Choosing Supply Voltage to Minimize
Power Dissipation
The MAX6954 drives a peak current of 40mA into LEDs with a 2.2V forward-voltage drop when operated from a supply voltage of at least 3.0V. The minimum voltage drop across the internal LED drivers is therefore (3.0V - 2.2V) = 0.8V. If a higher supply voltage is used, the driver absorbs a higher voltage, and the driver’s power dissipation increases accordingly. However, if the LEDs used have a higher forward voltage drop than 2.2V, thesupply voltage must be raised accordingly to ensure that the driver always has at least 0.8V of headroom. The voltage drop across the drivers with a nominal 5V supply (5.0V - 2.2V) = 2.8V is nearly 3 times the drop across the drivers with a nominal 3.3V supply (3.3V - 2.2V) = 1.1V. In most systems, consumption is an important design criterion, and the MAX6954 should be operated from the system’s 3.3V nominal supply. In other designs, the lowest supply voltage may be 5V. The issue now is to ensure the dissipation limit for the MAX6954 is not exceeded. This can be achieved by insert ing a series resistor in the supply to the MAX6954,ensuring that the supply decoupling capacitors are still on the MAX6954 side of the resistor. For example, consider the requirement that the minimum supply voltage to a MAX6954 must be 3.0V, and the input supply range is 5V ±5%. Maximum supply current is 35mA + (40mA x 17) = 715mA. Minimum input supply voltage is 4.75V. Maximum series resistor value is (4.75V - 3.0V)/0.715A = 2.44Ω. We choose 2.2Ω ±5%. Worstcase resistor dissipation is at maximum toleranced resistance, i.e., (0.715A) 2 x (2.2Ω x 1.05) = 1.18W. The maximum MAX6954 supply voltage is at maximum input supply voltage and minimum toleranced resistance,
i.e., 5.25V - (0.715A x 2.2Ω x 0.95) = 3.76V.
ZitierenZitat von Razer
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