bright at even 2.5V, but current increases exponentially with brightness as voltage rises.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: January 14, 2018, 05:06:57 am
content: Quote from: SuzyC on January 14, 2018, 03:56:54 amcharge=5mA*.5 seconds=2.5e-3 coulcharge/delV=2.5e-3/(5.85-5.6)=0.01 farad=10mF=10,000 uFNot sure it's a good idea to not use a resistor but this is the capacitance to do that.Andy

author: rs20
author_rank: Super Contributor
timestamp: January 14, 2018, 05:39:12 am
content: Quote from: ahbushnell on January 14, 2018, 05:06:57 amThat calculation gives a capacitor than delivers the same amount of charge as is required to light the LED for 0.5s, but if one actually constructs the circuit without a resistor, it might instead discharge much faster, possibly destroying the LEDs with a very brief high current surge (still the same amount of charge though)Also, the capacitor ends up being huge because you're only swinging it 0.25V. That's like carrying around a huge water bottle and only drinking the first 50mL. You might be OK either this though, depends if this thing is portable.If you arrived at the 5.85V figure by observing which voltage the LEDs consumed their rated current at, please also be aware that LEDs have a forward voltage that depends on temperature somewhat -- so what seems to work now might be very bright or very dim in different temperature conditions -- this is why current control is preferred over voltage control.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 11:24:59 am
content: Thanks for the help, ahbushnell and rs20.As you can see, I am creating the most-efficient LED power usage because no series resistance is being used, so this gives me thegreatest possible battery life.My idea to save more energy is to switch off the switching power supply regulator once the upper voltage limit is reached to savepower losses due to inefficiencies and device power usage in the switcher power supply IC circuit. Once turned off, the switcher circuit draws <66uA.The capacitor would be charged quickly as possible, peak charge current .1A.I am charging the storage capacitor in this way because I can see from the switching regulator IC spec sheet that the regulator efficiency is much higher at higher currents. Efficiency is show to be only about 70% to 88% at <1mA, but >90% at 100mA with a single 1.5V battery.From your calculations, I see that:Charge = I x TCapacitance Required: Charge/(deltaV)But deltaV seems too simple a calculation, it just doesn't seem to take into account the change in current due to the top and bottom limits of voltage.Since the current drawn by the LED's varies exponentially, does this calculation really work, considering the apparent change in internal resistance of the LEDs?Note, the peak capacitor voltage applied is set precisely by a regulated switching boost converter, so no problem with any surge damaging the LED's. Once the peak voltage is applied to the storage capacitor, the charging voltage is switched off until the LEDvoltage droops to the lower voltage limit.Since the overall current is low, there shouldn't be much heating of the LED's, but the temperatures that the LED's are working with varies between 40 deg F to 28 Deg F during a 24-hr day. This means the LED's are running in a very cool environment.Is it safe to assume that the temperature coeff. of voltage across the blue-white LEDS is approx. -2.2 mv/Deg C ?In conclusion, am I making sense with this pulsating blinky capacitor storage idea?I do like the pulsating brightness effect, but is my idea fun or folly?Does it really look like I am increasing battery usage life by charging a capacitor instead of steady and boringly holding voltage steady for lighting the LED's at 5mA approx?

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 11:44:05 am
content: Thanks blueskull, but doesn't your equation assume a constant current, which it isnot, due to voltage change across the LED's?

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 12:01:48 pm
content: Exactly bluskull, but I there is no danger of the LED's being damaged by over current since the voltages are set precisely and LED's don't change their internal resistance over time unless they have been subjected to damaging excessive current.To keep efficiency highest,no constant current circuit is usedto set current to the LEDs, only Vf changes the current.The temperature range of operation is fixed between approx 28 to 40 deg F. The variation in brightness can be compensated for by usinga thermistor across the top voltage feedback setting resistor of the switching reg. IC.The Vf and my perception of balance of brightness of the LED's (quan 24 in the batch) used seems to be by my measurments seem suprisingly very closely matched, even over temperature. The initial current can be trimmed just once to 5mA, if there is any change in batches.The LED's can quickly be easily sorted into pairs/quads/groups of closely matched If at Vf, yet the batch I got is remarkably uniform.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 12:44:44 pm
content: I have taken calculus I and II at the university level and even got an "A" in calc I and and a "B" in calc II, but I just don't see yet the equations to solve. That is why I posted my questions here, so some phD's here could give me a hand to get a grasp on the math/calc detail.

author: rs20
author_rank: Super Contributor
timestamp: January 14, 2018, 08:54:48 pm
content: I wonder if the efficiency gains you're going for are cancelled out by the fact that the vast majority of the energy you are putting into the capacitor can never be used?Constant-current does not necessarily imply "inefficient", switching constant current drivers do exist.You don't necessarily need to do this by hand using calculus (although by all means, give it a go if you want.) I'd just use LTSpice, or a spreadsheet with time divided into small chunks.The reason that switching converters are "inefficient" at low output currents is that the driving circuitry has a certain fixed current requirement of its own, which appears as a largepercentageof the input current at low output currents. But that's only a percentage, and a percentage of one particular component of your system. If you're driving the enable pin of a DC-DC converter with some sort of microcontroller to try to reduce this fixed current, it would be a mistake not to take the current consumption of the micro into consideration.In short, don't focus on the percentage of that one component, you're at risk of not seeing the forest for the trees if you do. Define a typical usage pattern for your device, and calculate its overall battery life for various solutions. Only use the percentage efficiency of the converter to figure out how much current it uses, and be sure to integrate over time.With all this in mind, I wouldn't be entirely surprised if a simple resistor turned out to be most efficient (sure, you're losing some energy as heat, but you're not wasting anycurrent.) Or, if you want to have a bit of fun, and you already have a microcontroller on the board, a GPIO pin tied to an inductor + LED with the micro doing some simple switching (5mA is nothing to most micros, so most likely don't need any sort of H bridge) will make a perfectly adequate little switching converter for the LED, with practically zero extra current consumption (I mean, you already had the micro on the board, so it's "100% efficient" in some contrived sense.)

author: jmelson
author_rank: Super Contributor
timestamp: January 14, 2018, 09:01:23 pm
content: Quote from: rs20 on January 14, 2018, 08:54:48 pmYES!!QuoteYes, the Digi-Key catalog actually has over 10,000 product listings of LED regulators. (Yes, that actually has 3 items for each chip: bulk reels, cut tape and digi-reel, so it really is only 3333 actual different components.) Still, PLENTY for you to choose from.) Most of them are VERY efficient.Jon

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 14, 2018, 11:11:04 pm
content: Sure, there are many, many available LED driver chips, but I wanted to work on my own having the following characteristics maybe not offered by other chips.1)As close tp 100% complete discharge of a 1.2 V NIMH AA cell as possible.2) Maximum efficiency, no series resistor or constant current source that must waste energy.3)Ability to be give a pulsed-brightness effect to make it more interesting to the eye.4)Regulated LED current/brightness using a switchmode boost converter chip to charge storage cap to give a fixed brightness range over the full discharge of the NIMH AA battery used.No constant current device driving a LED can be ever close to be being 100% efficient, only no series LED resistor is 100% efficient.RS20, the higher the Vdd voltage driving the MCU onboard, the higher the current used by the MCU. A 1.5V (NIMH 1.2V nominal) battery cannot possibly light a blue-white hi-effic. LED because the threshold voltage for getting bright light or any light is always >2.4V.Although "5mA is nothing for most micros", but a GPIO pin requires a current limiting resistor, the MCU Vdd must be >LED Vf, and the MCU needs to use power itself.

author: rs20
author_rank: Super Contributor
timestamp: January 15, 2018, 12:25:26 am
content: Quote from: SuzyC on January 14, 2018, 11:11:04 pmI should have explained my idea more clearly -- GPIO pin + inductor + LED constitutes a little buck converter, removing the need for any limiting resistor.I didn't realise you needed a boost converter -- the same idea holds though, MCU pin + inductor + diode = a lovely little low-current boost converter. You can also play with those capacitor doubler things.Also, you're still obsessing over percentage efficiencies. Stop it! There's more to your system than one resistor or one converter. Even if this is not one of them, many many design problems are most efficiently solved using a resistor, by people not obsessing over a tiny I²R loss in the resistor. Again, even though a resistor dissipates some heat, at least it's not shunting current to ground. It's "100% efficient with respect to current". Yes, buck converters can output more current than they take as input, but in situations where that isn't true (low output current), a resistor (or LDO) would be more efficient.Your design requirement for having no series resistor is not your actual requirement -- your actual requirement is long battery life, so stick to solving for that, and considering options that achieve that goal through simplicity and low quiescent/shunt currents rather than throwing DC/DC converters all over the place. (Or, at least calculate the performance of such a solution to use as a benchmark for your other solutions.)