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I'm trying to understand the datasheets for several shift-register constant-current LED drivers, such as TLC59281, TLC5916, CAT4016, IS31FL3726.

All of them have something similar to this in their electrical characteristics:

enter image description here

In this example the "BLANK" pin is an ENable pin. Am I reading it correctly that it will consume 1mA even when disabled?

Or does the Vout = 1 imply that it would not consume current if Vout were 0V? Or is the only way to have a very low power mode would be to put the whole thing in front of a big mosfet?

example datasheet: https://www.ti.com/lit/ds/symlink/tlc59281.pdf

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    \$\begingroup\$ Please edit your question to provide a link to the datasheet you extracted that table from. It may take some sleuthing to figure out the answer to your question, and I don't want to hunt through too many datasheets. Initially it does look like the lowest power state of the IC is around 1mA. But if you link the datasheet maybe you will get a more complete answer. \$\endgroup\$ – mkeith Aug 15 '20 at 18:01
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In your example, BLANK is not an "enable" pin. It turns all outputs off, or allows the on/off states as set in the registers. It does not enable or disable the rest of the chip.

Since everything else is enabled, the logic sections of the chip and the reference current will still be active.

The logic sections don't seem to consume much current, but the reference current can.

Depending on how bright you've set the output (smaller resistor for Rref,) the higher the reference current will be. That current is consumed even when the lights are all off.

In the table, there are four examples with various logic states and reference current resistor values.

The current ranges from 1mA (dim lights) to 40mA. That tracks very well with the reference current as given in the chart on page 13 of the datasheet. That chart is the output current, but the output current tracks the reference current.

The chip itself consumes a few milliamperes in addition to the reference current.

Other ICs may work differently. In any case, BLANK does not shut off this IC.

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