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?
Cathode and anode mean the same to a diode as they do to everything else in electronics, you can research that. The electrons do not "follow" the direction of the arrow for a forward conducting diode. The electrons will flow against the arrow. I understand your confusion. Look up electron flow and conventional current. To conduct the anode needs to be more positive than the cathode. The cathode is marked with a stripe. The anode is the arrow.
Brings back memories of the good ol' days at Nuc School. The enlisted students were taught "electron flow" and the officers "conventional flow." What was HGR thinking?!
Conventional current is positive to negative. Electrons don't really flow, they sort of ooze slowly along at the blazing speed of inches per minute. And, in some conductors, the charge carriers aren't electrons at all! Don't worry about it. Current is + to - and goes in the direction of the "arrow" on the diode.
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?
_________ | | | |----(+ RED)-----(----->|-----)------------ | P.V. | | | (charging circuit) (mirror image with another PV cell and diode) | |----( - BLACK)--------------------------------- _________
The arrows in transistor and diode symbols point in the direction of "conventional" (positive) current flow - opposite to the direction of electron flow, so the arrow points to the more negative terminal of the diode if you wish the diode to conduct.
Note that Zener diodes are used in a reverse breakdown mode - in that case, the arrow points to the more positive terminal.
--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
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|.........................................................(charging circuit).....(mirror image with another PV cell and diode)
Yes, with the caveat that you'll need to be sure to use diodes with the proper ratings. Probably Schottky types, since PV systems don't have a lot of spare power to waste on diode drops.
Thanks. I will order up some schottky types which are not carried at radio shack... Top posting was unintentional, probably a result of my changing the subject line. I am attempting to bolster the charge on my walkway lights which work fine in the summer, but do not get enough juice in the winter months. I will let you know how it goes. Again, thank you.
It is an unfortunate fact that Benjamin Franklin guessed wrong when he declared the direction of electric current flow. Before anyone figured out that he was wrong, the scientific world had adopted Franklin's "convention."
Everyone knows that--at least in metallic conductors--it is the negatively charged electrons that flow, but we say that current flows from positive to negative because we have agreed to use Franklin's convention.
All the world's electrical engineers are taught to use conventional current flow. Unfortunately, a lot of technician-level courses and books teach electron flow. This leads to a lot of confusion for people coming out of such courses or reading such books.
If you change your way of thinking to assume current flows from positive to negative, then you will find it much easier to interpret schematic symbols and schematic diagrams. Electrical schematic symbols assume conventional current flow.
For example, think of the diode symbol as a funnel that lets the current flow into it, but the bar across one end blocks current from flowing the other way.
Also, transistors will make more sense since, using an NPN transistor, a current entering the base will follow the arrow and flow into the emitter, thus turning the transistor on.
Learn to think in terms of conventional current, not electron flow, and your understanding will be enhanced.
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.
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.
yes. You are referring to the anode of the diode, the side that does not have the line(band) on the body.
The (+) lead of the panel connects on that side and the (+) voltage from the panel will come out the other side of the diode which does have the line (band) on it.
Hope that was simply enough.
P.s. you'll give ~ 0.6 voltage loss through the diode.
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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|----+------+
+----------+
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...
JF
If both solar panels are about equally insolated (amount of light applied), then this, or the other approach with both in parallel, will be OK. The problem is that the current from the panel (or cell) with the least amount of light will limit how much the others can supply. A persistent shadow can be a real show-stopper. But I have not had much experience with PV, so maybe it is not as much problem as I think.
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.
What's wrong with that? It is the favorite game of the Usenet community. ;-)
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.
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