sampling 240VAC

Sep 11, 2009 34 Replies

I'm looking for the best way to sample the 240VAC waveform from a powerline well staying within my design constraints. I would like to go transformerless because of size requirements and I would like to have a fairly high bandwidth, so I am looking into other options other than a high impedance resistive ladders. Does anyone have any suggestions on how to do this in a modern fashion? What about using a hall effect IC and a load resistor to convert the current back to a low enough sampling voltage?



Thanks, Thomas


Hack a 'scope and use its attenuator?

Good Luck! Rich

I looked into a few schematics of scopes and looked at their attenuators. The circuity looked a little tweeky with the use of communication capacitors to overcome the RC input capacitance. I'm kinda looking for an 'outside of the box' modern approach (if it exsists).

Thomas

Use a couple of capacitors as a voltage divider?

Thanks John,

I hadn't really thought of capacitors as voltage dividers and you got me thinking, which lead me to determine that the reactance and frequencies would make it difficult to design anything that would be stable. First off you would be AC coupled which would prevent you from seeing any low frequency or DC offsets. Secondly, the large capacitor values needed to divide the low frequencies would be subject to inductive reactance at the higher frequencies and also subject to thermal tolerances.

It would probably be wise to steer clear of any reactive components, when trying to design a higher bandwidth analyzer.

Thomas

I once bypassed a large resistive divider to stop an HVPS "hunting". It didn't get approved, because every arc on the HV wrecked the input circuit, but if you can protect your input from LARGE glitches, and/or don't expect arcs, you might be able to get away with it.

Good Luck! Rich

You can get some farily small

Actually, the capacitor values required are quite modest. For example, if you had a divider configured to give you 2.4V out with a 240V 60Hz input and a 1000 ohm output impedance it\'d look something like this: (View in Courier) 240VAC>---+ | [26.8nF] Xc = 99000R | +----->2.4V | [2.65µF] Xc = 1000R | 240VAC>---+----->COM Regardless of the spectral content of any transients on the line, (well, within reason) the division ratio would stay constant and transients would show up faithfully overlaid on the background sinusoid. Admittedly, the values of the capacitances are zany, but if you\'re going to build a zillion of your widgets you could probably find someone in China who\'d build them for a song. Well, a symphony, maybe. But, I agree with that if you\'re trying to measure DC offset on AC mains AC coupling probably won\'t work Are you free to say what is it, exactly, that you\'re trying to do?

"Thomas Magma is a Pisshead "

** Don't expect to see frequencies below 50Hz on the AC supply and there are no " DC offsets " as such - but the average value may sometimes become non zero due to uneven negative and positive peak values.

There is nothing wrong with using a resistive divider to monitor the mains active relative to ground.

** ROTFL !!

You are talking straight out your arse - fool.

** The caps would be at the same temp - fool.
** You have no idea how to design anything.

There is STILL nothing wrong with using a resistive divider to monitor the mains active relative to ground.

..... Phil

"John Fields"

** The OP is a trolling fool with no clue about anything - consequently he has not considered the safety angle at all.

But YOU should have !!!

What you have drawn is potentially lethal.

.... Phil

Yeah, it was a bad idea. Thanks. :-)

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Hmm, no resistors or transformers. (why not?) How about an opto- coupler? Your still going to need a big R to limit the current to the LED... and I have no idea if it would be at all linear at lower voltages... well of course it's going to turn off when the voltage gets below 2 V or so....

George H.

So the problem is with a current measurement, not with the AC potential? Hall sensors are poor choices unless your fields are large (and large fields are the province of folk who can make room for thick wires and have large box dimensions available).

Transformers are cheap, rugged, hold calibration well. Use 'em. In fact, you can use a current-limit resistor into a current transformer circuit to pick off the line voltage, and a second (high attenuation) current transformer to sense the current. Heck, instead of going for the awkward high turns ratio in a single current-sense transformer, use two in cascade, instead!

"Well, there's transformer, transformer, eggs, bacon, and transformer, that's not got much transformer in it..."

"> "Well, there's transformer, transformer, eggs, bacon, and transformer,

Hee Hee. Yeah why not use a transformer? It's got everything going for it. It's safe, you can pick the turns ratio, it's safe....

George H.

I take it the risk is with the 26.8nF shorting and putting line voltage on the output? How about replacing the 26.8nF with 4 series 107.2nF 1000 Volt caps, with a

5.1V zener on the output and a suitable fuse? If one 107.2nF shorts the output voltage goes high causing the zener to conduct and opening the "suitable fuse". Heck, let's put a crowbar on it. Running for cover :-) Mike

What's wrong with resistors? 10M scope probes work into the hundreds of MHz. At MHz sorts of frequencies, a purely resistive front-end will work fine.

John

ine

Power circuits with 240 VAC (in the common US fashion) aren't suitably ground-referenced. Even 120V has a neutral return wire, which must be allowed to deviate significantly from local ground potential. So it takes true-differential high voltage input, which is going to have at least one amplifier in addition to a 'resistive front-end'.

Sell, sure, it has to be differential if your circuit's not riding on neutral. Two HV resistors, two regular resistors, one opamp. I've also done it with three or four surface-mount resistors in series as the high-side resistor.

John

"amdx"

** NO !

.... Phil

quently

age

Phil, If I ask nicely will you tell us what is the problem with the series capacitors?

Please,

(I'm only a physicist and can be ignorant about some practical sides of electronics)

George H.

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