electricpete
Electrical
- May 4, 2001
- 16,774
We have recently been using AEMC current probe model 651 together with a digital recorder to measure motor starting current waveforms. The current probe is identified as an ac current probe - more details here:
When we look at the current waveforms, they look the way I would expect them to look if our measurement was properly seeing the true current. Specifically, I see an exponentially decaying dc component on each of the three phases.
But it raises a question: do we expect this type of probe to correctly indicate a motor starting current waveform which includes a decaying dc component?
I have thought through the same question before for a CT and the answer was yes. I know how to analyse a CT based on textbook equivalent circuit (but I don't know if the same applies to a clamp-on). Here is the analysis for a CT:
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When we look at the current waveforms, they look the way I would expect them to look if our measurement was properly seeing the true current. Specifically, I see an exponentially decaying dc component on each of the three phases.
But it raises a question: do we expect this type of probe to correctly indicate a motor starting current waveform which includes a decaying dc component?
I have thought through the same question before for a CT and the answer was yes. I know how to analyse a CT based on textbook equivalent circuit (but I don't know if the same applies to a clamp-on). Here is the analysis for a CT:
So... the clamp-on probe is a little different device than the CT. The output of the clamp-on probe is voltage, not current. And what is inside of the clamp-on probe I don't know how to model. Is it reasonable to expect that the waveform we see on our scope fed from this clamp-on is a true representation of our actual current, including exponentially decaying dc offset?electricpete's analysis of CT response to signal containing dc:
Assuming a linear magnetizing branch (L), purely resistive burdren (R), with all quantities referenced to the secondary:
I1 = Im + I2 = V2/(L*s) + I2 = (R*I2)/(L*s) + I2 = I2* [1 + R/(Ls)]
I2/I1 = 1 / [1 + R/(Ls)] = s / [ s + R/L]
This transfer function I2/I1 has a zero at the origin and a pole at s = -R/L. For typical installation L > > > R, and the pole -R/L is very close to the origin.... perhaps at 0.001 sec^-1. The pole and zero being so close together means that their effects will cancel for all (complex) frequencies except those very close to the pole zero pair.
H(s) = s / [ s + R/L] transforms into a quirky "impulse response" h(t) which itself includes an impulse (delta). It's a little easier to look at the step response which is (1/s) * H(s) = 1 / [ s + R/L] or step_response(t) = exp(-Rt/L) So the system responds perfectly to the high-frequency components in the rising edge of the step, but then decays away as the steady-state (dc) response starts to kick in.
As long as the rate of decay of the motor dc component is much faster than the R/L of the CT circuit (which appears to be the case R/L of CT ~ 0.001/sec, R/L of motor decay ~ 1/sec or greater), the dc component should pass through the CT intact
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