If an LED is used in a circuit, the current for that LED is normally set with a limiting resistor. The LED can then be switched on and off by means of a transistor.
The following discussions explains a transistor based configuration which ensures a constant current drive to the conected LEDs, under all circumstance.
However, the method shown in figure l does not take into account any variations in the supply voltage. A small variation in the LED current can be very conspicuous especially when high efficiency LEDs are used. The addition of just one transistor can transform the circuit of figure 1 to a onstant current source which can be switched on an off (for instance, with TTL levels). The circuit of figure 2 shows that resistor R1 has been moved to the emitter of T1. When a drive voltage is applied to the input of T1, this transistor conducts which causes a current through R1. Transistor T2 controls the base current of T1 such that I the voltage drop across R1 remains at 0.6 V. The current, l, through the LEDs and R1 is calculated by l= 0.6/R1. lf, for instance, R1 is 12 S2, the current through the LEDs is 50 mA. Bear in mind that the dissipation of T1 is somewhat higher than in the circuit of figure 1, but against that, the dissipation in R1 is not as high.
The following discussions explains a transistor based configuration which ensures a constant current drive to the conected LEDs, under all circumstance.
However, the method shown in figure l does not take into account any variations in the supply voltage. A small variation in the LED current can be very conspicuous especially when high efficiency LEDs are used. The addition of just one transistor can transform the circuit of figure 1 to a onstant current source which can be switched on an off (for instance, with TTL levels). The circuit of figure 2 shows that resistor R1 has been moved to the emitter of T1. When a drive voltage is applied to the input of T1, this transistor conducts which causes a current through R1. Transistor T2 controls the base current of T1 such that I the voltage drop across R1 remains at 0.6 V. The current, l, through the LEDs and R1 is calculated by l= 0.6/R1. lf, for instance, R1 is 12 S2, the current through the LEDs is 50 mA. Bear in mind that the dissipation of T1 is somewhat higher than in the circuit of figure 1, but against that, the dissipation in R1 is not as high.
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