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I have a BTS142D "smart" FET low-side switch driving a 12v load, all works great except when the logic (gate) side of the circuit is powered down but the 12v load is powered (Vbb = 12v), it back-feeds through the device (protection circuit presumably), pulls the gate pin (Vin) up to ~5v and turns itself on.

Circuit in question

Is this as simple as just adding a stiff enough pull-down resistor to the input or have I missed some design criteria?

The data-sheet / appnotes say nothing about this behaviour.

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2 Answers 2

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The data-sheet / appnotes say nothing about this behaviour.

The data sheets tells you that the on-state input current might be as high as 30 uA. It also tells you that the input threshold voltage for activation might be as low as 0.8 volts (worst case). So, I would naturally want to choose a parallel resistor that was lower than 26.7 kohm. I'd use a 10 kohm pull-down resistor. In the absence of any other information you have to make of the data sheet what you can.

If in doubt, consult the manufacturer.

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  • \$\begingroup\$ I've tried pulling Vin down, when I got to 1Kohm and it was still turning itself on I thought I'd stop and check myself because that doesn't seem right. \$\endgroup\$
    – John U
    Aug 20, 2020 at 11:07
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    \$\begingroup\$ No, that doesn't seem right to me. Have you tried open circuiting the connection to the logic that drives the gate input in case there's anything strange going on with that? Then repeat with 1 kohm and 10 kohm pull-down resistors. \$\endgroup\$
    – Andy aka
    Aug 20, 2020 at 11:08
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Just to follow up and say I found the cause of the problem and it's nothing to do with the BTS devices - it was in fact power back-feeding through a somewhat convoluted path through the testing rig.

However, per Andy aka's advice I will be putting 10k pulldowns on the next revision of board just to be safe.

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