1,2,3 on D7

1,2,3 on D7

Postby EpicAbrewFail » Fri Jul 23, 2010 7:00 am

I'm trying to create something that uses D7 to say three different numbers. I'm trying to make numbers 1, 2, and 3. But not with separate circuits. I have a lot of grids to build on.

I have four switches in this circuit, the main switch, and three switches that when activated, will say either 1, 2 or 3. This is a huge job but I'm up for the task. But I'm not sure if I have enough parts. I think I need twelve diodes, not including the seven digit dispay.

But do I?
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Re: 1,2,3 on D7

Postby jpseymour » Wed Aug 04, 2010 10:23 am

How goes the project?
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Re: 1,2,3 on D7

Postby EpicAbrewFail » Wed Aug 04, 2010 3:36 pm

Can't even come close.
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Re: 1,2,3 on D7

Postby DaleDe » Wed Aug 15, 2012 6:08 pm

Here is how I would approach this problem. You need to reference experiment 644 for the Nor gate in SC-750. (Download the manual if you don't have it.) The idea is to drive each input to ground on the appropriate LEDs in D7. You need 7 transistors to drive all 7 elements of the gate but we don't need to drive F for the first 3 so you really only need 6 NPN transistors. The example shown uses resistors on the inputs but that is not enough for any real use since we need to isolate wires going from one transistor to another from effecting each other so you will likely need diodes in series. The idea is that there can be more that two inputs for the transistors. We will need three inputs.

For a 1 wire the switch to the transistors connected to B and C.
For a 2 wire the switch to the transistors connected to A, B, G, D, E.
For a 3 wire the switch to the transistors connected to A, B, C, D, G.

This can, of course, be extended as needed for other numbers. Four will need and additional transistor for F plus B, C, and G. This is a fourth input to B.

This is a typical logic problem. There really needs to be a digital oriented set but the current digital support in the 500 and 750 are a sham. I can give more details if you are interested.
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Re: 1,2,3 on D7

Postby DaleDe » Wed Aug 15, 2012 7:59 pm

After thinking about this some more I think it can be done without any transistors. It is possible to build an low true or gate using only diodes and resisters. Here is how to go about it. Instead of using a transistor for each LED in D7 use a resister (10K should be ok) from each letter to the power grid (i.e. the + side of the battery). Then attach the anode end of a diode to the other end of the resistor / LED pin. The cathode end of the diode is attached to the switch which is subsequently attached to the ground side of the battery. Attach addition diodes the switch as needed in the chart shown above in my previous message (12 diodes are needed). For example 1 would be two diodes on the switch, one for B and one for C. Using diodes in this fashion should isolate the various inputs from each other so the switches will work. If the switch is turn on the diode will allow current flow to go through the diodes that are attached and then to the LED and finally to the power connection on D7.
The connect . . . . . . . . . . . . . . 10K . . .B . . diode switch
Each circuit would look like + --- /\/\/\----|----->|---- \ ------ -
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