Optimizing Circuit Design: Voltage Range, Internal Resistance, and Wire Sizing

[KarenRei]

Hi all. In a project I'm working on I'm needing to model a DC circuit to find the required power sources/storage to keep the system voltages up to a sufficient level. There's a solar cell array, a reversible fuel cell stack, various DC "appliance" loads, and wire runs with resistances. I dug up some Kirchoffs' law code that I wrote and tested ages ago, but I'm mainly thinking now about the "appliance" loads at the moment

Often when you see a device it'll be rated for a certain voltage range and will have a specified current or power draw. But what I need for Kirchoffs' laws is internal resistances. Now, I can just get that from V=IR. Say we had a device rated for 10-15V DC with a power draw of 150W. For 15V we get the following:

150W: 150 = V * I = 15 * I; I = 150/15 = 10
V = I * R; V / I = R; 15 / 10 = 1.5. So 1.5 ohms.

Now for 10V:

150W: 150 = V * I = 10 * I; I = 150/10 = 15
V = I * R; V / I = R; 10 / 15 = 0,667. So 2/3 of an ohm

In short, we find double the internal resistance for the device if I assume the upper end of the voltage range rather than the lower end! But surely that's not going to happen in the real world, right? Would the safe bet be to assume that the power rating is only valid for some "nominal" voltage? Or should I use the upper or lower bound in calculating the internal resistance?

I'm also pondering how I should determine the wire diameters/gauges. Right now, all that comes to mind is picking some initial values and then either manually or automatically iteratively converging to diameters that keep all "appliance" loads above their minimum voltage rating at all times and don't exceed their current ratings. An iterative refining approach is also all that comes to mind for determining the sizing on the solar cell array and fuel cell stack.

Am I thinking about this wrong?

 

[Tom.G]

In short, we find double the internal resistance for the device if I assume the upper end of the voltage range rather than the lower end! But surely that's not going to happen in the real world, right? Would the safe bet be to assume that the power rating is only valid for some "nominal" voltage? Or should I use the upper or lower bound in calculating the internal resistance?

Short answer, It depends.
Specifically it depends on the type of the "appliance." For instance a space heater, an electric range, an incandescent light bulb have close to a constant resistance (although the light bulb is a little weird, at the instant of turn-on it draws ten times the expected current for a small fraction of a second, and while operating its resistance goes up a little bit with increased voltage). Things using a switching power supply, a computer, computer monitor, most flat screen TVs, LED lamps, etc. have a constant power characteristic, and some motors also approach constant power (except at startup where they draw roughly 5 times the expected current). Flourescent Lamps are constant current appliances due to their required ballast.

You should also consider if it is required that all appliances be able to operate concurrently. For everyday 'normal' usage probably not, but which ones? If there is a big party going on you might have to run all the lights, fans, air conditioner, dishwasher, and the pump for the water well.

If this is a significant sized project I recommend modeling this in an electronic simulation program, an electrical simulation program, or possibly a spreadsheet program but that would be almost as tedious as doing it all manually.

WIre size will be determined by how much voltage drop you are willing to accept at the end of the wire and the upper temperature limit of the wire and its insulation when the maximum current is being drawn. If this is a real-world problem you also have to meet the building and the electrical codes for your location. In the U.S. almost all local electrical codes are based the National Electrical Code (NEC) published by the National Fire Protection Association (NFPA). Local jurisdictions may pile on a few additional requirements.

If this is an exercise, it seems whoever created it gave you a good real-world mix of loads to worry about.

 

[KarenRei]

Thanks for the reply. Quite useful, but it doesn't really answer the problem. A variable load draw is easy enough to model, as I can just use the resistance at the lowest voltage, since the goal is to make sure that everything stays above its minimum required voltage in a peak load scenario. But for fixed-load hardware that is rated for a given power and a given voltage range - the sort of data you get from looking at the label - would the internal resistance most reasonably be calculated from the lower, nominal, or upper end of the voltage range? Because I can't do kirchhoffs laws without having some sort of resistance figure.

On the other part, you wrote, "Wire size will be determined by how much voltage drop you are willing to accept at the end of the wire and the upper temperature limit", which is pretty much what I wrote in my initial post. But every part of the circuit affects every other part; if I have higher resistance in a given wire then it will affect the voltages everywhere else, not just at the hardware at the end of the wire. So is the proper answer what I assumed, that the best way to do it is to try to iteratively converge on a solution? A little annoying if so since there's multiple things converging at the same time, but I should be able to do it without too much of a headache, I think.

As for your other comments:
* Surges are not an issue for this application, it's fine to assume whatever "buffer" is needed
* The software is just a minor cog in a much larger modeling project; it wouldn't be reasonable to impose a simulation program as a dependency along with the integration work that would entail
* The fact that a limited percentage of the hardware will be in use at a given time is taken into account. But I still need the resistances to model a realistic peak load.

 

[Baluncore]

But I still need the resistances to model a realistic peak load.

You have a fixed source voltage system. You do not need resistance for loads, you need only maximum surge and steady current requirements.

Make the fuel cell stack the centre of your wiring network, it sets the supply voltage and is the only source of current during the night. Loads radiate out independently from that terminal pair. Do not worry about iteration of cable size selection effecting other things. The star network will reduce that effect and make the problem tractable.

The PV will need an MPP controller to efficiently pass energy to the fuel cell stack. It can be considered independently.

Start by specifying a minimum of 1 mm² for each wire. For greater currents use 0.1 mm² of copper wire for each amp of DC load current. (Use 0.2 mm² of copper wire for each amp of AC power current). The length of the cables needed has not yet been specified. You will need to use a greater wire section if the load is far away. Remember that each device has two wires, the return wire will double the resistance and voltage drop.

Tabulate the range of data for each load and for the line to that load.
Load n: Isurge, Icontinuous, Vmin @ Isurge, Vmin @ Icontinuous.
Line n: mm², material, resistivity, length.

Your first attempt at a solution will be a learning exercise.

 

[KarenRei]

Fixed voltage? In a no-load condition, but as soon as you attach a load there's immediate voltage drop (I can say this from experience).

Calculating voltage drop in a nontrivial circuit requires Kirchoff's laws, which requires resistance figures.

 

[Baluncore]

Something, such as the fuel cell stack, must set the source voltage of the distribution system.
This is not Kirchoff's laws. It is power distribution, with voltage drop along the lines to the loads.

 

[KarenRei]

The solar cell is a source of EMF. The reversible fuel cell stack is either a source or sink of EMF, depending on which direction current flows through it, which in turn depends on the voltage differential across it. The solar cells operating at a higher voltage than the stack, but there are alternative paths (both series and parallel) which current can take. How is this not a Kirchoff's laws problem?

 

[Baluncore]

Charge is sourced, not EMF. Current is rate of flow of charge. EMF is voltage and is set by the chemistry of the process in batteries, fuel cells or PV panel bandgap voltage. Fixed chemistry is fixed voltage, with a little acceptable variation due to the complexity of reality.

The solar cell is a source of current proportional to light level. Charge flows into, or out of, the fuel cell depending on the PV current available and the total current demand of all loads.

Voltage drops along wires should not be greater than 5% of supply. Internal resistance of supply should be very low, so power consumption of different loads is determined primarily by load resistance.

 

[KarenRei]

Charge flows because of EMF; with no EMF, there's no current. But this is getting offtopic. The rate of flow (in either direction) in a fuel cell depends on the voltage. But the voltage will vary depending on the rest of the system. To an extent determined by Kirchoffs' laws. So I'm still failing to see how Kirchoffs' laws don't apply. Kirchoffs' laws are the solution to the case where there's more than one possible loop. Which is obviously applicable here. Heck, in a way a solar cell is a loop in and of itself, as it has a shunt path parallel to the EMF source...

It seems like you're picturing some sort of single loop system where you have everything wired in series. That's the only way I can picture that Kirchoff's laws would not be relevant.

The simplest possible case: switch wired in parallel with a solar cell and a fuel cell.

((removed ascii-drawing that the forum messed up))

Picture what happens to the fuel cell based on whether or not the switch is closed or not. If it's closed (near-zero resistance), power shunts through it. The fuel cell discharges. If it's open (infinite resistance) all of the power of the solar cell instead charges the fuel cell. The fuel cell is affected by what happens elsewhere in the system. You can't just say "the voltage is X" as if it's some simple fixed value; the voltage at any point needs to be determined by Kirchoffs' laws.

Maybe you're seeing something that I'm not. But I'm just not getting your argument that Kirchoff's laws aren't needed.

 

[Baluncore]

Maybe you're seeing something that I'm not. But I'm just not getting your argument that Kirchoff's laws aren't needed.

So long as you argue that you must use KL, you will be unable to proceed.

Attached to the two terminals of the fuel cell is a positive rail and a negative rail in the distribution box. Each load will be independently connected to those two rails by two wires. The PV panel also connects to those rails through two wires.
That is a star topology with everything in parallel, not series.

When the sun shines on the PV panel, it's voltage will rise above the fuel cell voltage and charge will flow into the FC, (and support some load). The PV panel voltage will be selected as sufficient when illuminated to exceed the FC voltage, the PV voltage will then be held down by the fuel cell as a PV panel is a current source and the FC is a fixed voltage device.

 

[Baluncore]

 

[KarenRei]

So in your diagram above R1 and R2 must be the internal resistances, and vary with load. Because they're absolutely essential to modeling a realistic system, as without them a tiny solar cell would be the same as a thousand wired in parallel, and a fuel cell stack with 1cm² layers would be the same as one with 1m² layers. There's no shunt path in the PV panel in the diagram, which is a problem with real world photovoltaics, but that's at least something that can be worked around (with the assumption that this is reflected in the generation curves and that there's some hardware somewhere to prevent discharge through that path at night)

How are you suggesting that R1 and R2 be calculated without Kirchoffs' laws? You have a multi-loop circuit in the above, with two sources of EMF.

Let's calculate a simplified version - let's remove R7 from the diagram, and everything to the right of there, so that there's only one load resistor (R3) connected to the panel and fuel cell (and their internal resistances). So we have two junctions (the ones drawn above and below the fuel cell), presenting:

i_pv + i_fc - i_load = 0
i_load - i_pv - i_fc = 0 (identical)

We also have two loops:

emf_pv - (R1 * i_pv + R3 * i_load) = 0
emf_fc - (R2 * i_fc + R3 * i_load) = 0

Set up as a series of equations:

1 * i_pv + 1 * i_fc + -1 * i_load = 0
R1 * i_pv + 0 * i_fc + R3 * i_load = emf_pv
0 * i_pv + R2 * i_fc + R3 * i_load = emf_fc

i_pv = (R3 * (emf_pv - emf_fc) + emf_pv * R2) / (R3 * (R1 + R2) + R1 * R2)
i_fc = (R3 * (emf_fc - emf_pv) + emf_fc * R1) / (R3 * (R1 + R2) + R1 * R2)
i_load =(emf_pv * R2 + emf_fc * R1) / (R3 * (R1 + R2) + R1 * R2)

And thus:

v_pv = R1 * (R3 * (emf_pv - emf_fc) + emf_pv * R2) / (R3 * (R1 + R2) + R1 * R2)
v_fc = R2 * (R3 * (emf_fc - emf_pv) + emf_fc * R1) / (R3 * (R1 + R2) + R1 * R2)
v_load =R3 * (emf_pv * R2 + emf_fc * R1) / (R3 * (R1 + R2) + R1 * R2)

Simplifying gives us:

v_pv = (R3 * (emf_pv - emf_fc) + emf_pv * R2) / (R3 + R2 + R3 * R2 / R1)
v_fc = (R3 * (emf_fc - emf_pv) + emf_fc * R1) / (R3 + R2 + R3 * R1 / R2)
v_load =(emf_pv * R2 + emf_fc * R1) / ((R1 + R2) + R1 * R2 / R3)

Or:

v_pv = emf_pv / (1 + R3 / R2 + R3 / R1) + (emf_pv - emf_fc) / (1 + R2 / R3 + R2 / R1)
v_fc = emf_fc / (1 + R3 / R1 + R3 / R2) + (emf_fc - emf_pv) / (1 + R1 / R3 + R1 / R2)
v_load = emf_pv / (1 + R1 / R2 + R1 / R3) + emf_fc / (1 + R2 / R1 + R2 / R3)

So, what can we conclude from this?

* Changing the effective internal resistance of the solar cells (aka, increasing the number of cells in parallel) alters the voltage drop and current across both the load and the fuel cell.
* Changing the effective internal resistance of the fuel cell stack (aka, increasing the number of layers in parallel) alters the voltage drop and current across both the load and the solar cells.
* Changing the load resistance changes the voltage drop and current across both the solar cells and fuel cell stack - independent of each other.

Is this relevant? Absolutely. The voltage drop across a fuel cell determines its overpotential, which determines its efficiency and effective internal resistance. The voltages across the fuel cells and solar cells determine their resistive losses. And obviously the voltage and current across R3 is of great importance. To determine the constraints on the system, I have to be able to determine how big to make the solar cell array (aka, how low R1 needs to be), how big to make the fuel cell stack (aka, how low R2 needs to be), and how low the wiring resistances need to be (aka, part of R3... more likely constrained by joule heating, but either way it has an effect on the required size of the solar cell array and fuel cell stack)

Again: how can you possibly get all of this without iterative application of Kirchoff's laws?

Note that what was described above was the simplest possible scenario one could envision. Simplier than your example diagram. Which was in turn simpler than my scenario (I have two distribution buses, an upper and lower bus, which are connected both to the power source and to each other... and technically there's two separate solar arrays and fuel cell stacks, but they're mirror images of each other, so there's no need to consider them separately here)

 

[Baluncore]

Is this relevant? Absolutely.

The voltage drop across R1 is irrelevant because a PV panel is a current source. R1 causes slight energy inefficiency.
The voltage drop across R2 is irrelevant as the load resistances are so much greater than R2.

* Surges are not an issue for this application, it's fine to assume whatever "buffer" is needed

That assumes a low impedance, fixed voltage supply.
How is power distributed in your country 115VAC or 230VAC, +/-5% ? Why is the distribution voltage said to be fixed ?

 

[KarenRei]

The voltage drop across R1 is irrelevant because a PV panel is a current source. R1 causes slight energy inefficiency.
The voltage drop across R2 is irrelevant as the load resistances are so much greater than R2.

The voltage drop across R2 is absolutely not irrelevant, as the overpotential is a key factor in the efficiency of a fuel cell. Less than the minimum and the reaction doesn't proceed. More than the minimum and it's wasted to heat. It also affects the internal resistance (aka R2 itself)

Both R1 and R2 affect the voltage drop across R3, as you can see in the above.

You can't just say "the load resistances are so much greater than R2" (or R1 for that matter) when determining the maximum acceptable values of R1 and R2 is one of the main goals . You're begging the question.

Please read the first complete sentence written in this thread, right at the very top:

Hi all. In a project I'm working on I'm needing to model a DC circuit to find the required power sources/storage to keep the system voltages up to a sufficient level.

The whole point is to size the solar system, fuel cell stack, and wiring. You're ignoring the first two parts of that and focus only on the wiring. I mean, well, clearly if you can just assume that the internal resistance on the power side is irrelevantly small then of course you don't need Kirchoff's Laws. But that's not the issue at hand. You're not going to add a square kilometer of solar panels and a fuel cell stack the size of a house to power an iPhone, or whatnot. You have to size things appropriate to the task at hand. And that requires solving for the appropriate size.

That established, can we get back to how one best determines the internal resistance of a piece of hardware based on its rated power consumption and voltage range (minimum voltage, nominal, or maximum? My thought is that they'd base that on the nominal voltage - right?) Also, trying to resolve whether an iterative multivariate refinement is best for sizing the solar system, fuel cell system, and wiring, or whether there's a better approach.

That assumes a low impedance, fixed voltage supply.
How is power distributed in your country 115VAC or 230VAC, +/-5% ? Why is the distribution voltage said to be fixed ?

230 VAC, but I'm not sure how this is relevant - we're discussing a non-grid connected DC system. If you're just asking that as a hypothetical, it's because the grid supply presents a trivially small effective internal resistance versus the loads an individual consumer may present. Unlike the solar array and fuel cells in this scenario, which are to be sized to the minimum required size to prevent an unacceptable voltage drop across the loads.

 

[Baluncore]

I am sorry but I can't help you.

 

[KarenRei]

It's okay. I'll assume the rating is for nominal voltage, and that an iterative approach to multivariate optimization is best.