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Re: Measurements using field probe (fwd)
---------- Forwarded message ----------
Date: Wed, 8 Jul 1998 06:44:49 +0000
From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Measurements using field probe (fwd)
DR. RESONANCE -
The AC peak should equal the DC peak? This would be true when the AC is
rectified and stored. However, that would depend on the probe and the
instrumentation. I have not been able to find in my literature where a field
mill measures the potential.
What are you using for a probe? What is the equation used?
John Couture
-----------------------------------------------------
At 04:30 PM 7/7/98 -0600, you wrote:
>
>
>---------- Forwarded message ----------
>Date: Tue, 7 Jul 1998 08:24:08 -0600
>From: "D.C. Cox" <DR.RESONANCE-at-next-wave-dot-net>
>To: Tesla List <tesla-at-pupman-dot-com>
>Subject: Re: Measurements using field probe (fwd)
>
>Hi John:
>
>The field mill meter actually measures the potential difference and is
>usually used with a calibration voltage. We use 150 kV DC to calibrate our
>units, and then the potential is directly proportional to the distance from
>the measured terminal. Wouldn't the AC peak be equal to 1.4 x Erms? This
>would mean the ACpeak should be equal to the DCpeak. Example, if you use a
>common 1N4007 to rectify the AC line which is 120 VAC Erms, then the DC
>reading is around 170 volts. This is the same as measuring the AC line
>with a meter than is not a true RMS meter and gives only the peak AC
>reading --- again around 170 volts.
>
>DR.RESONANCE-at-next-wave-dot-net