Nonrelativistic particles radiate like dipoles, i.e. the pattern has a null along the direction of motion and a doughnut wrapped round it. Relativistic particles do the same in their co-moving frame, but in the lab frame, the radiation is concentrated near the direction of motion. It's an effect of Lorentz transformation.
However, there's still a null in the actual direction of motion.
Cheers
Phil Hobbs
Classical fields have momentum too.
Cheers
Phil Hobbs
No. Immagine a hall sensor with a permanent magnet.
ellipse by
time.
meaning.
Thimking.
?-)
handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?
Yes. And there are a handful or so of new versions tailored specifically to that purpose. Search for "positive rail current monitor".
dn apisumop
?-)
I think it's one of them quantum thingies!
Maybe zero point energy!
Jamie ;)
Positive rail current monitor? Like a high side shunt monitor? Does this apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?
Most likely just for DC.
The simples high side current monitor would be shunt between a PNP transistor emitter and base and a high resistance from collector to ground. Split the resistance into two resistors and get a nice voltage (say 0..5 V) from that voltage divider proportional to the current through the shunt.
For AC only, simply use a current transformer.
For AC+DC, use a isolated power supply to feed a V/f chip sitting on the shunt and send down the frequency modulated signal through capacitors and/or isolation transformers and then apply f/V conversion.
For real high common mode voltages, say 400 kVac, use a laser injected into one fiber to power the current measurement device sitting on the EHT line, then use an other fiber to digitally transmit down the measurement.
Not the safest thing to do but with the appropriate gain, the high common mode voltage is no problem. If you have a gain of 1/50, the common mode voltage scales inside a single 5V rail. Of course the differential signal is also divided by 50 and common mode rejection is dependent on resistor tolerances so there may be other problems.
can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?
specifically
apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?
Only for DC rails so far as i have seen.
?-)
ellipse by
of time.
meaning.
It actually works for classical physics as well. This is one of my problem areas going through college.
?-)
Really? The common mode voltage scales down?
Isn't common mode voltage (Vin,inv + Vin,non-inv)/2? so regardless of the gain the input of my opamp would see all 110V right? or maybe I'm misunderstanding something here....
WoW ! That's an impressive way to measure HV current. Hope I never have to do that, quite.
1000V is high enough for me.boB
The best solution to high common mode voltage is, of course, an isolation amplifier (input circuit has its own isolated power supply, output circuit is by transformer or optocoupler, or by voltage-frequency conversion and acoustic coupling...).
In an AC system, of course, you'd need 'positive rail current' and 'negative rail current' sensing, or the job's only halfway done! For wye (three phase) systems, that'd become 'only one third done'.
Look at this way, you have two voltage dividers, one before the shunt, the other after ther shunt. Assuming the shunt will generate a 1 V voltage drop and you are using 1/50 _ideal_ voltage dividers, you are going to measure a 20 mV voltage differences.
Completely doable with low voltages and realizable with existing resistor tolerances for a few tens of volts, but after that, no hope.
Really! Honest injun!
No, the non-inverting input resistor and the resistor from the non-inverting input to the reference form a voltage divider. If your gain is 1/50, you have a similar voltage divider between the shunt resistor and the opamp. As long as the opamp's output is between the rails, the inverting input is at the same voltage as the non-inverting, so the same holds on the inverting input.
This is trivial with current transformers.
At EHT (above 100 kV) things get a bit more complicated.
The required precision of the resistors gets crazy as the ratio of bus voltage to shunt drop goes up.
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