I posted this as a response in another thread but I think the issue merits a stand alone thread.
The discussion of driving a LED or LD with a voltage source regularly arises. Due to the electrical nature of LEDs and LDs the preferred method of driving them is with a constant current source. There are two reasons for this:
1. The Voltage-Current (VI) Curve is very steep (i.e. a very small change in voltage results in a very large change in current)
2. The VI curve shifts with the temperature of the device.
I will use an image I snagged off the net and marked up. It cannot be applied to specific situations but it is representative of the concept.
1. Check out the Green markings on the drawing. Using the 25C curve and tracing the corresponding current for voltages V1 and V2 you get:
V1= 1.17V, I1 = 4mA
V2= 1.34V, I2 = 20mA
So for a 0.17V change in drive voltage the current went from 4 to 20mA.
Expressed in percentages that's a 500% increase in current for a 15% increase in voltage!.
2. There are three Voltage-Current (VI) curves for the same LED for three different temperatures (-40C, 25C, 100C).
Let's pretend that the LED is driven by a regulated voltage (indicated by the red arrow).
At -40C the current through the LED would be about 1.5mA.
At 25C the same voltage would result in a LED current of about 4mA.
At 100C the same voltage would result in a LED current of 15mA.
Note that the 10x difference in current between a cold (really cold) LED and a really warm LED.
In practice, the temperature will not swing that much but it should be extra warning for people using TECs. A room temperature LED will heat up in proportion to the power pumped into it and the amount of cooling it has.
Also recall that the LED/LD die will be much hotter than the module it is housed in.
You can extrapolate from this graph that the current for a LED/LD supplied by a fixed voltage could easily double as it heats up.
To check the change in resistance vs temperature use R=V/I
For -40C curve, V=1.17V, I=1.5mA, therefore R=780 ohms
For 25C curve, V=1.17V, I=4mA, therefore R=292.5 ohms
For 100C curve, V=1.17V, I=15mA, therefore R=78 ohms
I hope this example clarifies things for people with questions. Good luck and happy modding.
The discussion of driving a LED or LD with a voltage source regularly arises. Due to the electrical nature of LEDs and LDs the preferred method of driving them is with a constant current source. There are two reasons for this:
1. The Voltage-Current (VI) Curve is very steep (i.e. a very small change in voltage results in a very large change in current)
2. The VI curve shifts with the temperature of the device.
I will use an image I snagged off the net and marked up. It cannot be applied to specific situations but it is representative of the concept.
1. Check out the Green markings on the drawing. Using the 25C curve and tracing the corresponding current for voltages V1 and V2 you get:
V1= 1.17V, I1 = 4mA
V2= 1.34V, I2 = 20mA
So for a 0.17V change in drive voltage the current went from 4 to 20mA.
Expressed in percentages that's a 500% increase in current for a 15% increase in voltage!.
2. There are three Voltage-Current (VI) curves for the same LED for three different temperatures (-40C, 25C, 100C).
Let's pretend that the LED is driven by a regulated voltage (indicated by the red arrow).
At -40C the current through the LED would be about 1.5mA.
At 25C the same voltage would result in a LED current of about 4mA.
At 100C the same voltage would result in a LED current of 15mA.
Note that the 10x difference in current between a cold (really cold) LED and a really warm LED.
In practice, the temperature will not swing that much but it should be extra warning for people using TECs. A room temperature LED will heat up in proportion to the power pumped into it and the amount of cooling it has.
Also recall that the LED/LD die will be much hotter than the module it is housed in.
You can extrapolate from this graph that the current for a LED/LD supplied by a fixed voltage could easily double as it heats up.
To check the change in resistance vs temperature use R=V/I
For -40C curve, V=1.17V, I=1.5mA, therefore R=780 ohms
For 25C curve, V=1.17V, I=4mA, therefore R=292.5 ohms
For 100C curve, V=1.17V, I=15mA, therefore R=78 ohms
I hope this example clarifies things for people with questions. Good luck and happy modding.
