i want to buld my own CT

[studyete]

My instructor thought that as i'm trying to make law cost energy meter which also can give many other service in future like adding the remote metering to it, so i should use this 2 uC based energy meter, but as CT is very costly, like 30 dollar, why not i use a shunt sensor.

any way, I've little idea what my CT give as output. As you describe earlier, if maximum 10A goes through the phase, 1/40 of that current will come through the CT which is 0.25A right? Also as the avr465 uses 1:2500 Ratio CT which mean i need a shunt which can give .004A at the output when maximum 10A current goes through primary line? if these are correct, then my understanding with the shunt is zero. So, pleas help me in this part.

Finely, I want to built shunt with resistor at my home, so pleas tell me what are the are the elements do I need to reform this circuit . if possible, pleas add some calculation

 

[Reloadron]

Attached is an image of a 50 Amp current shunt. It explains the terminals. The small terminals are the voltage output terminals. A current shunt is no more than a very low value resistor for practical purposes. When we pass a current through a resistor we get a subsequent voltage drop across the resistor. Current shunts like this are normally used for DC current! However, they will work for AC.

I suggest you read this link in detail and note where it mentions:

Since the shunt is a resistor which operates in a varying ambient temperature environment and generates heat whenever current flows through it, we must design the shunt resistance portion with a special metal alloy which has, essentially, a constant resistance value through its operating range. In addition to using a precision resistance alloy with a negligible temperature coefficient of resistance, the alloy must have a low resistivity in order to keep the self-heating to a minimum, and it must also have a low thermal E.M.F. against copper. A special shunt manganin alloy meets all these criteria, therefore, all our shunts are designed using this special alloy.

I would not use a current shunt for this but whatever your instructor wants I guess.

You will still need to amplify and signal condition the milli-volt output of the shunt. You also want the shunt on the low side of the load. I suggest you ask your instructor about high and low side current sensing.

Ron

 

[studyete]

I've question about old way (using ct) as i've change the given CT using 5:200 instead of 1:2500, I've change the shunt or burden resistor to 1 ohm, but the question is, do i have to change the whole analog current front i.e capacitor value and resister value? if so, then how do i know the value for these components? capacitor and Resistors? pleas do have a look of the analog current front of the avr465

 

[Reloadron]

Back to the basics again. The CT you had if I recall correctly was a 200:5 CT. Therefore the actual ratio is 200 / 5 = 40 so in reality it has a ratio of 40:1. The fact that it has a 200 amp rating means it will handle 200 amps of primary current. Therefore 40 amps through the primary yields 1 amp through the secondary. 200 amps through the primary is 5 amps through the secondary. Therefore 0 to 200 amps results in 0 to 5 amps.

So we add lets say a 1.0 Ohm shunt or burden resistor across the secondary. There will be a voltage drop across that resistor. With 0 amps through the primary of the CT the output will be 0 amps through 1 ohm for 0 volts. With 200 amps through the CT primary the secondary current will be 5 amps through 1 ohm for 5 volts. V = I * R so 1 * 5 = 5 volts. Thus the span is 0 to 200 amps = 0 to 5 volts. This is basic ohms law.

The capacitors you see in the original Atmel drawing are used for coupling. They are 47 uF electrolytic capacitors. They block any DC component level. The value is chosen based on the 50 Hz. line frequency. All of that is well covered on page 8 of the AVR pdf. The formula is shown. Everything is explained in detail.

The CT used in the drawing has a max current of 275 amps. It should be noted that in the circuit they never exceed 10 amps. That is it! The circuit is designed around a max current of 10 amps. The fact that the CT is rated for a max of 275 amps means nothing! With a 2500:1 ratio and 10 amps the CT secondary would be 10 / 2500 = .004 amps or 4 mA. They use a 68 Ohm shunt so .004 * 68 = .272 volts. The amplifier at this point has a gain of 1.25 so we get 1.25 * .272 = .34 volts RMS. That becomes .34 * 1.414 = .48076 volts peak and .48076 * 2 = .962 volts peak to peak.

The micro controller is a 10 bit ADC using a 1.1 volt reference. The ADC input needs to be kept below 1.1 volts which is what is done.

OK, so the max current is 10 amps. You want 100 amps? Then change Rshunt to .1 * 68 = 6.8 ohms. You want 200 amps? Then change Rshunt to .5 * 6.8 = 3.4 ohms.

You need to start understanding the basic fundamentals and apply them.

You have a current that is converted to a small voltage proportional to that current. Matters not if that small voltage comes as a result of a CT or a plain current shunt. You work with that small voltage and signal condition it.

The AVR in this case uses a 1.1 volt reference. The A to D is a 10 bit A/D converter. Ten bits means 2 ^ 10 or that an input of 1.1 volts = 1024 bits. When 1.1 volts is applied to the A/D input we get an analog input of 1024 bits. The input to the analog channel can not exceed 1.1 volts or the input will over range.

If the instructor suggest using a shunt I suggest you read up and understand high side and low side current sensing with a basic shunt.

I am not about to do this assignment. I did all my homework over 40 years ago. More to your question:

I've question about old way (using ct) as i've change the given CT using 5:200 instead of 1:2500, I've change the shunt or burden resistor to 1 ohm, but the question is, do i have to change the whole analog current front i.e capacitor value and resister value? if so, then how do i know the value for these components? capacitor and Resistors? pleas do have a look of the analog current front of the avr465

First, the ratios are 200:5 which is 40:1 and 2500:1 they are not 5:200 or 1:2500. Please get it right! May seem silly but it is very important. The value of Rshunt has nothing to do with the cap value or the resistors that determine the gain of the amplifier.

Ron

 

[studyete]

Halo again, Can you help me on power supply part for the meter pleas? My current circuit consumes (+/-)30mA. But the power supply can serve about 24mA, so the circuit is getting heated. I've managed to increase some mA by adding a .68uF capacitor and 470ohm resister parallel way. Resistor i'm using is 10Watt, still it is getting hot. So, I'm thinking to have transformer less power supply which can give me 50 to 100mA 5V clean output. Can you give me some good circuit design on that? U've searched google, but most of them are not good enough. so i thought it would be good to discuss it here.

 

[Reloadron]

I am not quite understanding what you are saying? If the PSU is capable of 24 mA and you need 30mA nothing will change that. Why are you placing a resistor across the power supply? A capacitor opposes voltage change, they are used for filtering at a power supply output. They filter and that is all they can do. They do not and will not increase the power supply output current. If a power supply is capable of 24 mA you can't place a 30 mA load on it. Can you post a schematic of this?

Ron

 

[studyete]

https://www.daycounter.com/Circuits...Supplies/Transformerless-Power-Supplies.phtml This site suggest how I can increase output current at the output. If you put the value at calculator box, you will get the result. My main problem is heat, 470ohm resistor is getting hot. I've added LCD another uC, so the power consumption got increased.

here is the schematic for this **broken link removed**

 

[Reloadron]

I didn't know you were talking about a transformerless supply. Personally, and I know the AVR circuit uses one, I dislike them as they have no isolation from mains power. Supplies like this deliver very little current and while the value of the cap contributes you can only get so much. I assumed you were using a standard transformer supply. The supply used in the AVR circuit should supply the AVR and that is about it.

Ron