I have a water distiller that draws 6amps I have a solar voltage inverter that barely puts out 6 amps at 12 noon if there are no clouds. So, I put a diode in line thinking that it would reduce the current in half. But it still draws 6 amps, according to the gauge on the inverter! Any suggestions?

 

Is the inverter output AC or DC?

 

Is the inverter output AC or DC?

It is 120 V AC. The inverter is powered by batteries charged by solar panels.

 

Putting a diode in series with the AC is generally a no-no, since it adds a large average DC component to the output current and most AC supplies don't like that.

The easiest would be to use a Variac whose output can be adjusted to reduce the voltage to give a tolerable current level.

Another way would be to use a large AC motor run (not start) non-polarized capacitor in series with the load to reduce the current.
Try about 100µF to start.
That will reduce the current to about 3.5A.

 

I have a water distiller that draws 6amps I have a solar voltage inverter that barely puts out 6 amps at 12 noon if there are no clouds. So, I put a diode in line thinking that it would reduce the current in half. But it still draws 6 amps, according to the gauge on the inverter! Any suggestions?

Sorry to bother everyone; I had three diodes and I had one in backwards! Ha ha on me!

 

Putting a diode in series with the AC is generally a no-no, since it adds a large average DC component to the output current and most AC supplies don't like that.

The easiest would be to use a Variac whose output can be adjusted to reduce the voltage to give a tolerable current level.

Another way would be to use a large AC motor run (not start) non-polarized capacitor in series with the load to reduce the current.
Try about 100µF to start.
That will reduce the current to about 3.5A.

Pardon me for the bother. I had one diode in backwards! The Variac is a very good idea!

 

Pardon me for the bother. I had one diode in backwards! The Variac is a very good idea!

Sorry to bother everyone; I had three diodes and I had one in backwards! Ha ha on me!
[/QUOTE thanks for the idea of the capacitor! I will try that next. I wonder if the fan will run at half speed?

 

I have a water distiller that draws 6amps I have a solar voltage inverter that barely puts out 6 amps at 12 noon if there are no clouds. So, I put a diode in line thinking that it would reduce the current in half. But it still draws 6 amps, according to the gauge on the inverter! Any suggestions?

It would help with finding a solution if you will explain what the problem is. All that you have told us is that the water distiller is using all of the power delivered by your solar power inverter. You have told us what does not work.
Now tell us what you want to achieve.

 

It would help with finding a solution if you will explain what the problem is.

I have a water distiller that draws 6amps I have a solar voltage inverter that barely puts out 6 amps at 12 noon if there are no clouds. So, I put a diode in line thinking that it would reduce the current in half. But it still draws 6 amps, according to the gauge on the inverter! Any suggestions?

 

NO, a diode in series with an AC line will not cut the current in half. It may reduce the power to one half, but it may also damage the connected equipment.
IN addition, the TS has not told us why that is a problem or what they hope to achieve. That makes suggesting a solution difficult.
Repeating the original statement does not help explain what the actual need is. " thinking that it would reduce the current in half. " is not an adequate explanation of why the current draw is a problem.
And without any description of how the water distiller works we have no means of arriving at a suggestion, unless we make guesses. And guesses are not what we should be providing.

 

" thinking that it would reduce the current in half. " is not an adequate explanation of why the current draw is a problem.

No, but it does provide insight into what the TS wants to do. His PVP's barely produce 6 amps. The TS wants to limit the current draw of the distiller. I agree, a diode is not the right approach. But it seems to be clear the TS wants to limit the current to 3 amps. I don't see why there's confusion over this.

 

I think the water distiller will not work at low power. It probably boils water, and the water will not get hot enough.
I thought about storing power in the battery all day so you can distill for 2 hours.
You just don't have the power you need.

 

Fun fact. Boiling water draws 10x more energy density than lead acid batteries can store.

It takes at least 1700 kJ/kg to boil water from 25'C.
Car batteries might hold about 180 kJ/kg of energy.

So the battery might weigh 10x more than the water to boil it. Is that Ok?

I thought @crutschow 's solution of a large run cap sounded good until you compare 120V/6A= 20 Ohms with a 10 uF Run cap at 60 Hz is 265 ohms. But Princess Auto (CDN) has big 50uF caps for $20 which might still not boil water.

A 10A diode might work and drop the nichrome heater current 50% but the controller might not work and the inverter might make some bad noises.

Max Power Transfer occurs when your inverter impedance matches to the load.

 

I think the water distiller will not work at low power. It probably boils water, and the water will not get hot enough.

If the distiller is reasonably well insulated, it should still boil water, albeit at a slower rate.

 

Max Power Transfer occurs when your inverter impedance matches to the load.

No, max power to a fixed load occurs when the source impedance is zero. Do the math.

 

Max Power Transfer occurs when your inverter impedance matches to the load.

That's a common misconception when using the maximum power theorem.
It doesn't work both ways.
You adjust the load to the source impedance, but if the source impedance is adjustable, then you make it as close to zero as possible.

 

I have a water distiller that draws 6amps I have a solar voltage inverter that barely puts out 6 amps at 12 noon if there are no clouds. So, I put a diode in line thinking that it would reduce the current in half. But it still draws 6 amps, according to the gauge on the inverter! Any suggestions?

Yes, you will need a bigger power source. Your distiller is designed for 120 VAC operation drawing 6.0 Amps. Anything less and it won't work. You could always call the distiller manufacturer and ask them about running on a lower voltage to get a lower current but pretty sure they will tell you that you need to provide the current and voltage called out on the unit name plate data.

Ron

 

Fun fact. Boiling water draws 10x more energy density than lead acid batteries can store.

That depends on the volume of water. You can boil a teaspoon full of water MUCH faster than you can boil a swimming pool full of water. OK, WAY WAY WAY over the top. But the point is made.

A few decades ago I was in a classroom where they were discussing physics. The teacher asked which will boil faster, a pot of hot water or a pot of cold water? Same heat source, same volume, same container. The students all agreed the hot water would boil faster. Then the teacher set one of the pots to boil. When it was finished he then took another pot full of cold water. Then turned them on at the same time. The hot water didn't boil for an hour while the cold pot boiled in the normal time. The students were amazed. I was there servicing emergency lighting equipment but the teacher had caught my attention. So I asked him if he disabled the thermal switch that shuts the hot pot off. For a moment he looked puzzled. I then said at some point the cold water is going to be hot but not yet boiling. Why would the cold started pot boil faster than the pot of hot water. He claimed it was physics. As far as I know to this day he probably still thinks it's true that cold will boil faster than hot. But I still contend that if both pots of water were put over a flame, the hot would always boil quicker.

OK, way off track. But the point is that the volume of water is as critical as the amount of energy being put into it. A larger volume of water will boil slower. So you can't say "Fun fact". You are on the right track with "Energy Density" but all things considered, if it takes "J" amount of energy to boil "W" amount of water and you double W then (assuming a perfect world) it would take 2J to boil the water from 25C to 100C in the same time. If you use 2J to boil 1W it will boil in half the time. Relatively speaking.

 

Yes, you will need a bigger power source. Your distiller is designed for 120 VAC operation drawing 6.0 Amps. Anything less and it won't work. You could always call the distiller manufacturer and ask them about running on a lower voltage to get a lower current but pretty sure they will tell you that you need to provide the current and voltage called out on the unit name plate data.

Ron

Just an assumption on my part, but if you drop the VAC to 1/2 the normal supply, the result should be four times slower - if at all. Feel free to correct me.

6A @ 120VAC would be 720VA. (or Watts)
120VAC @ 6A would assume a working resistance of 20Ω on a hot heating element.
So at 20Ω, 60VAC would deliver 3A and 180VA. One fourth the power.

 

Just an assumption on my part, but if you drop the VAC to 1/2 the normal supply, the result should be four times slower - if at all. Feel free to correct me.

Yes, but that is not what a diode does. It provides the same voltage for half the time.