AMS1117 LDO voltage regulator question

For a design I would need to include an AMS1117 3.3V regulator (non-adjustable) which at times may be used without input voltage, yet receive 3.3V at its output and have the GND pin grounded.

I can't find anything in the datasheet about this; what is the board's opinion?

Edit: I cannot use a diode protection at it's output.

From the data sheet. Most regulators require this if Vout>Vin. Read the data sheet in this area.

ronsimpson said:

From the data sheet. Most regulators require this if Vout>Vin. Read the data sheet in this area.
View attachment 109906

Click to expand...

Thanks ronsimpson; now in the text with this picture this diode must be used when large output capacitors are used and when the input may be shorted to ground. Yet in my case the input would have no potential at all (unconnected): so by the same token may I then assume that this IC does not need protection with a correct output voltage (here 3.3V) present (other source) yet no potential on the input?

I don't know your part. I think it will work well.
Older parts that I have used many times; require the diode because the voltage at Vout needs to flow backwards through the part and charge up the capacitors at Vin. (Vout = 3.3V and Vin is not connected but will be pulled up to about 3.0V by current flowing backwards) The old parts could only handle (don't remember maybe) 1A for a short time.
I think your part might have a power MOSFET+diode not a transistor so it is strong against this problem. (inside the IC)

Hi ronsimpson, I did the actual test. 3.3V on output, input floating, for a while. Then regular operation (5V on input), I get output 3.3V, under load. Then 5V on output, input floating; again for a while. Again Vin 5V and under load there still is 3.3V output. I will do the test on a prototype board for a number of weeks and see what that gives.

You say Vin=5V. So you do not need the "low drop out" function.
If you want another idea; Here I added a diode (inside the feed back loop). This causes Vout=3.3V and "OUT" pin = (3.3+0.6=3.9V) The dropout voltage will be more by 0.6V. I would use a Schottky diode, or any low voltage diode that can handle the current.

I can not see your entire circuit but most likely you do not need to do this.

ronsimpson said:

You say Vin=5V. So you do not need the "low drop out" function.
If you want another idea; Here I added a diode (inside the feed back loop). This causes Vout=3.3V and "OUT" pin = (3.3+0.6=3.9V) The dropout voltage will be more by 0.6V. I would use a Schottky diode, or any low voltage diode that can handle the current.
View attachment 109907
I can not see your entire circuit but most likely you do not need to do this.

Click to expand...

Actually I do need the LDO: this circuit has to be able to be solar panel powered and still operate when the battery runs low.