I have a problem that may lead to deeper problems very soon. I have a board with two isolated DC-DC regulators (for 3V3 and 5V):

DC-DC regulators

The outputs share a common ground, as seen. Both regulators have an LED on their respective output rails.

As I am soldering the board and testing step-by-step along the way, I realised that if I power the 3V3 rail, the 5V status LED also lights up. On the contrary, if I power the 5V rail, the 3V3 status LED does NOT light up.

I believe that this is caused by some level-shifter circuitry:


It explains this behaviour I suppose.

Between the 5V rail and ground, I measure several million ohms of resistance, which is expected. Between the 3V3 rail and ground, I only measure 1.2k.

The reason it is 1.2k is because there are six of these level-shifters on the board. The parallel configuration of six 10k resistors more or less conforms to this.

I suppose it's not going to be an issue since the current is limited by the resistor. Is there anything I should be concerned about in terms of the regulators themselves?

NOTE: The regulators have NOT been populated yet. I am simply powering the pads externally.

  • \$\begingroup\$ @R.Joshi Yes ;) but I'm just wondering about its consequences. \$\endgroup\$
    – Shreyas
    Commented Apr 25, 2018 at 13:04
  • \$\begingroup\$ Depends on what else is on the 5V net, it may not like getting high impedance 3.3V where it is expecting 5V. If there is no situation where 3.3V is on while 5V isn't you might want to sequence them to make sure that 5V turns on first just to be sure. \$\endgroup\$
    – PlasmaHH
    Commented Apr 25, 2018 at 13:18
  • \$\begingroup\$ Yes, the FSIG1 is going through MOSFET diode (forward biased when no 5V ) and through some protection diode to 5V \$\endgroup\$
    – Dorian
    Commented Apr 25, 2018 at 13:49

2 Answers 2


You would not be having these problems if you used a level-shifter IC such as this one (SN74LVC1T45):

enter image description here

• Fully Configurable Dual-Rail Design Allows Each
  Port to Operate Over the Full 1.65-V to 5.5-V
  Power-Supply Range
• VCC Isolation Feature – If Either VCC Input Is at
  GND, Both Ports Are in the High-Impedance State

As well as properly handling the translation and isolation functions, it is considerably faster than the sluggish pullup of the transistor method (sub 16ns in most cases, with a 15pF load). 10K and 15pF has a 150ns time constant.



simulate this circuit – Schematic created using CircuitLab

This is the actual path from 3.3V to the 5V led.

From FSIG high signals and, where FSIG is in high impedance, through R1 the power is going through the forward biased zenner where it looses ~0.6V then through internal protection diode of the 5V circuit (also through R2) where it loses another ~0.6V to the 5V rail. Around 2.1V are on the LED and R3, enough to light up.

Dangerous? Not quite, CMOS outputs are resilient to some short overdriving, we are talking about hundreds of ms at most, if you didn't fried anything by now it's very unlikely to happen when both regulators are starting almost together.

If still uncertain put some limiting resistor on pin 1. Tough , in your place, I woudn't bother.

If it's possible you could also configure FSIG and (1) to start low at power on

  • \$\begingroup\$ "This is the actual path from 3.3V to the 5V led [...] From FSIG high signals [...]" The OP hasn't mentioned any signal applied to the FSIG net - the only voltage mentioned is via the 3.3V rail. Surely a simpler option exists of 3.3V supply, via R1, the MOSFET body diode, then R2 and onto the 5V net (with obviously a reduced voltage & current, but enough to light the 5V LED as described by the OP)? Notice how I'm not downvoting your answer, even though I disagree with it, as your suggestion might help - yet you downvoted mine (now deleted) from yesterday with no comment. Disappointing :-( \$\endgroup\$
    – SamGibson
    Commented Apr 26, 2018 at 18:15
  • \$\begingroup\$ @SamGibson I mentioned the R1 path. Since the 3.3V is active presumably some FSIG signals might be active, high, low or Hiz. It is not a 5V led, it's the led from 5V powered line, a red led has down to 1.8V forward voltage. I'm sorry, I'm guilty, the answer to this question seemed to obvious that's why I posted a comment hoping that someone else will make an answer from it which didn't happened. Your post contained some useful information but the actual answer to the question was very hard to find because it was to long. I was hoping that you will notice that and take back the downvote. \$\endgroup\$
    – Dorian
    Commented Apr 26, 2018 at 19:11
  • \$\begingroup\$ "It is not a 5V led" I know, I just copied your phrase. I agree the reason for the lit LED on the 5V rail is obvious - it's just which path(s) are powering it which is unknown. I don't see why ESD diodes have to be involved. Regarding my answer: It seems you were looking for a specific answer and found it too long. Instead I was explaining the issues with that level-shifter, with references & alternatives, allowing for multiple "answers" to be found in what I wrote, depending on exactly which part(s) people were unclear about. Short answers are often not welcome. Anyway, that's all from me. \$\endgroup\$
    – SamGibson
    Commented Apr 26, 2018 at 19:43
  • \$\begingroup\$ @SamGibson Obviously you don't welcome any criticysm. To bad, all your other answers are concise and well documented. This was not. ESD diode might have an influence, cannot tell if the current flows through R2 or through it, that depends on the 5V load. Please undelete your answer, why rely on my opinion? \$\endgroup\$
    – Dorian
    Commented Apr 26, 2018 at 20:20

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