confusion on diode polarity

[Paul E. Schoen]

"rabiticide" <rabiticide@gmail.com> wrote in message
news:ac850a05-6f1b-4009-93fa-bfb516cbffdd@o4g2000pra.googlegroups.com...

okay I'm going to try them in parallel with the schottky diodes and if
that does not work ... series connection will be okay? I don't know
what a "buck switcher" is... it's a cheap set-up from Kmart which
charges 3, AA NiMH batteries...

You can put a small resistor, maybe 1 ohm, in series with each diode to
monitor the current in each panel to see if it is balanced. If so, all is
fine. If there is really not enough light to produce the voltage needed to
charge the batteries, then a series connection will help, but there could
be a problem if there is too much sunlight. PVs are pretty much current
limited, however, so I doubt you will cause overcharging or damage.

I'm fairly sure the charger device that connects the PV panel to the
battery is very simple, and perhaps nothing more than a diode to prevent
reverse connection which can damage the cells. You can measure the voltage
drop across the charger to see just how much it drops. If it is more than
0.5 volts it may be a silicon diode, and you might get a little better
performance if you replace it with a Schottky.

Good luck!

Paul

 

[Jasen Betts]

On 2008-12-31, John Fields <jfields@austininstruments.com> wrote:

On Tue, 30 Dec 2008 18:26:24 -0500, "Paul E. Schoen" <pstech@smart.net
wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:jmpkl4tiej4avd7492j3vbs985h2ofnk5s@4ax.com...
On Tue, 30 Dec 2008 09:47:06 -0800 (PST), rabiticide
rabiticide@gmail.com> wrote:

On Dec 30, 9:30 am, rabiticide <rabitic...@gmail.com> wrote:
Okay, so I'm trying to hook two solar cells in parallel and I don't
want current to try and go reversi through it. So I take the red wire
from the photovoltaic cell (+), hook it to the tail of the arrow, and
connect the head of the arrow to the electronics I want to power,
right?


+--[-PV1+]---[DIODE>]---+
| | +----------+ ____
+--[-PV2+]---[DIODE>]---+--|+IN +OUT|----+-O O-+
| | | |+ |
| | CHARGER | [BAT] [LOAD]
| | | | |
+--------------------------|-IN -OUT|----+------+
+----------+

The problem with this is that only the PV array with the higher output will
contribute to the charging process. It is more efficient to use two
chargers, the outputs of which will be closely matched, and they can be
designed for parallel output without additional steering diodes. The
chargers should be boost converters (assuming Vbat > Vpv), and they can be
designed so that they extract the maximum power from the photocells by
monitoring input voltage and current.

---
Good point but, since two chargers are twice as expensive as one and as
the OP stated that his lights suffered only in winter, I'd recommend:


+----------+ ____
+--[PV2+]---|+IN +OUT|----+-O O-+
| | | |+ |
| | CHARGER | [BAT] [LOAD]
| | | | |
+--[+PV1]---|-IN -OUT|----+------+
+----------+

to paraphrase:

the problem with this is that only the PV array with the higher output
current will contribute to the charging process.



i think in real life it depends on which parameter the arrays are
most closely matched.

Real arrays behave like they have non-trivial internal resistance,
so they if they procude similar voltages they can be paralleled and will
share the load quite well.

 

[Eeyore]

"Paul E. Schoen" wrote:

"rabiticide" <rabiticide@gmail.com> wrote

I'm trying to make a basic circuit with a diode. Had it not been
marked "cathode/anode" I would have not been confused - I would've
just had the electrons follow the arrow. But the arrow points away
from the anode and towards the cathode, and this is the opposite of
what I'd expect.

What does "cathode" and "anode" mean on a diode? To have the electrons
flow with the arrow my "anode" will be a higher voltage than my
"cathode" which is the opposite of what I would have thought. How am I
being confused?

Conventional current flow allows for a more intuitive concept of having the
highest positive voltage represent the source of current, and then there
are voltage drops around the circuit for each element as current flows
through them. The fact is that the more negative point has the most
electrons, but we have chosen their polarity to be negative. It is easier
to follow conventional current flow through the arrows of semiconductors
(when forward biased), and show a higher (more positive) voltage at the
anode as compared to the cathode.

An 'accident of fate'. It was a 50/50 chance. Doesn't seem to cause any trouble
though unless you're dealing with device physics.

Graham

 

[John Fields]

On Tue, 30 Dec 2008 22:21:33 -0600, krw <krw@att.bizzzzzzzzzz> wrote:

In article <grdll4dnr0mv5jof76qqcvrt7hg6gec9oa@4ax.com>,
jfields@austininstruments.com says...

---
Good point but, since two chargers are twice as expensive as one and as
the OP stated that his lights suffered only in winter, I'd recommend:


+----------+ ____
+--[PV2+]---|+IN +OUT|----+-O O-+
| | | |+ |
| | CHARGER | [BAT] [LOAD]
| | | | |
+--[+PV1]---|-IN -OUT|----+------+
+----------+

Wouldn't that presume the PVs are identical? As I read it, he
wanted to augment the built in PV panel with an external. It would
seem difficult to match them.

---
I think that would be true in the parallel case, where the PV with the
highest voltage would cut the other one off by back-biasing its diode.

In the series case, the current into the charger would be limited to the
current available from the lower-current PV, but the voltage into it
would be the sum of the PVs' voltages, making the power into the charger
greater than that available from either PV alone, I think.
---

Which will get rid of all of the diodes and, if the charger is a buck
switcher,

Well...

Now we venture into the land of conjecture...

What's wrong with that? It is the favorite game of the Usenet
community. ;-)

---
Well, then, to continue the game, , if the charger is a buck
converter the extra power supplied to the input will allow the power put
into the battery to increase, charging it more quickly and, probably,
for a longer time than with a single PV.

Just what one needs for those short winter days.

OTOH, using a linear regulator will just burn up the extra power in the
regulator instead of transferring it into the battery.

However, since the voltage into the regulator will be higher, its
dropout voltage will take longer to get to, allowing the battery to be
charged for longer than it would be with a single PV.

JF

 

[John Fields]

On 31 Dec 2008 10:57:43 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2008-12-31, John Fields <jfields@austininstruments.com> wrote:

---
Good point but, since two chargers are twice as expensive as one and as
the OP stated that his lights suffered only in winter, I'd recommend:


+----------+ ____
+--[PV2+]---|+IN +OUT|----+-O O-+
| | | |+ |
| | CHARGER | [BAT] [LOAD]
| | | | |
+--[+PV1]---|-IN -OUT|----+------+
+----------+

to paraphrase:

the problem with this is that only the PV array with the higher output
current will contribute to the charging process.

---
Not true, since the current into the charger will be limited to what the
_lower_ current PV array can supply. However, the _voltage_ into the
charger will be the sum of the array output voltages, allowing the
battery to be charged for a longer time until the charger's input
voltage falls to the point where current can no longer be forced into
the battery.

JF

 

[John Fields]

On Wed, 31 Dec 2008 12:29:09 +0000, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

"Paul E. Schoen" wrote:

"rabiticide" <rabiticide@gmail.com> wrote

I'm trying to make a basic circuit with a diode. Had it not been
marked "cathode/anode" I would have not been confused - I would've
just had the electrons follow the arrow. But the arrow points away
from the anode and towards the cathode, and this is the opposite of
what I'd expect.

What does "cathode" and "anode" mean on a diode? To have the electrons
flow with the arrow my "anode" will be a higher voltage than my
"cathode" which is the opposite of what I would have thought. How am I
being confused?

Conventional current flow allows for a more intuitive concept of having the
highest positive voltage represent the source of current, and then there
are voltage drops around the circuit for each element as current flows
through them. The fact is that the more negative point has the most
electrons, but we have chosen their polarity to be negative. It is easier
to follow conventional current flow through the arrows of semiconductors
(when forward biased), and show a higher (more positive) voltage at the
anode as compared to the cathode.

An 'accident of fate'. It was a 50/50 chance.

---
Not really, since one of Murphy's law states something like:

"If you have a 50/50 chance of being right, 90% of the time you'll be
wrong."

JF