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jcverive
modified 8 years ago

Constant current LED driver

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04:24:51
A basic circuit for driving a light-emitting diode with constant 20mA current. A zener is usually used in place of the green LED, but this circuit shows how a LED's forward voltage is fairly constant over its current range. The circuit has the benefit of showing how although the current through the green LED changes considerably over an input voltage range of 6 to 12V, the current in the red LED is fairly constant. We can measure these values directly, but we can also tell visually that the green LED's brightness varies much more than that of the red LED.
published 8 years ago
diego92
8 years ago
Where is the base resistor? Now there are 20 mA on base of pnp ...this circuit will burn. The current must to be uA . And base resistor of the + 100Kohm.
rich11292000
8 years ago
1.82 uA @ base
jcverive
8 years ago
diego92: the 330 ohm resistor is actually the base resistor, though the 60 ohm resistor also helps limit base-emitter current. This circuit is different from basic transistor switches commonly used to drive LEDs, because here the transistor is operating in its linear region (whereas when used as switches we usually operate transistors in the saturation region), so the method of limiting base current is somewhat different. If you simulate or build the circuit, you will see that base current is only around 0.2mA.
jcverive
8 years ago
An interesting characteristic of this circuit is the negative feedback due to collector current. If we disconnect the collector from the red LED, base current is the only source of emitter current (which is approximately 19 mA). With the collector connected, the emitter current increases to around 20mA, but now most of the emitter current is supplied by the collector, and base current drops significantly.
jcverive
8 years ago
A final note: the transistor (2N3906 or equivalent) is a PNP device, and it will not burn up with 20mA of base-emitter current: forward voltage of this junction will be somewhere around 0.7V, so with 20mA current the power dissipated by the junction is only 14mW. This is hardly enough power to raise junction temperature by more than a few degrees centigrade.

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