7805 error in electrical rules

Hi.
I made this scheme.


When I check the electrical rules I get an error on the GND pin (which should be marked as ADJ). The error is only exceeded if it is specified as PWR_FLAG. Because?
Thanks in advance.

The GND pin of a voltage reguator typically is tied to ground. So the pin type of the symbol is “power input”, requiring a connection to a “power output”. The node between the resistors is no power output, so a power flag is required.
Or you can ignore the error.

As far as I know, a 7805 is not meant to be used with a rheostat in the GND lead like that. Also why do you have 100nF and 10nF capacitors in parallel? You could do with just a 100nF one, since the other one is just 10% of it, and it’s not a critical capacitor anyway. The 470µF capacitor on the output is probably wrong too as it will prevent the regulator from responding quickly to load variations. Normally only a 10µF one is used if at all. Check the datasheet for a typical application circuit.

Using the 7805 as an adjustable regulator is quite common, AFAIK.

There are regulators that are designed to be adjusted such as the LM317. Using a 7805 this way requires calibration and you are relying on the ground pin current to be constant for the loads encountered which may depend on the internal circuitry. Does the OP really want more than 5V given that they’ve put a 5V zener across the output (which is also cutting it fine since you would use a rating somewhat above 5V to protect the load).

It is a 6.4V zener… stupid manufacturers numbering system.
The LED may be a bit dull with a 1K resistor.
Not sure of the purpose of D4.

Excerpt from a 7800 Series datasheet…

If it’s a red LED you’ll have 3+V across 1k which will give you 3+mA. Modern LEDs are bright enough at that current.

The purpose of D1 would be to dump any overvoltage at the output to the input voltage. Not necessary here as there is the Zener.

@jmk Sorry, I replied to the wrong guy…

Even 1 mA would suffice. For red, that is.

I still wouldn’t use that circuit. as it depends on Io, and you still have to calibrate it. If you want a variable voltage source, a variable regulator doesn’t cost that much more. In fact I would prefer the LM1117 series, which includes a variable model, as that has a lower drop voltage than the 78/79 series. That series is ancient, I remember using them in the 70s.

Whatever you choose…
Btw, unless a VERY clean voltage is required, I would prefer a buck converter. That’s what I do in all my MCU supplies.

Yeah, if I had a buck for every phone charger in my junk^Wspares box, I could have a nice dinner. :rofl:

The joke is moderately funny :slight_smile: , anyways buck converter is the correct term.

I know, I know: he who is dead earnest is dead already.

Right guy, wrong diode. :slightly_smiling_face:
D1 dumps back to the input.

Lool, all wrong.

@rmoggia D4 is reverse polarity protection.

@rmoggia And D4 should be on the other side of the fuse or a reverse polarity will eventually vaporize the diode. Maybe something more sturdy than a 1N4148. SS34 maybe or similar.

I’d probably use a more robust diode for both D1 & D4.

I have seen reverse schottky diodes on balanced psu outputs to prevent nasty start up latchups when one supply gets pulled into reverse.
The diode across the regulator is to prevent the IC getting destroyed by an input short.
1N4004 is a common choice