Whirlpool Induction cooktop only half works on inverter

[brightontrader]

I have talked to technical support at the Victron distributor here (UK) and he was also mystified as the 1.6 phoenix and load of only 1200w was well within specification, as was cabling battery to inverter although he did think it could be a ripple effect on the cabling not carrying the load from the batteries enough to inverter, there is a 5 metre distance and I hav enever tried this much load before at once, but have overloaded inverter many times before and normally it just cuts out and needs resetting, which its not doing on this occasion.

[transmitterdan]

5 meters is a long cable run for a high power inverter. For this distance the wire should be 11mm dia. What size wire did you use?

[brightontrader]

I do need to look a where inverter is mounted vis a vis batteries. Cable from memory is definately not 11mm, more likely 5mm.

boat is overloaded with electronics that would need an electrical engineer on board to maintain on a ocean passage, I am reviewing and simplifying over next month.

I am almost at the point of looking at gas which I had on last boat. It worked fine and boat never blew up..In the long run maybe a lot simpler.

Wondering if I can hump 5kw Mastervolt on board and try out, it weighs 47kgs...can barely lift it.

[TeddyDiver]

Originally Posted by brightontrader

I do need to look a where inverter is mounted vis a vis batteries. Cable from memory is definately not 11mm, more likely 5mm..

Seek no more faults untill you have changed appropriate cables

[transmitterdan]

Electrical fires onboard are more common than propane fires I think.

[brightontrader]

Yes on cable changes......however:

Is there such a thing as ripple effect, pushing a charge up a too thin a cable causes a ripple, I would have thought would heat up, but bounce charge back. It sounded kind of plausible when the inverter guy told me that and the symptoms of on/off clicking would seem to support that sort of thing however electricity electrons bouncing...

I have another question I may ask on another thread, how is cooking on a non gimballed cooker work in practice at sea? (my last was a gas force 10 mounted gimballed, this cooker is athawtship, or whatever the word is for the wrong way if the boat is heeled....)

[MarkSF]

Originally Posted by MarkSF

Have you calculated the losses in the cables between battery and inverter, when you're trying to take 150A or so? What gauge are they?

If I assume you have 6 gauge cable, based on the estimated diameter above, at a 5m cable run and 150A you would have a voltage drop of 2V. This dips the voltage below the minimum acceptable of 10.5V.

If you try to put a 5KW inverter on that cable, I suspect you won't get much more power out of it, as the cable is your limiting factor. It will probably also catch fire.

To power the 5KW inverter with a 10% voltage drop along the cable, you're going to need at least 2/0 cable, or 4/0, ideally.

"he was also mystified as the 1.6 phoenix and load of only 1200w was well within specification, as was cabling battery to inverter"

This is strange as I just read the Pheonix 1600 installation manual, and it states that the minimum cable cross section is 50 square mm, while if you really do have 5mm diameter, your area is 19 square mm.

[transmitterdan]

Originally Posted by brightontrader

Yes on cable changes......however:

Is there such a thing as ripple effect, pushing a charge up a too thin a cable causes a ripple, I would have thought would heat up, but bounce charge back. It sounded kind of plausible when the inverter guy told me that and the symptoms of on/off clicking would seem to support that sort of thing however electricity electrons bouncing...

Yes, there is such a ripple effect. The issue is that the current from the battery to the inverter is not pure DC. There is an AC ripple on top of the DC. This ripple is quite high in amplitude.

I have tried to explain this problem on other threads and was ridiculed for my explanation. I don't know if I can make it any simpler but I will try. Below is the output from a "simulation" of a theoretical ideal 1.6kW pure sine inverter with 80% efficiency driving a resistive load (like a water heater). The waveform is the 160A DC current from a 12.5V battery. You can see that the battery current it is not DC at all but it is 160A "DC". It peaks at nearly 250A and goes almost to zero. This is what the engineer means by "ripple effect". If the wire is not large enough to handle 250A then the voltage drop will cause problems for the inverter right at the moment it needs maximum power from the battery. You have to make the wire big enough to handle these peaks.

[txg]

Originally Posted by transmitterdan

Below is the output from a "simulation" of a theoretical ideal 1.6kW pure sine inverter with 80% efficiency driving a resistive load (like a water heater). The waveform is the 160A DC current from a 12.5V battery. You can see that the battery current it is not DC at all but it is 160A "DC". It peaks at nearly 250A and goes almost to zero. This is what the engineer means by "ripple effect". If the wire is not large enough to handle 250A then the voltage drop will cause problems for the inverter right at the moment it needs maximum power from the battery. You have to make the wire big enough to handle these peaks.

Actually, that's not the case. I'm an electronic engineer currently working on a sine inverter project. Current inverters are working with a much higher frequency than 50 or 60 Hz (somewhere in the range of 20-200kHz), and there is much filtering involved which is a lot easier (less space needed) due to the higher frequencies. Yes, there will always be some ripple, but it's a lot less than what you simulated and it will not be a sine waveform. The battery current will be mostly DC with some ripple (let's say somewhere in the range of 10%) on top of it.

Your simulation would be correct for an inverter working directly on 50/60Hz without an intermediate circuit and without any filtering. I don't have any personal experience with the victron units but i can't believe that they are working like that, as this is heavy, inefficient and expensive technology from the 70's or maybe 80's.

[Garrettw]

Everyone who is flipping out about his cable size needs to reread the first post in the thread. You missed something.


Sent from my iPad using Cruisers Sailing Forum

[transmitterdan]

You're right he has a 24V battery so the current is 1/2 of what I thought. So instead of 11mm cable it needs to be 7.5mm.

[transmitterdan]

Originally Posted by txg

Actually, that's not the case. I'm an electronic engineer currently working on a sine inverter project. Current inverters are working with a much higher frequency than 50 or 60 Hz (somewhere in the range of 20-200kHz), and there is much filtering involved which is a lot easier (less space needed) due to the higher frequencies. Yes, there will always be some ripple, but it's a lot less than what you simulated and it will not be a sine waveform. The battery current will be mostly DC with some ripple (let's say somewhere in the range of 10%) on top of it.

Your simulation would be correct for an inverter working directly on 50/60Hz without an intermediate circuit and without any filtering. I don't have any personal experience with the victron units but i can't believe that they are working like that, as this is heavy, inefficient and expensive technology from the 70's or maybe 80's.

The internal switching frequency isn't the issue. The power delivered to the load varies from 0 to maximum each 1/2 cycle of either 50 or 60Hz. So for example with a 1.6kW heating element as the load the power varies from zero to 3.2kW at each peak of the 50/60Hz sine wave (i.e. twice per AC cycle). Assuming the inverter is 90% efficient at the peak that means the inverter is consuming about 3.6kW at each sine wave peak. Where does this 3.6kW come from? It either has to come from the battery or from energy storage (capacitors and inductors) inside the inverter. There is no other place it can be.

The lower you try to make the battery ripple current the more energy that must be stored inside the inverter. This increases complexity, size, weight and cost. Plus it's really hard to get capacitors that are better than a battery at storing and delivering energy when you need it. So why would a cost conscientious engineer add lots of redundant energy storage inside the inverter when his customer has to supply a battery anyway? Yes, they will try to reduce the peak current especially at the switching frequency. But it takes huge capacitors and/or inductors to put a big dent in the 50/60Hz ripple current.

When you finish your high frequency inverter design please come back and show us how much you are able to reduce the 50/60Hz battery ripple current without huge capacitors and/or inductors whilst keeping the efficiency at 90% or better. It ain't easy.

[smac999]

with heater off, measure the battery voltage.


then turn the heater on and measure the following.


dc voltage at battery
dc voltage at inverter input
dc current on inverter wire.
ac voltage input to stove
ac current feeding stove
ac freq. at stove.


post all your results

[FlyingCloud1937]

OR

just hook a scope up to it.

lacking a scope?

Hire one ME for 1 hr @ $85.00, he will bring said scope, and be able to tell you what said problem is.

Or ask the "cruisers forum" for help, you might even be able to derive said problem.

But it may take months, and you may also question, the answer for many more months to come. Even if it seems to solve the issue??

smac, what are you smoking?

Originally Posted by smac999

with heater off, measure the battery voltage.


then turn the heater on and measure the following.


dc voltage at battery
dc voltage at inverter input
dc current on inverter wire.
ac voltage input to stove
ac current feeding stove
ac freq. at stove.


post all your results

[BigBeakie]

Brightontrader,

Any joy getting the induction cooker to run properly? I have since found out that as soon as you put the pot on the coil, it draws maximum current for a short time as part of it's sensing routine.

So perhaps there is a spike of current draw that is affecting the inverter?

It will be good to hear if you get a successful diagnosis and resolution of this issue!