Is this a normal behaviour of a diode?

So I have a 18650 charger circuit with solar panel, and I used a 1n5819 schottky diode. However I can measure a 7.-ish voltage (there is a little drop through the step down converter I guess) on the marked point.

But when I am trying to put a load there (a fan) the voltage almost disappeares. I have also tested the diode with a multimeter and it worked properly.

Is this a normal behaviour of a diode?

You measure voltage between 2 different places, not in one place as per your picture. Where is your other meter probe connected?

PerryBebbington:
You measure voltage between 2 different places, not in one place as per your picture. Where is your other meter probe connected?

It's on the ground of course.

Where is the fan connected? I assume it's supposed to be connected to the batteries?

Are the batteries fully charged?

Does the fan run when just directly-connected to the batteries?

Or is the solar supposed to be able to drive the fan?

What's the voltage & current rating on the fan?

Have you got any specs on the charger module?

Have you got any specs on the solar panel?

Is this a normal behaviour of a diode?

When it's charging and current is flowing from the solar panel to the charger you should get about a 1/4V "forward voltage drop" across a Schottky diode. That is, the solar panel side of the diode should read slightly higher than the charger-side.

That is not behaviour of the diode, that is the fan drawing more current from the solar panel than it can supply, so the voltage at the marked point drops.

I think I could not explain what I meant, so I simplified the circuit. So if there is no resistor I can still measure voltage backwards of the diode. But when I put load there (resistor or anything), that voltage disappeares.
Is this normal from a diode?

tosoki_tibor:
It's on the ground of course.

There's never any "of course".

Oh! Leakage current.
What kind of diode? Silicon junction diodes have very low leakage and I would not expect to see anything much, Schottky diodes maybe something as they have higher leakage.

[EDIT]
I’ve just tried this for you with a Schottky diode and get pretty much the PSU voltage on the DMM. With a silicon junction diode I get almost nothing. I’m not surprised, Schottky diodes are leaky. The reason it goes away when you connect a resistor is that the leakage is tiny, as soon as you connect a resistor there isn’t enough current available to develop any appreciable voltage across the resistor, so you see little or nothing on the DMM.

++Karma; // For experimenting and trying to learn :slight_smile:

It is a 1n5819 schottky diode. I just want to protect my solar panel from current going back to it (i dont how how it's called), and I chosed this as I have read that schottky diodes have less voltage drop. I compared it to a 1n4007 and it seems to be true.

Schottky diodes have a much lower forward voltage drop than silicon diodes, they are also a bit leaky. Check the data sheet for details of how much a particular diode leaks (look for the reverse bias current).

I don't know if solar panels need protecting like that, solar panels are basically diodes, so maybe they don't. However, I don't know, consult the data sheet. As the title of this question does not hint at a solar panel maybe to get the interest of someone with the correct knowledge you should edit the title to mention solar panels.

A diode is required to prevent the solar panel from discharging the batteries when it is dark. To see why, take a look at the equivalent circuit of a solar cell:

You don't gain much by using a Schottky diode instead of an ordinary silicon diode.

PerryBebbington:
I don't know if solar panels need protecting like that, solar panels are basically diodes, so maybe they don't.

It is not a matter of "protecting" the solar panel, it is essential to protect the battery from discharge!

Solar panels are indeed diodes - whose forward voltage goes up during illumination so that they pass current in the reverse direction (remember that is what a photodiode does!) in order to charge the battery. But when they are not illuminated, the voltage drops and without the diode, the battery discharges back through the panel. :roll_eyes:

Of course, that is why you use an actual charge controller which incorporates the diode or "active diode". :grinning:

Thanks Paul, never used one.