Post by thread:[OT]Solar Panels?

Wagner, A solar panel power delivery can be likened to a Low pass Bode plot. It you unload the cells they tend to rise in voltage - it is a natural physical response from being srtuck by light. So, at this point the Array voltage rises and there is an effective higher voltage output. As you load this with a circuit the draws power the array voltages drops again. But if you pulse load this then the average voltage delivered is some higher. The regulation unit must look for the maximum product of V panel to I delivered to battery / load. It now is a simple proportional drive to track a hi - right on - low point that allows the maximum power to be tracked and delivered. There is a need to understand that PV cells are not linear they a bowed - the V*I max in the bow can be tracked. I have PVs on my roof and collect 1/3 of all my electrical needs. If I can help folks witht his write... Trace Engineering makes some nice $value products... Eric Borcherding

Hi Eric, thanks for the explanation. I am just learning about solar cells, and it looks like you have much more experience. So I need to ask you: What you are saying, is that the *delivery power* of a solar cell changes with the current drained, is that right? Nothing to do with internal impedance, is that correct? Note that *power* means voltage x current, so according to what you are saying, suppose: 1) A pack of cells can supply 100V with open output. 2) You connect a load and the current goes to 10mA, while the voltage drops to 10V. The power delivered is 100mW. 3) If you increase the load to drain only 5mA, the voltage will be higher than 20V (more than 100mW supplied). It is natural that *any* voltage cell increase the voltage at the output connections when load is removed since the voltage drop at the internal impedance is reduced. If you connect a capacitor in parallel with any battery with high internal impedance, it will delivery a higher peak of power at the initial connection since the cap has an impedance lower than the battery. But according to what you are saying, this is not what happens to a solar cell, and that there is an "optimum" point in the power curve (current versus voltage) where it is higher. Is that right? > But if you pulse load this then the average voltage > delivered is some higher. It works the same way if you apply a capacitor in parallel to any power cell, reducing the internal impedance. In real, a solar cell may act somehow as a capacitor, since the large surface area. In the analogy to the accelerating car it doesn't work; A car with the traction tires lifted in the air, then accelerating at high rpm, creating mechanical inertia, and then releasing the car into the track, it will jump a little bit because the mass inertia of the rotating mechanics, but it has no torque, just inertia, and the acceleration process will reduce the rpm according to the friction so it needs to create the torque and at the end of the track it consumed more fuel/distance/time than if not doing that. The fuel consumed to create the initial inertia (in the air) has a terrible low productivity. In a videogame for the Nintendo64 has a SuperMario race game where this effect allows you to speed up more than your competitors, jumping the car continuously along the race, this is not true in the real world. If yes, a simple reduction in the gear box would have the same effect. I have a friend that once told me that a hammer-drill can penetrate the concrete not because the hammering effect (that breaks the crystallized concrete surface), but because when the hammer retracts it gains more speed... can you imagine? -------------------------------------------------------- Wagner Lipnharski - UST Research Inc. - Orlando, Florida Forum and microcontroller web site: http:/http://www.ustr.net Microcontrollers Survey: http://www.ustr.net/tellme.htm Eric Borcherding wrote: {Quote hidden}> > Wagner, > > A solar panel power delivery can be likened to a Low pass Bode plot. > It you unload the cells they tend to rise in voltage - it is a natural > physical > response from being srtuck by light. So, at this point the Array voltage > rises > and there is an effective higher voltage output. As you load this with a > circuit > the draws power the array voltages drops again. But if you pulse load this > then > the average voltage delivered is some higher. The regulation unit must look > for > the maximum product of V panel to I delivered to battery / load. It now is a > simple > proportional drive to track a hi - right on - low point that allows the > maximum > power to be tracked and delivered. There is a need to understand that PV > cells > are not linear they a bowed - the V*I max in the bow can be tracked. I have > PVs on my roof and collect 1/3 of all my electrical needs. If I can help > folks > witht his write... Trace Engineering makes some nice $value products... > > Eric Borcherding

Eric Borcherding wrote: > There is a need to understand that PV cells are not linear > they a bowed - the V*I max in the bow can be tracked. Where can we find a solar cell data sheet, with informations like that? Remember that the best power transfer happens when the external (load) impedance is equal to the internal (cell) impedance, so at that point, the voltage-drop at the internal (cell) impedance is the lowest possible with the maximum possible output current. Lots of folks doesn't understand that, this is culture, so here we goes: 10Volts-----INT.IMPEDANCE(10 Ohms)----x---LOAD(15 Ohms) Circuit current V/R: 10 / 25 = 0.4 A Power transfered to load RI^2: 15 x 0.4 x 0.4 = 2.4 W 10Volts-----INT.IMPEDANCE(10 Ohms)----x---LOAD(5 Ohms) Circuit current V/R: 10 / 15 = 0.66 A Power transfered to load RI^2: 5 x 0.66 x 0.66 = 2.178 W 10Volts-----INT.IMPEDANCE(10 Ohms)----x---LOAD(10 Ohms) Circuit current V/R: 10 / 20 = 0.5 A Power transfered to load RI^2: 10 x 0.5 x 0.5 = 2.5 W As you can see, when get the max transfered power when both impedances are equal. This is applied to sound, video, RF, power, pressure, including mechanics. Look what happens when the equalization changes a bit: 10Volts-----INT.IMPEDANCE(10 Ohms)----x---LOAD(11 Ohms) Circuit current V/R: 10 / 21 = 0.476 A Power transfered to load RI^2: 11 x 0.476 x 0.476 = 2.49 W 10Volts-----INT.IMPEDANCE(10 Ohms)----x---LOAD(9 Ohms) Circuit current V/R: 10 / 19 = 0.526 A Power transfered to load RI^2: 9 x 0.526 x 0.526 = 2.49 W The general formula is: PowerLoad = Voltage x LoadImpedance -------------------------- (SourceImped + LoadImped)^2 or PowerLoad = Voltage ------------------------- (SI^2 / LI) + 2xSI + LI I wonder if the bowed curved you said is not related to optimum power transfer when both impedances are equal, that a pwm current can do, caused by the current slack raising time. Wagner

Hi, At 10:45 AM 3/8/99 -0500, you wrote: >If you connect a capacitor in parallel with any battery >with high internal impedance, it will delivery a higher >peak of power at the initial connection since the cap >has an impedance lower than the battery. Won't you actually achieve max. power transfer to the cap when the product of the voltage across it and the current thru it is maximum? This won't occur when you first hook it up because the voltage across the cap is initially zero,so the power transfered is initially zero. Sean | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 spam_OUTshb7TakeThisOuTcornell.edu ICQ #: 3329174

At 10:45 AM 3/8/99 -0500, you wrote: >If you connect a capacitor in parallel with any battery >with high internal impedance, it will delivery a higher >peak of power at the initial connection since the cap >has an impedance lower than the battery. |Won't you actually achieve max. power transfer to the cap when the product |of the voltage across it and the current thru it is maximum? This won't |occur when you first hook it up because the voltage across the cap is |initially zero,so the power transfered is initially zero. The initial transfer of energy to the cap will not be terribly efficient, and if the cap is substantially discharged its repl- enishment won't be terribly efficient either. Suppose, however, that you have a device powered by a "9 volt" battery which needs to drive a 100-ohm solenoid for 100ms with at least 60mA. If you have a 4,700uF cap in parallel with the battery it will be able to supply that demand while dropping less than two volts even if the battery is not in good shape and has an internal resistance of 100 ohms (in which case connecting a 100ohm load without the cap would cause the voltage to drop by half, failing to meet the 60mA requirement). Note that recharging the cap after each pulse will not be 100% efficient (since the cap's starting voltage will be around two volts less than the battery voltage) but it will be much more efficient than would be the battery's efforts at driving the solenoid directly.

Hi John, At 05:40 PM 3/8/99 -0600, you wrote: [SNIP] >|Won't you actually achieve max. power transfer to the cap when the product >|of the voltage across it and the current thru it is maximum? This won't >|occur when you first hook it up because the voltage across the cap is >|initially zero,so the power transfered is initially zero. > [SNIP] >Suppose, however, that you have a device powered by a "9 volt" >battery which needs to drive a 100-ohm solenoid for 100ms with >at least 60mA. If you have a 4,700uF cap in parallel with the >battery it will be able to supply that demand while dropping >less than two volts even if the battery is not in good shape and >has an internal resistance of 100 ohms (in which case connecting >a 100ohm load without the cap would cause the voltage to drop by >half, failing to meet the 60mA requirement). > I see,I misunderstood the original post,then,I should have followed the thread more closely. I thought that the original post was talking about simply using a battery to charge a cap,nothing else,and was saying that the power delivered to the cap was greatest when the cap was discharged completely. Thanks, Sean | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 .....shb7KILLspam@spam@cornell.edu ICQ #: 3329174

Yes, the main function of a low FSR cap in parallel with a high impedance power source is to be a kind of intermediate low impedance element, to supply high peak current demand when it is necessary. The main reason here is to allow a high current =with= high voltage for a brief moment, what is very difficult to acquire from a solar cell, for example. Wagner. Sean Breheny wrote: {Quote hidden}> > Hi John, > > At 05:40 PM 3/8/99 -0600, you wrote: > [SNIP] > >|Won't you actually achieve max. power transfer to the cap when the product > >|of the voltage across it and the current thru it is maximum? This won't > >|occur when you first hook it up because the voltage across the cap is > >|initially zero,so the power transfered is initially zero. > > > [SNIP] > >Suppose, however, that you have a device powered by a "9 volt" > >battery which needs to drive a 100-ohm solenoid for 100ms with > >at least 60mA. If you have a 4,700uF cap in parallel with the > >battery it will be able to supply that demand while dropping > >less than two volts even if the battery is not in good shape and > >has an internal resistance of 100 ohms (in which case connecting > >a 100ohm load without the cap would cause the voltage to drop by > >half, failing to meet the 60mA requirement). > > > > I see,I misunderstood the original post,then,I should have followed the > thread more closely. I thought that the original post was talking about > simply using a battery to charge a cap,nothing else,and was saying that the > power delivered to the cap was greatest when the cap was discharged > completely. > > Thanks, > > Sean > > | > | Sean Breheny > | Amateur Radio Callsign: KA3YXM > | Electrical Engineering Student > \--------------=---------------- > Save lives, please look at http://www.all.org > Personal page: http://www.people.cornell.edu/pages/shb7 > shb7KILLspamcornell.edu ICQ #: 3329174 -- -------------------------------------------------------- Wagner Lipnharski - UST Research Inc. - Orlando, Florida Forum and microcontroller web site: http:/http://www.ustr.net Microcontrollers Survey: http://www.ustr.net/tellme.htm

Sean, If you try to draw current from a battery with a high impedance, the internal resistance of the battery will limit the maximum current you can draw. If you place a capacitor in parallel with the same battery, the battery will charge the capacitor up to the full voltage within a certain time period determined by the product R*C. The capacitor will initially charge according to the standard charge curve. Assuming that the capacitor has no leakage, the power system as a whole still has the same total energy capacity it had before. However, since the impedance of the capacitor can be MUCH lower than that of the battery, it can supply high currents for short periods. *After* being discharged, there will be a time delay before the capacitor fully charges again. That is the tradeoff. Every time you use a decoupling cap from an IC power pin to ground, you are doing the same thing. This techniques is also useful when driving relays: they have an initial need for a good stiff voltage source *until they are closed* then, they only require a certain minimum holding current. After releasing such a relay, there is a certain minimum recharge time that must be observed before you can re-activate the relay. With relays that can tolerate this wait time between activations you can actually add a resistor between the battery and the capacitor that will power the relay. This greatly reduces the chance that activating the relay will produce noise on the power bus. Hope this helps. Fr. Tom McGahee ---------- {Quote hidden}> From: Sean Breheny <EraseMEshb7spam_OUTTakeThisOuTCORNELL.EDU> > To: PICLISTspam_OUTMITVMA.MIT.EDU > Subject: Re: [OT]Solar Panels? - Why PWM Works > Date: Monday, March 08, 1999 5:31 PM > > Hi, > > At 10:45 AM 3/8/99 -0500, you wrote: > >If you connect a capacitor in parallel with any battery > >with high internal impedance, it will delivery a higher > >peak of power at the initial connection since the cap > >has an impedance lower than the battery. > > Won't you actually achieve max. power transfer to the cap when the product > of the voltage across it and the current thru it is maximum? This won't > occur when you first hook it up because the voltage across the cap is > initially zero,so the power transfered is initially zero. > > Sean > > > | > | Sean Breheny > | Amateur Radio Callsign: KA3YXM > | Electrical Engineering Student > \--------------=---------------- > Save lives, please look at http://www.all.org > Personal page: http://www.people.cornell.edu/pages/shb7 > @spam@shb7KILLspamcornell.edu ICQ #: 3329174