# Best way to reduce AC voltage for water heater.

#### Nicholas Howell

Joined Jun 15, 2018
6
I want to reduce the power being sent to a heater element in my hot water tank. It currently takes about 12 amps at 240 V AC (~3000 Watts). What I want is to drop that to about 1500 Watts, what is the best way to do this? At first I thought resistors but I don't want to dissipate all the energy as heat as that will defeat the purpose of the whole exercise, that being to use less energy. So then I thought about using a transformer. I currently switch on the heating element using a smart plug when I have at least 3000 Watts of solar going spare. Ideally if I only had 2000 Watts spare, I want to drop the heating element to just 2000 Watts so that it only uses the excess solar and not draw anything from the grid. Ideally I could change the single smart plug for 2 smart plugs, each being able to deliver 1500 Watts so that I could then switch one or both on depending if I had less than 3000 (but more than 1500) Watts or more than 3000 Watts. Does that make sense? Anyway could I do this with a couple of appropriately rated transformers?

Joined Jul 18, 2013
28,515
Are you in N.A.? If so connect it to 120v.
Max.

#### SLK001

Joined Nov 29, 2011
1,549
Dropping the voltage to 120V will give you 1/4 of the original wattage.

#### Dodgydave

Joined Jun 22, 2012
11,243
Use a light dimmer to adjust the voltage,. .
138V is 1Kw, 195V is 2Kw.

Last edited:

#### Ylli

Joined Nov 13, 2015
1,083
OEM might have a 2000 or 1500 watt element available.

#### ebp

Joined Feb 8, 2018
2,332
[edited to fix some bungling]

You could do it with a transformer arranged as a bucking autotransformer. You would need a transformer with a 72 volt secondary rated at 8.5 amperes (about 600 VA). This would be connected so that the phase of the secondary was opposite of the mains phase (hence the term "bucking"), thus reducing the voltage to the element to 240 - 72 = 168 V. This would give you (168/240)^2 x 3000 = 1500 W. A double throw relay could be used to switch the heating element between the bucked voltage and full line voltage.

As you can see, the required transformer is less than half of the rating you would need if you ran the full power through it. You may be able to find a "control transformer" (so-called because they are often used in industrial machinery control systems) that would be suitable. Hammond Manufacturing and (I think) Acme come to mind, if they are available where you live.

Phase angle control would certainly work, but may make for some problems with your inverter.

If you can obtain and fit a second heating element, Ylli's suggestion makes it easier.

#### oz93666

Joined Sep 7, 2010
739
No need for a transformer ... wastes power ...all you need is a diode !

1000v 3A ,,, available on eBay ...20 for $1.... best to wire at least 7 in parallel for the current you need with a safety margin 7x3 =21A (your heater pulls 15A), or perhaps you can find some with higher amperage rating . Just cut one of the power leads feeding the heater and put the diodes so the power goes through them The diodes will block half of the waveform , halving power in the heater. Last edited: #### dl324 Joined Mar 30, 2015 16,696 wire at least 7 in parallel for the current you need with a safety margin 7x3 =21A (your heater pulls 15A), or perhaps you can find some with higher amperage rating . It will just block half of the waveform , halving power in the heater. That's just going to be a diode tester. One diode will hog current and die, starting a cascading failure until all have failed. #### oz93666 Joined Sep 7, 2010 739 That's just going to be a diode tester. One diode will hog current and die, starting a cascading failure until all have failed. That's why I recommended at least 7 in parallel , that gives plenty of lee way .... if nervous put all 20 in parallel ... with 20 the average current in each is 0.75A , even if there is some hogging it will never be so great as to cause failure . or better find a single diode that can handle the required current ... diodes must be a cheaper and more sensible option than anything else. Last edited: #### Dodgydave Joined Jun 22, 2012 11,243 No need for a transformer ... wastes power ...all you need is a diode ! 1000v 3A ,,, available on eBay ...20 for$1.... best to wire at least 7 in parallel for the current you need with a safety margin 7x3 =21A (your heater pulls 15A), or perhaps you can find some with higher amperage rating .

Just cut one of the power leads feeding the heater and put the diodes so the power goes through them

The diodes will block half of the waveform , halving power in the heater.
Wrong!! Half the voltage is a 1/4 of the power !

#### oz93666

Joined Sep 7, 2010
739
Wrong!! Half the voltage is a 1/4 of the power !

Certainly in DC if the voltage is halved and the resistive load is the same , then V=IR so the current is halved and P =IV or I2R ... then the power is a 1/4

But putting a diode in the AC circuit does not halve the voltage ... the voltage is the same ,delivered for only half the time.... we have just removed half the cycle , so the power delivered is halved

#### SLK001

Joined Nov 29, 2011
1,549
But putting a diode in the AC circuit does not halve the voltage ... the voltage is the same ,delivered for only half the time.... we have just removed half the cycle , so the power delivered is halved
You're wrong. Cut out half the cycle and your RMS voltage drops by 1/2. You calculate the power using the DC equivalent voltage, or RMS voltage.

#### SLK001

Joined Nov 29, 2011
1,549
@Nicholas Howell Does your water heater have two elements or only one? If two, you could wire them in series (they should be parallel now) to get half the power. If a single, you can go to Home Depot and buy an element with less of a power rating.

#### MrAl

Joined Jun 17, 2014
11,272
Hello,

The diode idea is interesting. A rectifier diode at the rated current perhaps. The power is halved not quartered because we are only powering it for 1/2 the time then.

But the better way i think is to use a transformer, in fact a Variac. You can then adjust the power to whatever level you like. For example, full power, three quarters power, half power, 1/3 power, 1/4 power, however you find you like it best.
I say this because the heater is designed for a certain voltage, and that nominal voltage assures you of a certain top temperature level with a given ambient temperature. So if you dont have full voltage and full power, you may not be able to get the hot level temperature you need. Reducing power means it takes longer to heat the water, but it also limits the top temperature, so that's something to think about too.
Also, using a diode will cause 1/2 wave conduction and that means one half of the cycle gets loaded a lot more than the other. That may be a problem too.
The variac is a little expensive though so you have to think about this. It does ensure complete variability however in case you need to adjust the power to some level other than one-half. Because of the current levels involved here i would shoot for a 20 amp rated variac.

You're right, series resistors eat up the extra power which is reflected on the electric bill

#### ebp

Joined Feb 8, 2018
2,332
As I understand the problem, Nicholas has a PV array feeding an inverter feeding his AC mains and depending on his local power requirements he wants to optimize his use of power from his PV system to avoid paying for power he doesn't need. In such a situation a water heater can be thought of as a very large, highly reliable battery, at least in terms of where to put some power that would otherwise go to waste. He hasn't mentioned if he gets paid for power put back into the mains.

If half wave rectification is used:
When insufficient power to deliver the conducting half cycle requirement is available from the PV system, the balance of the power for that half cycle comes from the utility company power. During the other half cycle the power from the array is put onto mains. Exactly how the power from the PV array is utilized over the course of a full cycle depends very greatly on the amount of capacitance at the DC input of the inverter and how the inverter manages PV MPP. It would take a massive amount of capacitance to reasonably average the current from the array with this scheme. Exactly how the utility company handles and feels about getting feed into the mains on a half-cycle basis is hard to say. There are just a whole lot of questions that are hard to answer if half wave rectification is used. If the utility company isn't paying for the power it receives and his inverter doesn't have huge capacitors on the DC, the half wave rectification will waste half of the limited power he does have available - he gets half(ish) and the utility company gets the other half(ish) which it may not want, given the nature of it.

Using a bunch of low-current rectifiers in parallel when a suitable single rectifier with adequate rating costs a couple of dollars at most, is easily attached to a heatsink and is far more reliable is silly. The forward voltage of a silicon PN junction diode will increase by about 60 mV for each decade increase in current - which would help with sharing. But a temperature rise of less than 30 °C will cancel that increase. With diodes paralleled and not thermally coupled, runaway is probable. Even with monolithic dual diodes, derating by 10 to 15 % is recommended.

#### ebp

Joined Feb 8, 2018
2,332
This can be done with an open-loop PWM regulator with adjustable duty cycle and very little filtering.
• full wave rectify the AC; don't filter except for a little at switching frequency and a little for RFI management
• PWM at a fixed duty cycle using a FET or possibly and IGBT; the switching frequency need not be high and could probably even fall into the audio spectrum without making anything sing annoyingly
• filter the output only sufficiently for RFI management
• feed output to the heater
The duty cycle could be controlled with anything from a potentiometer to an computer that talked to the rest of the PV system to optimize the power put into the water tank.

I think it should be easy to accomplish at least 97% efficiency at 3 kW (100% efficient if the lost heat could be captured in input water). Power factor would be very near unity. Components cost would be a very few tens of dollars.

I implemented a similar scheme about 15 years ago, but working directly from the DC of the PV array with MPP tracking. I think the prototype controller is still in one of my boxes of "why the hell am I keeping this" junk.

#### crutschow

Joined Mar 14, 2008
34,054
Don't use a diode as oz93666 suggested to give a rectified half sinewave.
The resulting large DC current can damage a transformer or whatever else is driving the load.
In some places it's illegal to use a half-wave recitifier with loads above a certain wattage for that reason.

Does the solar system have battery storage?
If you could use use a solid-state relay AC with a zero-phase turn-on, and control it with a 50% duty-cycle signal (or whatever duty-cycle you want) from a 555 timer or similar.
That will give bursts of an integral number of full-wave power cycles to the heater, with the average power reduced by the duty-cycle with the battery averaging the power from the panel.
The duty-cycle frequency can be low, such a 1Hz.

#### MrAl

Joined Jun 17, 2014
11,272
Hi,

I tend to agree with the condemnation of the diode solution because of the heavy current involved.

I also tend to agree somewhat with the non linear regulation method over the linear because the linear means the hottest temperature may not be able to be reached and so an on'off scheme would still be better in that case. If i had a variac already though then that is what i would try first because it's so easy to try. Buying one just for that though may be a little bit of a risk unless you can easily return it because of the top hot temperature level issue with reducing total power.

#### crutschow

Joined Mar 14, 2008
34,054
.... Reducing power means it takes longer to heat the water, but it also limits the top temperature,......
I don't see that reducing the power will limit the maximum water temperature, only increasing the time it takes to get there.
The only way it would limit the maximum temperature is if the heat loss through the insulation equals the heater power, and that's not likely with the typical water heater insulation.
The power would need to be limited to much less than 1/2 power for that to happen.