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I've spent a lot of time designing a simple circuit for my car. Here's what it does: it has 2 inputs, and whenever one of them is triggered, the output switch is shorted, connecting power supply to some load. So it's a switch with 2 inputs that implement logical OR. Now, here's the trick:

  • input 1 is connected to the ignition key (ACC pin), so it must accepts 11-15 volts and must tolerate whatever interference there is when the engine's running (hint: there may be a lotta interference). Labeled ACC_IN on the schematic;
  • input 2 is triggered by 3.3V from an MCU's output port. Labeled MCU_PWR_CONTROL_OUT.

So I've searched for some automotive-grade parts, and came up with this circuit:

enter image description here

VNP14NV04 is an automotive-grade smart protected power switch with logic-level input that should be triggered by the 3.3V signal from the MCU. TLE4264G is 5V output automotive-grade voltage regulator that I'm using to convert the noisy 12V input to something more predictable.

Thoughts? Will it work? Did I miss something? I did miss the R2 resistor initially, so figured I'd better ask before stuffing my car with this stuff, especially since debugging it on site, in the car, will be less than pleasant. I need it to work reliably.

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  • \$\begingroup\$ You have from International Rectifier and Infineon new devices for automotive high or low side switches. They have protected input, so you can eliminate the TLE circuit. Also I think it would be cheaper and better to use a resitor divider and trisil instead of voltage regulator. HINT: In a car there is no such large interference and spikes anytime as you think and worry (from yours previous posts). \$\endgroup\$ – Marko Buršič Mar 15 '16 at 0:06
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Your MCU may put out 3.3V, but D1 drops ~0.5V so the FET Gate will only get ~2.8V. According to the datasheet it doesn't fully turn on until the Gate reaches ~3.25V. In the graph below I have marked where you will be on the transfer curve - not even half turned on (and this only shows the typical response. Some units will be worse).

enter image description here

To efficiently switch a high current load the FET must be fully turned on. Your 3.3V MCU output is barely enough even without a diode in the way. You should add a level shifter that increases Gate voltage to at least 4V.

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  • \$\begingroup\$ One thing I don't understand, though: how can there be voltage drop with so little current flowing? the gate's input current is very small. So I thought no current means no voltage drop. \$\endgroup\$ – Violet Giraffe Mar 15 '16 at 5:46
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    \$\begingroup\$ The protection circuit inside the VPN14NV04 draws ~100uA. \$\endgroup\$ – Bruce Abbott Mar 15 '16 at 6:52

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