My solar power system contains a lead-acid battery but as soon as I use the inverter to power some load, the voltage drops instantly by 1 volt.

Why does this happen? And is it proportional to the load (bigger load = bigger voltage drop)?

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    \$\begingroup\$ Did you try to search for this information on your own? For example reading electronics.stackexchange.com/questions/9183/… \$\endgroup\$ – pipe Jul 28 at 11:11
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    \$\begingroup\$ Thank you but I already checked both before posting : no and no. Check "Andy aka" post for an amazing answer. \$\endgroup\$ – bob dylan Jul 28 at 11:33
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    \$\begingroup\$ Just to weigh in on that, it's not the first time I've seen that in stack "Electrical". Often people are voting to close / put on hold without reading or even knowing anything about the subject of the question. (Just copy pasting the question in their search bar ?) Maybe it's a recurrent problem that should be addressed. \$\endgroup\$ – bob dylan Jul 28 at 11:55
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    \$\begingroup\$ I’ve nominated for a re open \$\endgroup\$ – Andy aka Jul 28 at 14:59
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    \$\begingroup\$ So my question is a duplicate of a closed question about Lipo batteries with no accepted answer. Nice ! \$\endgroup\$ – bob dylan Jul 29 at 11:57

Take a look at this graph from here: -

enter image description here

From All About Batteries, Part 3: Lead-Acid Batteries.

It's a typical 12 volt lead-acid battery discharge characteristic and it shows the initial drop from about 13 volts to around 12 volts occuring in the first minute of a load being applied. Thereafter, the discharge rate doesn't unduly affect the output voltage level until the battery gets quite depleted of stored energy.

This site explains in detail why that initial drop of terminal voltage is steep compared to the much slower drop in voltage that occurs afterwards: -

When a current is being drawn from the battery, the sudden drop is due to the internal resistance of the cell, the formation of more sulphate, and the abstracting of the acid from the electrolyte which fills the pores of the plate. The density of this acid is high just before the discharge is begun. It is diluted rapidly at first, but a balanced condition is reached between the density of the acid in the plates and in the main body of the electrolyte, the acid supply in the plates being maintained at a lowered density by fresh acid flowing into them from the main body of electrolyte. After the initial drop, the voltage decreases more slowly, the rate of decrease depending on the amount of current drawn from the battery.

  • \$\begingroup\$ Amazing. Thanks ! \$\endgroup\$ – bob dylan Jul 28 at 11:34
  • \$\begingroup\$ An additional question... Do you know if this voltage is proportional to the Ah of the battery or unrelated ? For example if I put several batteries in parallel will it change the voltage drop ? \$\endgroup\$ – bob dylan Jul 29 at 11:55
  • \$\begingroup\$ @bobdylan Sorry but I don't know that answer. You might have to go on battery supplier's websites and make comparisons. \$\endgroup\$ – Andy aka Jul 29 at 12:22
  • \$\begingroup\$ It is, according to the answer of electronics.stackexchange.com/questions/57508/… . If you can't do a full capacity test, simply measure the instantaneous voltage drop with a known large load, and again graph the results. \$\endgroup\$ – bob dylan Jul 29 at 14:08

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