Solar panel is hooked up and working!

[jwcolby54]

Originally Posted by P. Raz

Interesting. I have an old charger that uses PWM. I've always assumed the off time was to provide the sensing circuitry with a "disconnect" to check the battery voltage and the lowering of the duty cycle was to slow the charging down to avoid overheating the electrolyte. I'm not sure what you mean by throttle.

RAZ, PWM is used because the current applied to the circuit is directly proportional to the on time if the wave. Assume that the pulse width is zero (always off) - no current delivered. At 50% the current delivered is 50% of the available current. At 100% (always on) the current delivered is 100% of the available current.

Usually the width is varied based on some "algorithm". The current may be applied across a very small resistance (a shunt). The resistance causes a very small voltage across the shunt which is amplified by an op amp and then that voltage is measured. Probably the "wave" is averaged (using a capacitor or inductor) in the op amp circuit so that you get a filtered voltage which can then be measured. That voltage can be used to feed back into the PWM circuit to vary the width as required.
That control circuit is not going to be extraordinarily precise but "good enough". Turn the thing on most of the way for a bulk charge (very low battery charge), cut it back when the current into the battery starts to drop below a certain threshold, and then at some point cut it almost off for the float charge.

[Brian B-P]

Originally Posted by jwcolby54

2) "The current" is NOT being delivered. The current is being wasted as heat (power loss).
To make the math easy to understand assume that the battery is a 1 Ohm load and the solar panel is delivering 12 V. Put 12 volts across a 1 ohm load = 12 amps. I = E/R. Now add .1 ohm resistance in the wire as loss (just an example). The wire is placed in series with the original load and so the resulting load is 1.1 ohms. 12V / 1.1 Ohm = 10.909 amps. You have just "lost" .9 amps. Where did it go? To heat loss.

No, current is not lost, as heat or otherwise. There is just less current flowing, because the available voltage is not capable of pushing as much current through the larger resistance. Certainly some of the power being put out by the panel is being wasted as heat due to that resistance:

12 volts into 1 ohm load:

• 12 volts across 1 ohm means 12 amps of current flows
• 12 amps at 12 volts is 144 watts delivered

12 volts into 1 ohm load through 0.1 ohm wire resistance:

• 12 volts across 1.1 ohms of resistance means 10.9 amps of current flows
• 10.9 amps at 12 volts is 131 watts total, consisting of
  • 10.9 amps through 1 ohm is 119 watts into the battery, plus
  • the same 10.9 amps through 0.1 ohm is 12 watts wasted as heat from the wire

In practice, the panel itself is not an ideal zero-impedance voltage source. If the load has more resistance, the panel output voltage will be higher, and vice versa. In the example, the voltage would rise a bit when the resistance was added, so the delivered power would not drop so much.

The example is still a good illustration of why it is undesirable to have significant conductor resistance.

Originally Posted by jwcolby54

http://et.nmsu.edu/~etti/fall97/elec...htmlElectrical

My guess is that this was intended to be http://et.nmsu.edu/~etti/fall97/electronics/solder.html

[jwcolby54]

Well of course your math is absolutely correct.

And of course, even if you raise the voltage, you increase the current pushed through the resistance, which delivers more current to the battery but also wastes proportionately more watts in the wire. In any event, the less resistance in the wire, the more current delivered to the battery.

Someone mentioned using a 3 wire extension cord to deliver a two wire solution. If we simply short (the same) two of the three conductors at each end, then the resistance in one leg is halved - assuming equivalent gauge in all three wires. Thus the total resistance is 1.5 R instead of 2 R.

All of which shows why we try to use the right wire for the job, but as the OP pointed out, simply getting the panel in the sun over a 50' lossy extension cord will almost certainly be more efficient (deliver more current anyway) than the same panel in the shade over lossless wire.

Also I am looking at higher current fold-able panels which I certainly wouldn't want to hookup over a 16 gauge wire.