My knowledge on this is a little rusty. What kind of resistor do I need to put in there? There are 10 LEDs. I'm pretty sure they are in parallel.
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\$\begingroup\$ That depends on the current drawn by those LEDs. \$\endgroup\$– Jonathan S.Commented Mar 2, 2023 at 20:35
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\$\begingroup\$ If you are only ever driving a fixed load like 10 LEDs, then a resistor is ideal. Compute approximate value as 2 V / load_current. If you want to drive a load that varies from use to use, then a resistor is totally the wrong thing, use a regulator. \$\endgroup\$– Neil_UKCommented Mar 2, 2023 at 20:45
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\$\begingroup\$ I do not know what the current is, can you get away with an educated guess? If yes, do you have an educated guess? Thanks for your replies!! \$\endgroup\$– BiomedicalEngineeringStudentCommented Mar 2, 2023 at 20:47
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\$\begingroup\$ Need a lot more information to answer this quesition \$\endgroup\$– Voltage Spike ♦Commented Mar 2, 2023 at 20:48
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\$\begingroup\$ An educated guess for LEDs is that 20mA is pretty darn bright (like those way too bright blue LEDs that light up the ceiling and keep you awake at night) with modern ones. They used to design on/off LEDs for 20mA, then LEDs got brighter and people still kept using the 20mA rule of thumb. You want it not so bright, decrease it down to more like 2mA. That's per LED. \$\endgroup\$– Criticizing Israel not allowedCommented Mar 2, 2023 at 20:52
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3 Answers
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Avoid connecting LEDs directly in parallel!!!
The issue arises because LEDs, and diodes in general, exhibit slight variations in their properties. Each LED has a different forward voltage curve, leading to some LEDs receiving excessive current. This can result in damage to those LEDs. If one LED fails, the current will then pass through the remaining LEDs, causing a chain reaction of failures until all LEDs are non-functional.
How to mitigate this issue?
The best solution is to connect all LEDs in series using a single current-limiting resistor (or a current-limiting circuit). Aligning LEDs in series reduces the voltage across the resistor, minimizing energy losses. Your power supply voltage must be sufficiently high for this arrangement. If it falls short, you can create multiple series of LEDs, each with its own resistor.
Ensure that the voltage across the resistor is sufficient to prevent significant variations in current. This can occur because variations in supply voltage or the variations in LED properties.
The best thing to do is to keep a safety margin in mind. Let's say the max current of the LED's is 20mA, keep the current at 15mA or so. This is also better for the lifespan of the LED's
A circuit
To complete my answer, a little schematic. I am assuming your LEDs are max 20mA and have a forward voltage of about 1.8V. Using this schematic, the current through the LEDs is a little under 14mA.
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The LEDs seem to need ~16.33 mA. Along with @Neil_UK's answer, this gives that this situation needs a 2 V / 0.01633 A =
~122.45 Ohm resistor
answered Mar 2, 2023 at 21:18
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Does your battery box include some circuit? I bet it has a resistor to set the overall current for the string, probably at about 20mA (2mA per LED). Or it may have an actual current limiter IC to do this.
Either way, without knowing what's in the box, you could just limit the gross current delivered to the whole string to the same as what a fresh set of batteries delivers.
How to make that? Well, here's an answer (simulate it here):
Design procedure: Measure the current off the battery, adjust the limiter current sense to achieve that target value. The design shown delivers about 18mA at 5V, you could use it just like that.
Or play around with a series resistor, like they did here: Calculating forward voltage for a string of fairy lights: multimeter shows "1" Back-of-the-book answer: they chose 85 ohms.
answered Mar 2, 2023 at 23:38