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Lici2013-01-09 09:41:10
Do it yourself
Lici, 2013-01-09 09:41:10

How to power the diode and not burn the logic?

Task: a decoder + a self-segment indicator to assemble a small decoder of a binary signal (set by switches) into a decimal one. I already assembled a small version habrahabr.ru/qa/27337/

And now the seven-segment indicator will be large, which lights up at a voltage of about 12V. And the logic can only be supplied with 5V. How do you put all this stuff together?

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4 answer(s)
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Eddy_Em, 2013-01-09
@Eddy_Em

Supply low-power field keys .

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Yuri, 2013-01-09
@ploop

I did this on the watch: dl.dropbox.com/u/22991016/03_BigDigit.png
Indicator with OA. For each segment of the field. The indicators are 35mm, under it the register and 8 smd transistors are easily placed on the board
. There are specialized driver microcircuits.

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DejmosSm, 2013-01-09
@DejmosSm

If it’s very powerful, then you can switch through Darlington assemblies ULN2003 / ULN2004

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slpdmn, 2013-01-09
@slpdmn

You have not specified the required output currents for the indicators. Different currents require different solutions.
- or just a block of powerful inverters is enough
- or integral keys can be
- or still you need to hang a transistor on each segment

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