0
\$\begingroup\$

i try to make simple wire heating using solar panel. Problem is the wire not heating at all. I assembly solar cells 52mm*26mm in parallel 11.38 volt, amperage should be at least 2 or 3 amps.

Some experiments that i have tried using Solar Panel assembly as power source, 11.38v enter image description here 775Motor: 0.09V, 0.11A, motor not running. enter image description here enter image description here ToyMotor: 0.34V, 0.1A, motot not running. enter image description here enter image description here SMD LED: 2.83V, 0.1A enter image description here enter image description here Cree LED: 2.56V, 0.14A enter image description here enter image description here Single 30AWG SS304: 0.74V, 0.1A, 7.1Ohm, not heating. enter image description here enter image description here enter image description here Multiple 30AWG SS304: 0.05V, 0.11A, 1,4Ohm, not heating. enter image description here enter image description here enter image description here

|improve this question
\$\endgroup\$
  • 2
    \$\begingroup\$ That's a lot of pictures. What is the short circuit current? You say "amperage should be 2 to 3 amps", but why? It looks like the short circuit current is probably ~0.1A, and therefore the total wattage is much less than you expect. If you want to use the panel as an efficient power source, you need a maximum power point tracking converter. \$\endgroup\$ – uint128_t Sep 18 '17 at 4:08
  • \$\begingroup\$ I forgot to measure it. \$\endgroup\$ – Jonni Chan Sep 18 '17 at 4:14
  • \$\begingroup\$ Working right now, will measure it after work. Thanks for your fast respon. \$\endgroup\$ – Jonni Chan Sep 18 '17 at 4:15
  • 1
    \$\begingroup\$ It looks to me like this is sitting on floor tile. That suggests to me room light, not full sun. If so the cells won't be putting out anything like their rated power. \$\endgroup\$ – Loren Pechtel Sep 18 '17 at 4:25
  • 1
    \$\begingroup\$ "amperage should be at least 2 or 3 amps" You mean current. Don't assume, measure! \$\endgroup\$ – winny Sep 18 '17 at 5:32
1
\$\begingroup\$

THe maximum power will be about 80% of the short circuit current and 80% of the open circuit voltage, ( more or less) at whatever ratio of resistance that works out to be. Since motor resistance (DCR) starts at 10% of full power impedance, its a poor match without a battery. An impedance matching MPPT optimizes the charge current according to input solar power by tracking or some other algorithm based on Voc and/or Isc.

Consider solar power as a current source, limited by some Voc voltage just the opposite of a zener which acts like a voltage sink. Both have a knee in their curve and ideal for PV is 72% to 82%Voc from morning to mid-day full direct sun, with some variations for PV chemistry etc.

Looks like a smoggy day or overcast at least. Try a battery like with a Schottky diode to protect the PV so you can run 3 White Power LEDs in series with a series R or LM317 current limiter at night. WHite LEDs are like 3V zeners. The higher the power rating, the lower the Zzt where Zzt or ESR ~1/Pd rated.

|improve this answer
\$\endgroup\$
0
\$\begingroup\$

You show two different wire configurations, one at .74 V and .11 A, and another at .05 V and .10 A. The fact that the current is essentially the same in both cases is an indication that your measurements are correct, since photodiodes (of which a solar cell is an example) act as current sources when shorted.

Now calculate the power dissipated in the wires. .74 V x .11 A equals about 81 mW, while .05 V x .1 A equals 5 mW. The ratio, 16 should be equal to the ratio of the resistances, which is 5 in this case. The error is probably due to resistance measurement error using your meter. I suspect you failed to zero out the probe resistance.

Now. Look at the actual powers. 80 mW should be detectable, but 5 mW - not so much. So, how are you "detecting" your temperatures? I doubt you're using a thermal camera, and any physical thermometer may well be too big a heat sink to allow easy measurement. Assuming your power source is well-isolated, you might try letting the wire run for a few seconds, then touch it to your (dry) lips, which are very sensitive. Note that, if you do this, the temperature will almost immediately seem to cool to neutral temperature - again, heat sink effects. Of course, this requires absolute certainty that your power supply is isolated, and is not remotely a good idea if you don't know what this means.

|improve this answer
\$\endgroup\$
0
\$\begingroup\$

You have built a panel with 23 solar cells in series, and 6 in parallel. For each cell of 52x26mm in size, you have total of 0.187 m^2 of area. The cells you seem to be using are coming from eBay China, they claim 17.8% efficiency of their products (which seems to be reasonable in 2017 technology).

Now, assuming that the panel is under direct sunlight in absolutely clear sky day and oriented perpendicular to the direction of light, the incoming power will be about 187 W, and, after the assumed efficiency of 17.8% this panel should produce about 33 W at about 0.5*23 = 11.5 V.

Since there are serious indications in your pictures that your panel is not under direct sun, assumptions about how much sun does your panel get can vary wildly. It can be minimum 1/10th of direct sun, so you should expect no more than 3 W out of this conditions. Your measurements with LED indicate that you have about 0.35W, which is pretty normal for indoor light conditions (1/100th of direct sunlight).

Try to get your panel into afternoon direct sunlight, maybe you will get a better results, assuming that your soldering is good, and wires are think enough (which doesn't look true from the pictures).

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.