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MikeMike said:I believe you're going to need "source" drivers for the LED anodes. The STP16C596 and the ULN2803A are both "sink" drivers.
If you can exchange your matrix displays for common anode versions then you could use the STP16C596 16-bit "sink" driver for the LED cathodes along with a UDN2981A "source" driver (instead of the ULN2803A) to drive the common anodes.
Mike
Amen!Nigel Goodwin said:Well, this has sparked some lively debate!
OK, I'll explain further.
My original plan was to use source and sink drivers in order to push more current through the LED's - and the original circuit I started drawing incorporated that - but I thought it was getting a little away from the 'simple' philosophy of the tutorials.
While doing mental calculations of suitable resistor values I realised it's quite plausible using just the PIC to drive the displays, so I dropped the drivers and increased the values of the current limiting resistors to 150 ohms.
Due to the modular nature of my tutorials, a further thought was to optionally add driver boards between the PIC and the display - I was thinking of two seperate boards, a 'sink' board and a 'source' board - both of which could also be used for other purposes (again reinforcing the modular nature of the boards).
BTW, I've just assembled a quick program that puts all the LED's ON, and measured the battery current - it was only 62mA? - which seems rather low?, but the LED's seem plenty bright enough. This was on my meter at work, I wonder is the current ranges aren't reading properly? - it very rarely gets used on current, I'll measure again with my meter at home.
philba said:that does seem low. Are you certain the duty cycle for each row is 1/8 (minus a little for blanking during row change)? based on your numbers, it should be about double that. with some voltage drop from the pic, I'd expect more like 100-110 mA.
philba said:that does seem low. Are you certain the duty cycle for each row is 1/8 (minus a little for blanking during row change)? based on your numbers, it should be about double that. with some voltage drop from the pic, I'd expect more like 100-110 mA.