powering electronics by current flow

I would suggest delaying the conversion from A to D until after it had been transmitted and received.

For example, the current from the transformer could be applied to a coil whose movement in a magnetic field was controlled by, say a spring/torsion device plus a bit of mass damping. A mirror could be attached to the coil and a remote light source shone on it. An array of photosensors could be positioned on a cricular arc and able to detect the degree of deflection of the mirror and hence the coil.

What do you think? It could be called a photo-reflective galvanometer... or something ;)

-- Sue

| I would suggest delaying the conversion from A to D until after it had | been transmitted and received. | | For example, the current from the transformer could be applied to a coil | whose movement in a magnetic field was controlled by, say a | spring/torsion device plus a bit of mass damping. A mirror could be | attached to the coil and a remote light source shone on it. An array of | photosensors could be positioned on a cricular arc and able to detect | the degree of deflection of the mirror and hence the coil. | | What do you think? It could be called a photo-reflective galvanometer... | or something ;)

But how do you make sure the transmission path is accurate?

Have you explored the laser gyro that Kearfott builds. They have doing this kind of thing for about 50 years. Try:

Oh, that's just mechanics. You could split the source light beam and use the two differentially, with a coaxial unenergised coil and mirror assembly and an equal length light path as the reference.

Or use a couple of different wavelength lasers as the source and pass the light through a pair of gratings, instead of mirrors. Then count the nulls and peaks in received intensity.

Or simply put a rare Earth alloy in the coil and use (linear) Faraday rotation of polarisation with magnetic field strength to change the polarisation of a laser beam incident on it. No moving parts at all, that way. Just remotely measure the degree of polarisation shift of the reflected laser light.

-- Sue

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Piece of cake! Of course you need some numbers to see which ball park you are in. As an example, lets assume that the AC current you want to measure is between one amp and 100 amps, 60Hz. Lets further assume that you allow

Let the electronics on the secondary side of the current transformer work on

The sense resistor on the secondary of the current transformer is in series with the electronics power supply and has a value of .05 volts at.02 amps (1 amp divided by the ratio) = 2.5 ohms. The electronic power supply has a shunt regulator (maybe a 2 Amp zener) that absorbs the up to 2 amps on the secondary of the current transformer so the electronics is not overvoltaged at sensed currents up to 100 Amps.

A micro processor with an ADC senses the voltage on the series sense resistor and converts it to digital then drives an Infrared LED with a serial pulse train representing the sensed current. Low baud rate RS 232 or other serial transfer is used to operate the LED and is generated in the micro code and internal UART.

Another processor with a phototransistor and internal UART is used to receive, detect and decode the infrared beam. You do whatever you want with this digital output like drive a display, etc.

This is just an example. Much lower power than 100 milliwatts in the electronics is possible with the right processor and judicious use of LED power at low rep rates.

| Or simply put a rare Earth alloy in the coil and use (linear) Faraday | rotation of polarisation with magnetic field strength to change the | polarisation of a laser beam incident on it. No moving parts at all, | that way. Just remotely measure the degree of polarisation shift of the | reflected laser light.

I like this one best.

On Wed, 9 Apr 2008 13:56:32 -0700 Bob Eld wrote:

| Piece of cake! Of course you need some numbers to see which ball park you | are in. As an example, lets assume that the AC current you want to measure | is between one amp and 100 amps, 60Hz. Lets further assume that you allow | 100 milliwatts to power the electronics and that the sense resistor on the | secondary of the current transformer produces .05volts per amp, 5 volts at | 100 amps. | | Let the electronics on the secondary side of the current transformer work on | 5 volts at up to 20mA = .1 Watts. Let the current on the primary of the | current transformer create that required .1 watt plus a little for the | sensing at one amp sensed. Therefore the voltage on the primary must be | .1W/1A = .1 Volt. plus a little for sensing. The current transformer must | step up the .1 volts to 5 volts or a ratio of 1:50. | | The sense resistor on the secondary of the current transformer is in series | with the electronics power supply and has a value of .05 volts at.02 amps (1 | amp divided by the ratio) = 2.5 ohms. The electronic power supply has a | shunt regulator (maybe a 2 Amp zener) that absorbs the up to 2 amps on the | secondary of the current transformer so the electronics is not overvoltaged | at sensed currents up to 100 Amps. | | A micro processor with an ADC senses the voltage on the series sense | resistor and converts it to digital then drives an Infrared LED with a | serial pulse train representing the sensed current. Low baud rate RS 232 or | other serial transfer is used to operate the LED and is generated in the | micro code and internal UART. | | Another processor with a phototransistor and internal UART is used to | receive, detect and decode the infrared beam. You do whatever you want with | this digital output like drive a display, etc. | | This is just an example. Much lower power than 100 milliwatts in the | electronics is possible with the right processor and judicious use of LED | power at low rep rates.

While it certainly would not be the same model used in both cases, consider the extremes:

1. 240 volts, 100 milliamps to 400 amps working range
2. 765000 volts, 50 amps to 5000 amps working range

Without giving it too much thought, how about a full-wave bridge with a resistor and zener on the DC side and the current transformer on the AC side?

The 'trouble' with current transformers are they are usually rated for a specific 'burden'. If your resistor has too high a value, the voltage would rise across it to dangerous levels. That's where the zener comes. So size the resistor to give you enough voltage to power things at the lower limit of current including the measuring circuit current. BTW, the lower limit of current has to be large enough to power your circuit at that voltage. Size the zener to be able to carry full current from the CT.

I'm sure there are other ways, but this is one idea that comes to mind...

daestrom

You might want to do a little more research on this. You can actually do this without any power at the CT. You simply do not use a conventional CT. There are optic materials whose properties change in the presence of magnetic fields. By controlling the geometry (how close to the conductor you install the sensor) you can determine the current by these changes. Of course this involves using optic fiber and lasers. At my office I believe I still have some materials on this. Look in T&D World magazine. There have been articles on optic CTs.

As for using a normal magnetic CT and communicating down the information, it is doable. We have played with different designs for powering devices from current flow in our lab. If you need very low power levels you can do ok with 10s of amps flowing.

Charles Perry P.E.

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The voltage really doesn't matter because it is a current transformer situation. With infrared coupling, the extreme high voltage can easily be isolated while at 240 volts, transformer isolation should be sufficient. In otherwords, there are two places for isolation. so getting the required isolation level and common mode rejection should not be impossible.

One issue is the dynamic range. In my above example, the dynamic range is

The other voltage should be easier, however, the dynamic range is much greater, 4000 to 1. If you allow a one volt drop on the 240 volt line across the transformer primary, 100 milliwatts will be available for the electronics. The rest is similar to the example, but, depending on required accuracy which must be at least one in 4000, a minimum 12 bit system is in order and 16 bits would be better. The minimum signal to noise ratio will be about 72 dB, that's doable.

Are you sure the high voltage values are AC? This kind of voltage is typical of DC transmission lines. Obviously you can't use a simple current transformer on DC.

A number of years ago there was some experimentation using lasers which may have been something along this line. I have no idea of how it turned out. Possibly weather disturbances were a problem, accuracy another problem. -

| Are you sure the high voltage values are AC? This kind of voltage is typical | of DC transmission lines. Obviously you can't use a simple current | transformer on DC.

I'm only considering AC. There are AC transmission lines at that voltage. DC would be a different challenge.