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Any EEC-V/Tweecer/CalEdit/CalCon knowledge here?


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Not pretty, but here's what I was thinking. R1 is always 2X R2, but the values can vary to decrease the current. Again, the current needs to be at least an order of magnitude more than what the ECU pulls, although that is probably very, very little.

Thanks for the visual, Gary. I thought that was what you were saying but the graphic confirms it.

I'm curious about the part where the current (amps?) needs to be 10x what the ECU wants. Can you explain that a bit more?

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Thanks for the visual, Gary. I thought that was what you were saying but the graphic confirms it.

I'm curious about the part where the current (amps?) needs to be 10x what the ECU wants. Can you explain that a bit more?

The current is what sets the voltage. So if the current pulled by the ECU is an appreciable portion of that pulled by the resistors the voltage will change as it only pulls through R2.

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The current is what sets the voltage. So if the current pulled by the ECU is an appreciable portion of that pulled by the resistors the voltage will change as it only pulls through R2.

Ah, gotcha. So if the current available from the battery/resistor stuff is only a bit more than the EEC draws, it will drag the voltage down to next to nothing. But if it's 10x more than the ECU needs, it will only drag it down about 10%.

Makes sense. Thanks!

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The current is what sets the voltage. So if the current pulled by the ECU is an appreciable portion of that pulled by the resistors the voltage will change as it only pulls through R2.

Sidebar question - would I need 1 3.3 ohm resistor and 1 6.6 ohm resistor? Or would 3 3.3 ohm resistors work, one on the + side and 2 in series on the - side?

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Sidebar question - would I need 1 3.3 ohm resistor and 1 6.6 ohm resistor? Or would 3 3.3 ohm resistors work, one on the + side and 2 in series on the - side?

You can use three resistors of the same value, which is a good idea as it keeps it simple. I have ordered from Digikey for many years and have this search set for the kind of resistor I think you want, through-hole, and 1/4 watt. And there I see 10 ohm resistors as an option.

So let's do the math. 3 x 10 = 30 ohms total resistance. I (current) = E (voltage)/R (resistance) = 1.5/30 = .050 amps, or 50 milliamps. Wattage, meaning the power that the resistors will dissipate, is I x I X R = .05 x .05 x 30 = .075 watt. So you'll be fine with a resistor rated at .250 watts. In fact you could go smaller if you wanted, but 1/4 watt resistors are probably the cheapest as they are quite common.

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