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Current detector switch

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cobra1

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Ok

So i have this idea of what i want but not sure what to look for or how to start.

heres the idea.

i have 4 switches/relays, labelled 25%, 50%, 75% and 100%

what i want is these switches/relays to activate depending on on a load being taken at that time.

the 240v output will be coming from a 1kw continuous inverter. so if for instance i am drawing 500watts, the 25% and 50% relays become active, if i then plug in another device and am now drawing 500watts or more the 75% will also activate, i hope you get the idea.

thing is i dont know where to start, can anyone help, maybe point me to an existing circuit?
 
Don't call them switches if you don't physically push a button or throw a switch to turn them on it's confusing. It's just a relay, most people should know what a relay is.

The first place you need to start is with the current/voltage sense end of things as before you can do anything you need to figure out how to measure the power or you have nothing to work with from the getgo. Are there any existing shunts on the inverter that you can tap so you have a measure of the power it's using?
 
there is a 12volt source to the input of the inverter, there is nothing else there.
i could put a shunt in there though should it be needed,

how would this help, would it just measure the input current on the 12v side and use a circuit that works out how many relays need to be active??

are there any existing circuits that do this, if so what do i look for??
 
240v @ 1kw = ~4A
So with all relays in series, the first one pulls in at 1A of coil current, the next at 2A and so on.
You could use identical relays and shunt each coil with a resistor of value such that the relay only sees a portion of the current.

The total voltage drop across all relays/resistors should be <<240v.

The relays should look like the current regulator relay in vehicle electromechanical voltage regulators. I have not been able to find anyone who sells these relays as standalones.
 
Why not use 4 independent relays?

Use a current transformer to detect the amount of current passing through. Use a ADC converter block and connect it to a 1X4 MUX circuit. The MUX output energizes the relay coil.

Connect all the relays in parallel to the main line. However you have to make sure that 2 dont operate at the same time or you will see a lot of smoke:D.

I can't really help with the actual specs as I'm not very good at that sort of stuff. Just my 2 cents.
 
at some points all 4 relays would be on at the same time so this may not be a good idea. cheers for the input though.

i have had an idea, someone told me to use a resistor this way depending on the current drawn the resistor will measure between 0.5-4.5v

i think i have seen circuits that use this input before but cannot find them now as im not sure what there called.

basically depending on the input voltage, depends on what outputs are on, maybe a comparator using an op-amp would work??

any help or pointers ??
 
@cobra81

Well double switching can be avoided if the circuit is planned properly. Theoretically of course, it shouldnt happen at all. There should be some amount of grey area between the 2 relay swithings.

This grey area can be achieved by a proper ADC. The MUX is a discrete device anyways so it should work. I think. Any other comments?
 
While it might be better to use a current transformer on the secondary, it might be simpler to use a simple current shunt on the primary to measure the current. Amplify the shunt voltage and feed it into some comparator having appropriate reference voltages. See attached schematic for details (note the 1p capacitor should be more like 100n, it's 1p just for simulating).

As the input 12V is unlikely to be regulated, you should attach a better reference voltage where indicated on the cct. R12-15 are the relay coils (put freewheeling diodes across the coils of course).

The shunt 100uR can just be a section of the power cable (e.g. just tap 2 points in the cable to connect the 1k resistors to. You may need a noise-decoupling cap across the inputs also.

As an improvement, you could reduce the gain of the differential amp and follow it with an adjustable gain amp; this will allow the shunt to be most any arbitrary value and improve amp stability.

Hope this helps.
 

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Your max output is 4 amps, so here is a much better idea: Simply buy a 5 amp panel meter (make sure it includes a shunt!) and use it in series with your load. Since your output is AC, you will need to place an appropriately rated diode and capacitor between the shunt and one terminal of the panel meter. This will allow the voltage drop of the shut to be DC while not interfering with your AC. Although, I'm not entirley sure if the voltage drop of the diode will cause calibration issues or not...
 
i can suggest you to use a shunt resistor to measure voltage upto a certain range as required proposanel to current, amplify it and feed to an IC used for LED bar graph display (work with AC or DC). i remember some 40XX series available with option of bar display or dot display of level of input. ie: two modes of display.
instead of LED feed your driver transistor for relays from the out puts.

you may need to select propper shunt ressitor. also may need to use potentio meter to adjust required level of input.

EDIT: the IC is LM3914, the accuracy can be achived by propper configuration and settings, to be used for above application
 
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A 10mΩ shunt resistor in the +12V input line, a ZXCT1009 High-side current-monitor IC wired across the shunt, which feeds an LM3912 used as a four-channel comparator, driving four small relays directly, or driving four transistors which drive bigger relays...
 
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