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Originally Posted by shahidali55
How will you use 12C508A to drive 4 segments?
You need 8 pins for LEDs and 4 pins for the commons; Totally 12 pins.
Will you use a serial 4x7 LED display?
If yes, it won't save you much money.
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"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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(not possible in $0.59 so i've updated my previous message)
Total cost will be atleast $1.49 (upto $1.79).
I will be using a shift register in combination with a BCD 7segment driver IC
(74HC595 = $0.2 , CD4511 = $0.2)
(12C508
Last edited by shahidali55; - 12th November 2006 at 05:19.
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a single pic will cost you less and will take less PCB room.. let's say 16f630 or grandpa 16F628
Steve
It's not a bug, it's a random feature.
There's no problem, only learning opportunities.
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Total cost using PIC16F628 = $2.79
Total cost using PIC12C508A = $1.79
Total cost using PIC12F629 = $2.39
Because of better accuracy with interrupts,
I will have to consider using the 12F629 . . .
There is a lot of space behind the displays, so space is not a problem . . .
Last edited by shahidali55; - 12th November 2006 at 16:05.
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