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Registered Member #347
Joined: Sat Mar 25 2006, 08:26AM
Location: Vancouver, Canada
Posts: 106
cowana wrote ...
I think I'll stick with the dimmer switch. It is very small and compact, and means I can turn down the voltage to make it a variable heater. I will set it by making the input just over 1A - that will equal a power output of around 250W.
Andrew
Setting the input to 1A will get you only about 60W. You need to set the dimmer to get 120Vrms on the output. Since the dimmer is in series with the input, its input and output current will be identical. With the dimmer set to 120Vrms ouptut, driving a 250W heater from a 240V input, the input apparent power (VA) will be about 500VA and input current will be about 2.1A, but the input real power will be 250W. This is due to the poor power factor of the waveform you're drawing from the mains.
Because of the power factor, you can't determine the heater power just by multiplying RMS voltage and RMS current. You must either calculate it based off the Voltage or current and heater resistance, or use a true power meter.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Bored Chemist wrote ...
Some jobs might need more than half the power. A quartet of heaters...
Two 110V heaters in series running off 220V give twice the power of a single one. Proof: Each heater gets 110V and delivers its rated power, and there are two of them.
Four heaters in series would give the same power as a single one: each heater gets 55V and delivers one-quarter its rated power.
Tesla500: If you set the dimmer switch to give the correct RMS line current (250W/110V = 2.3A) as measured by a true RMS meter, then I bet you'd get the correct heater power.
I imagine there would also be 110V RMS across the heater at this point, since Ohm's law holds even for odd-shaped waveforms. In other words, if you consider the RMS voltage across the heater and the RMS current through it, the waveforms are identical and the power factor is unity. You only see poor PF if you look at the line voltage and the heater current, in fact you can say straight away that PF = 110/240 = 0.46.
This would not work for a meter that wasn't true RMS.
Finally: What would connecting a diode in series with the heater do? This trick is often used for half power mode in hair dryers and the like, but is it correct for running a 120V heater off 240V? A cookie for the answer. (ie, I don't know.)
Registered Member #27
Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058
Some quite expensive meters marked "True RMS" can give the wrong result since they are only true under certain conditions like the signal being symmetrical around 0.
In the wiki we have some correction factors for a typical multimeter:
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Yes, it's quite common for the true RMS circuit to ignore the DC component of the signal. After all, the RMS function is usually engaged on the AC ranges only. On the DC ranges you expect a DMM to read average, like a moving-coil meter.
And on the AC ranges, you don't expect it to respond to DC, which it would if it really measured the true RMS value of asymmetrical waveforms.
This is the explanation we were given in our EE labs anyway, as to why the multimeters measured "RMS without DC". I wouldn't be surprised if AC coupling made the RMS circuit cheaper, too.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Oops! I got bits of the arithmetic the wrong way up. 4 in series would work in that they wouldn't be over-run and the total output power would be corect. If the diode works then it's a great solution. Dirt cheap and nigh exactly half thepower.
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Steve McConner wrote ...
Finally: What would connecting a diode in series with the heater do? This trick is often used for half power mode in hair dryers and the like, but is it correct for running a 120V heater off 240V? A cookie for the answer. (ie, I don't know.)
I think that won't work, because a 110V heater on 220V will get 4x its rated power, and the diode only cuts it to half, so it still gets 2x its rated power.
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
> Finally: What would connecting a diode in series with the heater do? This trick is often used for half power mode in hair dryers and the like, but is it correct for running a 120V heater off 240V? A cookie for the answer. (ie, I don't know.)
Diode half wave rectifies input voltage waveform. RMS input voltage of half-wave rectified 230VAC is equal to 230/sqrt(2) = 162.6 Vrms. Heater resistance is V*V/W = 110*110/250 = 48.4 Ohms. RMS heater current is 3.36 Arms. Average heater power is calculated from either averaging V*I over once cycle, or from Vrms²/R
Average heater power is 546 Watts. Mmmmm, Toasty!!!!
Hence.... Use an autotransformer. Preferably with isolation, then the 110V heater gets the correct voltage and doesn't see the 230V line voltage at all!
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Making 110V out of 240V
