half wave rectification, question about diode power loss

[JeanF]

Hello everyone,

I'm stuck on the following problem for a while now... Forgive me if it's almost too easy for many of you guys ! Also I tried to search in the Forum whether or not the question was already asked : again, no luck maybe I've been using the wrong keywords.

Anyway, here is the context : I have some mains-powered electric heaters that are too power hungry for my taste. I read that it's possible to use a chunky diode in series with the heater to divide their power by a factor of 2, by doing half wave rectification. (Also reducing the cos(Phi) as well, but that's not my main concern). As someone opportunely pointed out, I also need to be careful with the power dissipation in the diode, and I will probably need heatsinking.

I would like to calculate the power dissipated in the diode, so I dug out these old formulas from my textbooks :
Sine wave operation :
V_rms = V_max/sqrt(2)
I_rms = I_max/sqrt(2)

Half wave rectification:
V_rms = V_max/2
I_rms = I_max/2
V_av = V_max/Pi
I_av = I_max/Pi (Equation 1)

Then we have P_heater = V_rms * I_rms = (V_max * I_max)/4, which is half of what it would be without the diode, so I'm ok with that.

How can I calculate the power loss in the diode ? I read several times that I need to multiply the forward voltage drop V_F by "the current flowing in the diode". That's where it's not clear for me. For half a cycle, the current is (almost, if we ignore the ~1V drop, which is 0.4% of the 230V mains voltage) the same as it would be without the diode. Then for the next half of the cycle, the current is 0. (please correct me if I'm wrong)

Then what expression of "I" should I use ?  I_max/sqrt(2) or I_max/2 ? I am quite lost. Also, what about the average current ?

Someone on a French forum suggested me to use the "approximate" (he said) equation P_diode = (V_Fmax * I_max)/(2*sqrt(2))     (Equation 2), but he gave no further explanation. Could someone please elaborate ?

Also, some diode datasheets refer to "Average Forward Current, I_f(av)". Do they refer to the real average current (which is, if I'm not mistaken, the equivalent DC current) or to its RMS value ?

Some datasheets even come with a power loss plot, such as this one (fig. 8, page 4).

To see how Equation 2 would work, and to have something to compare it to, I imagined a load which would draw an average current of 10A. Then, using Equation 1, I calculated I_max, and then using Equation 2, I got a result of P_diode = 11.7W, which is a bit higher but not too far off what I can see in fig. 8 of the datasheet for this current.

Many thanks in advance to anyone who could clarify this for me ! 
Cheers 
JF

PS : I also found AN 604 from ST, "Calculation of conduction losses in a power rectifier" , here but I'm not sure I want to go into this level of detail and take junction temperature into account, I guess I would just need a ballpark figure to choose a heatsink. Besides, according to fig. 8 of AN 604, the conduction losses decrease with temperature ; the worst case (ambient) may be just fine for me. 

[ocset]

strictly speaking, you need to do a complex calculation taking into account the vf at each If.
In any case, you can simulate it to a degree in LTspice and check there.

[David Hess]

Rectifier diodes are usually rated for current instead of power because of how they are used.  Calculating the RMS current and selecting a diode based on that while following the thermal recommendations will be sufficient.

[danadak]

Consider using this tool -

https://www.semikron.com/service-support/semisel-simulation.html


and their handbook -

https://www.semikron.com/dl/service-support/downloads/download/semikron-application-manual-power-semiconductors-english-en-2015/


Regards, Dana.

[jbb]

I have some mains-powered electric heaters that are too power hungry for my taste. I read that it's possible to use a chunky diode in series with the heater to divide their power by a factor of 2, by doing half wave rectification.

Don't do it.  This kind of kludge results in DC current flow in the AC network, which can cause saturation of the transformer in the street.  This in turn could result in inferior power quality (making all your equipment age faster than it should), a blackout or even make the transformer in the street catch fire.  They're full of oil and burn really well once they get going.

Also, your power meter may rat you out to the power company.

[David Hess]

I have some mains-powered electric heaters that are too power hungry for my taste. I read that it's possible to use a chunky diode in series with the heater to divide their power by a factor of 2, by doing half wave rectification.

Don't do it.  This kind of kludge results in DC current flow in the AC network, which can cause saturation of the transformer in the street.  This in turn could result in inferior power quality (making all your equipment age faster than it should), a blackout or even make the transformer in the street catch fire.  They're full of oil and burn really well once they get going.

Also, your power meter may rat you out to the power company.

The effect just is not that big unless you are talking about something considerably larger than a 24 kilowatt heater.  A web search for "power transformer DC imbalance half wave rectifer" turns up discussions about this.

A really big TRAIC phase controlled dimmer could be used but will generate considerable electrical noise.

[JeanF]

Thank you for your replies !

Yes I know that strictly speaking I would need to take into account the variation of Vf when If changes, but I'd rather not go into this if possible (as well as the influence of junction temperature). I'm not familiar with LTspice but I may give it a try one day.

Thanks for the tip about Semikron documents, I just put them in my bookmarks

Thank you for the warning about DC flow into AC network, although as David Hess said, I don't think this is a major concern here. I want to use this trick with one or two 1500W heaters to drop them to 750W ; the rest of the installation is about 10kW of mainly resistive equipment. Besides, I don't know about NZ but here in France, we don't have small pole transformers like in the US ; here the transformers are quite robust, they are rated between 100KVA and 1000KVA depending on where you live, so I highly doubt such a small mod could start a fire (hopefully !)

I also heard about Triac dimmers, but here I don't need precise variation, a "divide by 2" is enough for my needs. And adding a diode will be much easier to implement

I've been suggested later on, for this rough power estimation in the diode, to use the formula Pd = (Vfmax * Imax)/Pi, thus using the Mean current in the diode instead of the RMS current. I tried the formula with the same parameters I mentioned in my first post, and it matches pretty well the power chart in fig 8 of the datasheet for this particular diode (even better than the previous formula)

I'm still a little bit confused about how RMS values should be used in calculations, as opposed to Mean values ; could someone please help me on this ? Well, I'm aware that Mean value is 0 for non-rectified sine waves, so not very useful for calculus, but I heard that RMS value was equivalent to a DC value that would result in the same power in the load ; so why here (and elsewhere), the Mean value is called "DC value" ?

Thanks for your help, looking forward to reading you again ! Cheers

[Sparker]

The 'DC value' of your signal indeed means exactly the same as 'mean', 'constant' and 'average', that's why they said that 'mean is DC value'.
The mean value of a function, according to wikipedia, is:

Now if we have the fourier series of a periodic function, the coefficient with 0 index represents the amplitude of a cosine with zero frequency... or just a constant value offset! You can see that A0 is defined the same way as the mean value in the formula above:

Allright, sorry for bringing this stuff here, but i guess it explains why 'mean' and 'DC' are the same.

As for your initial question,

I read several times that I need to multiply the forward voltage drop VF by "the current flowing in the diode".

It's instantaneous power. But what you want to calculate is the mean power. You wrote before that for a half-wave sine the average current is:
Iav = Imax/Pi
The instantaneous power dissipated at your diode will follow current's waveform, being Pinst = Iinst * Vfwd, Vwfd~0.7V; Pmax = Imax*Vfwd. So for average power we can also write:
Pav = Pmax/Pi.

I hope I helped you!

[JeanF]

Hello Sparker,

Thank you very much for these details. You were extremely helpful indeed  that makes more sense now ! 

[David Hess]

It really is not that complicated.  The datasheet for a part like the $7 VS-80PF80 DO-203 stud mount 800 volt 80 amp rectifier shows forward power loss versus conduction angle and average current.  These curves represent to total power lost do to the current times forward voltage drop.  In your case, the power and power loss is halved because only half of the line cycles are conducting making a rough estimate of 1/2 watt per amp which makes sense because the forward voltage drop of a power diode is about 1 volt.

https://www.mouser.com/ProductDetail/Vishay-Semiconductors/VS-80PF80/?qs=sGAEpiMZZMtbRapU8LlZDzhFJYKIcxrrUbBIQP%252bCzS0%3d
http://www.mouser.com/ds/2/427/vs-80pfrseries-86949.pdf