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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?

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  • \$\begingroup\$ You're at risk of human frailty. Forgetting that the jumper is "in", and applying the 9-12v supply is the dangerous scenario. \$\endgroup\$ – glen_geek Oct 16 '16 at 14:47
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Well, I decided to simply breadboard it. See below for the Summary.

I fully recognise @glen_geek's comment regarding overlooking the jumper - but I figure that's my lookout!
Yes, I realise that this is a single experiment and not to trust this empirical result. However...

Experiment

For those who are thinking of doing the same, I implemented the following schematic: LM7805 schematic

  • TP1: Current measurement (mA)
  • TP2: Voltage measurement (V)
  • LM7805: Regulator temperature (Tr)
  • 20R 5W: Dissipation temperature (Td)
  • JP1: When Batt was 9V, the jumper was OFF.
    When Batt was 4.5V, the jumper was ON.

1) 9V with JP1 OFF:
mA = 250 mA
V = 4.99V
Td = 36C
Tr = 39C

2) 4.5V with JP1 ON:
mA = 200 mA, descending to 180mA over 1 hour
V = 4.08V, descending to 3.51V over 1 hour
Td = 36C
Tr = 22C (ambient)

3) Back to 9V with JP1 OFF:
mA = 250 mA
V = 4.95V
Td = 36C
Tr = 39C

4) 4.5V with LM7805 removed! (JP1 ON):
mA = 180 mA
V = 3.52V (same old batteries as before)
Td = 36C
Tr = N/A

Summary

In short, with a 200-250mA load:

  • The regulator got hot when I expected it to;
  • The regulator was cold when it was jumpered over;
  • The source current didn't change regardless of voltage supply;
  • The regulator did its job again after being abused for an hour;
  • Removing the regulator did nothing for the low-voltage scenario.

I'm going to risk it!

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  • \$\begingroup\$ By taking a quick look at Functional Block Diagram on the datasheet. I predict that there will be a maximum voltage with this configuration. Perhaps the internal zener diode D1 will blow up or something. If my guess was right, that'll probably around 35 Volt, which is the absolute maximum DC input voltage rating. With the jumper inserted and a proper load, try to observe the regulator while increasing the input voltage towards 35 Volt, towards 40 Volt slowly, and then dozens and hundreds volts of DC. \$\endgroup\$ – Unknown123 Mar 30 at 23:54
  • \$\begingroup\$ If my guess was right again, then probably the ground pin which is still connected to the ground circuit were the culprit. Thus, The configuration given by @JIm Dearden answer and Rohat Kılıç comment is better than this, since isolating the ground pin would also require minimum two jumper or switch. \$\endgroup\$ – Unknown123 Mar 31 at 0:06
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Better to isolate the 7805 altogether using two links.

enter image description here

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  • \$\begingroup\$ Nice symmetry! Alas I have room for exactly one jumper pair: not four as you've shown. \$\endgroup\$ – John Burger Oct 16 '16 at 16:27
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    \$\begingroup\$ A DPDT switch can be a nice alternative to jumpers. \$\endgroup\$ – Rohat Kılıç Oct 17 '16 at 11:39

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