| Electronics > Beginners |
| Phase Angle Conventions |
| (1/1) |
| eev_carl:
Hi, I have a simple series RC circuit: R=1k, C=.22, E=5Vpp 1kHz. I was wondering if someone could explain the difference between the phase angle on my scope and what I've computed as arctan(Xc/R). For arctan(Xc/R), I get -36 degrees and my scope shows 54 (Phase 1 ->2, where Es=1 and Ec=2). Hand calculating the scope's value w. cursors, I get 54 (td/period * 360). Update: Looking at this on paper, I think the scope is just using a different location for Ec. Rather than pointing down at 270 degrees, it's appearing at 0 for the reference. I'm still wondering which value is typically used when talking about phase. Thanks, Carl |
| bson:
The AC phase angle is between voltage and current. Positive means the voltage leads (inductive), negative that the current leads (capacitive). |
| eev_carl:
I read a little more on this. I wired my scope to measure Et and Ec and the displayed 54 degree value (td / period * 360) referred to that angle. I had expected the -36 value because my book examples showed the angle between Er and Et. I can get Er using a MATH function and get my -36. |
| schmitt trigger:
--- Quote from: bson on April 27, 2018, 03:00:10 am ---The AC phase angle is between voltage and current. Positive means the voltage leads (inductive), negative that the current leads (capacitive). --- End quote --- A simple way to memorize this: ELI the ICE man. |
| bson:
--- Quote from: eev_carl on April 27, 2018, 01:33:00 pm ---I read a little more on this. I wired my scope to measure Et and Ec and the displayed 54 degree value (td / period * 360) referred to that angle. I had expected the -36 value because my book examples showed the angle between Er and Et. I can get Er using a MATH function and get my -36. --- End quote --- That's interesting! All I can think of is that because C has a negative reactance its current is stated as flowing out of it rather than into it, which means Ic = -Ir, and if you measure Ir (by proxy, via Er) you need to reverse the sign for Ic. (More formally, you calculated the angle of the conjugate impedance, as if it were an inductor.) Hence, atan(Ic/Ec) = atan(-Ir/Ec) = atan(-(Et-Ec)/(R*Ec)). But that reasoning feels incomplete and kind of ad hoc, in particular because it seems to violate KCL. |
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