Inverter(s) tripping

[paulcheung]

Never mind.

[bcroe]

I believe the intent here is to add a series diode or 2 to the circuit, to reduce the peak voltage
a bit. Such diodes could be shorted when not required, to improve efficiency. Shorting could
be done with a robust switch or relay. Just how you'll control them isn't obvious.

One approach I have used in designs up to 60V and 90A, is to use N channel MOSFETs as
switchable diodes; the control power could be a 9V battery which will last shelf life.

If you short the MOSFET gate to the source, the source to drain wil appear as a diode, with
conduction in the reverse direction from normal linear operation: the source more positive
than the drain. This diode may be shorted out by turning on the MOSFET, by connecting the
neg of the 9V to the source and the pos to the gate, instead of the short. The only voltage
drop will be across the S-D resistance. Paralleling MOSFETs works since they have a positive
temp coefficient.

Of course an on MOSFET becomes a short in either direction between S-D. That may not
matter here. You could also use external diodes to be shorted, seems like a waste. Bruce Roe

[jdupa]

Right, I was not correct in saying "as opposed to a simple diode" instead of "in parallel" - this shows how limited my ele knowledge is. Nevertheless, could you please advise on a design involving the diode and MOSFET (or other better semiconductor) in parallel to turn the voltage dropping diode on and off (preferably adjustable)

Thanks a lot.
Joe

[inetdog]

The proposed solution with the MOSFET (or other better semiconductor type available now) was not to try to regulate the voltage using a series pass element but simply to turn the voltage dropping diode on and off as needed. You would still need a high power diode in parallel with the MOSFET.

[jdupa]

I followed this older discussion on inverter tripping problems, unfortunately, I made an uneducated mistake and invested in a DC/AC 1000W inverter (Go Power) never realizing that it's max input power is limited to 15V. I charge my batteries with MPPT-60 Morningstar with temperature sensor and commonly see well over 15v coming out from the controller on colder mornings. This trips the inverter and requires a manual reset of the unit - this is not convenient at all as the equipment is located outside.
I noticed the last comment on use of MOSFET as opposed to a simple diode to avoid power losses. There was another earlier poster proposing the use of high voltage LDO: http://www.solarpaneltalk.com/showth...ll=1#post42154
Would you please be so kind and let me know more about implementation of such solution? My electrical/electronic knowledge is quite minimal and I wouldn't be able to design this on my own. Unfortunately, I am not in a position to invest in another, better quality inverter at this time.

Thanks a lot!!
Joe

[bcroe]

Just to give a restart/review for this extensive thread:

1. If you cannot charge your batteries fully if you limit your charging to the current and voltage limits the battery manufacturer recommends, you are screwed. :People with very low sun hour numbers (high latitude, for example) will need to use batteries such as AGM which will accept charging at a higher rate.
2. If you can stay withing the battery manufacturer's parameters but end up tripping the inverter as a result, then a voltage dropping diode may work for you, but an elegant solution will use a relay to switch the diode out of the circuit when the batteries are not being charged. The extra power to actuate the relay (Normally Closed contacts in parallel with diode) can easily be spared during charging. No diode drop and no relay drain when pulling down the batteries after the PV has cut off.

Another approach is use an MOSFET transistor which looks like a diode when biased off. When the
diode is conducting (backward for the FET), you can apply turn on bias, and the drop will become
the FET resistance. Power to turn on essentially zero, resistance can be managed. The bias will
need to be removed if current reverses. I have equipment working this way at 90A. Bruce Roe

[inetdog]

Just to give a restart/review for this extensive thread:

1. If you cannot charge your batteries fully if you limit your charging to the current and voltage limits the battery manufacturer recommends, you are screwed. :People with very low sun hour numbers (high latitude, for example) will need to use batteries such as AGM which will accept charging at a higher rate.
2. If you can stay withing the battery manufacturer's parameters but end up tripping the inverter as a result, then a voltage dropping diode may work for you, but an elegant solution will use a relay to switch the diode out of the circuit when the batteries are not being charged. The extra power to actuate the relay (Normally Closed contacts in parallel with diode) can easily be spared during charging. No diode drop and no relay drain when pulling down the batteries after the PV has cut off.

[russ]

Hi gmturner - Welcome to Solar Panel Talk!

That is what the old threads are for - reference and learning - you did well to find this.

Russ

[gmturner]

sorry to dig up an old thread, i have a similar issue im trying to find a solution for and need 1 post to view the diagram for the LDO posted earlier.

[Sunking]

OK I am going to tell you how to fix most of the problems you have.

First let's ignore the fact you bought a cheap Chi-Com POS inverter that does not conform to any accepted standards. That was your first mistake. However it not much of a problem if it trips at 15 to 15.5 volts

The fix is real easy. You claim to be using an Outback FM 80 on a bank of Trojan batteries configured for 12 volts right? The solution is simple if you simply follow instructions from the battery manufacture. Set your OUTBACK Bulk and Absorb voltage to 14.8 volts,the Float to 13.2 volts, and your problem goes away. If you will take the time to read the Trojan Battery User Manual on page 9 Table 4 you would already know this. The only time it will become a problem is when you need to Equalize because the set point voltage needs to be 15.5 volts.

If you placed a diode between the battery and inverter is a band-aid but a horrible fix because it waste precious battery power. I say this because you also complain your batteries are never able to get 100% fully charged each and every day. You stated there is not enough hours in the day correct? That is a cop out statement. It clearly indicates the system is not properly designed to do the work it is intended for.

Any properly designed off-grid battery system should be able to fully recharge the station batteries on even the shortest winter days with daylight to spare.So when you say there is not enough sun hours in a day is BS, it means you have a poorly designed system, or I should say it was not designed. You should be able to get to the FLOAT stage with the sun still shining. If that is not happening you are screwed. It means your batteries are less than 100%, and you are not replacing what you used in a day. That means 1 major thing. Your batteries will be toast in a very short time, because when they are below 100%. When below 100% lead sulfate crystals are being formed on the plates, and that destroys the battery if allowed to continue. Instead of the batteries lasting 3 to 5 years, you only get a year or less out of them.

OK I think you said you have 6 Trojan T-105's configured as 12 volts correct? You are using a Outback FM-80 correct? What is the panel wattage connected to the FM 80?. Assuming you have it maxed out to 1250 watts, living in Central Florida only gives you 3.3 Kwh of usable energy per day. At 12 volts would requires a battery amp hour capacity. of 1375 Amp Hours. If all you have is 6 Trojan T-105's gives you an AH capacity of 660 AH, or only 1.5 Kwh of usable energy per day based on 20% DOD/day. Since you complain of not ever being able to fully charge your batteries each day tells us you have a mismatch which you need to get lined up or else you are in a world of hurt buying and replacing destroyed batteries from abuse.

What it all boils down to is you failed to design the system properly and used cheap non conforming equipment trying to cut corners for that FREE ENERGY you claim to have. You got what you asked for; Problems. The fiddler has finished and now is coming to collect his dues.

EDIT NOTE:

That diode you added between the battery and inverter is costing you an automatic 5.8% constant loss in power transfer efficiency. Math does not lie. For example if the cheap Chi-Com inverter efficiency is say 80%, that diode you added turned it into 74.2%. 0.7 volts / 12 volts = 5.8% loss

[BritishPete]

Read this back if it's wrong, I did ask first...

You wish to fully charge your bank, but your inverter/charger was tripping off due to feed overvoltage.
In order to allow the inverter charger to do it's thing internally and cycle up the voltage to the bank, you dropped it's input feed voltage via diode to allow the full cycle to complete.

This reads to me like the gist of it, and the rest is miscommunication and rapid assumptive conclusions.
That is, the extra wattage was not wasted because it never had a path to collection - rather the opportunity was wasted to collect it.
The diode allows that opportunity to get full charge to your bank by maintaining a non tripped inverter/charger to rock on.

Yes? No?

Correct except the inverter and charge controller are separate units so only the inverter is receiving reduced voltage and even then only when charge controller output is increased. The biggest misconception seems to have been that the battery charging is also connected through the diode which it is not. ONLY the inverter(s). There was also intentional mis-reading on the part of some too. (sad).

So my options are;
1) use the diode (works great)
2)Sit in the dark (figuratively) while on Equalize
3) Go out for several beers while on Equalize
4) Replace my i(otherwise perfect) nverters with different units.

I choose the $18 diode.

[tandrews]

Read this back if it's wrong, I did ask first...

You wish to fully charge your bank, but your inverter/charger was tripping off due to feed overvoltage.
In order to allow the inverter charger to do it's thing internally and cycle up the voltage to the bank, you dropped it's input feed voltage via diode to allow the full cycle to complete.

This reads to me like the gist of it, and the rest is miscommunication and rapid assumptive conclusions.
That is, the extra wattage was not wasted because it never had a path to collection - rather the opportunity was wasted to collect it.
The diode allows that opportunity to get full charge to your bank by maintaining a non tripped inverter/charger to rock on.

Yes? No?

[Sunking]

To me it is very simple but I guess I am not expressing myself well enough.

You are right, everthing you have expressed and re-explained is MAKE BELIEVE VODOO. It is clear you have no understanding of very basic electrical principles. So when you come here trying to make yourself out as an expert and claiming to be an EE is clearly a lie and fraudulent. A first year EE student with ZERO experience knows more than you do.

You have been exposed, give it up, you have lost any and all credibility here. You are only making a fool out of yourself.

[BritishPete]

Britishpete there is something terribly wrong with your arguements in just about every area.
you wrote this..
I shall; When my PV array is capable of generating say 80 amps ant 14.8V but is only charging the batteries at 20 amps because they are 80% full where is the 60 amps going? Answer WASTED. Now install a diode that will permit me to increase charge current to 30 amps (at say 15.2V) without tripping the inverters and I have GAINED 10 x 15.2 = 152 watts not wasted.
It is not rocket science but some would have us believe it is.
Now im beyond interested because you said in a much earlier post the diode was between the battery/ies and the inverter. YES??So how the hell does it help if its there to increase the charge current to the battery/ies from the charge controller????????????
ANY DIODE PLACED BETWEEN BATTERY/IES AND INVERTER WILL ALWAYS WASTE POWER. IT IS A FACT OF LIFE. ITS ALSO A FACT THAT ANYTHING PLACED AFTER THE BATTERY/IES WILL NOT DO ANYTHING TO INCREASE/DECREASE THE CHARGE RATE TO THE BATTERY/IES
OK IM RUDE IN SAYING I THINK YOUR ENGINEERING DEGREE IS IN EITHER PLUMBING OR DENTISTRY OR MOUNTAIN BIKE CONSTRUCTION.BUT I AINT RETRACTING IT. IF YOU HAVE A DEGREE IN ELECTRICAL ENGINEERING ITS A PHOTOSHOPPED ONE OF SOMEONE ELSES. MORE RUDE BUT I THINK TRUE AND THE MORE POSTS YOU DO THE MORE I BELIEVE IT.AND EVERYONE ELSE DOES ALSO ON THAT FORUM.

I have stated this many times but will state it one last time.

There are not enough daylight hours in a day to bring my battery bank above about 85% at an aborb voltage of 14.8V because the charging current tapers off as battery internal resistance increases. Despite this the battery manufacturer says that they should be periodically charged to 100%.

I cannot do as the manufacturer requires without increasing charging voltage (and hence current) and I cannot increase charging voltage without the inverters tripping.

Now when I have say 80 amps charging a 50% SOC battery all is fine but when the battery goes above 85% SOC the charging current reduces to about 20 amps even though another 60 amps is available it is not used; Hence WASTED.

Now the diode stops the inverter from tripping by reducing the input voltage by 0.7V when I increase the charge voltage and am able to complete a 100% SOC.

Of course the diode is consuming power but it only power that would otherwise not be used anyway so which is the lesser of two evils?

To me it is a no brainer! Increase voltage as the manufacturer recommends to obtain a 100% SOC and "Waste" about 15Whrs (typical for my loads) of 10 amps) for 2 hours through the diode which otherwise would NOT have been accessed from the panels anyway so how is this a waste !!!

Now In order to protect a $1000 battery bank I am perfectly willing to use 30Whrs once a week. Watt hours whcih otherwisw would not have been produced anyway.

I hope that is clear now because it is truly my last comment on the matter.

To me it is very simple but I guess I am not expressing myself well enough.

[john p]

Britishpete there is something terribly wrong with your arguements in just about every area.
you wrote this..
I shall; When my PV array is capable of generating say 80 amps ant 14.8V but is only charging the batteries at 20 amps because they are 80% full where is the 60 amps going? Answer WASTED. Now install a diode that will permit me to increase charge current to 30 amps (at say 15.2V) without tripping the inverters and I have GAINED 10 x 15.2 = 152 watts not wasted.
It is not rocket science but some would have us believe it is.
Now im beyond interested because you said in a much earlier post the diode was between the battery/ies and the inverter. YES??So how the hell does it help if its there to increase the charge current to the battery/ies from the charge controller????????????
ANY DIODE PLACED BETWEEN BATTERY/IES AND INVERTER WILL ALWAYS WASTE POWER. IT IS A FACT OF LIFE. ITS ALSO A FACT THAT ANYTHING PLACED AFTER THE BATTERY/IES WILL NOT DO ANYTHING TO INCREASE/DECREASE THE CHARGE RATE TO THE BATTERY/IES
OK IM RUDE IN SAYING I THINK YOUR ENGINEERING DEGREE IS IN EITHER PLUMBING OR DENTISTRY OR MOUNTAIN BIKE CONSTRUCTION.BUT I AINT RETRACTING IT. IF YOU HAVE A DEGREE IN ELECTRICAL ENGINEERING ITS A PHOTOSHOPPED ONE OF SOMEONE ELSES. MORE RUDE BUT I THINK TRUE AND THE MORE POSTS YOU DO THE MORE I BELIEVE IT.AND EVERYONE ELSE DOES ALSO ON THAT FORUM.