Protection of high-power LEDs from excessive current. Encyclopedia of radio electronics and electrical engineering
Encyclopedia of radio electronics and electrical engineering / Radio amateur designer
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A common way to power multiple LEDs is to connect them into two independent strings that are connected in parallel. In this case, the control circuit turns out to be inexpensive and simple, and besides, the supply voltage of both garlands connected in parallel is required two times less than when supplying one large garland. However, the control circuit in this case must provide twice the current, and in addition, we also need a circuit that branches the current to power both garlands, and is not dependent on the direct voltage drop across the LEDs. The forward voltage drop tolerance of an LED is as high as 20%, and this voltage changes with temperature and LED aging.
The current mirror does a good job of splitting the current. But the failure of any LED leads to an increase in current by the current mirror, which leads to devastating consequences. A current mirror, however, can partially protect two parallel LED strings of any number of 350mA LEDs from incurring unacceptably high currents (Figure 1).
Rice. 1. The bodies of transistors Q1 and Q2 must be connected together to ensure thermal coupling
This circuit can equalize currents between strings with an accuracy of about 2%, this accuracy is due to voltage drops of 0,5 volts on resistors in the emitter circuits, R1 and R2 with a nominal value of 1,5 ohms, having an accuracy of 1%. The voltage drop across resistor R3 compensates for the mismatched voltage drops across the LED strings and keeps Q1 and Q2 operating points in the linear region. This voltage drop depends on how many LEDs both strings consist of.
If any of the LEDs of the second garland fails (break), then the base current of transistors Q1 and Q2 stops, and the transistors turn off, the current flow through the first garland stops. If the LED in the first garland fails (also a break), then now twice the current (700 mA) will flow through the second garland, which is guaranteed to disable it. Therefore, protection of the second garland is necessary. This problem is solved by adding only three components (Fig. 2).
Rice. 2. By adding a couple of diodes and a transistor to the current mirror, you can prevent the failure of the LEDs
During normal operation, transistor Q3 is in linear mode at an emitter-collector voltage of 0,7 volts because both diodes D1 and D2 are carrying current in the forward direction. The power dissipated by Q3 is only about 0,5W, so there is no need for a heatsink. The 700 mA power supply current from the Q3 collector is halved between the LED strings by the guide diode D2 and the current mirror. If any of the LEDs in string 1 fails, then diode D2 blocks the base current of transistor Q3, turning it off. The power supply current stops flowing to garland 2, which saves it.
It is necessary to compensate for the 2 volt drop across diode D0,7 by slightly increasing the resistance value of resistor R3. The current mirror can be adapted to any type of LED, the main thing is that the maximum current of the transistors, which is 1,5 A, is not exceeded. for example LM317, manufactured by National Semiconductor.
The circuit was tested using an integrated regulator operating as a 700 mA current source with five LEDs in each string.
Author: Luca Bruno, Italy
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