Following on from my previous question I'm trying to create a shutdown controller for my Raspberry Pi. The Raspberry Pi needs to be powered from the battery, but should power-down after the Pi detects that the ignition has been turned off.

The Pi will take a 3.3V feed from the ACC line (I have other components that will take 5V from the ACC line via a 7805, so I will step down to 3.3V using a voltage divider unless anyone has a better suggestion - I'll also be driving a uPD6708 which takes 5V CMOS I/O, so will have to step-down from 5V to 3.3V on another 2 lines).

Software running in the RPi will set one of the GPIO pins high, presumably when the RPi shuts down the GPIO pins will all go low. So Q1 should turn the relay on, keeping the RPi's power on as long as the ignition is on, or the GPIO pin is high.

I have 3 fuse kits with a 1000uF cap and some kind of transformer/inductor, so I may as well use one of them on each the 12V battery and 12V accessory line.

This shut down controller claims to draw only 50uA in standby - if I used a CMOS 4071 OR gate that would be a start, but from what I've read, you'd need more current from the OR gate to saturate the transistor - is that right?

Bearing in mind that I need to level-shift 5 lines from 3.3V to 5V and 2 from 5V to 3.3V in addition to the requirements of this sub-circuit, can anyone recommend components/alternatives for OR1, Q1, RLY1 and/or any modifications?


simulate this circuit – Schematic created using CircuitLab

Here's my attempt at following @Connor Wolf's suggestion.

  • R1 and C3 need to be be chosen to allow the RPi to shut down properly
  • I've added C1 because I image that it will take a brief moment before the relay switches after the ignition is turned off - I've got no idea how long that is, but I suppose the RPi is going to be drawing about 700mA from the capacitor, in addition to the 555 and relay


simulate this circuit

@Nick suggests it could be simpler - like this perhaps? I tried to remove the diodes so that I could just use an off-the-shelf 12V-5V 1A USB power supply (or a pair of them). The 555 datasheet says that it outputs 3.3V (max source 100mA? This page says 200mA). The RPi will take read the ACC line at 3.3V to determine when to shut down.


simulate this circuit

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    \$\begingroup\$ Your 7805 is going to get REALLY HOT as your circuit is drawn. With the 700 mA draw of the raspberry pi, at your vDrop of 7V (12v-5v), you're going to be dissipating 4.9 watts (7V * 0.7A) in the regulator. You really need to look into a DC-DC for that. \$\endgroup\$ Commented Mar 22, 2013 at 8:06
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    \$\begingroup\$ Also, your "fuse kit" thing is drawn incorrectly. Right now, the diode is simply shorting out the 12V input, and the inductor isn't doing anything. I would guess that the inductor is in series with the power line, and the diode is the other way around (which would let it prevent reverse-biasing the input). \$\endgroup\$ Commented Mar 22, 2013 at 8:08
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    \$\begingroup\$ Voltage dividers are not a good way of powering anything, as the load varies the voltage will vary. Just use a regulator of some sort, they're cheap & plentiful enough. \$\endgroup\$
    – John U
    Commented Mar 22, 2013 at 8:40
  • \$\begingroup\$ @ConnorWolf I've got a 7805CT which is rated at 1A. ...Does that only apply if you're coming down from 6V to 5V? When I Google for DC-DC converter many of the search results mention 7805. Unless the DC-DC converter outputs 5.7V I think I'd need to switch D2 and D3 for schottky diodes with a voltage drop of less than 0.25V \$\endgroup\$ Commented Mar 23, 2013 at 3:20
  • \$\begingroup\$ @NicholasAlbion - I didn't say it wouldn't work, just that you're going to need a very large heatsink, or a fan/smaller heatsink, to keep it properly cool. \$\endgroup\$ Commented Mar 23, 2013 at 3:31

4 Answers 4


While using a one-shot timer circuit will work, I think an easier solution can be used. Take a look at this circuit.

enter image description here

For clarification, "VBAT" is a 12V source that is always on as long as the battery is connected. However, "ACC" is a 12V source that is only on when the ignition is on or the key is set to "accessory." Rather than using a 5V relay just to control the power to the RPi, why not use a standard 12V auto relay as shown. This way, there is no wasted power (except for the coil current while the power is on) because everything will be disconnected from the battery.

One side of the coil is always connected to 12V. The opposite side is connected to ground (chassis) through an N-Channel FET (Q1). While a MOSFET is used in the diagram, any FET capable of sinking the coil current can be used. When "ACC" is powered ON, Q1 will switch ON, connecting the coil to ground and actuating the switch. This will in turn power whatever 5V regulation circuit you plan to use (a simple 7805 regulator with heat sink, a switching DC-DC converter, the USB supplies mentioned, etc).

The diode D2 is there to ensure the capacitor can only discharge into Q1 and can be regular or Shottky. Other methods should probably be used for over voltage and current protection from the battery.

The "ACC" voltage can be put through a voltage divider to create a 3.3V signal for the RPi. Be careful with this voltage level, considering a 12V auto battery can really be more like 14V DC. As long as this signal is HI, the RPi knows that the power is on. Obviously, this GPIO pin should be set as an input with any internal pullups disabled. When "ACC" is turned off, the RPi should see the LO signal on the pin and begin its shutdown.

When the "ACC" voltage is turned off, the capacitor C1 will retain the charge for so long, discharging through the resistor R1. Once the capacitor voltage drops below the gate threshold of Q1, it will switch OFF, disconnecting the relay coil from ground and removing power from the peripheral circuit. If a "logic level MOSFET" is used for Q1, it will remain switched ON until C1 voltage is fairly low. I tested this circuit using an NTD4960 (Datasheet), and it remained on for around 15 seconds - until C1 was around 2V. To increase the time, increase the capacitance value.

  • \$\begingroup\$ How do I select an appropriate FET? I have a 12V relay already - 621D012 (270ohm -> 44mA) The online Jaycar catalogue lists the following: 2N7000, PN100, VN10KM, IRF540N, IRF1405 and more \$\endgroup\$ Commented Mar 27, 2013 at 12:06
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    \$\begingroup\$ I believe the 2N7000 would be a good choice, but here is what you are looking for... Type: N Channel FET (not an NPN BJT); Forward Current (I_d): At least 100mA - (double of 44mA); Gate Threshold Voltage (V_gs or V_gth): No more than 3V; Gate Capacitance: Doesn't matter, we put more externally! "ON" resistance (R_ds): low is good, but your power dissipation will be low; Power Dissipation (P_d): 44mA ^2 * R_ds << at least double this \$\endgroup\$ Commented Mar 27, 2013 at 19:38
  • \$\begingroup\$ Can this be store bought? I am trying to do the same as the OP but I have not idea on how to create my own circuits. Are there any store bought solutions for this? \$\endgroup\$ Commented Nov 3, 2014 at 11:50

Honestly, I think you're overthinking this a lot.

Personally, I'd just use a one-shot with a period of a minute or two, triggered by the car being turned off.

When you turn off the car, the one-shot fires, holding the relay closed until it times out. All you would need to do would be ensure that your raspberry pi shuts down within a minute or two of the car being turned off. This should be easy enough by monitoring a input from the car's switched power.

The biggest advantage to a system like this is that when your software crashes (when, not if), it'll still shut down anyways, so you won't wind up with a dead battery. The one-shot should be plenty simple. You could use a 555, or a little microprocessor (like Olin will suggest).
Another nice thing is that, if you do the design properly, the system can disconnect itself from the car battery, ensuring the quiescent current draw is absolutely 0.

  • \$\begingroup\$ sounds great - how is it done? \$\endgroup\$ Commented Mar 22, 2013 at 13:05
  • \$\begingroup\$ Sounds a bit like the circuit used to keep cortesy lights on, after the car door closes. From memory, a capacitor is charged to 12V and holds on a FET or similar. Remove power, the cap slowly discharges - until the FET gets switched off. \$\endgroup\$ Commented Mar 26, 2013 at 7:05
  • \$\begingroup\$ The RPi draws 5V at about 700mA and I think I need about 10 seconds to shutdown. Using a capacitor was my first thought, but that would involve about $100 worth of capacitors \$\endgroup\$ Commented Mar 26, 2013 at 13:09
  • \$\begingroup\$ @NicholasAlbion, the cap wouldn't be used to provide power directly to the Pi, but a gate voltage to a FET (which would be controlling a relay). The gate resistance of a FET is very high, so you wouldn't need a big cap to last the 10 seconds needed. \$\endgroup\$
    – Pentium100
    Commented Apr 1, 2013 at 3:38

Any fixed delay method suffers from the problem of not knowing how much time the RPi really needs to shut down. It would be better to press a button that signals the Pi to shut down, it could then do what it needs for a clean orderly shutdown, taking as long as needed, then issue a GPIO signal back to the push button circuit that shuts down the power. That gives the RPi as much time as it needs to do things like safely shut down the SD card. The circuit doesn't have to be too complicated. You can see a simple circuit at


The web site describes the circuit's operation.

  • \$\begingroup\$ You make a good point, although I don't think anyone wants to have to go through such a process every time they shut off their car. It would make more sense for the RPi to sense the shutdown by monitoring the voltage of the ACC line (only on when the car is on), then kill its own power from the "always on" 12V line when it is ready to do so - automating the system. \$\endgroup\$ Commented Apr 22, 2013 at 21:14

Use 4 AA rechargeable batteries. Power the Pi from them and have them on charge from the car's battery.

Use 1 GPIO to tell the Pi if the ignition is on or off.

Shutdown when ready.

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    \$\begingroup\$ I think this answer needs more detail to be useful. Maybe post a schematic or desciption on how you think this would work and what sort of charging circuit / device/ power control you would propose. \$\endgroup\$
    – PeterJ
    Commented Sep 25, 2013 at 12:26

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