Simulate potency

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Nuno
Nuno 2011-7-15
Hi... I need to put something that simulates a load with power 1000W for exemple... I thought about putting the next circuit:
[Edit: Use '<<>>' to post pictures. -AU]
But, how do I choose the resistance value? I know that P=I^2.R If i have P=100W what the value of R i put? There are any way to make this?

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回答（3 个）

Doug Eastman
Doug Eastman 2011-7-15
Rather than a regular Resistor block, put a Current Sensor in series with a Variable Resistor. Square the measured current using the PS Math Function block, divide that by the desired power from a PS Constant block and then feed the result into the Variable Resistor.
To keep it well behaved, you may also want to consider adding a saturation block after the division to set some limits on the resistance value (1e-7 to 1e7 for example).
Here is an example circuit:
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Nuno
Nuno 2011-7-15
But as I do that?
Put another Current Sensor? Like this? Where i connect the other ports?
http://img34.imageshack.us/img34/904/75927358.png
Doug Eastman
Doug Eastman 2011-7-15
Hopefully the above image makes it clearer.

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Nuno
Nuno 2011-7-16
I already implement this... The image is this: http://img197.imageshack.us/img197/726/38245137.png
in fact, what is the difference between circuit and that it had before?
But the problem continues because i need that the resistence simulate charge with for exemple 50W. how do I do this?
  1 个评论
Doug Eastman
Doug Eastman 2011-7-17
You're missing several blocks in your circuit that I have shown in my circuit: PS Math Function, PS Divide, and PS Constant. The PS Constant is where you enter the power that you want that resistance to drain (for example, 50 for 50 W).
You circuit is simply feeding the current value in as the resistance, the idea is to use current to calculate resistance based on the forumla: R = P/I^2

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Nuno
Nuno 2011-7-19
But how i get control de current? Because in this case, the current is always very low (~0.02A)... This makes that resistence is very high... How i can controlled this? The normal current should be ~2 or 3A...
  2 个评论
Doug Eastman
Doug Eastman 2011-7-20
There will only be one value for resistance that will give you the correct power consumption - that will also dictate the current. If the current is not what you expect for a given power consumption, it probably means there's a problem in the rest of your circuit.
Doug Eastman
Doug Eastman 2011-7-20
Also - some of the errors you are seeing may be because it's simply not possible to draw that much power at this point in the circuit. It is probably a good idea to avoid other resistances in series with this element.
Imagine you have a simple circuit with a 10V source, a 10 Ohm resistor and your special resistor in series. There is no way that your resistor can use 100W of power, so in that case you'll get an error when you try to simulate.

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