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CONSTANT CURRENT SOURCE FOR LED
constant current sink over wide voltage range
If we have a DC source that does not change the voltage, we can use a resistor to limit the current for the LED.
[ Yellow LED] has a 30V source and needs a 1k2 ohm resistor for current about 20mA.
[ Blue LED] has a DC source of 10V. Three times smaller voltage source needs three times smaller resistor for the same current. 400ohm is enough. That's Ohm's Law.
We need an adjustable resistor that adjusts according to the voltage at the input and allows only the necessary current. The word transistor means trans-resistance/trans-resistor.
[Green LED] has a voltage source that varies between 10V to 30V. The transistor ensures that only the correct current flows.
– If the transistor is open, the current from the collector to the emitter "ICE" passes through the 100 ohm resistor to ground. The current that passes through the resistor is the same as that which passes through the green LED connected on the collector side. So the current through the 100 ohm resistor between emitter and ground is transferred to the side between source and collector where the green LED is connected.
– How do we set the current size? The reference voltage VREF is connected to the base of the transistor. Zener diode maintains 4V. The base-emitter junction inside the transistor has a drop of 0.65V to open and transfer the remaining voltage of 3.35V to the 100 ohm resistor connected in the emitter.
The amount of current that passes through a 100 ohm resistor is calculated from Ohm's law I=U/R I = (4-0.65) / 100
– The 1k resistor only protects the reference voltage source VREF, here the zener diode, from excessive current.
[ Red LED ] has the same source as the green LED, 10-30V. It is the wiring used in practice. The solution is not perfect, but it is usable in practice and very cheap. In practice, the source will not fluctuate that much. The point is that I can connect the LED to a voltage of 10V or 30V and it will light up and not burn out.
– The reference voltage source consists of two ordinary rectifier diodes 1N4004. Why two diodes? The voltage drop on one diode is 0.65V. The loss on the two diodes will be 0.65+0.65=1.3V
The transition of the base-emitter transistor needs 0.65V to open, so it transfers 0.65V to the resistor in the emitter.
– What current "ICE" will flow through the collector-emitter junction in the transistor? The same as flows through the resistor connected in the emitter at 0.65V.
We require a red LED current of 20mA
R = U/I
R = 0.65V / 20mA
R = 0.65 / 0.020
R = 32.5 ohms
– The 1k resistor only protects the reference voltage source VREF, here the two diodes, from excessive current.
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