Buck converter input drop

[ConfusedNewbie]

Sorry to be dim here but,
I have a PV supply going through a cheap, off the shelf buck converter (and on to a further circuit).
I get around 23V from the solar panels unconnected but as soon as I connect them to the buck converter, the input drops to 2.3V (so I can then only dream of the 15V I'm after).
At around 25V all is good again.
The PVs are only 0.5W (and I have them in series to achieve the 23 to 25V) but I'm wondering if I should up the power/drop voltage and split them into pairs in parallel.

Again, sorry to be dim.

[taste_tester]

Post at least a diagram of exactly what is connected where and how, or take some photos if that's not possible for any reason. We don't have enough information to give you any insight whatsoever at this time

[golden_labels]

Aside from providing what taste_tester asked about, provide some explanation of what kind of performance you expect. Because what you describe resembles a difference between open-circuit voltage and voltage under load.

500mW output is probably the power those cells could output under perfect conditions: maximum illumination they can take, at straight angle towards them, uniformly over the entire surface. Realistically that will not be reached even in full sun. And you did not say anything about the lighting conditions you tested the cells in.

Another thing is that a 500mW source, even ignoring all possible losses, can sustain 23V only at loads below 22mA.

[ConfusedNewbie]

I tried to add pics but for some reason it won't upload them.
Load is just the buck converter (the rest of the circuit is physically switched off). The only load I can see is the initial cap on the buck BUT still doesn't explain the drop in PD - the inputs to the buck shouldn't make any difference to the PD (at those points the +ve rail and -ve rail surely should show the full PD load or not - if I measure the PD at the terminals of a battery, what PD is shown? Doesn't matter if there's a load further down the circuit - I admit, over time that PD will drop in a battery but this isn't a battery - my circuit doesn't deplete the sun.

[ConfusedNewbie]

Not to worry, I'm going to take out the buck, send the full PVC PD through the cirsuit (I have a zener/Tip122 regulator later in the circuit which can cope with 24V. Perhaps the Buck was overkill in the unlikely event we had mega sunshine - not worried about power too much (it drip charges a Pb/H+ battery so V more important than W).

[golden_labels]

The reason you can’t upload pictures should be indicated when the forum refused to accept them. Did you read the message? Why did you not quote it, so you could get assistance? In this thread you are doing everything to discourage people from helping you.

The load is the buck converter itself. Not the capacitor specifically, but the converter. Possibly including its obligatory minimum load, which may be provided internally.

Charging a lead-acid battery with milliamps will be hard, considering that self-discharge is likely to be an order of magnitude larger. May depends on the exact scenario, but we still miss the actual picture.

The rest I can’t address, as it’s sprinkled with cryptic abbreviations. No idea why you used them, but — assuming a common trait  — if you thought it will make you look more “pro”, it has a strong opposite effect.

[ConfusedNewbie]

That's a bit rude! uploading the pics as attachments offered no explanation of failure (it just went to a new message everytime). Abbreviations, PVC (photo voltaic cells) PD (potential difference) W (Watts).
That said, clearly I entered the wrong place for advice and for that I appoligise (it did say "beginners" in the topic but one clearly needs to be more than that).

Problem solved BTW (linear regulator).

[ConfusedNewbie]

OK. Let's keep this REALLY simple for golden_labels' sake.

Photovolatic cells. Measure the unloaded potential difference (voltage) between + and - AT THE CELLS.
Now, add a load and measure the PD (voltage) AT THE CELLS (not anywhere else in the circuit, same place as before, between the + and - of the cells).

What is the difference? Surely it should be the same PD (voltage) - my input should remain the same.

I can't make this clearer!

[MrAl]

Post at least a diagram of exactly what is connected where and how, or take some photos if that's not possible for any reason. We don't have enough information to give you any insight whatsoever at this time

Hi,

If i understand you right, when the output of the panel is lower the buck does not work but when the panel output voltage is a little higher the buck circuit works ok.

If that is true then i would guess that you are working too close to the max power point of the panel/array.
The "max power point" is the operating point of the solar array that provides the maximum power possible.  It varies with insolation level so as the sun goes down the max power point moves and it then indicates that the max power you can get is less than with full sunlight.
Now before the max power point is reached the voltage output would be somewhat constant with any current within the range of the array for that max power point, but unfortunately once you draw enough current to get to the max power point the voltage drops off fairly fast.  So you can have 25v out at say 100ma and then as you go to 110ma the voltage drops down to 15v or even less.  Also, if you have 25v out at say 100ma and the insolation level decreases, the max power point moves to a point of less current and so now you draw 100ma but the voltage falls down lower than 25v because now the current is beyond the max power point.

There are circuits that help you maintain the max power point for any insolation level but if your circuit needs 100ma and the max power point is at a point where the current is only 50ma, the voltage still falls very low.
If that happens on a regular basis, then you need a larger solar array.

It does sound like you need a larger array unless you can limit the use of the circuit to times when the sun is brightest.