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Agreed - might be an error in the values Pat stated maybe? My resistors give nice even 0.4V steps: 270, 300, 390, 470, 560, 750, 1100, 1500, 2400, 4300, 10000.

TimH wrote:Agreed - might be an error in the values Pat stated maybe? My resistors give nice even 0.4V steps: 270, 300, 390, 470, 560, 750, 1100, 1500, 2400, 4300, 10000.

Thanks, I worked it all out with a spread sheet, will put up screen shot.
The one I am having trouble with is how this switch works in 3 wire mode with the 5v power feed through the 3300ohm in series to pin 12, the formula just does not work it out I must be missing some thing.

Tim do you run your switches are just two wire or both as Pat described? if its as 3 wire like mine could you make it clear how that works as i just am not getting it. I can get mine to work easy, yours as 2 wire Thermister no problem, but not the two together?????

The R1 side (pin 12) of the equation with Pats description would have the 3300 ohm plus the sum of the resistor values depending on switch position yes? I'm just not getting the right Vout

Yours


Pats

TimH wrote:Standard potential divider stuff.

Voltage on the switch contact in position 1 is (5*300)/(3000+300) = 0.45V.

In Position 2 it's (5* (300+360) )/ (3000+300+360) = 0.9V (not 0.83V).

A spreadsheet helps to work this out for all steps in the chain.

The values I use aren't the same as Pat's for what it's worth.

Position 2 and Position 3 you mean
Position 1 will be 0 as pin 1 is not going across any resistor on switch side in two wire set up

RICE RACING wrote:

TimH wrote:Agreed - might be an error in the values Pat stated maybe? My resistors give nice even 0.4V steps: 270, 300, 390, 470, 560, 750, 1100, 1500, 2400, 4300, 10000.

Thanks, I worked it all out with a spread sheet, will put up screen shot.
The one I am having trouble with is how this switch works in 3 wire mode with the 5v power feed through the 3300ohm in series to pin 12, the formula just does not work it out I must be missing some thing.

Tim do you run your switches are just two wire or both as Pat described? if its as 3 wire like mine could you make it clear how that works as i just am not getting it. I can get mine to work easy, yours as 2 wire Thermister no problem, but not the two together?????

The R1 side (pin 12) of the equation with Pats description would have the 3300 ohm plus the sum of the resistor values depending on switch position yes? I'm just not getting the right Vout

5V power feed through a resistor (PullUp) need to connect at switch common contact, just like ECUs inside Pullup.

As above - the pullup needs to mimic the ECU pullup so goes on the centre contact not position 12

I haven't made a generic switch that's 2 or 3 wire but do make up 3 wire 5V versions when needed (with a 3k pullup. or sometimes just with equal value resistors and no pullup - in this case 5V does connect to position 12 contact) - guess it would make sense for me to do a generic 3 wire version one fine day!

I also do versions that run on 12V but give steps only up to 5V. These get used on R35 GTR's runninb EcuTek software when the CAL switch is connected to the lambda connector or secondary air pump wiring (although usually goes to a MAF connector so runs on 5V).

ok that does not make any sense as the 5v option is to turn the pull up off.
If one of you can put up how you do the calculation on position 2 of the switch with Pats's values I'd appreciate it cause I just cant end up with his output voltage as per the 2 wire hook up with the pull up on in TH mode.

Pat's 300R between positions 1 and 2, with a 3k3 pullup gives:

5*300/(3300+300) = 0.42V

TimH wrote:Pat's 300R between positions 1 and 2, with a 3k3 pullup gives:

5*300/(3300+300) = 0.42V

Pat's 5*(300/(3000+300)=0.455v *the internal pull up value listed in literature is 3k? or are you saying its 3.3k? if 3.3k then its 4.170Vout I agree, (would like that clarified)
Yes that is easy I already worked that out and your set up too.
What I cant work out is this switch (his) in 3 wire mode, with the pull up on or off (presumably he meant off) as he typed using it on a 5v input channel. In this case I cant match his output voltage ~ 0.42 for position 2 of the switch IE: using the 300 Ohm across pin 1 & 2 as I graphically represented it in my spread sheet .....
Would love to see his calculation chain to derive that, or yours or the other member who posted, cause I am dumb and cant work it out

Not sure I'm understanding?

The pullup in the ECU is on the switch centre contact...the pullup when not using the ECU pullup is *also* on the switch centre contact, so behaves the same.

I'm probably missing your point