Strange voltage readings=Half-Wave Rectifier?

Sep 26, 2006 16 Replies

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'd guess 100 watts.

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

If that is your output of what you think is the rectifier, it does not seem to be a half wave rectifier.

Or your rectifier is shot.

Since it seems you have an AC device here is your calculation.

begin 666 6a46cd3f467c480c9386d7bc2e49fb87.png MB5!.1PT*&@H````-24A$4@```2 ````O" ````!FR4@]````"7!(67,```!X M````> "=]5I@```#%TE$051HWNV:Z[&K(!" MP5;H 5:H 5;H 5:L 5;L 5; MH 5:H 4.(L9'>"R:_+BYNS-)G ,^^ :6_9P#CJ(80 @.H1D(2X#R,3BG>@*T M1P.("TY-STTR #B'888'995>C7>P4:@%PF30$Z 3( M.*M(-0K[_.R4_55 "75J!S3,X\\_*:D*=VO=YI7[7YMD'!M Q\\[N $NK4OL^O M$V@41G W"RM :+'4UYH#DR] /83H%*(Z5QVPR:@&2'4E#I=AB/0%0. 6+ZYZL.'S74 M'/O7

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

How did you guess that, on the DC or AC

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

half-wave

call

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).

ground,

resistance.

output.

stands.

able

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.

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 G Wot\'s Your Real Problem?

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.

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

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 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.

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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.

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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

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

Ghostwriter

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