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Solar System for farm.

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This morning spent rewiring main DB board. Inverter now supplies plugs and old 32V DC lights (bulbs changed to 220V AC LEDs). At 8:30 the PVs were producing just over a 1 kW. Once I finished wiring the DB I started switching on appliances and tried to force the inverter to draw from the battery so that I knew what the PVs were producing. I eventually switched on the oven and it finally showed a draw of 31A from the battery but that we were getting 2.24 kW from the panels (Awesome!). Panels face true North.

  • 2 weeks later...

Does anyone know of an easy way to "monitor" fuses inside those 10x38 fuse holders, without having to take out the fuse and measuring it with a multi meter?

 

For DC: bridge rectifier, resistor and LED.  For AC: capacitor, diode and LED - diode reversed and parallel with LED.

 

Depending on the voltages, I can draw up the diagrams and suggest component values.  Tell me if you require it for AC or DC and what the maximum voltage across the fuse would be if it blows.

For DC: bridge rectifier, resistor and LED.  For AC: capacitor, diode and LED - diode reversed and parallel with LED.

 

Depending on the voltages, I can draw up the diagrams and suggest component values.  Tell me if you require it for AC or DC and what the maximum voltage across the fuse would be if it blows.

Thanx for the help!

 

I have 2x 265W Solaire Direct panels connected in series, and get about 60V through, though I want to add more panels and change to a 3-panel series connection. 

The fuses are 10x38 type, 10A fuses but I am thinking of rather changing them to 20A fuses. 

 

The thing is, the fuses are connected to a common busbar, which then goes to the inverter. So on the "bottom side" all the fuses are connected together, if that makes sense?

Thanx for the help!

 

I have 2x 265W Solaire Direct panels connected in series, and get about 60V through, though I want to add more panels and change to a 3-panel series connection. 

The fuses are 10x38 type, 10A fuses but I am thinking of rather changing them to 20A fuses. 

 

The thing is, the fuses are connected to a common busbar, which then goes to the inverter. So on the "bottom side" all the fuses are connected together, if that makes sense?

 

Not following.  :( You say you have 2x 265W Solaire Direct panels connected in series and want to add another panel in series - that is effectively 1 series string. Then you mention the "bottom side" on all fuses are connected to a bus bar. To me it sounds like the panels are connected in parallel rather that series?

Not following.  :( You say you have 2x 265W Solaire Direct panels connected in series and want to add another panel in series - that is effectively 1 series string. Then you mention the "bottom side" on all fuses are connected to a bus bar. To me it sounds like the panels are connected in parallel rather that series?

My bad... there's 2 strings at the moment, of 2 panels in series each. So 4 panels right now. Later on I'll add more, and have 4 strings with 3 panels each. So the Volt, per string would be about 108V

My bad... there's 2 strings at the moment, of 2 panels in series each. So 4 panels right now. Later on I'll add more, and have 4 strings with 3 panels each. So the Volt, per string would be about 108V

 

OK, so you'll have 3 strings, each with a fuse in series. Short-circuit current on a 265W panel is about 9A and to avoid nuisance-blowing you should install at least a 12A fuse in each string, but not more than 15A per string. 

Fuse Monitor


 

To monitor the fuse in each of the PV strings you need to construct 3 of the "DC circuits". Components required are as follows:

Calculation: Forward voltage drop across LED is about 2V and voltage drop across the diode bridge is about 1.5V. The voltage across the resistor will then be 108V-2V-1.5V => 104.5V

The maximum current through the circuit must not exceed the LED forward current. If a LED with 10mA forward current is used, then the current flowing through the circuit (all series connected components) will be 10mA, thus would the current through the resistor also be 10mA. Calculate the resistor using ohms law R = V/I, R = 104.5 / 0.01, R = 10450 ohms. Resistors are not available in every possible value and the E12 series is the most popular and most frequently available range. The closest values in the E12 series would be a 10K ohm or 12K ohm resistor, but by using a 10K ohm resistor the current through the LED might be too high. You can either use a 12K ohm resistor or use a 10K ohm and 470 ohm resistor connected in series to get closer to 10450 ohm.

The power dissipated by the resistor will be P = VI, P = 104.5 x 0.01, P = 1.045 watt. Although this is only slightly more than 1W, a 1W resistor might get very hot and I recommend rather using a higher watt resistor. The next option would be a 2W resistor.

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