1
\$\begingroup\$

Some devices have a smart pull-down resistor. It is active until the user drives the pin, and as soon as a threshold is reached, the pull-down is deactivated.

How is this implemented, e.g. if I wanted to design a smart pull-down or pull-up for a load switch myself, having low power in mind, so that during startup the load switch has a known state, and then an MCU controls it, how would that be done?

I'm not talking about a simple resistor (permanently consuming quiescent current while the Gate of the load switch is being driven).

As an example, for a device with smart-enable functionality, the TPS7A20 LDO:

This device has a smart enable circuit to reduce quiescent current. When the voltage on the enable pin is driven above VEN(HI), as listed in the Electrical Characteristics table, the device is enabled and the smart enable internal pulldown resistor (REN(PULLDOWN)) is disconnected. When the enable pin is floating, the REN(PULLDOWN) is connected and pulls the enable pin low to disable the device. The REN(PULLDOWN) value is listed in the Electrical Characteristics table.

How does it detect whether the enable pin is being driven or not? With low power.

\$\endgroup\$
1
  • \$\begingroup\$ trevor - Hi, Please note the site rule which requires that when a post includes content (e.g. text, image, photo etc.) copied from elsewhere, that copied content must be correctly referenced. The source webpage or PDF etc. should be linked as a minimum (references for books / articles should include title, author(s), publisher, edition, page numbers etc.). I found what I believe is the source for the copied content & added it for you this time. For the future, please note it's your responsibility to do that. Thanks. (Please see the tour & help center for the main site rules.) \$\endgroup\$
    – SamGibson
    Jun 3, 2023 at 12:26

1 Answer 1

0
\$\begingroup\$

First off, a disclaimer: I'm no IC designer, so there may be some pitfall that prevents this from working inside an IC. But this is how I'd do it on a PCB with discrete elements, if I had cause to.


It's pretty simple, really:

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor can be omitted on an IC, where you can control the size of M1. Just make it have a sufficiently low W/L ratio and its own on-resistance can serve as the pulldown resistor.

If you want the pulldown to be enabled when the chip doesn't even have power yet, it's slightly more involved, but it boils down to just using a depletion mode transistor for M1.

The required logic can be as low-power as you want; static CMOS logic doesn't need much power (doesn't need any power in the ideal case, when it's not switching).

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.