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'[OT]Solar Panels? - Why PWM Works'
1999\03\08@021015 by Eric Borcherding

picon face
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

1999\03\08@104611 by Wagner Lipnharski

picon face
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}

1999\03\08@140435 by Wagner Lipnharski

picon face
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

1999\03\08@173305 by Sean Breheny

face picon face
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_OUTshb7TakeThisOuTspamcornell.edu ICQ #: 3329174

1999\03\08@183951 by John Payson
flavicon
face
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.

1999\03\08@191352 by Sean Breheny

face picon face
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
.....shb7KILLspamspam@spam@cornell.edu ICQ #: 3329174

1999\03\08@211244 by Wagner Lipnharski

picon face
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}

--
--------------------------------------------------------
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

1999\03\09@003411 by Russell McMahon

picon face
From: Wagner Lipnharski <.....wagnerlKILLspamspam.....EARTHLINK.NET>


This is historically known as "the maximum power transfer theorum"

>Remember that the best power transfer happens when the
>external (load) impedance is equal to the internal (cell)
>impedance,

>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.


But is never applied to mains sockets for appliances etc :-)
(for reasons which will be apparent when you think about it).



Russell McMahon

1999\03\09@080332 by Thomas McGahee

flavicon
face
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}

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