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What is the simplest way to turn a regulated continuous analog DC +12 V (+/- 0.2 V) input signal into a 0 V output signal, and likewise turn a 0 V input signal into a +12V output signal?

Two separate sources are available for use, the first for driving the input signal at 12V+, or 0 V (when source is open circuited). The second for driving the output signal at 0 V (open circuit) or 12V+ nominal.

The first source is regulated to 12V +/- 0.2V or 0 V, the second source is unregulated 11.9 - 15 V.

The final detection circuit that will read the output signal considers 8.5V - 15V as a HIGH state. Anything below 8.5 V it registers as LOW.

Continuously-energized NC signal relays are undesirable due to duty cycle concerns (100% duty for months or even years), although some are rated for it and they may be the best solution. System is for safety purposes, so reliability is important.

I am looking for low power consumption, capable of withstanding -40°C to +85°C, reasonable amount of vibration tolerance, ideally a ready-made solution with minimal manual assembly and possible failure points.

Transistor inverter? Op amp comparator? Something else?

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    \$\begingroup\$ An inverter is actually pretty easy. But it would help a lot if you'd either focus the question on just one input and one output or else make a nice table of what you want. It would take perhaps a little less time to read, that way. Also, when you write "but that range is okay for .. that detects anything greater than 8.5 V" then you have already gone a little deeper into making decisions about design issues that, perhaps, you shouldn't -- or should be just a little less definite. Stay focused on what you want to achieve. If there is a list, then make the list and let others decide. \$\endgroup\$
    – jonk
    Commented Apr 28, 2021 at 2:13
  • \$\begingroup\$ It's such a specific system I was trying to generalize to make it more useful for others, but then I also didn't want to add complexity knowing that the final output was only monitored for > or < than 8.5Vdc. I'll try to clean it up a bit. \$\endgroup\$
    – VanGogh66
    Commented Apr 28, 2021 at 2:21
  • \$\begingroup\$ Okay. A lot of times someone will try and anticipate where to go and specify something. They might be right. I'm not saying differently. But it is better to say what really is needed. If it turns out that your thoughts are on-target, then others will just say so. That would be lucky. But it happens. Anyway, also think about hysteresis needs, too. And if this is just square wave to square wave, or something different. Details about what's driving the circuit and what the circuit will drive in return, though obvious to you perhaps, is better laid out in as much detail as you can. \$\endgroup\$
    – jonk
    Commented Apr 28, 2021 at 2:34
  • \$\begingroup\$ For a precise logic level you're going to want a comparator. There are many like the lm339N that work at 12V. \$\endgroup\$
    – Drew
    Commented Apr 28, 2021 at 2:42
  • \$\begingroup\$ Can 0V be used at an op amp inverting terminal and 12V at the non-inverting? Ideally 11.9-15V would be at the non-inverting, otherwise I need more voltage regulation to clean that up as well. Output doesn't need to voltage precision, just above 8.5V. \$\endgroup\$
    – VanGogh66
    Commented Apr 28, 2021 at 4:10

1 Answer 1

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You could build an inverter using a MOSFET that would look something like this.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you. I had gone through a tutorial for one like this but with any amount of load resistance (which admittedly I didn't specify) it appeared to divide the output voltage, and therefore require yet another component to keep Vout in range. Is that correct? I'll have at least a 150mA LED indicator on the output, and it'll feed an alternator regulator's feature-in port which I only know monitors for greater than 8.5V+. I'm trying to determine how much current it sinks. \$\endgroup\$
    – VanGogh66
    Commented Apr 28, 2021 at 14:15
  • \$\begingroup\$ I tried to create this circuit but the voltage across the 1kohm resistor was too low to trigger an IRF510N. I don't know if an IRF530N would be any different based on the quick spec search I did (admittedly may have overlooked something). Changing it to a 5K resistor worked great since that provided 4V which was the max turn on (2V min) for the IRF510N. If you can update the above to show 5K across the gate or help me understand why 1K works in that drawing, I'll mark it as accepted. It was a great help and has definitely provided the key to solving my BMS cutoff detection problem. Thank you! \$\endgroup\$
    – VanGogh66
    Commented May 10, 2021 at 1:11
  • \$\begingroup\$ Updated, of course you're right. Guess I didn't check the data sheet and just randomly picket a FET assumnig the gate treshold voltage is lower... \$\endgroup\$
    – po.pe
    Commented May 10, 2021 at 5:50

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