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I'm currently testing a PCB I designed, and found out the following circuit don't work properly:

enter image description here

the SYS_EN signal is connected to a 1.8V rail through a 10k pull-up resistor.

TXU0101DRYRU5 is a level shifter. Not having my 1.8V rail yet (this circuit is part of the logic for activating the 3.3V rail, which in turns generates the 1.8V rail), I had the clever (or rather bad) idea of using my SYS_EN signal also to power the low side Vcc (VccA) of the level shifter.

When I turn everything on at once, everything works fine (SYS_EN reaches about 1.78V, BY goes high, all my voltage rails go up). The current on SYS_EN is about 20µA (in circuit, about 2µA for U5, the rest goes in a 100k pull-down resistor not represented)

However, if I first turn on power, and only afterwards turn on SYS_EN, then SYS_EN never reaches 1.8V: it stays stuck at 0.7 - 0.8V, and the current is far bigger (about 100µA).

To make sure, I bought a TXU0101 in SOT-23 package to test it out of circuit, and observed the same behavior (excepted currents that are now a bit lower: 2µA when powering simultaneously, 80µA when powering first the 5V then SYS_EN from 1.8V (lab supply) through a 10k pull-up: this is because now there is nothing else on the SYS_EN signal (in particular not the 100k pull-down).

Tomorrow, we will cut the track for VccA and power it with a 1.8V obtained from 5V with a zener + resistor.

However, I'm curious why my circuit don't work as it? Did I miss something in the datasheet? Is it "normal" that ICs don't work properly if you power-up Vcc and inputs at the same time? Is the problem that SYS_EN isn't powered directly by 1.8V but through a 10k resistor?

PS: this is a follow up on this question (that didn't got any answer nor comment)

EDIT: at @Reinderien request: I added the complete schematics (for the breadboard test, which is enough to reproduce the behavior)

schematic

simulate this circuit – Schematic created using CircuitLab

Note that in the 0-85°C range, the max currents are:

  • -0.1 to +1.5µA for the input pin (A)
  • -1 to +2.5µA for the supply pin (VccA)

So a total of 4µA max, which is only 40mV drop trough the 10k resistor: so at least in steady state, it shouldn't mater. So why is it a problem during power-up? Big inrush current while the voltage is too low, and the voltage drop being to high to ever reach a voltage that allow nominal current consumption?

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  • \$\begingroup\$ It would help if you showed the circuitry (10k pullup, etc.) for the 1.8V source. But broadly, yes, feeding anything labelled Vcc with a high-impedance source is almost certainly an issue. \$\endgroup\$
    – Reinderien
    Oct 18, 2023 at 14:40
  • \$\begingroup\$ @Reinderien : could you explain why 10k resistor counts as "high" impedance for an IC with maximal current of 1.5µA input current (leakage) and max 2.5µA supply current, ie max 4µA total (which is only a 40mV drop out of 1.8V)? Or is there a huge inrush current in ICs? \$\endgroup\$
    – Sandro
    Oct 18, 2023 at 14:55
  • \$\begingroup\$ 10k is a high impedance for anything expected to act as a supply rail, speaking generally. And your new diagram certainly doesn't reflect reality: V1 is not an unconditional voltage source, but instead (I'm guessing) some line out of a microcontroller called SYS_EN. Is the microcontroller itself fed from a 1.8V supply? \$\endgroup\$
    – Reinderien
    Oct 18, 2023 at 15:05
  • \$\begingroup\$ For the bottom diagram, it's the out of circuit test I performed on breadboard to confirm that the TXU0101 is the problem. In this configuration, V1 and V2 are 2 outputs of my lab supply (if I turn on both channels simultaneously, then the voltage at VccA is 1.78V, if I turn on first V2 then V1, then VccA is stuck at 0.7-0.8V). There is literally nothing else connected (excepted the voltmeter or ampmeter) \$\endgroup\$
    – Sandro
    Oct 18, 2023 at 15:11
  • \$\begingroup\$ So... don't do that. Give the IC proper Vcc rails. Run the test with VccA being fed directly, not through a pullup. \$\endgroup\$
    – Reinderien
    Oct 18, 2023 at 15:13

1 Answer 1

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If there is any inrush current, then the resistor will prevent the voltage from rising.

A power supply must have a low-impedance connection to the rail (10 kΩ is certainly too much), and needs a decoupling capacitor.

You could use a level shifter that needs only a single supply, i.e., a comparator like the TLV4041.

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  • \$\begingroup\$ So if the voltage can't rise, the inrush don't stops? (unlike capacitors that cause inrush current only until charged to the supply voltage). good point for the comparator with fixed reference : I will go that way for the next version of my prototype (rather than adding a linear regulator as I planed) \$\endgroup\$
    – Sandro
    Oct 18, 2023 at 15:30

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