Re: Strange voltage readings=Half-Wave Rectifier?
- From: "Dana" <raff242@xxxxxxxxx>
- Date: Wed, 27 Sep 2006 08:45:22 -0800
"ghostwriter" <ghostwriter25@xxxxxxxxxxxxxxxx> wrote in message
news:1159372785.598977.68670@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
John Fields wrote:
On 26 Sep 2006 11:45:22 -0700, "ghostwriter"
<ghostwriter25@xxxxxxxxxxxxxxxx> wrote:
Here is the basics, I have a heating coil that I think has a half-wave
rectifier in front of it (I am still waiting for the sales guy to call
back). I get about 20Volts AC when measuring the hot wire to ground,
0.8Volts AC when I measure the COLD wire to ground and 19.2Volts
between the hot and cold wires.
DC voltage osillates between 1.2 and 1.8 volts. So whatever it is it
inst straight AC.
---
It's AC. If it was half-wave rectified AC you'd be reading a lot
more than 1.2 - 1.8VDC. That 1.2 - 1.8V is just your meter jumping
around on the DC scale with AC on it.
---
I want to be able to calculate the wattage that the system is running---
at, any help appreciated. The circuit havs 4.2 Ohemns of resistance.
It's "ohms".
E² 20²
P = ---- = ------ ~ 95 watts
R 4.2R
--
John Fields
Professional Circuit Designer
The sales man got back to me and stated that the coil has a phase angle
SCR that chops the sine wave in order to ramp the heat output of the
coil up and down. He also says that to get the wattage I would need a
watt transducer that will cost about $800 to purchase and get
installed. A found a supplier that can get me the transducer for $350
and walk me though the installation for free.
Is this necessary? And what is the difference between operating
resistance and the alternative?
Ghostwriter
Here is a better article for measurement applications.
Since there is no way of knowing what you are trying to do, take a look at
this to see if it answers your questions.
https://www.ohiosemitronics.com/pdf/tech_papers/Transducer-book-dec-04.pdf
.
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