Quadruple Digit Display resistance

[woodturner550]

I am working on connecting a Everlight Electronics quadruple digit display(led seven segment) to an Easy6 board using port D and E. I plan to use port D for the seven segments and decimal point. Use E for digit 1,2,3 and 4 and V+ . The display has a forward voltage of 2.0 (TYP), forward current of 25mA (TYP), power dissipation rating of 85 mW, reverse voltage rating is 5 V and peak forward current rating 160 mA.

I have several questions. I need more than just the answer, I want to know the formula to figure what the resistance is and do I need pull up resistors on the Easy6 or pull down and how to KNOW which is needed.

This display is the type that the microcontroller will turn on each segment many times a second so it will seem as if the displayed numbers are constant. The signal from the microcontroller will send a signals to four small transistor(2N3904) to turn on the digits as needed.

I think the resistance should be 150 ohms from looking at some schematics but I want to be sure and understand how to know what it should be. Similar to the old saying feed a person and it takes care of that day teach him to grow his food and take care of the tomorrows.

Thanks for your help.

woodturner550

 

[audioguru]

You forgot to say what is your power supply voltage.
If it is 5.0V then the current-limiting resistors will have 3V across them if they are driven from a driver with a low voltage loss. Then 150 ohms will create an LED current of 5V - 2V)/150= 20mA. The LEDs will be bright only if they are lighted continuously. Since your microcontroller pulses them then they will appear to be dimmed. The on-off duty-cycle determines the brightness.

 

[Pommie]

You should be aware that each port on the Pic chips is limited to 100mA and so your segments (+dp) can only draw 12.5mA each. With 25% duty cycle your display will be averaging 3mA and so will be very dim. You can increase the brightness by dividing the segments between 2 ports.

For 12.5mA the calculation is (Vsupply-Vled)/Iled or, assuming 5V supply, (5-2)/0.0125 = 240Ω.

You could use pnp transistors on the segments and pulse your display at 100mA to achieve full brightness. However, be aware that with this setup, stopping the system with the debugger will burn out your display as the current will be a constant 100mA.

Mike.