4033 to large CA 7-seg display. Do I need inverter before the darlington drivers?

[chriscross1966]

Hi folks, first post so please be gentle I'm knocking up a small veroboard to count events received as GPO pulses from some equipment I use at work. We want to be able to see it across the room so we're using 4" common-anode led displays (already got these). To keep it simple we're just using a pair of 4033's with the GPO's feeding the clock of the 1's digit and the carry out of that feeding the clock of the 10's digit. I've got a choice of ULN2003 or ULN2804 for the darlington driver (got them lying around) . I think as it's a common anode device I'll need the 2804, but I've never used them before and am a bit dense when it comes to logic stuff anyway so do I need an inverter between the 4033 and the input of the darlingtons, ie are they active low or active high?

Many thanks

chrisc

 

[jbeng]

Here's links to some datasheets: <ULN2003> <ULN2804> <CD4033>

Actually, either the 2003 and the 2804 would drive CA displays. The 2003 is used when the driving circuit is 5v logic (TTL) and the 2804 is used when the logic supply voltage is 6-15 volts. Also, the 2003 has seven outputs, the 2804 has eight.

You don't need the inverters between the counter outputs and the drivers. The drivers have open-collector outputs, so when the input of the driver is logic high, the collector of the darlington output is essentially connected to ground. When the input is logic low, the output floats.

 

[chriscross1966]

Many thanks I'd already got the datasheets, it was the fact that the symbol they use for the darlington stages was an inverter with a diode hung off it that was causing me angst....

Guess it makes sense to me now though, given that they're sinking the current....

chrisc