Amplifier for Rogowski Coil

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hiii guys

im doing my FYP on amplify the voltage from a rogowski coil

just to make it short
i have already design the coil
which can detect a 50k frequency and gives pick-to-pick = 10mV Ac
now, i have to design an amplifier that can give an actual value.

so, i have already got a circuit that can be use for this type of current measurements
and my problem is to recalculate it to make it suitable for my coil

and im using LT1630 Op amplifier

thanx alot
 

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  • #2
berkeman
Mentor
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hiii guys

im doing my FYP on amplify the voltage from a rogowski coil

just to make it short
i have already design the coil
which can detect a 50k frequency and gives pick-to-pick = 10mV Ac
now, i have to design an amplifier that can give an actual value.

so, i have already got a circuit that can be use for this type of current measurements
and my problem is to recalculate it to make it suitable for my coil

and im using LT1630 Op amplifier

thanx alot
The integrator first stage makes sense for a Rogowski coil, but I don't understand what the 2nd opamp in the feedback path is for. Can you explain the circuit to us?
 
  • #3
12
0
yeah even me i didnt get it well
but they explain to me and said
" The PI feedback adds a 2nd order high pass to the integrator characteristic. The capacitor sets the second high pass corner at
about 0.3 Hz, the first at about 1.4 Hz is set by the feedback gain. To get a suitable roll-off behaviour, both corner frequencies
have to be scaled "
that what i under stand from them

is there is another way to rearrange this circuit and make it easy for the calculation ????
 
  • #4
4,662
5
I think your major drift problem is going to be the input voltage offset temperature coefficient. If you have a 1 microvolt input voltage offset (temperature coefficient), this (with the 220-ohm resistor) will generate a current into the (-) input of 4.5 nanoamps, which across the 10 nf integrating capacitor is 0.45 volts per second drift. You might want to increase the 220-ohm series resistor to reduce this. What is the resistance of the Rogowski coil? The grounding resistor on the (+) input should be equal to the series resistor on the (-) input to minimize the input bias current offset. I have sometimes used a mult-megohm active feed back to the (-) input to zero the output voltage with a long time constant. How do you plan to discharge the integratig capacitor? How long is your integrating time? What is area and number of turns of the coil, and the B field?

Bob S
 
  • #5
12
0
you really make me blurrrrrrrr
so, you mean the above circuit is totally wrong ??

the coil contain 160 turns

i conntnect the function generator to produce AC Voltage ( 14 V ) through a resistor ( 100 ohm ) and i let the wire to pass through a resistor

when i increase the frequency , the output Voltage of the coil will increase

for example if i give

__________________________
function | Output |
generator | Voltage of the coil |
-------------------------------------
100K Hz | 20 mV |
200K Hz | 40 mV |
300K Hz | 60 mV |
400K Hz | 80 mV |
------------------------------------

but now
my problem is the amplifier doesnt work well :(

what should i do ??
 
  • #6
4,662
5
The feedback element in your input amplifier is a 10 nF capacitor, which has an impedance of =-318j ohms at 50 kHz, but is open at dc. The LT1630 opamp has an open loop gain of ~3500 volts per millivolt at dc, so a very small input offset voltage will produce a dc current that will be integrated by the 10 nF capacitor, causing the amplifier output to hit the rail in a few seconds. As one example, suppose the input voltage offset is V =1 microvolt. Then the current through the Rogowski coil and the 220 ohm resistor will be V/220 = I =4.5 nanoamps. The voltage slew rate across the 10 nF capacitor is then dV/dt = I/C = 4.5 microamps/ 10 nF = 0.450 volts per second. The amplifier output will hit the rail in less than 50 seconds. The easiest correction to the amplifier is to put a 50k-ohm resistor in parallel with the 10 nF capacitor. In this case, the dc gain is 50k/220 = 227, so a 1 microvolt input offset will produce a 0.227 millivolt output offset.

If this dc output offset is not a problem, then use the 50k resistor. A large ac capacitor (~1 uF) in series with the 220-ohm resistor will help, but not completely prevent the 10 nF capacitor from integrating any residual input voltage and current offsets.

What kind of signal are you trying to measure with the Rogowski coil? Do you have a specific application in mind? Rogowski coils are usually used to measure the dc magnetic fields around wires carrying large dc currents. Are you trying to measure an ac magnetic field at 50 kHz, or a rapidly changing magnetic field? Or is this an antennal for receiving a 50 kHz electromagnetic (transmitted) wave?

So, thoroughly evaluate the dc characteristics of your circuit, including all input offset currents, input offset voltages, and input voltage temperature coefficients.

Bob S.
 
  • #7
12
0
thanks bro for the information you gave me
but unfortunately im measuring Ac Voltage

im using a function generator to preduce an AC volt that pass through a ressistor , so, there is a current passed through a resistor

buy using the coil i can measure the current , by the relastionship between the frequency and the output of the coil
i can know the current
V=IR, so--> I=V/R

so, you advice me to put 50k in parallel with a capacitor
 
  • #8
4,662
5
so, you advice me to put 50k in parallel with a capacitor?
Yes. The 50k-ohm resistor is to control the dc behavior of the circuit.

Bob S
 
  • #9
12
0
What kind of signal are you trying to measure with the Rogowski coil?
answer --- > im using sin wave signal

Do you have a specific application in mind?
answer --- > measuring a frequency from 100k Hz to 1M Hz

Rogowski coils are usually used to measure the dc magnetic fields around wires carrying large dc currents?
answer --- > use to measure an AC Volt that produce from a function generator

Are you trying to measure an ac magnetic field at 50 kHz, or a rapidly changing magnetic field?
answer --- > measuring from 100k Hz to 1M Hz

i hope i give a clear idea about what im doing now :)
 
  • #10
12
0
Yes. The 50k-ohm resistor is to control the dc behavior of the circuit.

Bob S
but im measuring Ac not DC
 
  • #11
sophiecentaur
Science Advisor
Gold Member
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4,421
. . . Rogowski coils are usually used to measure the dc magnetic fields around wires carrying large dc currents. Are you trying to measure an ac magnetic field at 50 kHz, or a rapidly changing magnetic field? Or is this an antennal for receiving a 50 kHz electromagnetic (transmitted) wave?

So, thoroughly evaluate the dc characteristics of your circuit, including all input offset currents, input offset voltages, and input voltage temperature coefficients.

Bob S.
How does a Rogowski Coil work for DC? It's basically a form of transformer and I don't see how a flux change can occur for DC to induce a emf.
 
  • #12
4,662
5
but im measuring Ac not DC
You are MEASURING AC, but the amplifier has DC drifts due to the input offsets I mentioned, and without the 50k-ohm resistor, the dc amplifier gain is 3,500,000. If you put the 50k resistor in, the dc gain is only 50,000/220 = 227. Read the LT1630 datasheet, page 3:

http://cds.linear.com/docs/Datasheet/16301fa.pdf [Broken]

Read the line "Large signal voltage gain" = 3500 volts per millivolt..

(from sophiecentauer) How does a Rogowski Coil work for DC? It's basically a form of transformer and I don't see how a flux change can occur for DC to induce a emf.
Although the Faraday Law as normally written applies only to ac fields, it can be written in the following way:

V = -(d/dt)N∫B·dA volts (Faraday Law)

so by carrying the dt to the other side and integrating we get

∫V·dt =V·(t2 - t1) = -N∫B·dA = - N·(B2 - B1)·A volt-seconds

A common use is to put a Rogowski coil (N turns, area A) in a dc magnetic field, start a voltage integrator, turn off the field (or flip the coil over), and read the integrated voltage (volt-seconds) on the integrator output to get N·B·A (or 2·N·B·A). The integrator is an opamp with a good low-leakage capacitor in the feedback loop.

Bob S

[added] For the specific opamp integrator with a series input resistor R=220 ohms, and a feedback capacitor C=10 nF, the opamp voltage output is

Voutput = (1/RC)∫t1t2 Vcoil(t) dt = 455,000∫Vcoil(t) dt
 
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  • #13
12
0
thanks bro for a clear information you gave
i will try to do my expriment on this coming monday on the lab and see how its going to work

wish me all the best
and thanks again
 
  • #14
sophiecentaur
Science Advisor
Gold Member
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Bob_S
You mean that the DC current is measured, effectively, by what happens at switch on? It's a sort of 'dead reckoning' method.
It's interesting but why not use Hall effect - which wouldn't be subject to the sort of drift errors that an integrating method must suffer from?
 
  • #15
4,662
5
A) You mean that the DC current is measured, effectively, by what happens at switch on? It's a sort of 'dead reckoning' method.
B) It's interesting but why not use Hall effect - which wouldn't be subject to the sort of drift errors that an integrating method must suffer from?
A) If the switch (short across feedback capacitor) is opened and the Rogowski coil is not moving and the magnetic field is constant, there is no voltage to integrate, so the opamp output does not change. If you remove the probe from the magnet, then the opamp output voltage is = N·B·A/R·C.

B) Rogowski coils and integrators have been around for over 50 years. The first used vacuum tubes. They are very accurate (fractional %) and stable. The RC time constant can be measured electronically using a stopwatch and a precision voltage source. Rogowski coils were often calibrated against proton NMR (nuclear magnetic resonance) probes to get ~0.1% accuracy. (NMR probes are better than 1 ppm).

Hall Effect probes have only recently become available (last 10 or 20 years), and are only ~10% accurate, unless they are calibrated against an NMR probe. Calibrated Hall Effect probes are expensive. The Rogowski circuit in this thread was selected by the OP.

Bob S
 
  • #16
sophiecentaur
Science Advisor
Gold Member
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BoB S
I see now where your concern about DC drift comes in!
 
  • #17
12
0
hi
i have one question bro

now i have to simulate may expriment by a Pspice
but im facing a problem in getting the parameters of Rc, Lc and Cc

beacuse i found many equaction for them and im confusing which one is to use
and my second question is
what is the conductivity Value of copper ?????
 
  • #18
4,662
5
Hi abadi-

I have attached a dc (no ac signal) spice model for your Rogowski coil application. V1 is the Rogowski coil output (= 0 volts for this analysis), L1 and R3 are the coil inductance and resistance (place holders) and V2 is a 1 microvolt input voltage offset. C1 = 10 nF is the feedback capacitor. When the amplifier is turned on, the output voltage ramps up to the positive rail in 70 milliseconds.

Questions: Why does it ramp up to the positive rail, and what are you going to do about it?

Bob S
 

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  • #19
12
0
thanx bro

but what i know is
the equvelant parameters are Rc, Cc and Lc
which are coil resistance and coil capacitor and coil inductance.

but in your design , you just use Rc and Lc !!!!

what i know about the coil, it will increase rapidly when the frequency increase, and it will reach a point the voltage will be constant.

am i right :D ??

and about your question ...
Why does it ramp up to the positive rail, and what are you going to do about it?
ramp at the positive rail, becouse we are using only positive side,

what to do,,, i just know it in hardware , which is to use 2 batteries !!!!!
am i right ????
 
  • #20
4,662
5
but what i know is
the equvelant parameters are Rc, Cc and Lc
which are coil resistance and coil capacitor and coil inductance.

but in your design , you just use Rc and Lc !!!!

what i know about the coil, it will increase rapidly when the frequency increase, and it will reach a point the voltage will be constant.

and about your question ...
Why does it ramp up to the positive rail, and what are you going to do about it?
ramp at the positive rail, becouse we are using only positive side,

what to do,,, i just know it in hardware , which is to use 2 batteries !!!!!
am i right ????
Please look at the Rogowski coil in the picture in

http://en.wikipedia.org/wiki/Rogowski_coil

It is an air-filled toroidal coil that has only coil inductance, coil resistance, and perhaps inter-turn capacitance at very high frequencies. It is usually connected to the amplifier with no series capacitance. Variations of this design for picking up low frequency magnetic fields are flat multi-turn coils with inductance, coil resistance, and inter-turn capacitance at very high frequencies. It is usually connected to the amplifier with no series capacitance.

Please completely describe your concept of a Rogowski coil. Where is the capacitance?

Please look at Figure 2 in this paper:

http://homepage.ntlworld.com/rocoil/Pr7o.pdf

and note in particular the Rf in parallel with the integrator capacitor. (Rogowski coils are (usually) always used with integrators). What is the resistor for? Why is your circuit different than the one in Figure 2?

Bob S
 
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  • #21
12
0
you can see my equivalent circuit for rogowski coil at the attachment

i have tired that amplifier, and it doesn't work, thats why im using different circuit that i got it from different Forum.

and now im trying to modify it to let it suitable for my project
 

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  • #22
4,662
5
you can see my equivalent circuit for rogowski coil at the attachment

i have tired that amplifier, and it doesn't work, thats why im using different circuit that i got it from different Forum.

and now im trying to modify it to let it suitable for my project
Both integrating amplifier stages in your original post are open-loop dc, meaning that both dc amplifier gains are over 3,500,000, and this has to be fixed. You should calculate and simulate (Spice model) the effects of dc input offset currents and offset voltages, and the effect of the input voltage offset coefficient (see my previous post #18 on Spice model of output voltage ramp). The equivalent curcuit of your Rogowski coil is self-resonant at ~ 500 kHz. What is your application, and what frequencies are you trying to measure? The Rogowski coil L and C can be modified to match the application. Your first circuit should be a basic integrating Rogowski coil circuit with leakage feedback resistor Rf, and without the extra active feedback circuit. Once that is working, then you can add the extra active feedback.

Bob S
 
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  • #23
12
0
:(
bro, im really stuck
i have been trying for almost 3 days
to resolve my problem but i couldn't
how to solve the ][ open loop dc ????
can you give my an advice ???
 
  • #24
4,662
5
Hi abadi-

You first need to get the integrator opamp working without the feedback opamp connected.

1) In original circuit, remove the 100k resistor and the second opamp from circuit.
2) Change the 22 ohm resistor to a 220 ohm resistor.
3) Put a 50k ohm resistor in parallel across the 10 nF capacitor.
4) In your original circuit as posted, you need symmetric + and - supply voltages (+/- 5 volts minimum) to the integrator op amp.
5) Now, without the Rogowski coil connected, the output of the integrator opamp should be less than a volt or two.

6) If 5) doesn't work, then change 10 nF capacitor to a 100 nF (0.1 uF) capacitor and retry.

7) If 5) works, then attach Rogowski coil. The output should still be less than several volts.

8) Post a complete copy of your new circuit, and discuss results.

Bob S
 
  • #25
12
0
ooooo

my problem was, i test the amplifiers together

thanx alot,
i will work on it
 

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