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To start, I'm a complete layman on this topic. If I say something that sounds inaccurate please do correct me. I'm here to learn as much as I'm here to ask for help.

I have a truck with aftermarket wheels that can't accept tire pressure monitoring sensors. Without the signal from the sensors my tpms light constantly flashes on my dash. The blinking light is driving me crazy so today, I basically disabled the system and fed some power into the wire that controls the light. When the circuit sees +5v, the light turns off. If it sees no voltage, the light turns on.

The light is off and my goal is to keep it off without causing electrical issues in my truck. Currently, I'm feeding 12v into the circuit. Even with my complete lack of knowledge on this topic, I know it's a bad idea. I'm not even sure what I'm overloading. If it's just an led then fine, it'll burn out. But if it's my tpms module then I want to reduce the 12v down to ~5v to avoid a potential fire. This is a pretty common thing for people do to in these particular trucks and I don't think anything will happen. But for my peace of mind, I want to do this small project right.

I've poked around a bit and see that people regularly make their own little resistor arrays for things like this and was wondering if you guys might be able to point me in the right direction. My first thought was not to solder something together but rather to use something like this:

https://www.amazon.com/Step-Down-Waterproof-Miniature-Converter-Supply/dp/B07MYPCM73

Would that even work for my application? I have access to good ground points and I already have the wire I can draw off of. All I need to know is how to step that voltage down from 12v to 5v. Currently my thought is to:

  1. Get that little convertor
  2. Tap into my 12v wire and connect it to the input of the convertor
  3. Connect the ground on the output side to a ground point

This is where I get a bit confused. What do I do with the ground on the input side? Can I connect it to the same ground point as the ground from the output side? Is there a better way to do this?

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    \$\begingroup\$ it's an automobile, maybe 12V is fine. can you measure how much current flows into that input when 5V is connected? \$\endgroup\$ Nov 14, 2021 at 23:04
  • \$\begingroup\$ the wire colours on that converter worry me, it looks like it could be a negative regulator, but there's nothing in the description. \$\endgroup\$ Nov 14, 2021 at 23:11
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    \$\begingroup\$ A better way is to get the wiring diagram for your truck and check for where the signal comes and goes. Alternatively the 5v ground would still tie to the same ground points. I bet the 12V input ground of that regulator is tied to the 5V ground internally so no extra wire would likely be needed. But I'm not liable for your car if not. \$\endgroup\$
    – Passerby
    Nov 14, 2021 at 23:39
  • \$\begingroup\$ Bottom of this article I share a solution I’ve found doing a lot of research on this. Hope it helps!! toyotanation.com/threads/how-do-you-bypass-the-tpms.1677635/… \$\endgroup\$
    – Omie Mann
    Mar 26, 2022 at 19:44

4 Answers 4

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It expects a simple signal voltage, no current for a load is carried. Just a few microamps for the input circuit. A voltage regulator is overkill and they often have issues with very small loads. You can complicate it with zeners or other voltage regulation circuits but this is pretty much what resistor voltage dividers are for. Low current, static loads.

enter image description here

A simple pair of resistors would work. The specific values are not important as long as the output is what you need. You can use the 8k/6k in the picture, or 2K/1.45K or multiple resistors in series/parallel to get the right value. The output voltage also isn't super critical. As long as it's almost in range like 4.5V to 5.5V or maybe even lower.

Formula is simple. VOut = (Vin * R1) / (R1 + R2)

Or

5V = (12V * R1) / (R1 + R2)

Resistor wattage is minimal, 1/4 would be fine.

As for the ground/0V just use the ground on the same connector.

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As you mentioned feeding +5V to the input turns the lamp off and 0V turns it on. So you have an input which gets a control signal and not some power supply. That means this input has a relatively high input impedance. Assuming the input voltage should not exceed approximately +5V, the only (and simple) thing you need is a kind of voltage divider like in the image I added. Just connect the 470Ohms, 1/2Watt resistor in series with the 5.1V Zener diode (also about 1/2Watts) as shown in the image and take your +5V output from the cathode of the Zener diode (the side with the line marking). This should probably help protecting your input circuit.simple 5V regulator

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    \$\begingroup\$ It's a logic input. It doesn't need 15mA current capacity. This circuit may overload the input if the Zener diode were to fail open. I don't recommend using it at all. \$\endgroup\$ Mar 27, 2022 at 0:16
  • \$\begingroup\$ I cannot remember having seen any Zener diode fail open in the past 30 years. They always short which is no problem at all. A resistor divider would allow the input voltage exceed 5V when battery charges while the motor is working. \$\endgroup\$
    – Andreas66
    Mar 27, 2022 at 10:03
  • \$\begingroup\$ You're most likely right. I've never see any fail open either, but got a couple low-impedance ones that acted like a somewhat decent Zener paralleled with resistance in the tens-hundreds Ohm range. Still, an extra component that's not needed in this application. \$\endgroup\$ Mar 27, 2022 at 17:06
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All you need is a series resistor, loaded with the presumed clamping diodes on the digital input and a potential pull-down resistor.

Typically you'd want to start with say a 470kOhm series resistor, check if it works, and if not then reduce the value to about 2/3rds (33% smaller), and try again. Once the signal gets recognized, use the next smaller value (2/3rds of the current one), and that should do the trick.

A Zener diode is probably unnecessary since the logic input itself has some clamping action already, and the failure of the Zener would be an additional source of uncertainty. The input already has components that do the clamping.

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Since you already put 12V to it, 12V is probably OK. In other words, you could probably just protect it against wattage rather than voltage. As suggested, you could check how much current is being drawn.

If you measure current and find it to be 0, it's probably fine.

If you cannot, then I suggest you get a 1M potentiometer. This can then be used in several ways including tuning it to produce 5V as a Voltage divider. Since you already used 12V, the usage I suggest is as a tunable 0-1M ohm resistor in between 12V and where you connected it. You know it would work at 0. if it works at 1M, you protect against overpowering and burning out since any small current would cause voltage to drop significantly. Does not protect against every mode of failure, although overvoltage does not appear to be an issue based on what you tried.

If it is drawing current, there likely exists a sweet spot that results in about 5V, otherwise it will always produce 12V, proving that connecting 12V did not push power through something.

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