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I was studying the manuals of the Horizon 1000XP Fuel Cell and I came across this chart: enter image description here

First Question: Lets assume that I put a specific resistance to the fuel cell output. How can I find the operating point of the fuel cell? As I see it, neither the voltage, nor the current are stable. I also don't quite understand how can the voltage depend on the current being drawn. In classic circuits, one has a certain voltage and a certain resistance, and can then find the current (I=V/R).

(In the manual it says that when the load is zero (also current is 0) the voltage has its max value. When a load is connected, a current flows through it. This in turn should drop the voltage a bit, which would also cause a decrease in the current and so on..)

Second Question: I have read that to increase the power sent from the fuel cell, you need to increase "the load", or "the current drawn from the load". Assuming that I have only ohmic loads, how can I control the current flowing through my load? For example, lets say that I want 1000W to be given from the fuel cell. What should my load be in that case?

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    \$\begingroup\$ A resistor establishes a fixed, linear relationship between current and voltage. If you plot the current-voltage relationship for a given resistance on the above graph, you'll get a straight line. The place where the line representing the resistor intersects with the blue "voltage" curve will be the operating point. \$\endgroup\$ – Solomon Slow Mar 5 '18 at 15:39
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    \$\begingroup\$ Having chosen load R, simply plot it as the straight line V=IR, through the origin (where V and I are both 0) and see where it intersects the voltage curve. The X axis value of that intersection is the current. \$\endgroup\$ – Brian Drummond Mar 5 '18 at 16:07
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The graph is showing you that the load cell has a significant terminal resistance. That means you lose volts at the terminals of the load cell the more current you take.

schematic

simulate this circuit – Schematic created using CircuitLab

This is typical of any power source.

Based on about a 10V drop at 20A, that gives you a ballpark resistance of 0.5 Ohms.

As for your power requirement. You use the graph to find the crossover point at which the thing will nominally be operating at the power you are requiring.

enter image description here

So for about 1000W, the voltage will be around 35V and the current will be @28A. Your load therefore needs to be @1.25 Ohms.

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One way to find the voltage and current given a specific resistance is to estimate the approximate working point, linearise the voltage at that point and then solve for both voltage and current.

For example, let's assume your resistance is 50 Ohms, which will result in approximatively 1A of current. At 1A, the V curve looks like V = 47-I, which gives

I = V / 50
V = 47 - I

V = 47 - V/50 -> V = 47,9V
I = 47,9/50 = 0,96A
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