Inverter Dying?

[irwin37]

"Here might be a way to convince you. Let's assume a 1000W resistive load such as a water heater element. At 120V RMS the current is 8.333A RMS. The peak voltage times the peak current is 2000W."

This makes absolutely no sense, were are you getting a peak load with a resistor? at 120vac 1000watt heater rated for 120vac will draw 1000watts. now an inductive load may draw a much high peak current during start-up but that is momentary and the load will fall back within seconds.

[transmitterdan]

[QUOTE=socaldmax;1612445]The bold part is where I agree with you. Once a current is rectified, it is a direct current. However, like you pointed out, it is a varying sinusoidal current. Clamp on AC ammeters mistakenly take this varying current, with it's rising and falling magnetic fields, to be alternating, which it isn't.

/QUOTE]


The meter is correct. It measures the RMS of the time varying current minus the DC term. That is exactly correct. It is a mistake to ignore that current. That extra current is real and it contributes to the heating of the wires.

[rwidman]

[QUOTE=transmitterdan;1612476]

Originally Posted by socaldmax

The bold part is where I agree with you. Once a current is rectified, it is a direct current. However, like you pointed out, it is a varying sinusoidal current. Clamp on AC ammeters mistakenly take this varying current, with it's rising and falling magnetic fields, to be alternating, which it isn't.

/QUOTE]


The meter is correct. It measures the RMS of the time varying current minus the DC term. That is exactly correct. It is a mistake to ignore that current. That extra current is real and it contributes to the heating of the wires.

Are you just arguing an obscure point or are you claiming that the manufacturers are not accurately specifying the input current or cable ampacity needed?

[transmitterdan]

Originally Posted by irwin37

"Here might be a way to convince you. Let's assume a 1000W resistive load such as a water heater element. At 120V RMS the current is 8.333A RMS. The peak voltage times the peak current is 2000W."

This makes absolutely no sense, were are you getting a peak load with a resistor? at 120vac 1000watt heater rated for 120vac will draw 1000watts. now an inductive load may draw a much high peak current during start-up but that is momentary and the load will fall back within seconds.

Nevertheless it is true. A 1000W resistor fed by an sinusoid AC voltage draws sinusoid AC current. When the voltage is maximum the current is also maximum and the instantaneous power at that moment is twice the average power. When the voltage crosses zero the current and power are also zero. If you average that power over one cycle of the AC it averages out to 1000W. But at the peak it is 2000W. The thermal mass of the resistor averages the heating effect the same as if the power was a constant 1000W all the time. In AC circuits the voltage, current and power is constantly changing. An inverter therefore must constantly vary the power it draws from the battery. The battery voltage is pretty constant. So the battery current has to vary from near zero to twice the average value two times for each AC cycle (once for the positive half cycle and once again for the negative half cycle).

[Garrettw]

Originally Posted by Auspicious

You have at least two EEs and a marine engineer saying you are mistaken and yet you persist in describing our posts as based on "lay" information. Frankly I believe that any technical person who cannot responsibly describe an issue in terms any reasonably competent person can understand does not have sufficient grasp on the issues at hand.

You can make that three EEs although I gave up a while ago since the worst outcome of transmitterdan's advice is oversized wire which isn't necessarily a bad thing.


Sent from my iPad using Cruisers Sailing Forum

[transmitterdan]

[QUOTE=rwidman;1612479]

Originally Posted by transmitterdan

Are you just arguing an obscure point or are you claiming that the manufacturers are not accurately specifying the input current or cable ampacity needed?

Some manufacturers give vague recommendations. Look at the OP's manual. It's hard to know what cable size he should use in his installation.

I am arguing that the manufacturers almost always recommend cable size that exceeds normal boating ampacity tables. If you use DC formulas it isn't clear why they do that. But there is a reason. Don't ignore their advice. Also, due to the very high battery currents the connections must be perfect. There is no margin for error. Giving some actual current numbers may wake people up to the importance. But all it seems to so is generate controversy for which I am truly sorry.

[irwin37]

I did look at the manual,

"This illustrates that at 10 ft. away from the battery and
with large cables, you can expect voltage drop. Again,
try to keep cable length to a minimum and use the
maximum gauge cable possible. Xantrex recommends
the following cables for optimum inverter performance
(apply to both 120 V and 230 V versions).
1000/12: #0 AWG or 55 mm2
1000/24: #6 AWG or 13 mm2
1800/12: #4/0 AWG or 110 mm2
1800/24: #2 AWG or 34 mm"

Seems pretty straight forward to me, right from the manual Use #0 or bigger for runs upto 10 ft one way (20 ft of cable) for the 1000 watt inverter.

[Ex-Calif]

Disclaimer to the OP - sorry for thread drift.

1 - Follow the manufacturers recommendation
2 - Move your inverter closer to the batteries
3 - Use larger DC cables if possible
4 - Periodically tighten the connections.

For the scientists...

I enjoy a learned conversation. Often I learn something. However tossing theory back and forth is not useful.

How about we look at the attached white paper covering a Texas Instruments full sine wave inverter with battery charger. This paper describes in technical detail how this works. I could not find one for the Xantrax but I presume they operate the same way.

There is no magic DC that is AC. There are two steps. Amplify the low voltage DC to high voltage DC and then convert the high voltage DC to AC. I am not an electrical engineer but that's how I read it.

http://www.ti.com/lit/an/slaa602/slaa602.pdf

I would love to hear the discussion about how the scientists think this thing works.

[irwin37]

Works just like it says, Xantrex included. 12DC > approx 170 Volts DC > 120 Volts AC This white paper suggests the 12 to 120 volt transformer is the most popular method. This may have been true some years ago but today most inverters as as you described, 12 volt to high volt DC to 120 Volt AC

Think I'm done here, its been fun, night all.

[transmitterdan]

Originally Posted by Ex-Calif

Disclaimer to the OP - sorry for thread drift.

1 - Follow the manufacturers recommendation
2 - Move your inverter closer to the batteries
3 - Use larger DC cables if possible
4 - Periodically tighten the connections.

For the scientists...

I enjoy a learned conversation. Often I learn something. However tossing theory back and forth is not useful.

How about we look at the attached white paper covering a Texas Instruments full sine wave inverter with battery charger. This paper describes in technical detail how this works. I could not find one for the Xantrax but I presume they operate the same way.

There is no magic DC that is AC. There are two steps. Amplify the low voltage DC to high voltage DC and then convert the high voltage DC to AC. I am not an electrical engineer but that's how I read it.

http://www.ti.com/lit/an/slaa602/slaa602.pdf

I would love to hear the discussion about how the scientists think this thing works.

I agree with your recommendations.

Now those not interested in theory can stop reading.

You have a slight misunderstanding about how this thing works. This is a quote from page 5:

"The method, in which the low voltage DC power is inverted, is completed in two steps. The first step is the conversion of the low voltage DC power to a high voltage DC source, and the second step is the conversion of the high DC source to an AC waveform using pulse width modulation. Another method to complete the desired outcome would be to first convert the low voltage DC power to AC, and then use a transformer to boost the voltage to 120/220 volts. The widely used method in the current residential inverter is the second one and hence this reference design is based on this method."

The author's English is a little lacking. He is saying the reference design does not convert 12VDC to a high voltage DC. It converts 12VDC directly to an AC voltage that drives the primary of a transformer. On page 4 you see they label the transformer as a 50Hz transformer. That's ok because a 50Hz transformer can also handle 60Hz. The voltage applied to the transformer is an AC voltage. Transformers cannot transform a DC input.

I don't know if this design is representative of any typical marine inverters but I suspect it is not. They normally use a different topology to avoid the need for a heavy 50Hz iron core transformer. The one where they boost 12V to a high DC voltage and then convert to 60Hz without a 50Hz transformer requires much less iron and therefore weighs a lot less.

[transmitterdan]

Originally Posted by irwin37

I did look at the manual,

"This illustrates that at 10 ft. away from the battery and
with large cables, you can expect voltage drop. Again,
try to keep cable length to a minimum and use the
maximum gauge cable possible. Xantrex recommends
the following cables for optimum inverter performance
(apply to both 120 V and 230 V versions).
1000/12: #0 AWG or 55 mm2
1000/24: #6 AWG or 13 mm2
1800/12: #4/0 AWG or 110 mm2
1800/24: #2 AWG or 34 mm"

Seems pretty straight forward to me, right from the manual Use #0 or bigger for runs upto 10 ft one way (20 ft of cable) for the 1000 watt inverter.

I agree with this recommendation. In the ProWatt SW1000 manual on Xantrex web site I cannot find this text. What manual is this from?

[transmitterdan]

Originally Posted by irwin37

Works just like it says, Xantrex included. 12DC > approx 170 Volts DC > 120 Volts AC This white paper suggests the 12 to 120 volt transformer is the most popular method. This may have been true some years ago but today most inverters as as you described, 12 volt to high volt DC to 120 Volt AC

Think I'm done here, its been fun, night all.

I would guess Xantrex convert 12VDC to plus and minus 170VDC. That way they can create a pure AC sine wave without a 50/60Hz transformer. That saves a lot of weight and cost over the app-note design which requires a heavy 50Hz iron core transformer.

[Ex-Calif]

Originally Posted by transmitterdan

<snip>

"...The widely used method in the current residential inverter is the second one and hence this reference design is based on this method."

Thanks for that - I noted that on a reread but it was too late to edit my post.

So they convert to low power AC first and then amplify.

I am still wondering how battery alternating current comes into play.

[transmitterdan]

Originally Posted by Auspicious

*sigh*


Pardon? You're going to have to show me the math for pure 100A DC having an RMS current different from 100A. No superimposed AC...

You have at least two EEs and a marine engineer saying you are mistaken and yet you persist in describing our posts as based on "lay" information. Frankly I believe that any technical person who cannot responsibly describe an issue in terms any reasonably competent person can understand does not have sufficient grasp on the issues at hand.

I don't believe I ever claimed what you are quoting. You may have taken something slightly out of context. An inverter does not draw pure DC from the battery is my point. And many people assume that it does (as seen in earlier posts on this thread) which can be a dangerous assumption.

The Fourier series describing the battery current contains a zero frequency term (DC) and several harmonics of the AC output frequency. I contend each of these harmonic frequencies constitute an AC current although some call it ripple current but no meter has a "ripple" position on the selector switch. But a good true RMS meter's AC position will correctly measure that RMS current in spite of contention by some that the meter is in error in so doing. And the true RMS current is the sum of the RMS of the AC plus the DC. This value is higher than the pure DC term. If there are some EEs and marine engineers that disagree with that then you can easily prove they are wrong by pretty simple math.

What I don't understand is why it is important to anyone to try to prove I am wrong about it. The reason it is important is that this explains and should reinforce why we should not scrimp on the inverter battery wires. Do what the manufacturer says. Don't let some 3% drop ampacity table from a web site based on pure DC lighting loads lull you into thinking you can drop down a size or two in wire. That's what Sandero (the OP) did and another poster said it was alright. I see it all the time and think it's important enough to speak up and try to explain why the wire size is so important. I apologize again for thread creep but this post is exactly on topic IMO.

[transmitterdan]

Originally Posted by Ex-Calif

Thanks for that - I noted that on a reread but it was too late to edit my post.

So they convert to low power AC first and then amplify.

I am still wondering how battery alternating current comes into play.

They don't amplify, they just use a transformer with a turns ratio that transforms the low voltage AC to 120V RMS. This transformer is big and heavy which is why marine inverters don't use this topology.

All inverters regardless of topology need DC current plus some "ripple" current which is slang for AC. For our purposes we'll stipulate they are the same thing. To measure "ripple" you put the meter on the RMS AC position. Ergo, I call it AC current cuz that's what it is. Also, the inverter needs lots of peak current from the battery. 50-100% more than the DC value.

The inverter engineer could design in a "ripple" filter (aka low pass filter) between the battery terminal and the inverter proper. It would be a multi-stage inductor/capacitor filter with lots of capacitors and really big heavy inductors. Then the inverter would draw pure DC from the battery with low ripple and no high peaks. Wire and battery both could be smaller. But his boss won't let him do that because the size and weight of the inverter would just about double and the cost would go up substantially. So the inverter engineer says "ok" but I want to put the RMS and peak current specifications in the manual so the purchaser will know what to expect. Again, his boss says no just give the DC current value, they don't need to know that other stuff it'll just confuse 'em. So the engineer does the best he can and recommends a wire size bigger than anyone expects from the DC value published. Then the tech writer edits what the engineer wrote because he thinks engineers can't explain things well and then the poor customer has no idea what to do.

And that pretty much sums up this thread.