 # Does this very simple circuit look ok?

Hi,

I'm looking to build a circuit to power six white 3 watt LEDs. I have sketched up what I think will work: .

Does this look like it won't blow up/work?

For more details on the parts: http://www.ebay.co.uk/itm/321554308949 http://www.aliexpress.com/snapshot/6444469412.html

Why do you need the resistor?
You are using a constant current driver so you do not need the resistor.
Otherwise it looks OK.

Great thanks! The total of the 6 LEDs is 21v. Where will the extra 3 volts go?

There are no "extra" volts. The circuit is designed to drive 700mA of current through the load (LEDs) and will set the output voltage level, between the limits of 1.2 to 28 volts, sufficient to drive that level of current.

Fantastic, just want I wanted to hear. Thanks!

Maybe I am misunderstanding your labels but the led driver can support a total of 700 mA and you show 6 3W leds at about 3.3V each which is 3W/3.3W=0.909 A (900 mA) each x 6 = 5.45A total.

Can you please explain what the correct specs are since 5.45A >> 0.700 A. (6 * 700 mA =4200 mA)

raschemmel, when you put LEDs in series, you add voltages, not currents.

3.6V at 700 mA is 2.52W. Close enough.

You don ‘t know the half of it. Good thing I can’ t be pulled over for PUI (Posting Under the Influence)
So using in series does not change their Power dissipation. 6 * 2.52 W = 15.12 W

(6 * 3.5 V = .21 V
15.12 W/21 V = 0.720 A

V4.7 ohm R = 0.720 A * 4.7 ohms = 3.384 V

28 V - 3.384 V = 24.616 V

(6 * 3.5 V = 21V)

24.616 - 21 V = 3.616 V

6* 2.52W = 15.12 W
I = P/V = 15.12 W /21 V = 0.720 A ?
:

Raschemmel

You are missing the point - it is a constant current power supply source

Edit : added the following to clarify the point - you could connect 6 x 10w LEDs instead of 6 x 3W units but you'd still only get 700mA of drive current

ok got it.