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I'm using this gate driver IC. My question is, do I have to externally pull up the /RST pin? Because it can't be left floating, but I don't get the part that says "pulled down internally".

In the datasheet this is mentioned,

enter image description here

So is this correct to be connected like this?

enter image description here

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    \$\begingroup\$ Datasheet says: "RST/EN pin needs to be pulled up to enable the device; when the pin is pulled down, the device is in disabled status.With a 50kΩ pull down resistor existing, the driver is disabled by default." \$\endgroup\$ Mar 18, 2021 at 7:27

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Pulled down internally means there is a pull-down resistor inside the chip. If it was an internal pull-up then you don't technically need an external pull-up resistor but may want one anyways for noise immunity.

Since it is a pull-down internally, you most definitely need an external pull-up resistor. The internal pull-down is probably a very high resistance so if you pull-up with a lower resistance it form a voltage divider where the voltage is still high enough to keep it out of reset.

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  • \$\begingroup\$ Thanks for clarifying. So like Lars mentioned, if pull down resistor is 50 kohm, then I should put a higher like 100 kohm pull up resistance? \$\endgroup\$
    – SM32
    Mar 18, 2021 at 7:42
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    \$\begingroup\$ No. A divider with 100K on top and 50K on bottom will produce a voltage of 1/3 of the rail. Which is the very upper edge of a LO voltage and likely into the gray area. You are trying to Pull UP so you need a lower resistor. so the divider output voltage is much closer to the rail. You you want something more like 10K or lower so the divider output voltage is 5/6 of the rail voltage or higher. \$\endgroup\$
    – DKNguyen
    Mar 18, 2021 at 7:47

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