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# is this led array right?

+18V

+ -|>|- -|>|- -|>|- -|>|- -|>|- -/\/\/\- + R = 22 ohms

+ -|>|- -|>|- -|>|- -|>|- -|>|- -/\/\/\- + R = 22 ohms

+ -|>|- -|>|- -|>|- -|>|- -|>|- -/\/\/\- + R = 22 ohms

+ -|>|- -|>|- -|>|- -|>|- -|>|- -/\/\/\- + R = 22 ohms

+ -|>|- -|>|- -|>|- -|>|- -|>|- -/\/\/\- + R = 22 ohms

this lay out has the resistor on the negative side is this right

many thanks for helping

karl

## Comments

7 years ago

First ... each of these horizontal LED strings works the same way.

Answer 7 years ago

Now here is your LED string as calculated by bwrussel

7 years ago

Check out THIS led array wizard. Assuming white (3.3 V @ 20mA) LEDs you're resistors are a bit small. Really they're small for most any LED. Basically, sum the voltage drop across a single series of LEDs and then subtract that from the source. If it is negative you need more power. The remainder needs to be dropped by the resistor. Using V=I*R (Ohm's Law) we can calculate the resistor size as R= V(remaining) / I (current). Using my assumptions that would be R= 1.5V / .02 A = 82 Ohm (rounded UP to the nearest resistor.)

7 years ago

Doesnt matter whether the resistor is on the high (+) side or low (-) side in this circuit. Not sure what kind of led's your using (typically 1.7-2.1 voltage drop for standard red, green or yellow) but the 22 ohm resistor looks a little small. (5) 2.1 volt leds will add up to 10.5 volts of drop. The remaining voltage would be 7.5 volts. 7.5 volts divided by 22 ohms would be 340 milliamps which is very high for most leds. You may be using high current, high brightness led's that require that or have higher voltage drops. Just check your data sheet.

7 years ago

Doesn't matter what side of the LED you place the resistors. The common rule of thumb is to have the resitors on the negative side.