??? 12/10/04 04:25 Read: times |
#82875 - Solving in the frequency domain Responding to: ???'s previous message |
Prahlad said:
I have read it carefully and but I think I will need to read it more times to grasp the things properly. Maybe it helps if we analyze the problem in the frequency domain? Let's assume this equivalent circuit for the coil (Rogowski coil or current transformer) ![]() and let's take into account that the induced voltage 'Uind(t)' is ![]() where 'n' is the number of windings, 'A' is the cross section of toroid coil and 'k' expresses the proportionality between welding current 'Iw(t)' and magnetic flux density 'B(t)'. 'Ri' is assumed to be allowed to be ignored. By solving the according differential equation it can be shown, that if the welding current Iw(t) is a sinus ![]() then at output of above circuit, means across RL the following signal can be observed ![]() with ![]() If the welding current increases by an arbitrary factor, then Us(t) will also increase by the same factor. So, the quotient of amplitudes of both sinus ![]() defines something like the 'gain' of your coil. You will immediately notice, that this 'gain' will depend on actual frequency 'w'. But if the welding current is a function of arbitrary shape, means if it contains mains frequency and lots of harmonics, then you do NOT want, that 'gain' is frequency dependent! How to solve this problem? In the case of current transformer, where the inductance of CT can be very high, RL can be choosen in such a way, that w x L >> R. Then the quotient goes to ![]() Means, if RL is choosen very small in combination with current transformer, then 'gain' is no longer frequency dependent! For the Rogowski coil, where RL is choosen to be much greater than w x L, the quotient does not simplify so easily, unfortunately. Here an additional signal conditioning is needed, which integrates Us(t). After integrating of Us(t) the following quotient is valid: ![]() Now things are easier. If we choose R >> w x L the quotient goes to ![]() Means, if RL is choosen very high in combination with Rogowski coil, then 'gain' is no longer frequency dependent! You had asked, what error the introduce of poti will cause. Well the last two equations give you the answer: Introduce of your poti will result in a decrease of signal by the factor ![]() This factor assumes, that wiper of poti is at 100%. If wiper of poti is at 80%, for instance, then the signal is ADDITIONALLY decreased. Wiper at 80% means a multiplication of signal by a factor of '0.8'. Again, this is additional to the above factor! Kai |
Topic | Author | Date |
Rogowski Coil Interchangeability. | 01/01/70 00:00 | |
echo | 01/01/70 00:00 | |
re: Echo | 01/01/70 00:00 | |
Still missed it then ? | 01/01/70 00:00 | |
I guess, standard practice for interchg | 01/01/70 00:00 | |
Re: seeprom storing caliberation. | 01/01/70 00:00 | |
But I want something simpler. | 01/01/70 00:00 | |
stability | 01/01/70 00:00 | |
1-Wire = 3 Pins?! | 01/01/70 00:00 | |
Calibration is unavoidable. | 01/01/70 00:00 | |
Calibetaion through POT. | 01/01/70 00:00 | |
error budget. | 01/01/70 00:00 | |
exceeding error budget | 01/01/70 00:00 | |
OP07 amplifier | 01/01/70 00:00 | |
Re: Error Budget | 01/01/70 00:00 | |
CT versus Rogowski coil | 01/01/70 00:00 | |
Re: CT vs Rogowski Coil | 01/01/70 00:00 | |
Solving in the frequency domain![]() | 01/01/70 00:00 |