I remember reading that installing a rectifier diode in series with a resistive heating element saves energy. I understand that the heating element will receive half the power. Although this, if done correctly, will make the heating element last longer (lower temperature) and possibly safer (again, lower temperature), wouldn't the element have to be on for twice the time to produce the same amount of heat?
I don't think it's true because if it is, every electric heater will have a rectifier. (Some heating equipment do contain a large rectifier but that's for control.) I think I read it while googling for "true-rms" but I couldn't find it anymore.
I don't think it's true because if it is, every electric heater will have a rectifier. (Some heating equipment do contain a large rectifier but that's for control.) I think I read it while googling for "true-rms" but I couldn't find it anymore.
rectifier diode in series with AC heatin elementg
Star882
Probably not a good idea. It would take ages to heat up, and the element may even buzz. It may however, be worthwhile switching in the diode when the required temperature is reached.
A thermosat would be far better , though.
I am not sure how nearby A/V gear would like this method.
SandyK
Star882
Probably not a good idea. It would take ages to heat up, and the element may even buzz. It may however, be worthwhile switching in the diode when the required temperature is reached.
A thermosat would be far better , though.
I am not sure how nearby A/V gear would like this method.
SandyK
If a heater designer wants the heater element to last longer, he/she may make the element larger (but rated for the same amount of power as a smaller element) to distribute the heat, and thus thermal stress, over a larger area. Upon that, one might cause airflow around the element to prevent it from heating up as much.
A diode in the path to the element would induce very bad harmonic switching noises, and possibly EMI, into the environment immediate to the heater.
Diodes to handle the amount of current for most typical heater elements would probably be somewhat costly as well...
A diode in the path to the element would induce very bad harmonic switching noises, and possibly EMI, into the environment immediate to the heater.
Diodes to handle the amount of current for most typical heater elements would probably be somewhat costly as well...
I recall there was a device that effectively reduced the voltage to a water heater during off-peak times and restores full voltage during peak times. I forgot how it did that, but since it advertises something about half power, a rectifier diode can definitely be a way it did it.
I'll make a model using a small rectifier and a 5w power resistor. Then I'll experiment to find out exactly what the rectifier does.
I'll make a model using a small rectifier and a 5w power resistor. Then I'll experiment to find out exactly what the rectifier does.
Hi,
can we assume this heater runs direct off the mains voltage?
The diode half wave rectifies the mains.
Output power is halved.
The reverse voltage of the diode will need to exceed the peak voltage on the supply. 200V is a bit too close for 120Vac.
In the UK we would need 600V for 240Vac.
This power reduction is what was fitted to the trigger of two speed power drills. Massive reduction in torque made them run slow. Big drills then made them run even slower. Slow turning fan and very high half wave currents = burnt out motor.
can we assume this heater runs direct off the mains voltage?
The diode half wave rectifies the mains.
Output power is halved.
The reverse voltage of the diode will need to exceed the peak voltage on the supply. 200V is a bit too close for 120Vac.
In the UK we would need 600V for 240Vac.
This power reduction is what was fitted to the trigger of two speed power drills. Massive reduction in torque made them run slow. Big drills then made them run even slower. Slow turning fan and very high half wave currents = burnt out motor.
Isn't this idea commonly used in hairdryers?
I've never dismantled one to find out but assumed it to be true because I've read so many people complaining about hairdryers affecting the sound quality from HiFi.
I've never dismantled one to find out but assumed it to be true because I've read so many people complaining about hairdryers affecting the sound quality from HiFi.
See this post:
http://www.diyaudio.com/forums/showthread.php?postid=1222808#post1222808
With the series diode in place only half the power will be applied to the heater, and DC will be drawn from mains line. This is not a recommended practice.
http://www.diyaudio.com/forums/showthread.php?postid=1222808#post1222808
With the series diode in place only half the power will be applied to the heater, and DC will be drawn from mains line. This is not a recommended practice.
I have a diode in the power cord for my bench soldering iron. When I need more power, I engage a bypass switch. I guess it is not too bad on the line power at that lower power level. The set-up is simple, small, and lightweight.
I have seen it used in portable heaters, heat guns, and even coffee makers.
The experiment had some strange results. With a 1:1 isolation transformer powering a 3k9 resistor with and without a 1N4007 rectifier, adding the rectifier halved the voltage, current, and delta-T. A true-RMS meter was used to measure the voltages and currents. Based on voltage and current, the resistor was drawing only 1/4 its original power, but based on delta-T, it was producing only 1/2 the heat. This makes no sense so I think I made a mistake somewhere.
The experiment had some strange results. With a 1:1 isolation transformer powering a 3k9 resistor with and without a 1N4007 rectifier, adding the rectifier halved the voltage, current, and delta-T. A true-RMS meter was used to measure the voltages and currents. Based on voltage and current, the resistor was drawing only 1/4 its original power, but based on delta-T, it was producing only 1/2 the heat. This makes no sense so I think I made a mistake somewhere.
star882 said:
Looks like the voltage and current measurements are out to lunch, which might be the case if the "true RMS" meter has a tough time with nonsinusoidal waveforms. It makes sense for the resistor to dissipate half the heat instead of 1/4, because the diode takes away half the AC wave so intuitively one should expect to get half the heat.
I'd bet a Fourier transform of the half-waves would show the fundamental is a bit more than half the amplitude of the full wave, say around 63% at a guess, with the rest of the power contributed by harmonics.
Half the voltage and half the current means 1/4 of the power dissipation, not necessarily 1/4 the temperature change. Although the temperature change is usually pretty linear too.
so lets see.......
assuming (aarrrggg) that Vp,Ip,Pf, and VAR are perfect........
115 vac in
3900 ohm load
=.029 amps
=3.3 watts
=11.5 btus (~3.4 btu's per watt)
then thowing a half wave rectifier in the mix then
115 vac in
diode makes it ~57.5 vac
3900 ohm load
=.0147 amps
=.845 watts
=2.85 btu's
soo......
assuming (aarrrggg) that Vp,Ip,Pf, and VAR are perfect........
115 vac in
3900 ohm load
=.029 amps
=3.3 watts
=11.5 btus (~3.4 btu's per watt)
then thowing a half wave rectifier in the mix then
115 vac in
diode makes it ~57.5 vac
3900 ohm load
=.0147 amps
=.845 watts
=2.85 btu's
soo......
Along the lines of what jimbo posted, using the average values with 120VAC would mean 3Wavg dissipation with no diode and 0.75Wavg dissipation with the diode.
The RMS voltage of a half wave rectified sine wave is .707 times the RMS voltage of the unrectifed sin wave. Therefore, the power delivered by the rectifer circuit is .5 time that of the unrectifed circuit (for the same load resistance).
So for a heater if you want to cut the power in half double the resistance, rather than rectify the line voltage.
So for a heater if you want to cut the power in half double the resistance, rather than rectify the line voltage.
no,BWRX said:
it's all of the voltage and all of the current for half of the time.
That in my book means half power.
Think about it, and forget the maths!
Draw a sine wave (AC mains) and then remove everything below the center zero line (1/2 wave rectification).
You have just removed 1/2 of the "area below the graph" - remember that at school?
So 50% it is!
Draw a sine wave (AC mains) and then remove everything below the center zero line (1/2 wave rectification).
You have just removed 1/2 of the "area below the graph" - remember that at school?
So 50% it is!
AndrewT said:
When I said half the voltage and half the current I meant average voltage and current for the half wave rectifier compared to the regular sine wave. It now makes sense that you can't calculate it that way.
Here's a good picture for those of us who see things better graphically.
An externally hosted image should be here but it was not working when we last tested it.
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