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The PCB circuit I'm using expects 5V input voltage which it then converts to 3.3V using an AMS1117-3.3 voltage regulator. The power supply I'm using is 3.3V. What happens if I feed the 3.3V to the input of the 5 to 3.3 voltage regulator?

The input of the voltage regulator has an easily accessible jumper header. Whereas bypassing the voltage regulator would require the use of bodge wires. Can I get by with just feeding the 3.3V into the regulator? What are possible drawbacks if it even works? What would happen when the PSU voltage drops below 3.3V (as I doubt it will provide a precise and constant 3.3V)?

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    \$\begingroup\$ Dropout Voltage (V IN - V OUT) 1.3V means input must be higher by this much \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 29 '17 at 18:35
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The critical parameter is drop-out voltage, which I'll call \$V_{do}\$, listed in your datasheet on page 3.

The output voltage will generally not be more than \$V_{in}-V_{do}\$. (Obviously this breaks down when \$V_{in}\$ is very low. If \$V_{in}<V_{do}\$, for example, all bets are off)

For the AMS1117-3.3, the drop-out voltage is listed as 1.1 (typ) to 1.3 (max) volts. That means to be sure of getting 3.3 V out, you need to provide at least 4.6 V in.

And if you provide only 3.3 V in, you shouldn't expect more than 2 V out. In fact, that far below the nominal input voltage you shouldn't count on the listed drop-out voltage applying, so you can't even be sure of getting 2 V out.

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Read the datasheet! The dropout voltage is clearly specified as 1.3 V (page 3, fourth row, "Dropout Voltage"). Since you are using the 3.3 V output version, the input voltage must be at least 4.6 V for any of the other specs in the datasheet to apply.

So the correct answer to what will happen when you supply 3.3 V in is "You don't know". All other promises made in the datasheet are null and void since you are violating the minimum dropout voltage requirement.

Probably, the output will track the input minus some voltage, at least for input voltages that aren't "really low". Many chips that are meant to run from 3.3 V need 3.0 to 3.6 V. It is unlikely, and certainly not guaranteed, that the output voltage will be 3.0 V or more with only 3.3 V in.

Overall, bad idea. Solder a small wire across the input and output of the regulator. Or, remove it altogether and jumper across its input to output. Or, get a 5 V supply.

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In the Electrical Characteristics table in the datasheet, there is a line for "Dropout Voltage", with a typical value of 1.1 volts. That is the minimum voltage difference between input and output of the regulator at which the regulator is guaranteed to work.

So, for a 3.3 volt regulator, the input voltage must be above 4.4 volts for correct operation. I would expect that if your input voltage is only 3 .3 volts, the regulator output will be around 2.3 volts.

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It will likely not pass through 3.3V, as the are internal components that will use up some of that voltage. Even a simple diode drop will take up some of it. The regulator may also oscillate.

But this is simple enough to test. There would be little harm in attempting it and measuring the result yourself.

Or just avoid the issue and solder on a pass through wire. 10 second job really.

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