??? 02/11/05 16:18 Read: times |
#87144 - There's no 'corner frequency' Responding to: ???'s previous message |
Prahlad said:
Yes I could see inverse proportionality, by increasing 10k pot value the signal output from rogowski reduces. The swing obtained by using 10k resistor is exactly 18%. For example with 47k alone the RMS current output shown by my LCD is 21.4kA and with 47k+10k pot for same current the output on LCD is 17.6kA. This looks pretty good! Actually I was planning to use R=100k but then this stuff [National opAmp AppNote] scared me...
![]() In case of my integrator with C=47nF and R=47+10=57k Fc=59Hz If I use 100k as R and say 20k as Ra then Fc=31Hz and I am using this integrator for F=50Hz and of course a lot of harmonics. Don't confuse 'fc' with 'corner frequency'. Here it means 'center frequency'. If you apply a sine signal to input of above integrator, then after integration the term 1/w appears, which is to be multiplied by 1/(R1 x C1). Means, Vout is proportional to 1/(w x R1 x C1), or, differently spoken, Vout is inversely proportional to 'f'. In opposition to the performance of a RC filter you will not see something like a 'corner frequency' with above shown integrator, because if you draw a plot of Vout versus frequency there will not be any plateau. But you will find a certain frequency, where the sine signal at output (concretely spoken the cosine signal) shows the same amplitude as the sine signal at input. This frequency is called 'center frequency' and its value is as stated in the picture. But please remember, that your application totally differs from the situation above, because you don't apply a sine signal to input of integrator, but the output signal of your Rogowski coil. And this signal contains the time derivative of welding current. This is something totally different. From the calculation below, you will see, that an integration is needed to recover the welding current from its time derivative. After integrating the Rogowski coil signal there is no further frequency dependency (ideal components assumed, of course): If your welding current to be measured is Isig(t), then Rogowsky coil gives a signal voltage of Urog(t) = M x dIsig(t) / dt. If you apply this voltage to the input of following integrator ![]() then, assuming having an ideal operational amplifier without showing any input bias current and input offset voltage, at output the signal Uout(t) can be found, according to: ![]() The minus results from the fact, that integrator is built using an inverting circuitry. Provided that load resistance at output of Rogowski coil is properly choosen, what I discussed in another post, then you get at output of integrator an exact copy of welding current. No further frequency dependency and no corner frequency stuff! By suited choice of 'R' and 'C' output signal height can be adjusted. But it's always an exact copy of welding current you see at output of integrator (again ideal components assumed)! Kai |
Topic | Author | Date |
My Rogowski Coils | 01/01/70 00:00 | |
If it ain't broke | 01/01/70 00:00 | |
Lack of confidence. | 01/01/70 00:00 | |
Art of electronics | 01/01/70 00:00 | |
No I haven't. | 01/01/70 00:00 | |
OP AMP basic definitions | 01/01/70 00:00 | |
Kai- Please Check your mailbox. | 01/01/70 00:00 | |
Three things that scare me. | 01/01/70 00:00 | |
More concrete schematic | 01/01/70 00:00 | |
Re: Concrete Schematic. | 01/01/70 00:00 | |
Rogowski coil | 01/01/70 00:00 | |
Did I Miss Something -Kai. | 01/01/70 00:00 | |
You should see a change! | 01/01/70 00:00 | |
Yes I could see the change Thanks. | 01/01/70 00:00 | |
Alu foil, asbestos shield | 01/01/70 00:00 | |
Asbestos | 01/01/70 00:00 | |
There's no 'corner frequency'![]() | 01/01/70 00:00 | |
Integrating Capacitor. -Kai | 01/01/70 00:00 | |
MKP, select for low offset | 01/01/70 00:00 | |
How about these. | 01/01/70 00:00 | |
Are ok | 01/01/70 00:00 |