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Mick,<br>
<br>
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.<br>
<pre class="moz-signature" cols="72">Kent Osterberg
Blue Mountain Solar, Inc.
<a class="moz-txt-link-abbreviated" href="http://www.bluemountainsolar.com">www.bluemountainsolar.com</a>
</pre>
<br>
On 5/19/2012 12:35 PM, Mick Abraham wrote:
<blockquote
cite="mid:CABOJ=XR028VESviTt4it-ntXzBEptTGQ67jSbCxYYPfqTVvHzw@mail.gmail.com"
type="cite">
<div>Hi, Wrenchies~</div>
<div> </div>
<div>Kindly educate me regarding PV behavior when the load is a
resistor. Here's a hypothetical situation:</div>
<div> </div>
<div>* Eight 250 watt PV modules (60 cells per module), all
connected in series for "peak" ratings of 240 volts DC & 8.3
amps </div>
<div> </div>
<div>* Lab type cell temperature & illumination so that the
eight would truly pump 2,000 watts <strong><em>into an
ideal load</em></strong></div>
<div> </div>
<div>* 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</div>
<div> </div>
<div>* Connect the PV string to the heat element, with nothing in
between except a fused disconnect.</div>
<div> </div>
<div>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?</div>
<div>Would the heater actually get 2,000 watts to turn into heat?</div>
<div> </div>
<div>+++++++++++++++++++++++++++</div>
<div> </div>
<div>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</div>
<div>resistor--again with no intervening electronics, could we
count on 1000 watts of heat?</div>
<div> </div>
<div>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?</div>
<div> </div>
<div>Thanks & Jolliness,</div>
<div><br clear="all">
Mick Abraham, Proprietor<br>
<a moz-do-not-send="true" href="http://www.abrahamsolar.com"
target="_blank">www.abrahamsolar.com</a><br>
<br>
Voice: 970-731-4675<br>
</div>
</blockquote>
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