Re: [RE-wrenches] PV into a resistor: please advise

[Allan Sindelar]

This is a fun one... Kent, I would think that an LCB, assuming it could output the same current at half voltage, would not help at all. By Ohm's law, if you drop voltage by half, the power (wattage) drops by a factor of four. I *think* (the **'s mean I'm not certain at all here) that the available wattage would be dependent on only the resistance, not the voltage or the current. Allan Allan Sindelar al...@positiveenergysolar.com NABCEP Certified Photovoltaic Installer NABCEP Certified Technical Sales Professional New Mexico EE98J Journeyman Electrician Founder and Chief Technology Officer Positive Energy, Inc. 3209 Richards Lane (note new address) Santa Fe, New Mexico 87507 505 424-1112 www.positiveenergysolar.com On 5/19/2012 3:20 PM, Kent Osterberg wrote: Mick, In the first case you are selecting a resistor that matches the IV curve peak power point at 1000 W/sq meter irradiance. You will indeed get 2000 watts of power into the resistor. For the second case, at 500 W/sq meter irradiance, the PV module's IV curve has about the same peak power voltage, but only half as much current. The resistance that would get the maximum power changes from 29 ohms to 58 ohms. With a 29-ohm resistor connected, the power into the resistor will be a little more than half of the 1000 watts the PV module could deliver. The linear current booster is designed to boost the current when the voltage (and resistance) are less than optimum, so I think it would improve the power transfer. Kent Osterberg Blue Mountain Solar, Inc. www.bluemountainsolar.com On 5/19/2012 12:35 PM, Mick Abraham wrote: Hi, Wrenchies~   Kindly educate me regarding PV behavior when the load is a resistor. Here's a hypothetical situation:   * Eight 250 watt PV modules (60 cells per module), all connected in series for "peak" ratings of 240 volts DC & 8.3 amps    * Lab type cell temperature & illumination so that the eight would truly pump 2,000 watts into an ideal load   * A 240 volt AC heating element designed for 2,000 watt heat dissipation at 240 volts AC...that's about 29 ohms resistance for the heat element   * Connect the PV string to the heat element, with nothing in between except a fused disconnect.   In the above situation, would the resistance of the heating element be all that's needed to force the PV array to operate near the "peak" wattage? Would the heater actually get 2,000 watts to turn into heat?   +++++++++++++++++++++++++++   Now consider the same cell temperature but half the illumination. That's similar voltage but half the amps at peak wattage. If this is sent into the same 29 ohm resistor--again with no intervening electronics, could we count on 1000 watts of heat?   If the answers come up "no", would the power throughput be helped by a SolarConverters style MPPT pump controller (Linear Current Booster kinda thing), assuming that one could be found to operate in the 240 volt range?   Thanks & Jolliness, Mick Abraham, Proprietor www.abrahamsolar.com Voice: 970-731-4675 _______________________________________________ List sponsored by Home Power magazine List Address: RE-wrenches@lists.re-wrenches.org Options & settings: http://lists.re-wrenches.org/options.cgi/re-wrenches-re-wrenches.org List-Archive: http://lists.re-wrenches.org/pipermail/re-wrenches-re-wrenches.org List rules & etiquette: www.re-wrenches.org/etiquette.htm Check out participant bios: www.members.re-wrenches.org

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