I have a 5V hobby circuit that will be externally powered from a wall wart, with way less than 1 amp. For full flexibility, I'm thinking of using a simple LM7805 regulator – but that has a 7V minimum. Since 5V supplies are becoming ubiquitous, I'd like to allow that too – but then I'd need to bypass the regulator.
What would be the impact / effect / damage if I simply provided a jumper that shorted the regulator's input to the output for semi-permanent use with a 5V supply? Use a 9-12V supply: remove the jumper. Use a 5V supply: add the jumper. I'd leave the input and output capacitors to do their thing: it won't harm the 5V case.
Would the regulator be at minimum consumption because the output was already at 5V? Or would it struggle at full load with the less-than 7V input – or worse, short output to ground?
Best Answer
Well, I decided to simply breadboard it. See below for the Summary.
Experiment
For those who are thinking of doing the same, I implemented the following schematic:
TP1
: Current measurement (mA
)TP2
: Voltage measurement (V
)LM7805
: Regulator temperature (Tr
)20R 5W
: Dissipation temperature (Td
)JP1
: WhenBatt
was9V
, the jumper was OFF.When
Batt
was4.5V
, the jumper was ON.1) 9V with
JP1
OFF:mA
= 250 mAV
= 4.99VTd
= 36CTr
= 39C2) 4.5V with
JP1
ON:mA
= 200 mA, descending to 180mA over 1 hourV
= 4.08V, descending to 3.51V over 1 hourTd
= 36CTr
= 22C (ambient)3) Back to 9V with
JP1
OFF:mA
= 250 mAV
= 4.95VTd
= 36CTr
= 39C4) 4.5V with LM7805 removed! (
JP1
ON):mA
= 180 mAV
= 3.52V (same old batteries as before)Td
= 36CTr
= N/ASummary
In short, with a 200-250mA load:
I'm going to risk it!