I'm reading the data sheet for the ULN2803A Darlington array (TI: ) and its collector current is rated at 500mA, qualified by the phrase "single output". Does this mean that the 500mA rating applies only when a single output is on, or that each individual Darlington's Ic is 500mA regardless of whether the others are on? If the former, would I be able to assume that the real max Ic for each of the eight pairs is (something like)
500mA/8, given the possibility that they could all be on at once?
This data sheet seems to lack the graphs that would make this clear.
500 mA is the maximum from any one output - but as the bulb says, you can parallel sections for more current.
The maximum continuous TOTAL current from several sections combined is limited by the TOTAL dissipation in watts - which is limited by the maximum chip temp of 150C, the thermal impedance of the particular package and the ambient temp.
About 1.5 watts seems to be the safe upper limit - equates to 4 sections running at about 300 mA each, continuous.
Of course, this makes perfect sense...if I were to fix the max ambient operating temperature at 120F (49C) then the max dissipation would be about 1.6W for the DIP version. Thanks for getting me this far.
I'm really afraid I'm about to say something stupid, because I'm still in the process of developing intuition about how electricity in general, and transistors in particular, work.
So, the power being dissipated includes both base-emitter and collector-emitter currents, I would gather. If, as you suggest, I assume four pairs on at any given time, I'd expect the power dissipation to be P = n*( IiVi + IcVce ), where n is 4 in this case.
The intended circuit looks something like this:
+5V o---------------------| |\\ COM \\>-----| GND
If the input is low, then the LEDs should be off since their cathodes aren't grounded.
If the input is high, then Ic would be something like (5 - 3.2 - Vce)/100 times the number of LEDs. I'm having a tough time figuring out what Vce ought to be because I don't have a clear idea what it depends on.
Once that Ic is figured out, is the resulting dissipated power thereon actually IcVce?
The listed Ii is 1.35mA max for Vi = 3.85V. If the input is a 74HC supplied at 5V, couldn't Vi easily be higher than 3.85V?
So, I think I have a formula, but I'm not sure which figures I use to fill in the blanks. Suggestions?
The biggest problem you are going to have with this approach is the rather large (and variable) voltage drop of the darlington switches. If you used logic level mosfets or high gain transistors, instead, that saturation voltage could be so low that the resistors would much better determine the LED current.
I have eight distinct 74HC logic outputs. Each of them corresponds to the on/off state of 10 high-brightness LEDs of Vf = 3.2V, preferably run
15-20mA. The HC itself doesn't have nearly the sink/source necessary, so I'll be using transistors. I'd considered using discrete transistors, but I'd like to avoid the requirement for all the base resistors, and having all of the emitters connected together would be a help, since they're all going the same place anyway.
The topology of the ULN2803A seemed to fit the bill; there's a common cathode, internal resistors tuned for TTL-level input, and 8 inverters to a device. But apparently I don't know how to use it as well as I thought. Is there a different array I can use instead or am I stuck going the discrete route?
You can use the ULN2803, if the exact current through the LEDs is unimportant, so long as nothing is damaged. However, in your description, you did not include the Vcesat drop in the ULN2803. That's what John Popelish referred to. If you compute the LED current without considering that drop, you get about 18 mA with your circuit. However, the actual figure could be a lot closer to ~8 mA, due to that drop.
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