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PWM effect on heaters.. ?

#21
I don't completely agree with the above. The average RMS voltage will be cut in half but for each half cycle that the element is powered, full power will be dissipated, so cutting the number of cycles per second in half will cut the average watts in half.

I have a soldering gun somewhere around here that runs all the time it is plugged in with a 1N4007 (what else) powering the element. When the trigger is pulled the diode is shorted out and the gun heats up and you can start soldering.
 
#22
I don't completely agree with the above. The average RMS voltage will be cut in half but for each half cycle that the element is powered, full power will be dissipated, so cutting the number of cycles per second in half will cut the average watts in half
Ohms law dude .... Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations are equivalent to a DC circuit

P = (V^2)/R

If R is fixed to 10 Ohms

P at 120V = 1440 Watts
P at 240V = 5760 Watts .... if you cut 240V in half with a diode that's still 2880 Watts

In the case where we were stepping up the 120V to 145V
P at 145V = 2103 Watts ...

Notice using a diode to cut the 240V provides 2880 Watts which for us was within an acceptable 2103 Watt limit when running the machine at 145V
 
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dknguyen

Well-Known Member
#23
Ohms law dude .... Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations are equivalent to a DC circuit

P = (V^2)/R

If R is fixed to 10 Ohms

P at 120V = 1440 Watts
P at 240V = 5760 Watts .... if you cut 240 in half with a diode that's still 2880 Watts
The V^2 to power relationship is correct. However, your assumption that adding a diode to turn a full-wave to half-wave is the same as halving the voltage is not.

A full-wave has an RMS value that is 1.414 that of a half-wave, NOT 2. Conversely, a half-wave as an RMS value that is 0.707 that of a full-wave, NOT 1/2. Therefore, adding a diode cuts the power in half while actually halving the full-wave voltage reduces the power to 1/4.

Dick's explanation of half the number of cycles = half the power is basically the calculus explanation put into laymen terms which is as follows: Half the number of cycles means half the area under the curve. Half the area under the curve means half the power.
 
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#24
However, your assumption that adding a diode to turn a full-wave to half-wave is the same as halving the voltage is not.
This is not what I am saying .... in fact you and I are saying the same thing.

Halving the wave with a diode reduces the wattage by half
Halving the voltage reduces the wattage by a quarter
 

dknguyen

Well-Known Member
#25
This is not what I am saying .... in fact you and I are saying the same thing.

Halving the wave with a diode reduces the wattage by half
Halving the voltage reduces the wattage by a quarter
Oh, on closer inspection I suppose you are. Your wording for this part was really weird and threw me for a loop:

camerart,

SO simply adding a diode at 240 to cut the wave in half will only reduce the wattage from x4 to x2.
 

dr pepper

Well-Known Member
Most Helpful Member
#26
Slightly off subject so dont mind me.
I have a solder gun similar to the above that has a dual position switch, both are now full power, I was gonna bin it thinking the transformer is dead.
Sounds like maybe its just a shorted diode.
 

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