That would be fine.Soooo would there be anything wrong with connecting a reverse polarity prevention diode in line between the 18volt supply and the anodes of all the LEDs ?
How are you arriving at that number?My estimate would be max current with all LEDs on at about
1.4 amps so would need a 2 amp diode?
Let's estimate high and say the LEDs draw 25mA.
25mA is required per row of LEDs.
2 rows make up a segment.
7 segments make up a digit and there are two digits.
25mA x 2 x 7 x 2 = 700mA
If you have spare 1N4001s, it wouldn't be a bad idea to put on the anode of each digit, thus reducing the current going through each diode to 350mA.
This is why you may need to use a smaller value resistor between the driver IC and a row. Now, if you can't "see" any difference in brightness with the 3mA drop, then no worries.I noticed with connecting a diode as mentioned in the first paragraph, it dropped the current draw by about 3 mills... to about 18-19 mills.