0
\$\begingroup\$

I would like to multiplex 8, 7-segment LED displays with two, 8-bit shift-registers. The required current for one segment is 20mA. I have no desire to overdrive the leds with 8 times the specified current, that will shorten their precious lifetime.

Since the displays are common cathode, the TPIC-6B-595 is suitable for sinking current, but I did not find any current-sourcing shift registers, so I need to use an octal buffer/line-driver. I prefer octal-buffers over Transistor/FET arrays.

The 74**541 and 74**244 datasheets are not clear about the current-sourcing parameters, to me it looks like they are only sinking and cannot handle 7+1 (dot) * ~30 mA = 240 mA.

This is the circuit diagram: Multiplexing 7 segment displays

Does anybody knows any solution to my problem?

\$\endgroup\$
  • \$\begingroup\$ I think you're calculating wrong. Each gate would only be sourcing the current for one segment, so only 20mA. \$\endgroup\$ – Majenko Mar 1 '15 at 21:19
  • \$\begingroup\$ Yes, but the whole device has to take 8*~30=240mA (worst case) which is far beyond any octal-buffer specs. \$\endgroup\$ – Gábor Dani Mar 1 '15 at 21:22
  • \$\begingroup\$ Majenko is right. You are wrong. The whole device only has to produce 30 mA at any given instant. That said, 74**S is not a good way to source current. You "prefer octal-buffers over Transistor/FET arrays", but that just means you need a high-side driver. \$\endgroup\$ – WhatRoughBeast Mar 1 '15 at 21:29
  • \$\begingroup\$ If all the 8 led, rated ~20mA each, (7 segment + 1 dot) is under power, how can the whole thing be only ~20-30 mA? \$\endgroup\$ – Gábor Dani Mar 1 '15 at 21:34
  • 1
    \$\begingroup\$ Ah you mean the whole device current - ok, yes that can be 240 max for your arrangement. \$\endgroup\$ – Majenko Mar 1 '15 at 21:36
2
\$\begingroup\$

In the "for what it's worth" category, this resource I found below lists quite a few latching constant-current LED drivers (8 and 16 bit). http://www.altera.com/literature/an/an286.pdf

Some of them actually have shift outputs so you can gang them together and write one long stream to the first device in the daisy chain.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.