sampling 240VAC

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

Reply to
Thomas Magma
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Hack a 'scope and use its attenuator?

Good Luck! Rich

Reply to
Rich Grise

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

Reply to
Thomas Magma

--
Use a couple of capacitors as a voltage divider?
Reply to
John Fields

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

Reply to
Thomas Magma

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

Reply to
Rich Grise

You can get some farily small

Reply to
David L. Jones

--
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?
Reply to
John Fields

"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

Reply to
Phil Allison

"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

Reply to
Phil Allison

--
Yeah, it was a bad idea.

Thanks. :-)
Reply to
John Fields

ne

gh

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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.

Reply to
George Herold

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..."

Reply to
whit3rd

"> "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.

Reply to
George Herold

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
Reply to
amdx

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

Reply to
John Larkin

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'.

Reply to
whit3rd

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

Reply to
John Larkin

"amdx"

** NO !

.... Phil

Reply to
Phil Allison

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.

Reply to
George Herold

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