Hello,
I have a 12v 5amp power supply. I am using it to drive 12v 20w COB LED. so when I am connecting my led with power supply with out any other ckt in between (other than multi meter which is used to measure current) I notice that my power supply is sourcing only 1amps and when I am connecting 2 COB LEDS they are drawing 1.5amps of current. so my question is how can I make my led to illuminate at maximum power ie 12v * 1.7amp = 20w.

 

You need to drive LEDs with a constant current circuit. The power rating is specified for a particular current value. The actual voltage across the LED varies between samples of the LED and with temperature.

Les.

 

You need a Constant Current circuit like this simple transistor one...R3 sets the current

http://www.instructables.com/id/Power-LED-s---simplest-light-with-constant-current/

 

An LM317 is even easier way to set current but I wouldnt recomend you use any form of linear regulator with a 20W load.

The simplest option at those voltages and currents would be a buck board with a current limit.

It sounds like the PSU isnt up to the job either. If it is genuinly capable of 5A then it isnt capable of it at a voltage high enough to drive the COB, which is probably just as well because you would have burned them out if it were.

You either need to find or measure the forward voltage of the cob.
You cant drive two LED's in paralell unless something in the circuit is having more effect on the individual LED current than the LED its self.

You can use a linear device or just a resister but you will get big losses - quite a bit of heat.

Assume the forward voltage is circa 14v, remember this will vary between COB's and with temperature.
Almost all LED's have a negative temperature coefficiant so as they get hotter the VF falls leading to more current and thus more heat... hopfully you can see where this is going.
A 20W COB at 14V will require about 1.4A (Round figures)
If you went with an 18V supply that leaves you 4V to drop at 1.4A which requires a little less than 3 ohms, 2.75 would be 1.45A
The resistor will be disipating 5.8 W which is toasty...
Worse still assume the VF falls to 13V when it is hot now you have 18V - 13V into 2.75 ohms which is 1.8A, almost 24W at 13V

The upshot is that you NEED a constant current supply with a maximum voltage greater than VF
and you need this for each COB or series string of simmilar COB's

Sorry, it is what it is.
Cheap buck boards are the quickest win.

Al