# Charging a battery using solar panels, and IV curve of the solar panel

I am aware that a solar panel has a fixed IV curve, and the load resistance can move the operating point of the solar panel along the IV curve. If a solar panel was used to directly charge a battery without MPPT(I know this is bad practice cuz we need charge controllers, BMS and voltage regulators, but IF!), where would the operating point of the solar panel be? Here, would the internal resistance of the battery be the load resistance? But the internal resistance of battery are very low, in 0-100 milliohms (for example for phone batteries).

Would this mean that the circuit (battery and solar panel connected) would be like a short circuit? Then, would that mean that the voltage of the solar panel is close to 0V and current will be close to the short circuit current, resulting in almost 0 power? If that is the case, then wouldn't adding a 5V buck boost converter not help charging the battery since despite the voltage being 5V (if the input voltage is higher than minimum input voltage), the current will be so low?

If my way of thinking is correct, then how are there so many tutorials on solar chargers which state using an appropriate power solar panel, 5V buck boost converter and a USB port is enough for charging a phone?

Please correct me!

• There are solar panels designed to be directly connected to car batteries to trickle charge them with no mppt. They don't cause the battery voltage to be zero either. Commented Jun 24 at 8:47
• The charge controller is built into the phone. It just needs a 5 V supply to operate from, which the buck/boost controller provides. Commented Jun 24 at 9:30

## 2 Answers

Almost all batteries act as a voltage source/sink, with (as you state in your question) very low internal resistance.

A plain solar PV panel acts as a current source over most of its IV curve, its output voltage only being the limiting factor at light loads or open circuit.

That means that a direct connection between the two is very simple to analyse. Assuming that the O/C panel voltage is higher than the battery voltage, the battery will pull the panel voltage down to its present charging voltage, and the panel will charge the battery at its output current, which will be light flux dependent.

If it's a tame battery chemistry like lead, then a shunt float voltage regulator is all that needs to be added to shunt away surplus current when the battery is charged. Lithium needs a proper charge controller of course.

The above assumes that the panel is driving the battery directly. If the panel is instead driving some sort of battery charger, then all bets are off. The battery charger will have its own programmable behaviour, and if it doesn't like the IV curve of the panel, it may shut down, or oscillate between drawing current and not.

• What do you mean by "its output voltage only being the limiting factor at light loads or open circuit."? Wouldn't the output voltage be lowest at short circuit and low load resistance? Commented Jun 24 at 10:13
• I mean that on heavy loads, the load determines the panel output voltage, by dragging the panel's voltage down until the lad current matches the panel's current. It's only when the load doesn't take much current that the panel voltage rises to its open circuit value. Commented Jun 24 at 10:53
• @siyunlee There's a bit of circularity in the definitions of 'heavy' and 'light' loads. A light load allows the panel voltage to be about the same as the unloaded value, a heavy load pulls the panel voltage down significantly. Commented Jun 24 at 12:23
• So just to clarify, a battery is a high load because it draws a lot of current to be charged from the solar panel. At high load, the current from the solar panel remains constant, and therefore the load determines the panel output voltage, which in this case would be the charging voltage of the battery? Commented Jun 25 at 4:46
• @siyunlee yes, you have it Commented Jun 25 at 8:07

But the internal resistance of battery are very low, in 0-100 milliohms (for example for phone batteries).

Would this mean that the circuit (battery and solar panel connected) would be like a short circuit? Then, would that mean that the voltage of the solar panel is close to 0V

No - batteries retain their terminal voltage unless fully discharged, and most phone batteries can't be recovered from that level of discharge. So the voltage ends up not really going below 3.6V, and any analysis should be done on the difference between that voltage and the applied voltage.

But! In your example:

how are there so many tutorials on solar chargers which state using an appropriate power solar panel, 5V buck boost converter and a USB port is enough for charging a phone?

(if there are so many tutorials, please link an example!)

That's a completely different scenario. You've not connected the panel to the battery. You've connected it to the boost converter, and the boost converter to the charger in the phone (which itself looks like a buck converter most of the time).

You've got two different fairly sophisticated bits of circuitry in there. Both of which will be running control algorithms based on the voltage and current passing through them. I suspect what happens is that most people doing this simply overpower it a bit; the input impedance of the boost converter will settle at a value that is not zero, the input voltage is nonzero, and power transfer happens as expected.

If underpowered, what you might have is a situation where capacitors in either the boost or buck charge until the controller decides the voltage is adequate, charging commences, the stored energy is drained, the controller "browns out" and stops charging, and the cycle continues. Inefficient and the phone will probably complain, but better than nothing.

• So does the voltage produced by the solar panel never go below 3.6V when charging a battery? Commented Jun 24 at 10:42
• If you connect them directly (NOT as described in your scenario), then no. Commented Jun 24 at 12:18