How it works
- On power-up Q1 and Q2 are off. There is no collector current so L1 is off.
- If the digital control input on the left is brought high (5 V) Q1 will turn on. Current will flow through L1, Q1 and R2.
- As the voltage drop across R2 increases to about 0.6 V Q2 will start to turn on and shunt some of the base current away from Q1.
- The result is that the circuit will settle at whatever Q1 emitter current will drop 0.6 V across R2.
- When the input signal drops to zero Q1 and L1 turn off.
- If an unswitched version is required then just connect R1 to Vbb.
Note that the usual series resistor is not present with the LED. Q1 acts as a variable resistor in this case adjusting to maintain the required current.
Example
Calculate R2 value to set L1 current at 30 mA.
\( R = \frac [V][I] = \frac [600m][30m] = 20\ \Omega\).
(We’re using ‘m’ as shorthand for ‘milli’.)
