VARISTOR A CHOICE?

[VARISTORS]

A 400 WATTS INVERTER (12 V DC TO 120 V AC, RATED 300 WATTS CONTINUOUS) RUNS A VIDEO & TV UNTIL SOMEONE GETS A GLASS OF WATER. THE TAP (ON A BOAT) TURNS ON A 12 V WATER PUMP... ON OCCASSIONS THE TV FLICKERS THEN CONTINUES ... BUT USUALLY THE INVERTER SHUTS OFF. YES I KNOW THE OBVIOUS NON ELECTRONIC ANSWERS. SOURCE TWO 12 V BAT IN PARALLEL. SO IS A VARISTOR A SUITABLE REMEDY? ... WHAT SIZE? ... WHAT ADDITIONAL INFO IS NEEDED? ... ALTERNATIVES?

[Chris Smith]

A varistor clips the voltage, when or if it should rise above the given. <p>Because it is a resistor, it can not "add" to a drop in voltage.

[Robert Reed]

How about this rule
"No water or refreshments served while movie is in progress"

[terri]

How about you "hefty-ize" the wiring to the inverter (and its connections !) Four hundred watts is 33 amps more or less from the battery. Betcha you've only got about 10-ga wiring to the inverter. And, if possible, shorten the feed lines to the inverter. When you're talking 33 amps, you're talking a significant supply line drop with even a little resistance in either the supply line itself or its connections.<p>Other solutions would involve adding an extra battery (as you mentioned) or supercaps and stuff like that near the inverter. Adding temporary energy reservoirs near the heavy-drawing appliances, like these supercaps and the like, are fairly standard solutions to getting enough current to heavy-drawing appliances like high-power audio amps and transmitters during either peak loads or source voltage drops.<p>Scan around the forum for discussions of stuff like this.<p>[ September 21, 2005: Message edited by: terri ]</p>

[philba]

You can't be the first boater/RVer/off-gridder/... to face this problem. <p>I suspect that the resistance of the wire has minimal impact - battery internal resistance will probably be the dominant factor causing voltage drop. I'm sure that the motor start-up has a pretty significant inrush. Single battery supply robots have a very similar problem. Basically, a motor's start-up voltage drop is enough to wreak havoc with sensors or even reset the microcontroller(s). <p>The solution there is to use a storage reservior cap (like terri mentioned) and a diode in series between the battery and cap bank (to prevent "draw-back"). Schotty diodes are often used for a lower voltage drop. The size of the cap bank is easy to calculate if you know how long the start up inrush is, the inverter's draw and the shut down voltage level of the inverter. I suspect we are talking about some pretty big capacitors. Given the size of the inverter's draw, a second battery is likely to be a more cost effective idea.<p>Another possible solution is to softstart the pump to mitigate the inrush. I'm not 100% certain it would work but may be worth a try if a second battery is out of the question. This could be accomplished by PWMing the motor in a steady ramp from 0 to 100% over a couple of seconds. The idea is to spread the impact over a longer period. There maybe be ready made solutions out there - a lot of hand held tools use this approach to soften the initial start-up jerk.<p>[ September 21, 2005: Message edited by: philba ]</p>

[terri]

That's why I emphasized the connections. I assume a fairly heavy gauge supply, but I couldn't be sure from the problem as stated, so I figured I'd throw that in about line resistance, too. As opposed to connection resistances.<p>[ September 22, 2005: Message edited by: terri ]</p>

[VARISTORS]

CHECKING OUT CABLE SIZES, CONNECTORS, START UP INRUSH, DRAW & SHUT DOWN V. CONSIDERING AC RELAY OFF INVERTER TO OPEN PUMP POS. FROM HOUSE BATTERIES WHICH RUN INVERTER; SWITCHING PUMP ONTO MOTOR BAT (ISOLATED FROM HOUSE BAT.) DURING INVERTER USE ... THE SUGGESTED SECOND BATTERY.
PHILBA - COULD YOU SUGGEST AN ON LINE CALCULATOR TO DETERMINE THE CAP BANK SIZE?
THANKS GUYS! - STILL OPEN TO POSSIBLE SOLUTIONS.

[Chris Smith]

Caps are expensive compared to batteries, farad for farad values. <p>Even a small motor cycle battery down stream with back flow diodes can stop this temporary outrush of current draw from the main battery.<p> The pump probably draws 5 to 10 amps at start up and thus any small 6 to10 amp motorcycle battery will last for 1/4 hour or more on its own.<p> Surge is all you trying to stop from the main battery, and although caps can do the job, their value and cost is not comparable to a inexpensive surplus MC battery.<p>You could also use a sensory relay that cuts out the main battery and taps off the small battery when the pump is turned on. <p>When it relaxes, it reconnects the two batteries back together for a recharge. <p>This would isolate the two, and still allow pump use for short bursts.<p>However, a by pass would be needed for things like a shower, if this happens to be a house boat.<p>[ September 22, 2005: Message edited by: Chris Smith ]</p>

[philba]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by JACK MCLEOD:
CHECKING OUT CABLE SIZES, CONNECTORS, START UP INRUSH, DRAW & SHUT DOWN V. CONSIDERING AC RELAY OFF INVERTER TO OPEN PUMP POS. FROM HOUSE BATTERIES WHICH RUN INVERTER; SWITCHING PUMP ONTO MOTOR BAT (ISOLATED FROM HOUSE BAT.) DURING INVERTER USE ... THE SUGGESTED SECOND BATTERY.
PHILBA - COULD YOU SUGGEST AN ON LINE CALCULATOR TO DETERMINE THE CAP BANK SIZE?
THANKS GUYS! - STILL OPEN TO POSSIBLE SOLUTIONS.<hr></blockquote><p>Would you mind using lower case?<p>Sorry, don't know of a calculator. the math is fairly straightforward. V = Vi*e^(-t/RC) where Vi is the initial voltage, t is the time in seconds, ^ is exponentiation, R is resistance in ohms of the load, C is capacitance in farads. You can plug that into excel. You can compute R by using ohms law with current draw of the inverter and the voltage. <p>An example:
300 watts computes to about .48 ohms (12V, 25 amps). A 10 Farad cap bank would drop by about 1 V (12 to 11) in 400 milliseconds. I don't know if that's long enough or is above your inverter's shut down voltage. But, that is one HUGE cap. Supercaps are usually rated at 6V or less. You'd need something rated above 12V - I'd stay at least at above 18V for 50% headroom. But this will be way pricey, not to mention pretty large. I'd use a second battery way before I'd do that. <p>But the more I think about it, the softstart approach could be very effective, simple and compact. The October 1997 issue of The Silicon Chip had a car headlamp softstart circuit that might be adaptable. I did a bit of looking on google and didn't see anything pre-made or in kit form.<p>[ September 22, 2005: Message edited by: philba ]</p>

[MrAl]

Hi there Jack and others,<p>
The solution *could* be as simple as installing
a single power resistor in series with the 12vdc
pump motor. A 2 ohm resistor would limit
pump current to 6 amps, which could be good
enough. It would help if we knew the pumps
running dc current. It may draw as little as
2 amps but have an initial surge of 30 amps
while it's trying to start.<p>Starting with a 1 ohm power resistor would be
typical too, and if that doesnt work try 2 ohms.
There will be some drop while it's running but
that's ok. If it draws 2 amps and you use a
1 ohm resistor then the drop is only 2 volts so
it still gets 10 volts. It might pump slower,
but that's it. A 10 watt 1 ohm resistor would
be ok.
If 1 ohm doesnt work try 2 ohms. If that doesnt
work then you can try current regulation using
a regulator circuit. The complexity of the
circuit will be determined by the normal run
current of the pump motor.
If 2 ohms does in fact work, it should be
rated at 15 watts or better.
If you want to get fancy, you can have a dc coil
relay connected to small cap and resistor to
short out that 2 ohm resistor after a delay of
say 2 seconds. This will limit surge but allow
the pump to work normally after it starts up.<p>Let us know what the pump draws normally while
it's running. This will help a lot.<p>
Take care,
Al

[terri]

There was a tricky little circuit I saw once for limiting the inrush of current to a high-power transmitter power supply, using a relay in series with the primary which switched in a resistance in series with the primary and then dropped out when the initial current surge dropped to running levels. This was about twenty years ago, and I remember thinking ten years later that the circuit concept might be adaptable to solid state circuitry, but I dropped the subject. Perhaps this is the kind of headlight circuit Philba touched on. It would be a struggle to look this one up again, my reference works being buried in storage, but the circuit was not as obvious as one would think right off the bat.<p>In the spirit of brainstorming, would it be better to put the supercaps or small battery near the pump motor rather than the inverter?<p>Also, would not a hefty L reactance in series with the pump motor do the trick? Maybe an old flourescent lamp ballast or the like?<p>Just brainstorming.

[philba]

Don't you want that power resistor to be a little higher wattage? <p>I'm not sure I understand the voltage drop calculations. If the motor pulls 6A at 12V (steady state), then its effective resistance is 2 ohms. Wouldn't the 2 ohm resistor in series with the motor have a 6V drop and limit the current to 3A in that case? and dissipate 18 watts. But the motor would only see 6V/3A as well, maybe that's enough to run it. <p>The inrush resistor/relay approach sounds like it might work though it's a bit brute-force in nature.

[terri]

Philba: What did you think of the flourescent ballast (inductance) in series with the pump motor?<p>Too way out? I've no idea what their inductances or DC resistances would be, it just popped into my mind.<p>[ September 24, 2005: Message edited by: terri ]</p>

[philba]

I think one could calculate the reactance of the inductor based on the rise time of the current (i.e. it's frequency) and the inductance. Probably easier to try it than to guess at the values to plug into the equation. what's the value of a typical ballast?

[terri]

That's sort of what I was asking, since I don't have any ballasts aroound to play with. But I also wondered about the general concept of putting any big L in series with the motor, of whatever nature, like maybe an old transformer or something cheap and reasonably available. <p>
The idea of using an old flourescent ballast, which should be dumpster-available, occured to me simply because I am a world-class cheapskate, always looking for the dumbest, cheapest, and simplest way of doing something. The old quote, "If you want to find the easiest way of doing something, give the job to a lazy man" applies to me. I'm eminently qualified on that basis, and have been well-rewarded for that kind of "resourcefulness."<p>[ September 25, 2005: Message edited by: terri ]</p>