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I am trying to mimick a sensor on a car that supplies 0.7 V to the car’s computer. So what I would like is a circuit diagram to build a unit that would use the 11 to 14 V input from the car battery and supply a constant 0.7 V output to the cars ECU. I would like to be able to adjust the oUtput between 0.3V and 0.8V. I do not know what current the ECU would draw at .7V from this unit. The idea is just that it must give a constant voltage signal to the ECU.Any help would be appreciated. Thank you

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  • \$\begingroup\$ Assuming the current drawn is low, which is usually reasonable given it is an input, you could use an op amp configured as a buffer, and feed its non-inverting input from a resistive divider, where the lower resistor is actually a potentiometer, so that you can adjust the output between 0.3 and 0.8 V. Because you need a voltage that's constant with respect to changes in the battery voltage, a voltage regulator is indicated to supply the resistive divider -- something like a 7805 should do. I believe a 47k upper resistor and a 10k potentiometer for the lower resistor would work. \$\endgroup\$ – swineone Mar 17 at 13:07
  • \$\begingroup\$ Thank you for that very quick response! Unfortunately I am only an electrician and a mechanical technician, so I am not familiar with the terms' op amp, buffer, and feed its non-inverting input from a resistive divider'. Could i buy some of these items or would it be possible to give me a diagram or a link to a diagram for the items I cannot buy? In other words if I could buy some of the items, how to connect them? Or am I asking to much here? \$\endgroup\$ – Redge Mar 17 at 13:25
  • \$\begingroup\$ Ok swineone, I found some very good info about op amps, I will see what I can find out about resistor deviders. \$\endgroup\$ – Redge Mar 17 at 13:45
  • \$\begingroup\$ Ok, I get the picture now. I have just one more question, Can you buy a op amp configured as a buffer over the counter? If you can, What rate or size should I ask for in my application? Thank you for your patience. \$\endgroup\$ – Redge Mar 17 at 14:00
  • \$\begingroup\$ Not really, you just make a circuit out of it. I’ll post the details later. \$\endgroup\$ – swineone Mar 17 at 14:02
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Here's a suggested circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Now for a quick explanation:

U1 is a voltage regulator. It outputs a fixed voltage of 5 V, regardless of fluctuation in the battery voltage. This ensures the output voltage stays fixed regardless if the battery voltage is 11 V, 14 V or anything in between.

R1 and R2 form a voltage divider. It takes the 5 V output from U1, and produces a voltage given by:

$$ V_{div} = \frac{R2}{R1 + R2} \times 5\ \textrm{V} $$

Note that R2 is not a fixed 10 kΩ resistance, but rather a potentiometer (a variable resistance). As R2 varies from 0 to 10 kΩ, the voltage at the middle node (connected to C3 and the + input terminal of the op amp) varies from 0 to approximately 0.9 V.

OA1 is an op amp configured as a unity-gain buffer or voltage follower. Whichever the voltage applied at the non-inverting (+) input terminal of the op amp, the same voltage will be replicated at its output. While that seems pointless, it "shields" the resistive divider from voltage fluctuations arising from variations in the current drawn by the ECU. It's possible the buffer is not necessary, but you'd have to test. If you want a guaranteed result (unless the ECU draws an excessive amount of current, say tens of mA, which I find unlikely).

Regarding the op amp model: the LM321 should be easy to find. If not, look for an LM358. It has two op amps in a single package, but you're only going to use one (I suggest configuring the second one with the same circuit, but leave the output voltage terminal disconnected.) If you can't find either one, there are other models that will work, but you need to ensure they're 1. capable of working from a 14 V supply and 2. are of rail-to-rail type, at least with respect to the negative rail. The datasheets will give this information, but if you don't have experience with electronics, it may not be easy to find. You could add a comment here if you need further help.

Note that I'm assuming your sensor indeed behaves as a voltage source. If it is based on a different principle (current source, resistive, etc.) then this circuit won't work. Some extra details on the actual sensor being used (possibly an O2 sensor?) would help determine if the circuit is adequate.

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  • \$\begingroup\$ Hi swineone, before you replied I did a lot of reading on the components you suggested, so I have a much better idea what you are talking about. This is a great explanation and I will definitely let you know if it worked or if I need more help. Thank you again \$\endgroup\$ – Redge Mar 18 at 5:43
  • \$\begingroup\$ Just to be 100% sure, all the down arrows represents earth which is the negative of the battery (which is earthed to the car body)? \$\endgroup\$ – Redge Mar 18 at 7:30
  • \$\begingroup\$ @Redge yes, ground is the battery’s negative terminal. \$\endgroup\$ – swineone Mar 18 at 11:44
  • \$\begingroup\$ Thank you swineone, your circuit is working great, I am just trying to connect it to the car's wiring. I am searching for a proper wiring diagram for a Suzuki Jimney. We have a lockdown now and I can't get to the car right now. Will let you know if it works! \$\endgroup\$ – Redge Apr 9 at 6:38
  • \$\begingroup\$ Could I use this circuit to make a usb charger to charge cellphones? \$\endgroup\$ – Redge Apr 9 at 6:42

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