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I have LDFM33PUR device in a circuit and it is responsible for generating stable 3.3V (for Micro-controller) from 5V input. I want not to use LDFM33PUR and rest of the circuit and just externally power the Micro-controller, which means that 3.3V would be applied to V_out of the LDM33PUR while its V_in and EN signals are 0 (LDM33PUR will be shut-off).

Do you think that can damage the device?

I am not sure since the data in LDFM33PUR datasheet is only specified for the device turned on, I suppose.

EDIT: What if V_in and EN signals are completely disconnected (only around 30uF capacitance will be present on V_in pin)?

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  • \$\begingroup\$ It depends on what is on the input side of the regulator. This regulator can't have the output higher than 0.3V more than VIN, which is approximately same with other regulators - some regulators can handle the situation with built-in diodes or relevant structures, some regulators need external diodes to handle the situation. \$\endgroup\$
    – Justme
    Commented Jun 14, 2021 at 20:52
  • \$\begingroup\$ Does it avoid the not to exceed max on interface signals while powered down? \$\endgroup\$
    – D.A.S.
    Commented Jun 15, 2021 at 1:04
  • \$\begingroup\$ @Justme I agree, the question is also what Tony is asking - also keeping in mind this particular device. Its interesting if it can handle it in non-working conditions (also when the V_in and EN are not connected at all and only some capacitance present- I edited the question to include that part) \$\endgroup\$
    – Vaso
    Commented Jun 15, 2021 at 14:49

3 Answers 3

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I want not to use LDFM33PUR and rest of the circuit and just externally power the Micro-controller

This won't work. The LDFM33 has a MOSFET between input and output, whose body diode will conduct when the output voltage is ~0.6 V higher than the input voltage. So the 'rest of the circuit' will be powered, just at a lower voltage.

enter image description here

Furthermore this operation is outside the absolute maximum ratings, which states that Vout must not go more than 0.3 V above Vin. It can probably handle this for a short time when the circuit is powering down, but if externally powered via the output and the 'rest of the circuit' draws a high current it could burn out the regulator.

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  • \$\begingroup\$ Makes sense, thanks for explanation. And what if the V_in and EN will not be connected at all? (One can argue that then there wont be any current flow - so it will be safe to use in that case?) \$\endgroup\$
    – Vaso
    Commented Jun 15, 2021 at 6:34
  • \$\begingroup\$ If there is no circuit on the input that can draw current then it will probably be safe. An example of this would be a power supply that goes directly into the regulator without powering anything else on the board, that is switched off at the board or unplugged from it. A power supply that is switched off at the wall but still connected could damage the regulator if it has a large smoothing capacitor in it. Inserting a diode in series with the power input will prevent this, as well protecting it from reverse polarity. \$\endgroup\$
    – Bruce Abbott
    Commented Jun 15, 2021 at 7:21
  • \$\begingroup\$ I will completely remove the original power supply, so there wont be any power coming at all to the input side of the regulator (V_in and EN wont be connected). I unfortunately didnt place diode in the circuit for protection. Only thing that can draw a bit of current is capacitor on the input of the regulator - 1uf, there is also 22uf capacitor at the output of the previous driver. \$\endgroup\$
    – Vaso
    Commented Jun 15, 2021 at 11:00
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Vout must not exceed 0.3 of Vin , so if Vin is 0 that will not work.

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I just wanted to see if this was possible:

Diode-ed Linear Regulator

If a diode is placed after the regulator (D1 to prevent it from being reverse-powered), then the output voltage will be lower due to the diode-drop. This could be overcome by also placing a diode on it's ground pin, which forces it to output a little higher than designed (Vout + D2 drop.) This over-voltage from D2, combined with the loss from D1, can be very close to 3.3v.

Unfortunately this presents a host of problems. The current through the ground connection is typically very small, so a different diode would need to be used there. The other main problem is the output will contain a temperature dependence. The forward voltage of D2 will change with temperature, introducing output voltage change, perhaps significantly. A tertiary concern is that all diodes are non-linear, resulting in less accuracy across the regulation and load range.

So I wouldn't recommend doing this. Might be something interesting to play around with on a breadboard though, for a non-mission-critical application.

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  • \$\begingroup\$ Thanks, interesting insights. One solution for that problems can be to place diode on the input pin of the regulator (if one is sure that the input voltage will not be smaller then around 4V - which will ensure that the regulator will be able to output 3,3V) \$\endgroup\$
    – Vaso
    Commented Jun 15, 2021 at 14:41

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