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I just bought a 2004 Explorer. I have a wireless phone charger and GPS plugged into the cigarette lighter port. The problem is this is always on even when the car is off so I'm forced to unplug everything when I shut the car off. Not a huge deal, but what about when I forget to unplug?

I have a "power strip" of sorts that plugs into the cigarette lighter port and expands to 3 ports and 2 USB charging ports. It also includes a voltage monitor on the jack that plugs in with a digital display.

The car is above 13 V (actually almost 14 V) when the car is running and quickly drops below 13 V. My idea is to modify this to cut power when it drops below 13 V.

I'm no engineer, but I tinker with Arduinos and am confident that I could write a sketch to monitor the voltage and trigger a relay to cut power but I'm thinking that is CRAZY overkill.

In my research, I came across this thread. I understand what is going on except the reference voltage fore the TL431. I understand that the reference voltage can be set with the correct resistor, but as the power drops wouldn't the reference voltage drop as well?

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  • \$\begingroup\$ for example if you put a 5V regulator (and appropriate capacitor) in, and regulator specs allow it to make 5V from 8-24 V, you get a 5V ref even if your voltage fluctuates. Does not solve your problem properly, though. Ideally you want system to be off unless voltage has exceeded a level. Consider a zener diode? There may exist dc brownout protection systems you can use. \$\endgroup\$
    – Abel
    Mar 17 at 13:16
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    \$\begingroup\$ IMO it would be far easier to trace the wire from the lighter port back to the fuse box , remove it from its current fuse, and hook it an ignition-switched one instead. \$\endgroup\$
    – brhans
    Mar 17 at 13:38
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    \$\begingroup\$ With your idea, is your voltage monitoring circuit liable to use more power than the things you want to disconnect? Would it disconnect its own power? \$\endgroup\$
    – HandyHowie
    Mar 17 at 14:01
  • \$\begingroup\$ Have you looked at the vehicle's actual voltage immediately after you turn it off? My car is pretty much always a steady 14.5 while the alternator is running, and up to about 13.1 with the engine off, so for my car, I'd use 13.5 as the threshold voltage. Your vehicle will differ. \$\endgroup\$ Mar 18 at 6:04
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    \$\begingroup\$ @brhans I agree but modifying the car isn't always that viable or easy. Adding a circuit to their outlet expander is easier. \$\endgroup\$
    – Passerby
    Mar 18 at 15:52

6 Answers 6

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Either change the supply from Batt+ to an ignition controlled supply, or fit a relay into that supply to the cigar lighter that controls it with ignition.

Changing the supply can lead to other problems if that Batt+ supply feeds other components as well.

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    \$\begingroup\$ This seems to be the simplest, most reliable answer...just add a relay that is controlled by the ignition circuit. \$\endgroup\$ Mar 17 at 14:31
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    \$\begingroup\$ @ElliotAlderson And to control that relay, you can normally pick up off a line to the radio. I've got a 2003 Transit and added 2 12V sockets plus 2 USB charging sockets switched that way, keeping the original socket which serves nicely to connect a solar trickle charger. \$\endgroup\$
    – Chris H
    Mar 18 at 16:14
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Depending on the vehicle, there may be a much easier way. This may not work for hdelien's Ford (if not, sorry hdelien) but there might be others with the same question who might be helped to learn this.

Check the owner's manual for the section about Fuses. There may be a note there saying something like "Fuse X or Y can be moved to the outer position to provide continuous power to the auxiliary outlets or cigarette lighter when the ignition is not on."

If you look at this photo of a fusebox (possibly of a 1998 Buick LeSabre Limited Sedan FWD), there are five small fuses at the bottom of the image, two yellow fuses marked 20 (but upside down) and three fuses marked 15. Notice those two yellow fuses have different sockets than the other small fuses. The special socket has three holes for the two fuse pins, and allows the fuse to go in to one of two positions. The left yellow fuse is to the right, and the right yellow fuse is to the left. This configuration is called "the inner position"; the alternative would be "the outer position".

Fuses positioned in the inner position as shown in the photo should disable the outlets and cigarette lighter when the ignition is off.

Again, this feature is not universal -- read the owner's manual, or ask a car dealership or car mechanic.

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    \$\begingroup\$ My Ford of similar vintage doesn't have that feature - but Ford are inconsistent across models \$\endgroup\$
    – Chris H
    Mar 18 at 16:16
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I understand that the reference voltage can be set with the correct resister, but as the power drops wouldn't the reference voltage drop as well?

No. That is what a voltage reference (reference diode, zener diode, etc.) gets you. Until the reference part is starved of current because the input voltage is lower than its reference value, it holds that voltage drop value over a wide range of operating current.

the current through the reference varies in direct proportion to the input voltage, per Ohm's Law. Since the FET requires almost zero gate current to remain on, the only current through R3 is whatever the reference needs. The datasheet gives the normal range.

Ohm's Law also determines the voltage at the R1-R2 node. This voltage drives an internal comparator that determines when the reference actually conducts current. The datasheet has design examples and equations.

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I've done this modification on my truck for the same reason, and on other vehicles to add a dash cam.

Use an "add a fuse" to add a circuit. You will want this to run off an accessory circuit. This means it will only have power when the key is in the accessory or on position. The easiest one is the radio main power fuse. Run that power to a new lighter socket.

Add a fuse: https://www.amazon.com/Types-Add-Circuit-Adapter-Fuse/dp/B085MGP931/ref=sr_1_11?crid=8C3CV92E7YPW&keywords=add+a+fuse&qid=1647609080&sprefix=add+a+fuse%2Caps%2C246&sr=8-11

Lighter Socket: https://www.amazon.com/Cigarette-Lighter-Waterproof-Replacement-Mounting/dp/B07XB7PBB4/ref=sr_1_4?crid=120244O51HYT0&keywords=mountable+lighter+socket&qid=1647609340&sprefix=mountable+light+socket%2Caps%2C116&sr=8-4

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The circuit of this allows for zero current draw when the voltage is low, and doubles as the on/off switch as well. It uses a TLV431 together with the MOSFET in a 'thyristor' configuration.

schematic

simulate this circuit – Schematic created using CircuitLab

The values shown are for about 11 V, but it's trivial to alter R1/2/4 to obtain a 13 V threshold.

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I'm no engineer, but I tinker with Arduinos and am confident that I could write a sketch to monitor the voltage and trigger a relay to cut power but I'm thinking that is CRAZY overkill.

You sound like an engineer. But no, it's not overkill. Sure you could do this with passive parts and a convoluted analog circuit but a 50 cents digispark attiny arduino clone and a voltage divider plus some code would be simple enough. Hell it might even fit IN your 3 outlet port expander. It's 2022, general purpose microcontrollers can be bought under a penny and used to replace all sorts of analog circuits. Don't be a luddite.

With deep sleep, it could use minimal current (disable any leds or at the least replace their resistors with higher ones). Just wake it up every few seconds to check the voltage and turn on/off the power to a small relay. This is a beginners sketch.

Basically you are just replicating an analog comparator circuit using human readable language instead.

As to your actual question

I understand what is going on except the reference voltage fore the TL431. I understand that the reference voltage can be set with the correct resistor, but as the power drops wouldn't the reference voltage drop as well?

As long as the cathode of the TL431 is above its internal reference voltage (2.5V) AND your external reference voltage is above that internal reference voltage, then it enables its internal transistor. Yes when the power drops the external reference voltage across the resistor voltage divider also drops, which is what you want. You set the external reference voltage so that when V+ is > 13V Ref is > 2.5V and when V+ is < 13V Ref is < 2.5V. You will have to play around with the resistor values in order to get that behavior. Keep in mind resistors vary in precision and their resistance also vary due to external environmental factors like heat. But a 5 to 1 divider will give you 2.6 V at 13V in and 2.4 V at 12V in which would be the perfect switch value for you.

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    \$\begingroup\$ +1 for “you are just replicating an analog comparator circuit using human readable language instead”. An MCU also gives you all sorts of benefits for future projects such as monitoring your engine RPM via the voltage spikes on the cigarette lighter port: youtu.be/t0ToYhjYV9I \$\endgroup\$
    – Navin
    Apr 11 at 19:58

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