Hello,

i have a transformer rated 230/12-0-12, 500mA. What is this 500mA?
Is it for the LV or HV?

 

Hello,

The current rating is for the Low voltage side.

Bertus

 

it about ur transfomer secondry lv 5ooma

 

The current rating is for the Low voltage side.

Bertus

then check the attachment.
Am i went wrong???
Can i use the arrangement for current measurement?
I have posted a similar thread earlier.... but my doubt was not cleared..


Check the attachments in detail..

 

then check the attachment.
Am i went wrong???
Can i use the arrangement for current measurement?
I have posted a similar thread earlier.... but my doubt was not cleared..


Check the attachments in detail..

You appear to be using a mains to low -voltage transformer "backwards" as a current sensing transformer. If that is what you are doing, I suspect that you may be infringing some safety regulations, but that aspect I will leave to others better qualified to comment.

As for the current sensing, you may not have calculated the output correctly. You appear to have used the low voltage winding resistance to predict the transformer output. If so, this will not give the desired result. You need to treat this thing as a current transformer. I'm afraid that it won't work very well - expect a pretty big error from magnetizing current.

Neglecting any errors, with a 12V:230V transformer, the nominal current ratio would be 1:0.0522, so 430mA input current would give 22.4mA output, or 22.4V into a 1kΩ load. Similarly, if you used the whole 12V-0-12V, the current ratio would be 1:0.104, so you might expect 44.8V into 1kΩ. Magnetizing current errors would reduce these values, but I have no idea by how much.

Bottom line: get a proper current transformer and you can hope for safe and predictable results - providing you don't forget to load the secondary!

 

A 12-0-12 volt transformer will give you about 17vdc after you rectified it and the 500mA means you won't be able to use a load exceeding 500mA.
500mA is also 1/2 amp and is normally use for low voltage diagrams

The 230vac is the primary side of the transformer and is usually plugged into the house mains outlet. This is the only side of the transformer that is high voltage. You need this input to enable the transformer to turn on.

 

Apparently you have made a number of assumptions about how transformers behave. Stick with the specifications given: your transformer provides 12 volts a.c. at 500 ma (maximum). This is determined, in part, by the ratio of turns (primary to secondary) and the a.c. voltage applied to the primary. The amount of current you can draw is determined largely by the size of the wire used in the windings.

 

Hello adjuster

You appear to be using a mains to low -voltage transformer "backwards" as a current sensing transformer.

Exactly

this will not give the desired result.

no problem

Magnetizing current errors would reduce these values

i need some voltage from the transformer which i can use it safely. I can calibrate it to work with a comparator(LM324) using a preset. see the attachment.[/QUOTE]

Bottom line: get a proper current transformer and you can hope for safe and predictable results

purchasing a small CT is not a big thing for me. Is it impossible to design with my plan. if not, what are the corrections i have to take.

You appear to have used the low voltage winding resistance to predict the transformer output

yes of-course, i know the inductance comes into picture. What ever, is it not possible to get at least a safe 5v DC.???

 

You could shunt the primary of the "CT" with lets say 0.1 ohm resistor, which should leave you with a voltage transformer sensing the current, with some amplification (or attenuation if you use it the other way around). The resistor will dissipate 10W at 10A, so this should be pretty safe and predictable, definitely more than using the transformer as a real CT.

Also you will need some low-pass filter after the diode so that you measure average current and not instanteneous.

 

Look at something like this:

http://iteadstudio.com/store/index....oducts_id=56&zenid=ifkcl1lihk66pr6v96t9bk6k22

You will not have to even cut your wires.

 

i need some voltage from the transformer which i can use it safely. I can calibrate it to work with a comparator(LM324) using a preset. see the attachment.

I think your basic idea will work, but I think the voltage you'll be getting will be too high for your comparator. I think both voltages (the measurement and the reference) need to be within certain limits, and those limits are related to the power supply for the comparator. I don't think they can be both purely floating, as you've drawn. You've shown a ground symbol, but it's not clear what that means. You'll need to work out how ground loops are dealt with.

You'll also want to filter (integrate) the peaks from your peak detector, or the comparator will flop at 60Hz. Maybe that's OK. A simple RC tank (a capacitor in parallel with the 1K resistor) will prevent the flopping at that speed.

 

You could shunt the primary of the "CT" with lets say 0.1 ohm resistor, which should leave you with a voltage transformer sensing the current, with some amplification (or attenuation if you use it the other way around). The resistor will dissipate 10W at 10A, so this should be pretty safe and predictable, definitely more than using the transformer as a real CT.

Also you will need some low-pass filter after the diode so that you measure average current and not instanteneous.

A shunt resistor across the input would certainly reduce the risk, but it would need to be very substantial to avoid going open in the event of a fault, with potentially disastrous effects for the connected equipment. A purpose-made shunt resistor would be preferable. The value of the shunt should be as low as possible, but the output voltage would be rather small.

With 0.1Ω shunting the 12V winding, he would get only 43mV rms input at 430mA input, so ignoring any losses a 12V:230V ratio would give 824mV rms or 1.166V peak. That sounds a bit low for straight rectification. Depending on what current level the OP needs to detect, he might need to amplify the signal or use an "active" rectifier, which amounts to the same thing.

There is not much doubt that something of the kind could be made to work, but is it worth all the effort compared to using something designed for the job? What would be the legal position if something went wrong with such a lash-up?

 

but I think the voltage you'll be getting will be too high for your comparator.

ok

I think both voltages (the measurement and the reference) need to be within certain limits, and those limits are related to the power supply for the comparator.

I can use a pot(voltage divider) to get in safe limits.

You've shown a ground symbol, but it's not clear what that means. You'll need to work out how ground loops are dealt with.

see the attachment[/QUOTE]

 

but is it worth all the effort compared to using something designed for the job?

of-course NOT. But, i wan2 try with this type....

What would be the legal position if something went wrong with such a lash-up?

what do u meant by "if something went wrong" ?
probably the 12v transformer may damage right????
is there anything more risk involved???

If it is that much dangerous..... then suggest me a suitable CT....

One more thing, instead of that transformer, if i use 0.1Ω, 1/4W resistor, can i directly connect it to the comparator....(see the attachment)

 

One more thing, instead of that transformer, if i use 0.1Ω, 1/4W resistor, can i directly connect it to the comparator....(see the attachment)

Definitely a bad idea. You really don´t want to get the mains voltage anywhere near the low voltage circuit. For example if the resistor for some reason goes open, you will have 230V right in the circuit burning everything in its way. And killing you aswell if you accidentaly touch anything.

 

Definitely a bad idea.

Ok.....then .... any suggestions about my transformer idea????

and, suggest me CT. Im first time using the CT. I dont know how they specified..

 

Ok.....then .... any suggestions about my transformer idea????

and, suggest me CT. Im first time using the CT. I dont know how they specified..

Real CTs are specified by their current ratio and a burden value. For example you have a CT with 5:1 ratio, max current 5A, burden <1ohm. That means if you have 5A through the primary, you will have 1A through the secondary which has to be burnt in a resistor of 1ohm or less so that the core doesn´t saturate. You then measure the voltage on the resistor (or current through that resistor if that suits you more, which is here not the case). The burden resistor has to be allways connected to the transformer or it will create very high voltage on the secondary as it is trying to force the secondary current through it.

The CT spinnaker suggested is not a "real" CT, because it produces some voltage for each amount of current, so the burden resistor is allready built in. That thing is more suited for portable measurement as you can wrap it around any wire without cutting the wire.

The shunted transformer will work for sure. It is the opposite of a current transformer, in CT you first transform the current and then convert it to voltage by the burden resistor.
The shunted transformer first converts the current to voltage by the shunt resistor, and then transforms the voltage.

Usage is about the same, but for a hobbyist I think it is easier and cheaper to get a small-power voltage tranformer that suits your needs than a CT.
What you choose depends on what you are actually building, your budget and how you expect it to perform.

 

ok see the attachment

This is making more sense now. I think you'll want to divide down the signal to the comparator, to avoid an overvoltage at its input.

You're using an op-amp as a comparator. I think you'd get a better switching by using a genuine comparator such as LM339, or by at least using a standard op-amp-as-comparator circuit (using positive feedback). As drawn, you'll be sending a varying voltage to the SG3524 shutdown pin. Its datasheet isn't real clear to me how it might respond to that. But sending it either 0 or 12v from a comparator output, and nothing in between, is sure to give a better result.