simple power controller

[johnh89]

Wonder if some of you tech heads could help me out . I am trying to reduce power to a heating element . The element is 2050W AC . If I place a 20A stud rectifier in series with the element then what will be the reduction in power of the element . I am thinking that by half wave rectification then I would be halving the power but my AC theory could do with brushing up . By placing a switch in parallel with the diode then I could run the element at full power when required . Also would it be possible to use one or two elements from a bridge rectifier in place of the diode as they are cheaper and more easily available . If a bridge rectifier is rated at 35A , what would each individual rectifier be rated at ? Thanks for any help you can give .

 

[ericgibbs]

Wonder if some of you tech heads could help me out . I am trying to reduce power to a heating element . The element is 2050W AC . If I place a 20A stud rectifier in series with the element then what will be the reduction in power of the element . I am thinking that by half wave rectification then I would be halving the power but my AC theory could do with brushing up . By placing a switch in parallel with the diode then I could run the element at full power when required . Also would it be possible to use one or two elements from a bridge rectifier in place of the diode as they are cheaper and more easily available . If a bridge rectifier is rated at 35A , what would each individual rectifier be rated at ? Thanks for any help you can give .

Hi John,
Look at this option when using a bridge rectifier.
The labels in BLUE are the markings on the bridge.

You will get two diode drops in the voltage out, but effectively have two pairs in parallel.

NOTE: the BLUE 'AC's are NOT connected

 

[tom_pay]

Hi

I would have it at a guess that the element would run at half power, because it is practically being run at 50% duty cycle, just like PWM control.

One diode out of a 35A bridge rectifier would be good for 35A. However if you were to put two in parallel they would need current limiters. So they equally share the current and one does not get more than the other.

Hope this helps

Tom

P.S. Eric, what program did you use to get the scope screen simulator?

 

[johnh89]

Thanks for the quick replies guys . I had not thought about using the bridge as you suggested eric . I presume that I would lose 1.4V at the output due to 2 voltage drops across the diodes but at 240v should not be a problem . I am only guessing that the power would be 50% but was wondering if there was a more scientific reason than my sometimes flawed common sense . I am also thinking that some sort of heatsink would be a good idea ?

 

[dougy83]

I presume that I would lose 1.4V at the output due to 2 voltage drops across the diodes but at 240v should not be a problem.

At 5A average, 1.4Vf, the power diss will be 7W, which will get pretty hot. I would recommend connecting both ~ (AC) terminals together as the anode, and using the + terminal as the cathode. This way you dissipate half that 7W, which may mean you can avoid a heatsink.

You can work out whether you need a heatsink by checking the specs of bridge; specifically the temperature resistance from junction to ambient (or at least case to ambient) - multiply it by the diode power dissipation (e.g. 3.5W, from above), and you have the case temperature without a heatsink.

I am only guessing that the power would be 50% but was wondering if there was a more scientific reason than my sometimes flawed common sense.

You get pretty close to half the power as the duty cycle is now 50%. It will be affected somewhat by the temperature of heater changing its resistance; it should be close enough to 50% though.

 

[MikeMl]

Showing how LTSpice can answer the original question.

First plot is Average Power in R1

Second plot is Average Power in R2

Third Plot is Average Power in the Diode.

Note that Nichrome heaters have a positive tempco, and the power with the diode will actually be less than half. The fact that the heater is immersed in water will make this effect small.

 

[cowana]

The new heater power will be greater than half the old power.

Logic:
Half the 'on time'
Wire will be cooler
Wire will have lower resisatance
More current will flow (per half-cycle)
More power will be emitted (per half cycle).

As there are half the number of half-cycles, and slightly more power per cycle, I'd imagine the new power to be between 50%-60% of the heater running on full AC. Obviously closer to 50% if running in water.

 

[MikeMl]

Yep, I should have said "slightly more than half".

 

[ericgibbs]

hi john,
I dont know how stable your local mains supply voltage is but if its say +/-5% on a nominal 230Vac this will give a variation in wattage, that will swamp out any any small change due diode drops or the ptc of the element.

I do agree with Dougy, just use two of the diodes in the manner he described.