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Subject: Re: Constant current sources to drive LEDs on long lines

Date: 02/26/03 at 1:36 PM
Posted by: Ron Harrison
E-mail: rmaxh@yahoo.com
Message Posted:

In Reply to: Re: Constant current sources to drive LEDs on long lines posted by David Ambry on 02/25/03 at 5:52 PM:

David,

If you are driving only one LED (i.e., not 2 or more in series) per current source, You can use this circuit. This circuit is efficient in the sense that all the current goes to the LED. You will have to adjust a resistor (I suggest R3) to get the current you want, which also depends on the LED voltage. The equation for the output current is (approximately):

Iled=(Vin-Vled-0.7)/R3+(((Vin-Vled-0.7)*R2)-0.7)/R1 (whew!)

The values shown will yield about 20ma, assuming vin=5v and Vled=1.9v.
Here's the circuit. I hope you can read a Spice netlist. If not, I can explain it.
********************
R1 vcc e2 200
V1 vcc 0 10
Q2 out c1 e2 0 NTE129P
Q1 c1 in e1 0 2N3904
R2 vcc c1 2k
R3 out e1 1k
********************
If you don't want to use the LED voltage as part of the reference current, you will waste a milliamp in each current source. Here's the netlist for that circuit:
********************
R1 vcc e2 200
V1 vcc 0 10
Q2 out c1 e2 0 NTE129P
Q1 c1 in e1 0 2N3904
R2 vcc c1 4.7k
R3 0 e1 3.9k
********************
This will give you about 20ma, assuming Vin=5v.
Change R3 to change the current.

I have shown NTE129P as the PNP. Any PNP that can handle the power dissipation and the voltage should work. If the current is 10ma, then a 2N3906 will work. If it's 20ma, then you will need something like the NTE129P (I don't know if these are available). You might get away with a heat sink on a 2N3906.

Ron

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