8051 - A simple demonstration
Uploader Comments (seemantadutta)
All Comments (17)
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nice work..........
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Segment 'e' has a lower resistor.
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@K1000PL 8051 is weak compared to Atmega.
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You said in your previous post that you only need one per digit; which implies that you are using more than one.
I guess in the rather odd instance that you had an application that uses only one digit and your goal was to save 6 or seven resistors at the expense of added code, then it can be done; but you could just as easily use a resistor network chip. Generally multiplexing displays is done to save ports on a controller.
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To clarify what I meant: If you mutiplex one digit, you only need one resistor as there is only one LED on at one time. If you have more than one digit then it is normal to use 7 (or 8) resistors, one per segment. Yes it increases code complexity but you just mask the byte you're outputting with a different single bit (one segment) and output it 8 times.
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That is incorrect; it's irrelevant whether you scan of not. When you multiplex, you are turning on one common at a time. If you put the resistor in the common, the current in that digit will be divided between each of the segments that are on during that scan. You will notice the number '8' is the dimmest number and '1' would be the brightest.
If you multiplex (as you should with more than one digit), there should be 7 resistors; one for each segment.
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You should only require one resistor per each digit if you're multiplexing that is!
Yes it is !
seemantadutta 3 years ago
is your xtal 11059200 ?
captncaveman81 3 years ago
Yes, it is !
seemantadutta 3 years ago
Programming on asm or c?
K1000PL 3 years ago
ASM !!
seemantadutta 3 years ago