How can I measure DC power with i2c?

Question

I am doing a project about a windmill. I have to make a controller system for this windmill The first thing I want to know is what is it generating?

So after the 3 phases there is a rectifier and after this I want to measure.

The windmill can generate at average about 50 Volts DC and 10 Amps of current. So this will make a power generated of about P = U * I > 50 * 10 = 500 Watts Right?

Now the problem is I want to measure voltage and current. And there are a lot of ways like a voltage divider with a voltage follower with a opamp that creates high impedance and will let little current through. That's for the Voltage.

But then I thought may be if there is a chip/board that can measure both and calculates this and then sends this data to a i2c line.

Now I found a LTC2946 chip
I actually don't understand how to set this up. The problem this chip needs a sense resistor. On that page they have chosen a 0.02 Ohm sense resistor.

So let's do a little math.

So I know the value of the resistor right? >> 0.02 Ohm Say my current is 10 Amps >>

Ohms law = V = I * R right? So my voltage will be 10 * 0.02 = 0.2 Volt or 200 mVolts So can I say then that I need a > 0.2 * 10 = 2 Watt High Power Resistor Is this right?

Say my current will be 15 Amps. That make 15*0.02 = 0.3 Volts This will make > 0.3 * 15 = 4.5 Watts of High Power Resistor right? I don't know if I am doing this right !

If so then the next problem will occur.

How can I measure voltages with this chip? I don't understand.

If you guys know anything, help is appreciated.

PART 2

Now I found something the LTC2947. The last one was LTC2946 and this one needed a external sense resistor. Take the LTC2946, which has a full-scale voltage of approximately 100mV. If the LTC2946 is used to measure a 30A rail, a 3.3 mili Ohm sense resistor is required. This is readily available but will have to dissipate 2.9W of power !!!!.

Now that is a lot of dissipation of power for a simple measurement. Furthermore the package is not "standard" and therefore expensive.

Then I saw the LTC2947 chip. Datasheet

This chip has a integrated sense resistor of 300µ Ohms. So this is a very efficient chip with high accuracy.

I have looked in the datasheet and found a the section "48V Bidirectional Power, Energy and Charge Monitor with Isolated I2C Interface and High Side Sense"

So I don't think I am gonna use the LTC2946 because of the high power loss.

But the next problem occurs. Does this chip measure voltages beyond 48Volts ? Because the windmill can generate higher than that, it depends on the wind.

If not how can I make this work? In which way?

Answer 1

To answer the parts of your question that deal with power rating for the current-sense resistor.

It's good that you are paying attention to such things. Yes, it's not uncommon for a current sense resistor to have a fairly small value and a fairly large power rating. That's a norm.

For an example, look at the development board for the LTC2946. The manual contains a schematic. The resistor in question is designated RSNS. The manual also contains an abbreviated BoM (bill of materials). RSNS is described as a 1.0W resistor in the BoM. 1.0W is comparable to the power rating which you have calculated for your application (don't forget to add safety margin, by the way).

Answer 2

You don't need a chip on a IIC bus or something. You must already have a microcontroller. You can use a analog high side current sensor, then feed the result into the A/D you already have. The high side current sensor usually creates a low side output current. With a resistor, you convert this to a voltage, which goes into the A/D.

As a example of a high side current sensor, see the INA169.

Answer 3

You can make this work, but you need to be very careful. Most importantly, you do not need to worry immediately about the average voltage and current - you must pay attention to the MAXIMUM voltage and current. Windmills are notoriously exposed to occasional high winds, and these form the stressing conditions.

In the 48 volt circuit which you are looking at, the components at lower left use Z1 to generate a floating ground, called FGND and indicated by the solid triangle, and the difference between +48 and FGND serves as the power voltage for the chip. More importantly, it also serves as the power for the optocouplers, which have a maximum recommended voltage of 5.5 volts. The 2647 itself can run up to 15 volts. The current measurement works normally, but the 48 volts is reduced by R10/R11 to 4.00 volts at an input of 48 volts.

Because this 4 volts is tied (effectively) to the Vdd of the 2647, at 48 volts the negative measuring point, VM, should be at about 1 volt above FGND. As a result, the circuit should work well up to about 60 volts, at which point VM will be nominally at zero volts WRT FGND. But this is where things get iffy on a windmill. If you get a gust of wind which causes the input to exceed 60 volts, VM will be below FGND and you will exceed the input limits, with no guarantee what will happen. Plus, I have not done and analysis on the level generator for FGND, so it's possible that one or more parts will die at higher than 48 volts. You would need to explore this.

In many respects the 2647 is clearly better than the 2646, but it is limited to a 15 volt input, which is why it needs this rather klugey workaround for higher input voltages. The 2646 will work up to 100 volts in, and is probably a better choice for something like your windmill, since it will handle transient high inputs better. Of course, you'll also need a voltage measurement device and you'll have to do the power multiplication yourself, but the overall system is arguably more robust.

And I'm not sure about why you are so obsessed with saving power. The 2646 (if you read the data sheet) works with a 100 mV input range on current sense, so measuring a 10 A current will only dissipate 1 watt with a .01 ohm resistor. Note that this is .2% of the total power produced, and you will lose a good deal more than that in your system wiring. If you're worried about dissipating power in your controller, don't be. You should use an external current shunt to generate your current measurement anyways. First, running 10 A around on a PC board is certainly possible, but getting into the range where if you are completely inexperienced you may get yourself into trouble. Furthermore, making a precise voltage measurement of the drop across a resistor is not difficult, exactly, but you need to be careful with layout and provide a kelvin connection. Using an external shunt with a simple twisted pair to connect to the controller makes for a more robust (although more mechanically complicated) unit. As a point in the favor of the 2647, the chip designers have done the hard work of producing a good shunt/sense setup right on the chip, where you don't need to worry about it. Running a small-diameter ground sense from the power ground allows a simple 3-wire connection to your sensor board without the need for running heavy wires into and out of it, which will simplify board mounting and sensor wiring.