I'm using 4 GPIO pins of my Raspberry Pi Model B 512MB Rev2.0 Board, to drive a single 7 segment LED display (1", Common Cathode, no part number on item) through a BCD to 7-segment latch/decoder/driver (HEF4511B).

Basically, VDD in the 4511 BCD Driver is connected to 5V pin in RPi. The BCD inputs to the 4511 (D0, D1, D2, D3) are from 4 GPIO pins (3.3V).

The output from the 4511 is passed through a 480 ohms resistor onto each LED segment. The circuit is set up like the first example here.

My problem is that the LED is too dim. How can I increase the brightness?

I know I could try swapping the 480 ohms resistor with a lower resistance or perhaps use transistors. Problem is, I'm afraid I might cause damage to my RPi.

I would prefer a solution with the 4511 IC itself, as I'm new to electronics and this IC is easy to understand.

EDIT: I have found this Relay driver ULN2803 which seems to be the thing to use, but I don't know how to connect it to my circuit and besides, it'll take 3-4 days for delivery.

  • You can't damage the Pi if you use smaller resistors. Did you measure the voltage across the LEDs? What does the data sheet say WRT voltage and current requirement? Feb 5 '14 at 21:48
  • The voltage across the LED is 1.7V. This LED doesn't have a part number, so I don't know where to look for the data sheet. All I know is it's a 1", red, common cathode 7 segment display.
    – Code Poet
    Feb 6 '14 at 5:57
  • This LED's part number is BAF056RCC. However, I can't find a data sheet for it.
    – Code Poet
    Feb 6 '14 at 6:03
  • If you want to run the 7-segment with an ULN2803, you need a common-anode display, not a common-cathode one. Feb 6 '14 at 19:33
  • The HEF4511B data sheet shows what to do (fig. 10 on page 13; the transistor is 1/8th of an ULN2803 or similar). Won't help you with your 7-segment, unfortunately. Feb 6 '14 at 19:42

The 4511 has internal buffer, decoder and latch. It will draw very little current from the GPIO pins. It can source up to 25mA to the display.

You will be limiting the current to ~3V/470Ω i.e. about 6mA. You could quite safely reduce these to 150Ω (or increase the voltage to the display).

  • The display is powered by output pins from 4511, which is connected to (VDD) the 5V line on RPi. How do I increase the voltage to the display?
    – Code Poet
    Feb 6 '14 at 6:05
  • If you only have 5V then use 5V. The 4511 can be connected to higher voltages. This was only a possibility; just reduce the resistors.
    – Milliways
    Feb 6 '14 at 6:56

Red LEDs typically have a forward voltage drop of 2.0V. So simply reduce the resistor's value until you get there.

Without a data sheet, you can either hook up the LED to an adjustable power supply and a current meter to figure out the LED's resistance at that point, or you can simply try smaller resistor values. Slightly higher voltages are unlikely to damage the LEDs if you just do a quick measurement.

  • Actually, I'm noticing an odd thing. If I light just one segment, the LED is bright enough to be seen in broad daylight. As the number of lit segments increase, the LED is getting dimmer and dimmer. I changed my resistor to 68Ω. Still no perceptible difference in brightness. These LEDs are not connected inside in series, are they? I'm powering them from the 8 output pins on the 4511. VDD on the 4511 is connected to the 5V pin on the Pi.
    – Code Poet
    Feb 6 '14 at 13:12
  • … how do you light just one segment with that chip? the minimum should be two. Anyway, it seems that the LED pulls more current than your driver IC can deliver, so you need to change your approach. I'll whip up a quick schematic. Feb 6 '14 at 16:38
  • OK, that quick schematic isn't that simple. Looking through the DigiKey catalog, there's no equivalent for an ULN2003 that works for common-cathode LEDs, so you'd have to use discrete PNP transistors. :-/ Feb 6 '14 at 20:02

You can use MIC5891 open source shift register to drive common cathode displays.


Only need 3 IO pins to control it.

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