Electronic Circuits and Projects Forum

4th March 2008 03:12 AM

Originally Posted by 3v0

String 4 1K resistors in series. Attach one terminal of each switch to a junction common to 2 resistors. Attach one end resistor to gnd and the other to +5. Tie all the unused switch terminals together and then to an analog input pin. Use a weak pullup on the pin say 20K.

Because 3v0 said to

Mondo Tech's Super Probe · 4th March 2008 03:15 AM

Also when all the switches are open the input to the analog pin will be floating/undefined (not driven) tristated..
You need a pull up or pull down to take care of that.

Mondo Tech's Super Probe · 4th March 2008 03:16 AM

Originally Posted by AtomSoft

Because 3v0 said to

Yes but you set it up as a current limiting resistor.
See the schematic I added to my XXXX second post. And I am not sure what value would be best. At least 10K. I would try the 100K if it works.

4th March 2008 03:18 AM

oh oh so it must return a value so we close the circuit no matter what and just care if the voltage is one of mines.
Like

found the switches:

4th March 2008 03:23 AM

Originally Posted by AtomSoft

oh oh so it must return a value so we close the circuit no matter what and just care if the voltage is one of mines.

Sort of. A pullup resistor is used to maintain a small, stable voltage on an input pin. If you left it off, your circuit MIGHT work correctly, but the pin's voltage may fluctuate, causing unwanted false inputs.

Mondo Tech's Super Probe · 4th March 2008 03:27 AM

Hey we went from 3 to 4 switches

You have the basic idea.

Mondo Tech's Super Probe · 4th March 2008 03:29 AM

Originally Posted by I²R

Sort of. A pullup resistor is used to maintain a small, stable voltage on an input pin. If you left it off, your circuit MIGHT work correctly, but the pin's voltage may fluctuate, causing unwanted false inputs.

I am not an EE. But I would have said steady voltage and small current?

4th March 2008 03:32 AM

ok cool. So there is no definite way to calculate the divided voltages right? Like i would have to place resistors and manually check each one to determine drop ?
All i know in ohms law is how to get 1 value from 2 others. I don't know how to subtract like 100k ohms from 5v to determine the drop on the resistor.

Hey we went from 3 to 4 switches

You have the basic idea.

yeah the concept is in the head but the head doesnt know how to draw lol

EDIT: Found this

4th March 2008 03:42 AM

It's simple. Consider a 20 kOhm resistor in series with a 5 kOhm resistor, with 10 V applied across them. You want to determine the voltage at the point between them. First thing you do is calculate the current through the resistors:

I = V / R

I = 10 / (15 k + 5 k)

I = .5 mA

Now that you know that, you can calculate the voltage drop across each resistor:

V = I x R

V1 = .5 mA x 15 k

V1 = 7.5 V

V2 = .5 mA x 5 k

V2 = 2.5 V

Reality check: the two voltages should sum to the total supply voltage. 7.5 + 2.5 = 10. If you have more than two resistors, you can just extend this process. Now, work out your voltage drops.

4th March 2008 03:45 AM

Originally Posted by 3v0

I am not an EE. But I would have said steady voltage and small current?

A negligible current. The inputs are high impedance and measure voltage, not current.

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