Strange voltage readings=Half-Wave Rectifier?

[ghostwriter]

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.

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.

Any help appreciated.

Ghostwriter

 

[Charles Schuler]

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.

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.

I'd guess 100 watts.

 

[Tom Biasi]

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.

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.

Any help appreciated.

Ghostwriter

How do you know the circuit has 4.2 Ohms of resistance?
If you measured it cold the reading will not apply to operating power.

Tom

 

[Tom Biasi]

How did you guess that, on the DC or AC

It doesn't matter. He squared the 20 volts and devided by 4 (rounded).
P=E^2/R
If the 4.2 Ohms was operating resistance then I agree with this "guess"

Tom

 

[Dana]

I'd guess 100 watts.

How did you guess that, on the DC or AC

 

[Tom Biasi]

Depending on what the original poster needs, it does. He is stating he
thinks he has a rectifier, well that implies he is expecting a DC output.
And with only 1.8v out, that would probably indicate an issue with the
rectifier.
I assumed he was measuring at the output of what he thinks is the
rectifier.
Hence that is where the values he gave come into play.
But if indeed it is just a simple AC heating coil, than yes about 100
watts
would be correct.




He squared the 20 volts and devided by 4 (rounded).

He stated a lot of things. But the question was of power consumption.
If the unit is getting 20 volts (ACorDC) then the power calculation stands.

If he really cares about the rectifier he shouldn't say " I want to be able
to calculate the wattage that the system is running
at, any help appreciated. "

I took all his extra readings to be just info we didn't need but he had no
way of knowing what we need.
If the voltage across the heating element is 1.8 volts then that's another
story.

Rectifiers before a heating element than has a stepped down supply of 20
volts makes no sense from a design standpoint.

Regards,
Tom

 

[Dana]

It doesn't matter.

Depending on what the original poster needs, it does. He is stating he
thinks he has a rectifier, well that implies he is expecting a DC output.
And with only 1.8v out, that would probably indicate an issue with the
rectifier.
I assumed he was measuring at the output of what he thinks is the rectifier.
Hence that is where the values he gave come into play.
But if indeed it is just a simple AC heating coil, than yes about 100 watts
would be correct.




He squared the 20 volts and devided by 4 (rounded).

 

[Dana]

He stated a lot of things. But the question was of power consumption.
If the unit is getting 20 volts (ACorDC) then the power calculation stands.

If he really cares about the rectifier he shouldn't say " I want to be able
to calculate the wattage that the system is running
at, any help appreciated. "

I can agree to that. But when he gave AC as well as DC readings, and
mentioned rectifier, I assumed that maybe he had a DC type element.

 

[John G]

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.

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.

Any help appreciated.

Ghostwriter

I think we need a much better description of this "Heating Coil"

You talk about Hot wire and Cold wire. Do you mean the houshold supply
wires or is there some other components between the supply and your
heater.
What is this heater for, just so we can guess what it might be to verify
the other "Facts"?
Does this thing have a nameplate?

 

[John Fields]

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

 

[ghostwriter]

---
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.
---

---
It's "ohms".


E² 20²
P = ---- = ------ ~ 95 watts
R 4.2R

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

 

[Dana]

---
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.
---

---
It's "ohms".


E² 20²
P = ---- = ------ ~ 95 watts
R 4.2R

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?

What is the Wattage you need, what is your source of power AC or DC, and
what is its limit.
And can your coil actually handle the wattage you think you need.

Must devices give operational characteristics, why not get a coil that can
do what you need.

Can you describe what you are trying to do.

Ghostwriter

 

[Dana]

---
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.
---

---
It's "ohms".


E² 20²
P = ---- = ------ ~ 95 watts
R 4.2R

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

Check out this link below.

https://www.ohiosemitronics.com/pdf/tech_papers/Single Element on 3-phase(D).pdf

 

[Dana]

---
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.
---

---
It's "ohms".


E² 20²
P = ---- = ------ ~ 95 watts
R 4.2R

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

 

[ghostwriter]

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?

What is the Wattage you need, what is your source of power AC or DC, and
what is its limit.
And can your coil actually handle the wattage you think you need.

Must devices give operational characteristics, why not get a coil that can
do what you need.

Can you describe what you are trying to do.

Ghostwriter

The coil works fine, I use it for burning high energy chemical
compositions, specifically thermites for the metalworking industry. I
want to measure the wattage in realtime so that I can find out the
energy input and output curves and correlate them to the material
performance in the foundries.

The tempature curve is a close approximation and I am getting very
close to the true energy curve of the reaction but I cant subtract the
input from the controller unless I know the wattage.

Ghostwriter

 

[ghostwriter]

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

I work in Columbus, Ohio and Ohio Semitronics is the company that is
helping me and will be suppling the transducer.

Ghostwriter