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Beginners / Re: The old "Replace blue LEDs with red LEDs" 7 segment question (74HC595)
« Last post by pcprogrammer on Today at 08:48:36 am »If possible change the 220 Ohm current limiting resistors to 470 Ohm when replacing the 7 segment displays. This will limit the current to about 6mA instead of 13mA the 220 Ohm allows. The 13mA won't kill the red ones but they might be to bright with it and it can kill the 74HC595A because with all segments lit the current will be above the max Icc of 75mA given in the datasheet.
Simple rule is to take the LED forward voltage and subtract it from the supply voltage and divide that by the current limiting resistor to get the operational current. So with a Vf of 3.8V and a supply of 5V and a 220 Ohm resistor it is 5 - 3.8 = 1.2 / 220 = 0,005454 Ampere. Will be a bit less due to the output voltage of the 74HC595A not being equal to the supply voltage.
The 74HC595A has an output enable. Controlling this with a PWM signal can dim the display. The outputs will only supply current when enabled and on average lower the current through the LED thus dimming it.
Simple rule is to take the LED forward voltage and subtract it from the supply voltage and divide that by the current limiting resistor to get the operational current. So with a Vf of 3.8V and a supply of 5V and a 220 Ohm resistor it is 5 - 3.8 = 1.2 / 220 = 0,005454 Ampere. Will be a bit less due to the output voltage of the 74HC595A not being equal to the supply voltage.
The 74HC595A has an output enable. Controlling this with a PWM signal can dim the display. The outputs will only supply current when enabled and on average lower the current through the LED thus dimming it.