I am Making an 8x8 Led Dot Matrix Display following this guide https://www.insidegadgets.com/2010/11/07/fun-with-8x8-led-matrix/ as you can see it requires a "ULN2803A Darlington Array (of Transistors)" I need to make this Project now and i dont have time to order the IC. Is it possible to make the required IC using Transistors. If so do i need special ones or can i use anyone Such as the 2N3904


2 Answers 2


The page you link above has this note following the last photo:

Edit: I’ve been advised that you don’t actually need a Darlington array for this, you could use the second shift register to become the ground. Shifting a 1 would disable the ground and shifting a 0 would enable ground.

Or, since the Darlingtons would only be driving one LED at a time, it should be possible to replace the ULN2803 with eight 2N3094 transistors, with a 1K or so base resistor for each transistor, like so:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ What do you mean? Can you explain more \$\endgroup\$
    – liljoey112
    Jul 19, 2015 at 23:16
  • \$\begingroup\$ Besides using less parts and easier construction does this method have any draw backs? \$\endgroup\$
    – liljoey112
    Jul 19, 2015 at 23:41
  • \$\begingroup\$ I can't think of any drawbacks. \$\endgroup\$ Jul 20, 2015 at 0:03

Sure, just copy the circuit inside the ULN2803 using (lots of) discrete parts.

enter image description here

For push-pull drive from the micro and LED drive you can probably leave out the diode (and the parasitic diodes with the dashed lines, of course), as well as the 3K and 7.2K resistors, so just the two transistors and the base resistor.

The maximum current will be set by the transistor Ic(max), for the 2N3904 that's only 200mA. The ULN2803 is good for 500mA, but whatever your requirements are.. you could use a 2N4401 and either another 2N4401 or a 2N4401 and a 2N3904.

Edit: Given the schematic in the link, you don't need to duplicate the current capability of the ULN2803. A single transistor with a 510 ohm base resistor (assuming 5V drive) should suffice, so 8 transistors and 8 resistors. The maximum collector current is about 120mA (at 12.5% duty cycle) so a 2N3904 or 2N4401 per channel will do.

  • \$\begingroup\$ So Basically Make this 8 times ? i.imgur.com/5R08oFz.png \$\endgroup\$
    – liljoey112
    Jul 19, 2015 at 23:15
  • \$\begingroup\$ Yes, just like that. \$\endgroup\$ Jul 19, 2015 at 23:18
  • \$\begingroup\$ Now I know you where talking about current Do you think theses can provide enough current for a 8x8 red led matrix \$\endgroup\$
    – liljoey112
    Jul 19, 2015 at 23:30
  • \$\begingroup\$ @liljoey112 - please provide a part number for the transistors you are considering. \$\endgroup\$ Jul 19, 2015 at 23:43
  • \$\begingroup\$ @WhatRoughBeast It would be the 2N3904. Also What do you think of using the Single Transistor \$\endgroup\$
    – liljoey112
    Jul 20, 2015 at 0:41

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