I have a bright white LED from a flashlight. Aproximatley How long will it light up with a 150 farad 2.5 volt capacitor. Do I need a resistor? And if so how many Ω? The capacitor is a maxwell 150 farad 2.7 volt boostcap here.
Electronic – How long would a 150 Farad Capacitor light up an LED
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Related Solutions
There are several circuits on the internet from noatable sources that show how LEDs can be powered directly from an AC supply without a transformer. Here's one from ON semi: -
And here is the document it came from. The document is entitled "High Current LED - capacitive drop drive application note".
Here is another interesting article, this time from EDN. It shows the following diagram, again transformerless: -
And another article and circuit here: -
Background:
I have designed a number of LED lighting products which are manufactured in China.
I have several cylindrical LED flashlights that have a large number of LEDs in them
... Are there ultra-bright LEDs that you can drive directly off of 4.5 volts without a current limiting resistor? Or are there special purpose ultra bright white LEDs made for 4.5 volt supply that have internal current limiting resistors?
No and no, unfortunately.
Many LED lights are constructed as you describe, with multiple white LEDs wired in parallel and connected essentially directly across the battery.
They are junk.
They are not "designed".
They build them this way "because they can" and they work well enough to be able to sell them.
When supplied with 4.5V + the LEDs are driven well above their maximum design rating and their lifetimes are greatly shortened. The LEDs used are typically low lifetime low cost devices.
Follow-up question: Does anybody know if the 12 volt LED bulbs that are in landscape lights have a voltage regulator in them?
The 12 volt LED strips usually use 3 LED die in series plus a series resistor.
Turn on / turn off time is liable to be sub `1 microsecond if capacitors are not used downstream of the switch.
Current is set to be "about right" at 12 Volts so will vary substantially if used in an automotive context where several volts of variation occurs. Many strips use individual LEDs but some use 3 die per package LEDs with all 3 independent die wired in series. It is possible but not certain that strips with individual LEDs will run somewhat cooler due to a lower concentration of Energy per package.
Lifetime of these LEDs may be better as the series resistor means that they are somewhat more properly driven. I have seen very substantial variations in output of similarly appearing strips. The brightness bears no obvious relationship to LED specifications and a brighter strip may simply reflect a manufacturers 'marketing decision'. You can get a range of LEDs per metre but current drain and number of LEDs are not directly related.
White LEDs are typically have a voltage drop in the 3.0 - 3.5V range at rated current.
Current increase tends to be exponential with voltage and at 4.5V almost any LED would self destruct almost instantly. The "saving grace" (if it can be called that) is that the combination of small batteries and many LEDs means that the batteries are unable to produce more than 'vastly too much' current when the batteries are new. Any light constructed in this manner demonstrates a total lack of concern and/or knowledge by the manufacturer.
Adding even a single common series resistor makes a substantial improvement in voltage/current profile and a resistor per LED would greatly assist current balancing between LEDs.
Added May 2016
Harper commented:
OP is asking about LED bulbs, not strips. Those are commonly made as screw-in replacements for incandescents. Some have a resistor, but many have a switching buck converter which will accept a range of voltages from 12-30V or higher. The LED series voltage is quite close to 12V actual, so if voltage drops much below 12V the buck converter will go to 100% duty cycle and simply pass the voltage through, causing the LEDs to dim rapidly.
My answer addressed LED strips as I noted, which the OP did not ask about, as Harper noted :-).
Harper's comments above are correct where applicable. I have not seen a bulb with a buck converter internally, but no doubt they exist. White LEDs have Vf typically in the range 2.8V - 3.5V. 2.8V is unusual and usually only seen in reasonably modern LEDs or ones operated well under full power. At 12V nominal, 4 LEDs have 12/4 = 3V each available. Allowing a small voltage drop in connectors and wiring 4 LEDs with Vf of 2.8V to 2.9V would be able to be operated at full power. In real world situations with Vin able to be somewhat below to substantially above 12V, 4 LEDs in series will often work but 3 x LEDs in series plus a series resistor is 'safer'. Bulbs may not match strips in configuration, but all 12V LED strips that I have seen use 3 LEDs in series plus a resistor.
Best Answer
Direct answer to the question
The direct answer to your question, assuming you intend to just connect the capacitor to the LED with a series resistor is no time at all. That is because a white LED takes more than 2.7 V to light. Check its datasheet. These things usually need a bit over 3 V.
There are two options. The simplest is to use a LED with a lower forward drop. Let's say you try this with a red LED that has a 1.8 V drop at 20 mA. That means at full charge, there will be 2.7V - 1.8V = 900 mV accross the resistor. If you want the maximum brightness at full charge, which we are saying is 20 mA, then you need a 900mV / 20mA = 45 Ω resistor. Let's pick the common nominal value of 47 Ω.
Now that we have a capacitance and resistance we can compute the time constant, which is 150F x 47Ω = 7050 s = 118 minutes = 2 hours. At full charge, the LED will be nearly at full brightness, which will then decay slowly. There is no fixed limit at which it will suddenly go out, so we have to pick something. Let's say 5 mA is dim enough to be considered not usefully lit anymore in your application. The voltage accross the resistor will be 47Ω x 5mA = 240mV. Using the first approximation of the LED having constant voltage accross it, that means the capacitor voltage is 2 V.
The question is now how long does it take to decay from 2.7 V to 2.0 V at a 2 hour time constant. That is .3 time constants, or 2100 seconds, or 35 minutes. The actual value will be a bit longer due to the LED having some effective series resistance too and therefore increasing the time constant.
A better way
The above tries to answer your question, but is not useful for a flashlight. For a flashlight you want to keep the light at close to the full brightness for as long as possible. That can be done with a switching power supply, which transfer Watts in to Watts out plus some loss but at different combinations of voltage and current. We therefore look at the total energy available and required and not worry about specific volts and amps too much.
The energy in a capacitor is:
$$E = \frac{C \times V^2}{2}$$
When C is in Farads, V in Volts, then E is in Joules.
$$\frac{150F * (2.7V)^2}{2} = 547 J$$
The switching power supply will need some minimum voltage to work with. Let's say it can operate down to 1 V. That represents some energy left in the cap the circuit can't extract:
$$\frac{150F * (1.0V)^2}{2} = 75 J$$
The total available to the switching power suppy is therefore 547 J - 75 J = 470 J. Due to the low voltages, the losses in the switching power supply will be quite high. Let's say that in the end only 1/2 the available energy gets delivered to the LED. That leaves us with 236 J to light the LED.
Now we need to see how much power the LED needs. Let's go back to your original white LED and pick some numbers. Let's say it needs 3.5 V at 20 mA to shine nicely. That's 3.5V * 20 mA = 70 mW. (236 J)/(70 mW) = 3370 seconds, or 56 minutes. At the end of that, the light would go dead rather quickly, but you will have fairly steady brightness up until then.