What to use for the high-side (anode) drivers of a common anode LED matrix?

Question

I'd like to multiplex rows of LEDs with a constant current sink driver. Should I just use BJTs to switch current into the rows of anodes, or is it better to use FETs, an array of FETs, or the same packaged into a convenient IC? It's easy to find current sinking ICs but I'm not sure what to search for on the sourcing side.

I would like to build something substantially similar to the Rainbowduino, http://nkcelectronics.com/rainbowduino-led-driver-platform.html . The Rainbowduino lights 24 LEDs at once with 24 current sinks and one current source using a Darlington array as the so-called "Super Source Driver", then it lights the next row of 24 and so on for 8 rows. Very standard. I would like to find a substitute for their unavailable-in-the-USA Darlington array. What do I need search for on digikey/Mouser to find an IC containing an array of current sources?

Answer 1

You can build your own common-anode driver out of discretes like this:
A forward-biased diode at the zener (D2) cathode will let you use 1 zener and a bunch of standard silicon diodes for lower cost and easier availability.

Remember that discrete devices are available in arrays, which will greatly simplify breadboard or perfboard construction. However, this will reduce your available power significantly. If you're doing it on a PCB, it's perfectly fine to do it with discretes (as long as you have room.)

With that circuit R1, D2, and Q1 all dissipate minimal power, so can be SOT-23 and 0805/0603, or low power arrays.

Answer 2

There isn't anything wrong with using discrete devices (BJTs or FETs). FETs will get you more bang for your buck (lower losses, easier driving) but BJTs are cheaper. All in one ICs are more convenient and take up less space, but are more expensive.

It's your call. 750mA across a 1.4V BJT drop is just over a Watt; I understand that you're duty-cycling the drive, but I'd suggest P-channel FETs with small NPN switching transistors driving the gates if I were to do it myself. You're probably going to want to drive those LEDs from a higher (12V?) supply in order to maximize brightness, and the transistor will give you a nice open-collector gate driver (although the pull-up resistor may end up drawing some current.) Without knowing more about your design it's hard to say what is best.

Answer 3

A Maxim MAX7219 can handle up to 64 LEDs as a current source. I wrote a short review about it here.

Answer 4

and one current source using a Darlington array as the so-called "Super Source Driver"

This is maybe just a terminology thing, but you don't want a current source, you need a voltage source. Current will be controlled by your constant current sink driver. A current source wouldn't be good anyway, since the sourced current will depend on the number of LEDs that's on.

Let's look at your requirements first. You need 8 outputs, each of which has to drive 24 LEDs. The outputs will be 1:8 multiplexed, so each LED will have to served 8 times its nominal current to have the same luminosity as when statically driven. I'm presuming common 20mA LEDs. That means each output has to supply 24 \$\times\$ 8 \$\times\$ 20mA = 3.85A. That's a lot.
I'll presume you have a low voltage power supply, like 5V, and that you want to control your driver using a logic level.

There are high side driver arrays. The AMIS-39101, for instance, looks nice, but isn't cut for the required current.

How about a discrete solution? If we can find a logic-level P-MOSFET that can supply 4A pulsed we'd be there. Let's take a look at the Si2377EDS I found at Digikey. \$R_{DS(ON)}\$ is 61m\$\Omega\$ at a 4.5V \$V_{GS}\$ and \$I_D\$ of 4.4A. Power dissipation is then (4A)\$^2\$ \$\times\$ 61m\$\Omega\$ \$\times\$ 12.5% duty cycle = 120mW. That's pretty decent!

The \$I_D-V_{DS}\$ graph shows that even at 2V \$V_{GS}\$ we have plenty of current.

So, just use 8 Si2377EDSs, possibly with a gate resistor, and you can drive the lot from 8 microcontroller outputs.