Hi. Can someone explain the directions of arrows labeled "I" here :






[ From:
Fig 9.3 at http://docplayer.net/21879535-Solar-cell-parameters-and-equivalent-circuit.html
Page 2 at http://icrepq.com/icrepq'11/339-rodrigues.pdf ]

 

In fig 9.3 a and b solar cell or panel is providing the current and it shows conventional current flow that is from +ve to -ve.

I might have to refer th ebook. Which book are you referring ?

Edit:

Ok. I saw the links now.

G is the solar cell.

 

Hello,

A possibility for figures 'a' and 'b' is that they are showing it in the fashion of a two port network.
For a two port network the currents are usually shown with the current flowing INTO the device from both sides even if the device is just transferring energy from left to right as usual and where we like to show the current flow from left to right for both input and output. Two port network theory is based on currents flowing only into the network from some unspecified sources so the currents will be shown flowing into the input and also into the output. It does not change the fact that current could be flowing out of the output, in which case the current shown as simply "I" (capitol 'i') would be negative.

This confusion regarding the sign of the current comes up now and then. We dont have any problem labeling the current I, A, B, C, x, y, etc., but as soon as we want to think of it as a negative current it gets a little confusing. For any of those mention like A,B,C x, etc., any or all of them can be negative but we dont show the sign until we actually calculate the actual current.
For example, if "I" in one of the drawings was actually 5 amps, then that would mean 5 amps was flowing into the output of the network, but if actually -5 amps then it would be flowing out of the output of the network. See, no problem really, one is I=5 and the other is I=-5, perfectly valid values.

 

No....your right.....one of them is lying to you.

 

Two more :

 

Abolish positive current/charge flow.........and all this confusion goes away.

One shouldn't use a mathematical equivalent for a physical event. ONLY negative flows are equivalent.

Is gravity plus or minus? It doesn't matter......because it always acts the same way.......just like electron current.

One never has to identify the polarity of current. It only has direction.....the polarity is constant.

A positive charge flow is never equal to a negative charge flow........only in anti-matter reactions.

In math......a +e charge and a -e charge are opposite, but equal. ONLY the amount of charge is equal. Every other property is un-equal in a large degree. Field density, mass/inertia, size, frequency, energy.......everything else is different.

 

No....your right.....one of them is lying to you.

Hi,

You should read more posts before you comment

I explained how this works.

But also, there is no reason why we cant power a solar panel at least as an experiment, and then current will actually flow INTO the output terminals. That happens at night if there is battery storage and no blocking diode.

But take a look at some two port network theory articles and see what you find out. The current is shown going in on both sides. Nothing wrong with that.

 

Hi. Can someone explain the directions of arrows labeled "I" here :

View attachment 133351
View attachment 133352



[ From:
Fig 9.3 at http://docplayer.net/21879535-Solar-cell-parameters-and-equivalent-circuit.html
Page 2 at http://icrepq.com/icrepq'11/339-rodrigues.pdf ]

You can flip a coin when deciding which direction to label any voltage or any current. You then just have to be consistent with that choice in your work with that circuit.

Depending on what you are doing, it is convenient to use one of a variety of conventions for choosing your directions, instead of resorting to coin flipping.

You might want to choose the directions so that the associated parameters are positive under normal conditions.

You might want to choose the directions so that they are consistent with the passive sign convention.

You might want to choose the directions so that they are consistent with the normal labeling of equivalent circuits or two-port networks.

These various conventions may not agree with each other for some of the directions, so you have to choose. There is no right or wrong choice, as long as you remain consistent with them.

 

The choice of arrow direction is completely arbitrary; it doesn't matter, as they are only a visual aid to understanding the circuit. What does matter, when analyzing a circuit (e.g., by nodal analysis), is that the math corresponds correctly to the chosen direction of the arrows.

For example, suppose I have a circuit node with three wires connected to it, carrying currents I1, I2 and I3.

If I choose to draw the arrows representing I1, I2 and I3 all pointing toward the node, the resulting nodal equation would be:

I1 + I2 + I3 = 0.​

If I chose to draw the arrows such that I1 and I2 are shown flowing into the node, and I3 flowing out of the node, the equation would be:

I1 + I2 - I3 = 0.​

And If I1 was shown flowing into the node, and I2 and I3 flowing out of the node, the equation would be:

I1 - I2 - I3 = 0.​

And so forth. No matter how you choose the arrow directions, provided those directions are correctly reflected in the resulting equations you will get a valid answer. (Note that if, after doing the analysis by solving the system of nodal equations to find the currents, you end up with one or more currents evaluating to a negative quantity, all that means is that the physical current is actually flowing in a direction opposite to that of your arrow, nothing more.)

The following graphic is an example of this in use, from a presentation done long, long ago in a job far, far away:


In the above, I simply chose, completely arbitrarily, to show I1, I2 and I3 flowing into the circuit node connected to the op amp's inverting input. But I could have chosen a different set of arrow directions, and the end result would still be the same.

Since the arrows themselves are just a visual aid to understanding the circuit operation, I usually choose them to they more-or-less "make sense."

 

Thanks, gentlemen.
Two things...
- The other current arrows in post #1 are shown in the same direction in every schematic. The "I" differ. Does it denote lack of consistency within some schematics ?
Al:
- It is now 24 hours am evaluating as experiment the behavior of a 3.77V lithium cell connected directly in parallel to a dark 4V solar panel. 'Reverse' current is 0.02mA without a blocking diode, the voltage staying solid at 3.77V. The lithium cell is not appreciably discharging trough the panel .

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If a dark solar cell behaves as a diode; what is the expected diode drop ? 0.7V ?
For a panel with -say 10 cells-; the lithium cell voltage would need to be greater than the 10 Vdrops = 7V in order for the cell to get discharged into the dark panel ?
Being the lithuim cell 3.77V; does not reach the 7V needed to 'overcome' the Vf of ten solar cells, thus, not discharging.
Is that the way it works ?

 

Thanks, gentlemen.
Two things...
- The other current arrows in post #1 are shown in the same direction in every schematic. The "I" differ. Does it denote lack of consistency within some schematics ?

No, not necessarily.

While the choice of direction is always arbitrary, some choices make more natural sense then others. It is not surprising that different people usually make the same choices for those currents.

Other currents are less so and one choice might make more sense than the other but that choice might differ depending on what they are trying to convey.

There's nothing wrong with analyzing the same circuit multiple times, under different situations, and making different choices for each one. You see this quite a bit with zener diode circuits, AC circuits, photocell circuits, charging circuits -- any circuit in which the actual current might flow in different directions under different conditions.

Al:
- It is now 24 hours am evaluating as experiment the behavior of a 3.77V lithium cell connected directly in parallel to a dark 4V solar panel. 'Reverse' current is 0.02mA without a blocking diode, the voltage staying solid at 3.77V. The lithium cell is not appreciably discharging trough the panel .

I wouldn't expect a 20 μA current to appreciably discharge any battery over the course of a day -- that's less than a half a milliamp-hour of charge. Even the smallest coin cells could tolerate that for a couple of months.