baileychic
Advanced Member level 3
Is this a good circuit that I have designed?
I am only concerned about RN8 (24R). It allow a max common anode current of 500 mA per digit and also a max 500mA can sink through each channel of the ULC2823.
ULN2823 can sink max 500mA per channel.
0.5A * 0.5A * 24R = 6W and so I should choose 10W resistors for 24R?
Actually my chosen 4 inch 7 Segment display (CA) draws 50 mA per segment and total 8 segments including dp.
50 mA * 8 = 400mA.
12V/400mA = 30R (ULN2823 pull up)
0.4 A * 0.4 A * 30R = 4.8W per resistor.
How can I avoid this large current sinking from ULN2823 common anode drive channels? If I use a 1k or 10l resistor between 24R and ULN channles then display doesn't turn On.
- - - Updated - - -
Issue solved. I replaced the common anode driver to UDN2982A and added a hex inverter.
- - - Updated - - -
Here is the final one (TD-2.pdf, TD-2.png)
I am only concerned about RN8 (24R). It allow a max common anode current of 500 mA per digit and also a max 500mA can sink through each channel of the ULC2823.
ULN2823 can sink max 500mA per channel.
0.5A * 0.5A * 24R = 6W and so I should choose 10W resistors for 24R?
Actually my chosen 4 inch 7 Segment display (CA) draws 50 mA per segment and total 8 segments including dp.
50 mA * 8 = 400mA.
12V/400mA = 30R (ULN2823 pull up)
0.4 A * 0.4 A * 30R = 4.8W per resistor.
How can I avoid this large current sinking from ULN2823 common anode drive channels? If I use a 1k or 10l resistor between 24R and ULN channles then display doesn't turn On.
- - - Updated - - -
Issue solved. I replaced the common anode driver to UDN2982A and added a hex inverter.
- - - Updated - - -
Here is the final one (TD-2.pdf, TD-2.png)
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