I'm so confused. How does this chip drive 8 LED's when there is no apparent power source? What am I missing?
New design based on comments.
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1\$\begingroup\$ It doesn't. That circuit won't work. \$\endgroup\$– Marcus MüllerCommented Dec 20, 2019 at 18:25
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\$\begingroup\$ What do I need to add? \$\endgroup\$– Kevin McQuownCommented Dec 20, 2019 at 18:26
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\$\begingroup\$ a sensible driving architecture, this is none. \$\endgroup\$– Marcus MüllerCommented Dec 20, 2019 at 18:27
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1\$\begingroup\$ Remove the ground on pin 10 and reverse the LEDs and it would work. \$\endgroup\$– Kevin WhiteCommented Dec 20, 2019 at 18:31
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1\$\begingroup\$ @KevinWhite ...and connect the other ends of the resistor networks RN3 and RN4 to some power supply voltage. \$\endgroup\$– Elliot AldersonCommented Dec 20, 2019 at 18:34
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2 Answers
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The circuit is incorrect.
The IC provides 8 x Darlington drivers which provide switched low side drives.
The load needs to go from the IC to a positive supply.
The LEDs should be reversed and the resistor pack common connections should go to a positive LED supply voltage.
Pin 10 should be connected to LED supply positive - or can be left open-circuit for a resistive LOAD such as LEDs.
While modern "indicator LEDs" can be driven from typical microcontroller output pins, if higher currents are required a driver of this sort is "useful". Despite its 'time honoured' design it still does the job it was designed to do.
Drive current is specified as 500 mA/channel but average drive currents are usually limited by package dissipation. Quite useful drive current on all channels is possible with due care to thermal issues.
At 100 mA/channel drive, dissipation is around 100 mW - which may be acceptable without heatsinking. For the olde DIP package dissipation total is specified as 2.25W and 1W max per channel. So at 100 mW x 8, temperature rise at 55 C/W = 44 C above ambient.
answered Dec 20, 2019 at 18:39
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ULN Darlingtons are low side NPN "hammer" drivers. You don't need a hammer.
For Indicators these days you don't need more than 1000mcd which can be 1 mA on a 20Cd LED rated at 20mA. However you could easily drive from the ports with 5 mA dissipating only < 2mW per port> .
The Current limiting R is just (5-Vf min)/ R.
You can choose either High side or Low side drive from CMOS.
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\$\begingroup\$ "If I had a hammer, I'd ... " :-) || He says he wants a wall display with about 20 mA/LED. \$\endgroup\$– Russell McMahon ♦Commented Dec 20, 2019 at 22:29
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\$\begingroup\$ I'd still bet, I could pick a brighter LED using only 5 or 10mA because he forgot to say anything about luminous intensity or brightness in the question which tells me alot.... @RussellMcMahon but I think I got hammered at last nite's dinner party. \$\endgroup\$– D.A.S.Commented Dec 21, 2019 at 12:12