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# Solution for measuring voltage of three phase supply

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#### lats

##### Full Member level 4
hi frnzs
need to know one thing.i want to measure the voltages of three pahse supply. i know the voltage between v1 and neutral and v2 and neutral.

for eg: say voltage between v1 and neutarl is 220volts and volatge between v2 and neutral is 215 volts .

now my question is how can i measure the volatge between v1 and v2?

3 phase voltage measurement

You just have measured the Va<->Vb voltage ..

For Va<->N=220V and Vb<->N=215V (against NEUTRAL)
Va<->Vb(of three phase system) ≈ {(Va+Vb)/2}*√3 ≈ 377V
(see picture below)

Regards,
IanP

### lats

Points: 2
phase voltage formula

that's really very helpful Ianp. Actually i'am trying to use ADE7754 for measuring 3-phase voltages and currents and frequency. Have u any idea about ADE7754? i've gone through it's datasheet and except datasheet no other information is available on net, Can u help me in this regard.

3 phase voltage

I have no experience with the ADE7754 IC, so I can't be of much help to you on this issue ..
Have you tried the evaluation board with documentation from Analog Devices?
Just in case, here is the link:

Regards,
IanP

3 phase voltages

Go to Analog site again and look under Application Notes. Look in AN-639 and AN-624. They contains info about current sensing and other things.

phase to phase voltage calculation

Ianp, someone also suggested me to go for evaluation board..but my financial conditions don't allow me to buy this board. I'm trying to do it myself, but just hoping for a small technical help on this chip....anyway, once again Thanks a lot.

I want to ask you Ianp , if some how i'm unable to succeed with ADE7754, can i measure the following parameters of 3-phase supply with normal components and how?:

1. Voltage (phase 1) and neutral.
2. Voltage (phase 2) and neutral.
3. Voltage (phase 3) and neutral.
4. Voltage (phase 1) and (phase 2). (i think this can be done using the formula u told).
5. Voltage (phase 2) and (phase 3).
6. Voltage (phase 3) and (phase 1).

7. current phase 1 using CT.
8. current phase 2 using CT.
9. current phase 3 using CT.

10. frequency

Thanks Borber too. I have gone throgh those applications. I need some coding example sort of help.

three phase voltage formula

1. Voltage (phase 1) and neutral.
2. Voltage (phase 2) and neutral.
3. Voltage (phase 3) and neutral.
4. Voltage (phase 1) and (phase 2). (i think this can be done using the formula u told).
5. Voltage (phase 2) and (phase 3).
6. Voltage (phase 3) and (phase 1).

7. current phase 1 using CT.
8. current phase 2 using CT.
9. current phase 3 using CT.

10. frequency

For 1), 2) and 3) you can measure Vmax (using peak detector, for example) and apply formula: Vrms=Umax/√2 = 0.707 Vmax ..
(Va = Vrmsa, Vb = Vrmsb and Vc = Vrmsc)

For 4), 5) and 6) {(Va+Vb)/2}*√3 ..

7), 8) and 9) can be measured with current transformers ..

10) from current transformers (as above) or independently with optocoupler(s) ..

To make calculations as simple as possible you will have to make one fundamental assumption: forget about wave distortions and treat all signals as pure sine waves ..

What accuracy are you after?
My gut feeling is that the above approximations/assumptions will keep you well below 5% mark ..

Regards,
IanP

### lats

Points: 2
voltage phase-phase equation

Thanks Ianp for such a good help. while searching on this site i found this discussion, in which u have given a peak detector and zero-crossing detector circuit. Can you please tell me four things:

1. is it compulsory to use transformers, cant i use directly a 10M ohm resitor at input (as i have seen in one readymade 1-phase circuit) (i can send this readymade meter's circuit diagram too, it's using 741, and 7107 for displaying, the value) .

2. what's the use of phase measurement, is it used to measure power factor?

3. what's the type of output of the peak detector circuit you gave.

4. which diodes have bee used in this peak-detector circuit, 1N4148 ???

3-phase phase voltage equation

1). It is not compulsory to use transformers, but one have to be aware of the possible exposure to dangerous voltages ..

2). Phase measurement, as you mentioned, is used for power factor calculations ..

3.) The output from this peak detector is DC (ripples determined by RC constant) ..

4). 1N4148 (or similar) is O.K.

If you could, please post the circuit of the single channel unit ..
Regards,
IanP

### lats

Points: 2
ct burden resistor calculation

ya i'm posting this circuit diagram, sorry for extremely bad drawing.

one thing more what would be the output of peak detector for 0-1000 volts AC

3 phase voltage calculation

Firstly, the 0-1000V ac voltage has to be reduced by a 2-resistor voltage divider (141.4:1 ratio, or different, if required) and then it can be fed to the input of the peak detector ..
Secondly, assuming that the original voltage was 1000Vac(rms) (peak ≈ 1414V) the output from the peak detector will be 10Vdc ..

Obviously you can choose different division ratio to produce, say, 0-5Vdc output ..

Regards,
IanP

### lats

Points: 2
3 phase voltage equation

Ianp, the peak detector circuit is amazing. It's working out damn perfectly, although i'm using OP07 instead of LM353 (it isn't available in my area).

Now, i want to troble you again, sorry .

I'm sending the pic of CT which i'm supposed to use (in case there are other types of CT's in market). Although i need to make the meter such that any CT (i.e CT's with different ratio's) can be installed. I've to use CT's from ratio:- 30A/5A to 900A/5A. that's a later on story. But first i want to know how to interface this CT (50A/5A) to read current. should i make some circuit such as i have shown in diagram, or how can i do it?

Thnks and sorry once again to disturb u !

formula for 3 phase voltage

While using XXX-to-5A current transformers you have to be aware that the output current from it at the maximum input current of 50A ( .. 900A) is 5A, so you need to provide adequate load to read this value ..
Your drawing suggests 10kΩ; well, this value has to much, much lower than that ..

A 20W load resistor of 0.707Ω would give you ≈5Vdc after peak detector at full primary current. However, 20W resistor would be impractical as far as its physical size and generated heat are concerned ..

If you decrease this resistance to, say, 0.2Ω, its power rating will drop to 5W, 0.1Ω --> 2.5W, and so on, and although they will convert current to lower voltages than 0-5V range, you can use single stage amplifier to correct this difference ..

Regards,
IanP

peak voltage in a three phase system

hi,
Can you please provide the concept and theory of this peak detector circuit.
and aslo the concept for selecting the wattage of resistors.

phase to phase voltage measurement

Can you please tell me the purpose of these CT's. These pictures are from a readymade 3-phase current measuring unit. I think the purpose of these small CT's is to reduce the current of external CT's (5 Amperes), so that small 1/4 W resistors (CT's current burden resistor) could be used. Am i right?

3 phase voltage measurements

These current transformers can have very wide current ratios, for example, 5A (1Turn - external) primary, 2mA secondary, with "burden" resistance 100Ω, and you can find several models on the market which can operate with quite substantial currents, yet producing "reasonable" outputs of, say, <100mA or so ..
Here are some examples:

Regards,
IanP

### lats

Points: 2
ct 3 phase convert current to voltage

The ratio of this CT is 1:2500, so does that mean if i feed 5 Amperes i'll get 2mA current?

and what does these mounting pin means....does it mean only ffor mounting on pcb and no interfacing with circuit?

3 phase voltage metering circuit

The 1:2500 ration will ensure that at 5A primary the output will be 2mA ..
Keep in mind, that the recommended burden resistor is only 30Ω, so the "output" voltage will be quite low .. (≈60mV at 5A primary) ..

Mounting pins are not connected to anything and can be treated as such - just for mounting purposes ..

Regards,
IanP

### lats

Points: 2
3 phase voltage formula

Is it necessary to use these types of small CT's along with external CT's. I think the designer of this circuit is using these ct's to eliminate the high wattage resistors. Can't i use directly (let's say) 0.1 ohm 5w resistor and get a particular voltage, which i will give to "single stage amplifier" (as u advised). Now for different CT's, the 5 volts achived (full value) would represent the CT ratio and the current can be displayed using simple mathematics.

ex:

for 50A/5A CT :- 5Volts would represent 50A

for 100A/5A CT :- 5Volts would represent 100A

etc.

this division ratio can be set usig simple micro-controller programming.

Am i going in right way?

ltc1966 ac voltmeter

If you don't mind a little bit of heat generated by directly connected low-ohm resistor(s) then, in my oppinion, this is the right way to go ..
The main adventage is this: without amplification you have 5V of signal in hand ..

Regards,
IanP

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