Load sensing circuit for inverters

[baggo]

Hi all,

I am looking to build a load sensing circuit to automatically switch my inverters on and off. At the moment I switch them on manually as required to save unecessary power drain. Have found a few designs for controling multiple items from one e.g. turning all your Hi-Fi gear on when you switch the pre-amp on. These rely on a permanent mains supply though which is not much use with an inverter as there is no mains until you switch it on!

Have any of you electronic Gurus out there any suggestions how it can be done?

Obviously the higher end inverters already have this built in but I am using the cheap MSW variety.

Thanks in advance for any help and suggestions,

John

[thunderhead]

I believe the simplest circuit is to provide a DC sense current on the line, which I guess would be at your invertor's input voltage.  The problem with this is that this sense current would need to be isolated when the AC comes on.

The other alternative is to run a tiny, tiny invertor (say 100 watts or less) using a tuned transformer, and drive it with short duration pulses at mains frequency.  Then you can check the voltage just before applying the pulse: when some sort of load is attached to the tuned transformer, this will result in that voltage dropping, which can be used to switch in more invertors.

I posted a description of this sort of circuit some months ago - but it's a design "in principle", and you'd need some fooling around with electronics to build a practical version.  I haven't done this because recently I've been more interested in charger design and microprocessor control - and getting a nursery room sorted for the son and heir!

Here is my old posting

http://www.fieldlines.com/story/2004/4/11/25436/6769

Hope this helps.

 

[RobD]

In order to sense a DC load on the battery side you would need the inverter to be on which would defeat your purpose.

You could design a circuit that turned on the inverter every, say, 100ms to see if there was a load on the secondary side (AC) at which point the inverter would stay on. I think this is how trace does it.

RobD

[Opera House]

I have a 1927 Kohler generator and it has this auto start technology!   Low voltage DC is placed on the AC line.  A load with some resistance causes current to flow and the electronics turns on the inverter.  A large inductance is placed in this DC sense line to block most of the normal AC current Inverters with transformer outputs (this would look like a load) must be isolated by an additional relay that has a capacitor in series with the coil to block the DC sense current.  

That turns everything on except in the case of switching power supplies unless you add a resistor to them.   Then a current sense in the AC line shuts everything off when current is no longer being drawn.   This can all be really complicated.   I wouls think using 3 way (for the first two) and 4 way switches (for additional locations) in every room to turn the inverter on would be simpler.

[RobT]

Hi John  You might try this.  It's not a real flash drawing, but the schematic does work!  If you have any problems, I may be able to clarify it a little.

Hope it's of some use!  Rob T

[baggo]

Thanks all for your suggestions - you've given me an idea which I am going to try.

I'm thinking of using a MAX472 current sensing IC across a low resistance shunt in the AC side of the inverter to sense the current when a light etc is switched on. Initially a small DC voltage will be connected across the AC output to provide the sensing voltage. The MAX472 will see the small current drawn from the DC by the load and then operate a relay to switch on the inverter. A second relay will connect the load to the inverter after a short delay so that the inverter does not start under full load. This second relay will also disconnect the DC sensing voltage. The current drawn by the load will keep the MAX472 relay and inverter operated until the load is switched off. The first relay will switch the inverter off and the second reconnect the DC sensing voltage until next time round. The MAX472 circuit will be fed from the 12 volt side via an isolating DC to DC converter for safety. Sounds simple enough but we will see!

Thanks again,

John

[Ungrounded Lightning Rod]

Please recheck your schematic.  Looks to me like the output is an input solely driving the LED and there's no connection from the current sensing circuitry to the output.

[RobT]

re checking the schematic:  It is as it should be. Please bear in mind that this is just the load control of a larger circuit. Power(+12v) comes in thru R1, and also powers Q1,2,3. R1 provides current to the forward-biassed 4004 diodes and produces a DC voltage of around 1.2v on the Load Neutral line.  If an external load is sensed, (something switched on!) then this voltage is applied to the anode of the 1N60 diode. This turns Q1 on, which turns Q2 on and provides current to the 8v2 zener. This gives  8v regulated power to the 8v output connector, which, in turn, is directly linked to the power pins of (in my case) the 555 timer/4017 combination, and presto!!

If no load is there, then the whole thing virtually shuts down, with only about 200uA being drawn from R1.

Now, if all this sounds like I know what I'm talking about, not necessarily so!  I painstakingly followed instructions from the schematic,(which worked) and am now passing said instruction on.

Rob T.

[Ungrounded Lightning Rod]

This turns Q1 on, which turns Q2 on and provides current to the 8v2 zener.

So far so good.

And the 8V zener drives emitter follower Q3 (to harden up the voltage with some current) which charges C4 (to smooth the output) through small resistor R6 (to protect Q3 from the inrush current into C4).

And then it goes nowhere.

Meanwhile the output is only connected to the indicator LED D5 through its ballast resistor R7.  This will never turn on unless an external voltage is applied to the output connection.

I think you're missing a wire from the + side of C4 to the output pin.

[Ungrounded Lightning Rod]

Also:  Missing or ambiguously placed labels:

 - Which diode(s) is/are 4004 and which are 1N60?  (I presume the 16 diodes are 4004s.)

 - Which pins are which on J2?  (I presume left is 1, right is 2, and "Tx" means the inverter.)

The inverter output must be floating because the ground bond is provided in this circuit through the diodes.  It might be nice to note this on the diagram.

About 2% of your power will be dissipated in the sampling diode stack when the inverter is running.  1N4004s are rated at 30A peak surge or 1A continuous rectified.  So your diode stack is good for about 2A of load.  You might want more (a LOT more) parallel strings of diodes and good air circulation.

[RobT]

Not sure where you're coming from here... With J1 and J2, all standard pinouts identify the square pin as pin one.(always!) As for the diodes, yes, we have 16 1N4004's with, as noted on the schematic, 1 1N60 sitting there on its own.

I can't follow your line of thought, but I do know that I have a working 600VA inverter sitting here on my bench, after building according to the above.(It pulls a lot more than 2 amps without any hassles.)If I get time, I'll recheck my drawing. It may be that I've missed a wire somewhere...

Rob T.

[Ungrounded Lightning Rod]

With J1 and J2, all standard pinouts identify the square pin as pin one.(always!)

Thanks.  I missed that the pins were one round one square.

As for the diodes, yes, we have 16 1N4004's with, as noted on the schematic, 1 1N60 sitting there on its own.

Sounds good.  (I didn't see the 4004s designated on the schematic, and the 1N60, while on top of that diode, was near the others.)

If I get time, I'll recheck my drawing. It may be that I've missed a wire somewhere...

Just take a quick look at the output wire.  Should it be connecte ONLY to R7, or should it also be connected to, say, the + side of C4?

I can't follow your line of thought, but I do know that I have a working 600VA inverter sitting here on my bench, after building according to the above.(It pulls a lot more than 2 amps without any hassles.)

600VA at 120V = 5A RMS.  You're only overloading your diodes by a factor of 2.5 from their rating when you max out your inverter, and the diode specs probably allow some slop - especially if they're in a relatively open place in your device, so they can dissipate the heat.  (I was assuming something like a 2.5 KVA average, 5 KVA motor-start surge unit.)  Even at 2.5A each they're only dissipating about 1.5 watts.

Nevertheless, I'd use five strings of diodes rather than two, to match the rating of the inverter, just to be safe.  More for a bigger inverter, of course.  1N4004s are cheap.

[dandober]

How about feeding in a small high freq sine wave and sensing current flow?? If you use a freq above 100khz a cap of 1 micofarad will allow the 100khz to pass into AC load...but the 1uF cap acts as high impeadance to 60Hz AC..thereby isolation from AC line when inverter on. Put a zener diode clamp acroos the cap on the 100khz side..that way any AC leaking thru from line( when inverter comes on) will be clipped. Rectify the 100KHZ AC current , drive base of transistor to power relay which turns on Inverter??

Dan

[commanda]

I was thinking along similar lines. Use a transformer to couple in audio. When no load, the secondary is open circuit, hence no load on primary. Switching on a load effectively shorts out the secondary, and the primary voltage drops.

Sort of like a shorted turns tester.

Amanda

[bill541]

I would have to agree on using an AC waveform to sense the load. The problem with using DC as a load sensing signal is that any appliance with a transformer in it that is always across the line will look like a short circuit to the DC signal.

If you only have non inductive loads on your branch circuit or have switches between the AC line and any inductive loads, then you DC sensing circuit would work fine.

I think most inverters with load sensing use a short pulse periodically to test for a load change. Every second or so should be fine.

-Bill-

[dandober]

hey I drew up a circuit that might be of interest to those seeking a possible solution to the above posted problem. I have not tested it, but I think it should work. Here is the possible snafus:

1. If internal impeadance is to low at 100khz than R4 will need to be reduced. If interal Z of inverter is real low ckt will not work.
   
2. I did not check values of 555 timer to insure that it osc @100khz, check data books/online stuff on 55 timer.
   
   

I up loaded my PDF schematic but cant seem to figure out how to stuff it into this window...how do i get my circuit diagram to post, help please..

Dan

[dandober]

hey amanda.. i posted another reply to that issue..i even drew up a schematic..I used a 555 timer to generate 100khz SQ wave then AC coupled to inverter output using 1uf caps.

If I had your email I can send the PDF of the circuit, its simpe and low cost.

It does require the output of inverter to have a high impeadance(AC and DC) in off state. If that is not the case a relay could be used to disconnect it. I have a 1kw inverter and it measures about 50k DC ohms ..AC impeadance not known..

anyway I posted to the site my pdf file but could not get to my post window??

dan

[commanda]

Hey Dan,

If you look real close you'll see my email address. It's in the header box of this article just after my name.

Also, when replying, if you scroll down the page, there's a box called User Files.

Just below the Post button.

Select the file you want & hit Insert Image.

Amanda

[TomW]

Dan;

Here is a link to the pdf you uploaded:

INVONOFF_CKT.pdf

T

[commanda]

I think it would be more reliable in starting up if the inverter secondary was connected through the relay. But then I don't see how it would shut down automatically.

How's about you build a prototype?

Amanda

[commanda]

Just found this on EDN asia. It's the second article on the page.

http://www.ednasia.com/EDNA/modules.php?name=News&file=article&sid=2011

Amanda

[Peter]

Please recheck your schematic.  Looks to me like the output is an input solely driving the LED and there's no connection from the current sensing circuitry to the output.

[kitestrings]

The older Trace inverters (2512, 2524) had a load sensing circuit, that put pulses across the AC line.  When you turned on a device like a drill, say, you would get an brief 'eh-eh-eh' (please don't make me repeat the sound ;>) before the thing sprung to life.  The nice thing with it, was that you could adjust the sensitivity, so you could select a lower or higher amount of load.  This is helpful when there is either a lot of long-distance wiring down-stream of the thing, or you have phantom loads that are trigger it on, or don't allow it to go off.  However you do it, this may be a feature that you want to consider.

~ks