Let's take the LT1763. The datasheet says that "The SHDN pin can be driven either by 5V logic or open-collector logic with a pull-up resistor." On the other hand the chart "SHDN Pin Threshold Off-to-On" says about 0.7V at 25*C temp suggesting 0.7V shall be enough to drive the SHDN pin high.

The question is whether 3.3V logic is enough to drive this line reliably? If not, which chart should I study?

I need the device to operate normally off and enable it from the microcontroller on demand. If the 3.3V cannot drive the SHDN pin correclty then I'll need to use an NPN and put the chip power supply between Vin and NPN's collector (for GND).


3 Answers 3


Yes, 3.3V should work. The datasheet says:


So as long as you can drive the \$ \overline{SHDN} \$pin lower than 0.25V when turning it off and higher than 2V when turning it on, it should work as expected.


With an Off-to-On threshold of between 0.8V and 2V, and an On-to-Off threshold of between 0.25V and 0.65V, that sounds very similar to a TTL level Schmitt Trigger.

So any voltage that rises above its Off-to-On threshold, which could be anywhere between 0.8V and 2V, will turn the regulator on. Any voltage which falls below 0.25V to 0.65V will turn it off.

Since 3.3V is greater than the upper Off-to-On threshold, that will happily turn the regulator on.

  • \$\begingroup\$ The datasheet says moreover: "The device will be in the low power shutdown state if the SHDN pin is not connected." - by that I'd expect no external pull-down resistor to be needed, right? \$\endgroup\$
    – eltomek
    Dec 23, 2014 at 10:41
  • 1
    \$\begingroup\$ Since it specifies a minimum current needed to activate the shutdown pin, that suggests an internal pull-down resistor that you have to overcome with your external drive current. \$\endgroup\$
    – Majenko
    Dec 23, 2014 at 10:43

It might not, but this can help:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Shouldn't you do that off of Vin and not Vout? \$\endgroup\$ Dec 23, 2014 at 11:23
  • \$\begingroup\$ @ScottSeidman Edited \$\endgroup\$ Dec 23, 2014 at 11:27
  • \$\begingroup\$ I think that in order for the device to be off the gate of the MOSFET would have to be kept high and the resistor would need to draw power continuously. I think the primary pull-up would need to be on the output side, but a push-button operated pull-up should be on the input. Further, a cap might be needed to ensure that even if the MOSFET turns off when the output voltage falls the regulator will stay shut down until the output is too low to pull the /shutdown pin high. \$\endgroup\$
    – supercat
    Dec 23, 2014 at 17:54

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