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I've got two solar panels set up in series as shown on the below schema. I connect positive on panel A to positive on battery, and negative on panel B to negative on battery. The two panels produce 9V each, and 18V in series on a good sunny day. That's what my multimeter says.

enter image description here

If I disconnect cabling between panels and battery, my multimeter shows that the panels produce about 15V (a bit cloudy today). However, when I connect the cables between panels and battery, my multimeter shows about 12.2V. I believe this is the battery voltage.

Why is it that when I connect my multimeter at point A and B it shows 12.2V (when solar panels are connected to the battery). As you can see on the schema, there's a diode in front of point B. Shouldn't that diode make sure that current from the battery not flows back to the panels?

If this is how it is supposed to work, how do I check the voltage my panels produce when the panels are connected to the battery.

enter image description here

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    \$\begingroup\$ You CAN get a reading of Vpanel when connected to the battery - it is Vbattery + 1 diode drop when charging. ie the panel voltage IS ~= the battery voltage. What you are trying to ask for (probably) is what the panel OPEN CIRCUIT voltage would be if you disconnected it from the battery. That is something like asking how fast a car engine would rev at this throttle setting if the trailer we are towing did not have an elephant in it. ie the load (or the elephant) is modifying the real voltage and the one you are asking about does not exist in this set of conditions.... \$\endgroup\$
    – Russell McMahon ♦
    Commented May 24, 2014 at 15:49
  • \$\begingroup\$ ... You can infer what Voc would be from Vloaded and current . BUT Voc is not very important in practice -as long as it is high enough to allow maximum current to flow to the load. \$\endgroup\$
    – Russell McMahon ♦
    Commented May 24, 2014 at 15:50

5 Answers 5

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When you connect the voltmeter between points A and B, you're connecting it directly across the battery, so if the solar panels put out less than the battery voltage the diode will be reverse biased and will, in effect, disconnect the battery from discharging into the solar panels and you'll be measuring only the battery voltage.

On the other hand, if the voltage from the solar panels is higher than the battery voltage and the drop across the diode, the solar panels will force current into the battery, charging it.

However, because the impedance of the battery is so low, it'll drag the voltage of the solar panels down close to the battery voltage, even though the solar panels will be continually pumping current into the battery.

As time goes by and the battery becomes more fully charged, you'll notice that its voltage will rise, but never to your solar panels' open-circuit full-sun value, because the battery chemistry won't allow it.

The proper way to monitor your battery's charging is to measure the voltage across it and the current into it, and never let either rise above the manufacturer's recommendations.

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  • \$\begingroup\$ Thanks for an easy to understand explanation. I'm a total beginners as this, but I get what you're saying. To check whether the battery is charging at all, could I somehow measure the current (before or after the diode)? \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 13:37
  • \$\begingroup\$ Current in a series circuit is the same anywherre in the circuit, so all that' \$\endgroup\$
    – EM Fields
    Commented May 24, 2014 at 14:30
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    \$\begingroup\$ Aarghh... Current in a series circuit is the same anywhere in the circuit, so all that's necessary to measure current is to open the circuit - anywhere - and connect an ammeter in series with the break. \$\endgroup\$
    – EM Fields
    Commented May 24, 2014 at 14:34
  • \$\begingroup\$ Actually, you don't even need to open the circuit. There are such things as clamp-on current meters, both AC and DC (The DC sensors are more expensive). A random example of a company which makes this sort of thing: automationdirect.com/adc/Overview/Catalog/Sensors_-z-Encoders/Current_Sensors_(AC-a-_DC) \$\endgroup\$
    – WhatRoughBeast
    Commented May 24, 2014 at 18:47
  • \$\begingroup\$ Yes, of course, but my take was based on that, - because of the OP's question - he was much more likely to have a DMM on hand than a clamp-on ammeter. Your comment, however, apprised him of another way to go; a good thing. :-) \$\endgroup\$
    – EM Fields
    Commented May 29, 2014 at 0:31
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If you connect the voltmeter leads to the terminals of the battery, you will read the voltage across the battery. There really isn't much more than can be said about that.

The battery is essentially modelled as an ideal voltage source in series with a relatively small resistance.

The solar panels, on the other hand, are more like a current source in parallel with a relatively large resistance.

When the solar panels are connected across the battery, a simple model of the circuit is

schematic

simulate this circuit – Schematic created using CircuitLab

The current \$I_{sc}\$ is the short circuit current produced (for a given illumination) by the panels when the panels are short circuited.

The voltage \$V_{oc}\$ is the open circuit voltage produced by the battery when the battery is open circuited.

Elementary circuit analysis tells us that the voltage across the battery terminals with the solar panel connected is

$$V_{bat} = V_{oc}\frac{R_p}{R_p + R_b} + I_{sc}R_p||R_b$$

Since, typically, the resistance \$R_b\$ is much less than the resistance \$R_p\$, the voltage across the battery is approximately

$$V_{bat} = V_{oc} + I_{sc}R_b \approx V_{oc}$$

In other words, unless the panels are providing a relatively large current and / or the internal resistance of the battery is relatively large, you will measure approximately the open circuit battery voltage with the solar panel connected.

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  • \$\begingroup\$ Alright, I just thought the Schottky diode would somehow "isolate" the panels from the battery and let me measure the panel's voltage "behind" the diode. However, I realise that the diode won't affect the voltage but only block the current from the battery. In my case, how would I go about to get a voltage reading from the panels when connected to the battery? \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 13:16
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    \$\begingroup\$ @sbrattla, if one connects two circuit elements in parallel, the voltage across each is identical. When you connect the solar panel across the battery, the panel and battery are in parallel and, thus, the panel voltage and battery voltage are identical; you are measuring the voltage across the panels. The panels are not voltage sources, they're imperfect current sources. When you measure the panel voltage when disconnected, you're measuring the open-circuit voltage given by \$I_{sc}R_p\$. \$\endgroup\$
    – Alfred Centauri
    Commented May 24, 2014 at 13:23
  • \$\begingroup\$ Alright, so bottom line is that I can't get a reading from my panels when connected with the battery. \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 13:32
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    \$\begingroup\$ @sbrattla, a reading of what? What are you trying to read from the panels? The power? The current? The voltage 'from' the panels is, as explained above, the same as the voltage across the battery. \$\endgroup\$
    – Alfred Centauri
    Commented May 24, 2014 at 14:37
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Solar panels have quite high output resistance and therefore any appreciable load will lower the output voltage of the panels. A lead-acid battery, on the other hand can supply a load of several amps without hardly changing its output voltage. This is largely the same when charging current goes into the battery so, in effect the battery will dominate and the voltage you'll read across it will be largely the same whether solar cells are connected or not.

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  • \$\begingroup\$ I've read that I should use a voltage regulator to make sure that I don't provide more than about 13.8V (trickle charge) from the solar panels to the battery. However, with what you're saying does that imply I don't have to use a voltage regulator as the battery's voltage will dominate? \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 13:05
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    \$\begingroup\$ Over time, as you pump more charge into the lead-acid battery, the battery's terminal voltage will rise and there is a recommended limit so, using a voltage regulator is a good idea generally. \$\endgroup\$
    – Andy aka
    Commented May 24, 2014 at 13:09
  • \$\begingroup\$ Alright, so it would be a good idea to put a voltage regulator in between the panels and the battery to make sure i "cap" the voltage at, say, 13.8V which will be a safe voltage? \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 13:11
  • \$\begingroup\$ I've definitely seen voltage regulators being used. \$\endgroup\$
    – Andy aka
    Commented May 24, 2014 at 13:30
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If you put a momentary, normally-closed switch in the loop to the battery side of where you connect the voltmeter, you can read the open-circuit panel voltage while you hold the button.

Edit:

enter image description here

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  • \$\begingroup\$ Could you elaborate a little on this? How would I do that? \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 14:18
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    \$\begingroup\$ (See edited diagram). \$\endgroup\$
    – JRobert
    Commented May 24, 2014 at 14:59
  • \$\begingroup\$ Ah, I see. Thanks, i'll try to implement that to be able to get a reading on the panels. \$\endgroup\$
    – sbrattla
    Commented May 24, 2014 at 15:27
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I've read that I should use a voltage regulator to make sure that I don't provide more than about 13.8V (trickle charge) from the solar panels to the battery

Depends on the panel... you don't show them in your photos, but I assume you are using the smaller ~12V panels, and not the more common 60-cell ~35V ones?

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  • \$\begingroup\$ 2 x 9v, so 18v in total \$\endgroup\$
    – sbrattla
    Commented Jan 18, 2019 at 18:15
  • \$\begingroup\$ Yeah you should be fine then. As the others have noted, the battery will basically overwhelm the panels. It does mean you only charge in bright sunlight, but that's not worth fixing. \$\endgroup\$
    – Maury Markowitz
    Commented Jan 18, 2019 at 21:23

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