Hi,

I am having problem with adjusting the current in this basic china "1500w" boost converter.

I want to power cob led with it and it needs 60-72 volts at 3.420 amps MAX.

The test setup contain 300w dummy load 8 ohms (originally for amplifier testing)

So my problem is that i cannot adjust the current low enough before the voltage starts dropping with the dummy load.

For example if i adjust
Voltage max but current low i get 2.1amps at 13,5 volts

When turning amps up, amps starts rising BUT so does voltage (for example 3.4a@33,5v) when reaching 60 volts, amps are too high.

Theres no way to do 0.5a and 60v because voltage drops immediately.

I see many people on youtube powering these cob leds with these boost converters.

Will the actual cob led behave differently than my 8 ohm test dummy load?

Dont want burn the cob

 

If you set the output current to be 3.42 amps then the voltage across an 8 ohm load will be 27.36 volts (3.42 * 8) Read your old school book from when you were taught ohms law at school.
LEDs do not obey ohms ohms law.

Les.

 

Hi,

I am having problem with adjusting the current in this basic china "1500w" boost converter.

I want to power cob led with it and it needs 60-72 volts at 3.420 amps MAX.

The test setup contain 300w dummy load 8 ohms (originally for amplifier testing)

So my problem is that i cannot adjust the current low enough before the voltage starts dropping with the dummy load.

For example if i adjust
Voltage max but current low i get 2.1amps at 13,5 volts

When turning amps up, amps starts rising BUT so does voltage (for example 3.4a@33,5v) when reaching 60 volts, amps are too high.

Theres no way to do 0.5a and 60v because voltage drops immediately.

I see many people on youtube powering these cob leds with these boost converters.

Will the actual cob led behave differently than my 8 ohm test dummy load?

Dont want burn the cob

If you want to test for 60v at 0.5 amps with a resistive load then you need a resistor that is 120 Ohms.
Also, because 60 volts times 0.5 amps is 30 watts, you should get a resistor that is rated around 60 watts. It will get hot if you leave it running so be careful.
If you cant find a 120 Ohm resistor rated around 50 or 60 watts, then you can use two or more in parallel to get the 120 Ohms and the proper wattage.

Some simple series and parallel resistor calculations...
For two resistors in parallel the wattage is two times one resistors power rating and the resistance is one half of one resistors resistance.
For two resistors in series the wattage is two times one resistors power rating and the resistance is two times one resistors resistance.
So two 240 Ohm 30 watt resistors in parallel result in one 120 Ohm 60 watt resistor, and
two 60 Ohm 30 watt resistors in series result in one 120 Ohm 60 watt resistor.

For N equal resistance resistors in parallel:
Rtotal=R/N
Ptotal=P*N

and for N equal resistance resistors in series:
Rtotal=R*N
PTotal-P*N

where R is the resistance of just one resistor and N is the number of resistors.

For example, ten 1200 Ohm 10 watt resistors result in a resistor of 120 Ohms and 100 watts.
Ten 12 Ohm 5 watt resistors in series results in a resistor that is 120 Ohms and rated for 50 watts.
Ten 12 Ohm 10 watt resistors in series results in a resistor that is 120 Ohms and rated fro 100 watts.

In any case of parallel or series power resistors there must be adequate space between resistors so they can get a reasonable air flow, and they must be mounted away from other components so the heat does not melt or damage anything else.

A fan can be used to cool them. With the lower voltages they can also be immersed in a non conductive container of water which will keep them cool for quite a while (although probably not indefinitely without a radiator and water pump to circulate the water).

 

V = I R

Learn what this means if you want to do anything with electronics.

Bob

 

Here's an example (the scheme is not optimized and it's an idea):