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### Topic: 120 VAC LED string to 12V DC (Read 1 time)previous topic - next topic

#### ryry

##### May 22, 2011, 10:16 pm
I have a string of 70 LEDs that run off 120 VAC. I need to run them off a 12V battery. Originally, they were wired in two groups of 35 in series. What I planned on doing was wire groups of 3 in parallel since 120/35 ~3.5 V/LED. That way each group of three would have 12 V across them. Is my logic correct here? The reason I ask is because one group of three burned out after being lit for a few minutes. Some of the lights in the string were bad, so I don't know if that was just a coincidence or if I need to add a current limiting resistor in there.

Any thoughts?

#### johnwasser

#1
##### May 23, 2011, 12:19 am
I don't know about your christmas lights but in mine, some of the LED's had resistors.  These resistors were included in the 120V circuit, meaning that the voltage drop across the LEDs was less than 120/35.

Also, 120V AC is +60V to -60V.  Since the LED's don't conduct backward the peak voltage across each chains of LEDs is 60, not 120.
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#### ryry

#2
##### May 23, 2011, 12:31 am
Right. Didn't think of that. I had the feeling that there were resistors in there somewhere. I tested a few groups of three and the current running through them ranged from 10mA to 30mA. I'll be using PWM to control the brightness, so I suppose I'll reduce the pulse widths until everything appears stable.

#### Magician

#3
##### May 23, 2011, 03:03 pm
Sorry for interrupting, but
Quote
Also, 120V AC is +60V to -60V.

isn't correct,
as 120V AC is 120 * sqrt(2) = +169.7V to -169.7V.

#### johnwasser

#4
##### May 23, 2011, 04:37 pm

Sorry for interrupting, but
Quote
Also, 120V AC is +60V to -60V.

isn't correct,
as 120V AC is 120 * sqrt(2) = +169.7V to -169.7V.

Oops!  My mistake.  I did not know that AC power was named after the DC voltage with equivalent Root-Mean-Square power.

Good thing I never interface stuff to the mains!
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