Can anyone please help me to find out the relation between KWH and the pulses generated in a chip used whose part number is WT7752B.

 

KWH is a large unit of energy, typically used in mains energy calculation, and is generally not relevant to chip operation.

What is a WT7752B?

 

it is a polyphase electrical energy measurement IC intended for use in any 3-phase distribution system.
m trying to design a energy meter using this IC n neuvoton n792844a microcontroller.please help.
m trying 2 find out if 1 kwh of power is consumed then what about the pulses that are generated as output n if 5 kwh of power is consumed then what is the difference in the shape n characteristic of pulses..i can not find out thru datasheet evn.please help.

 

Could you PLEASE stop with the text speak?

It would help if you post a link to the datasheet.

 

http://www.prolific.com.tw/US/ShowProduct.aspx?p_id=145&pcid=109

 

Here is a link to Prolific's contact page:
http://www.prolific.com.tw/US/contact.aspx
Fill in your info, and make certain to click the "Technical" button - then ask your question there. I suggest that this is the quickest route to obtain a good answer.

 

Could you PLEASE stop with the text speak?

It would help if you post a link to the datasheet.

this is the datasheet can u now help me solving my question?

 

That IC doesnt measure kWh. It outputs pulses at a rate proportional to instantaneous power (kW). You would have to feed those to a suitable counter to get kWh.
The pulse rate has no fixed relation to monitored power, since the IC relies on inputs from scaled voltage sources and from current-to-voltage transducers. You would need to calibrate the pulse rate in terms of known device-under-test voltages and currents. The IC has a CF output for calibration purposes.

 

That IC doesnt measure kWh. It outputs pulses at a rate proportional to instantaneous power (kW). You would have to feed those to a suitable counter to get kWh.
The pulse rate has no fixed relation to monitored power, since the IC relies on inputs from scaled voltage sources and from current-to-voltage transducers. You would need to calibrate the pulse rate in terms of known device-under-test voltages and currents. The IC has a CF output for calibration purposes.

thak u sir,so is this IC WT7752B suitable to be used in the project 3-phase Energy meter??as u said this IC measures instantaneous power.but in electric meter used at home we need to design it which would measure in terms of KWH?..please guide...tl me the suitable IC(if any) that we can use which would further be interfaced with microcontroller..(in my project).

please suggest me the ways to interface same IC WT7752B (if suitable to use)with microcontroller STM32F030 series...please guide??

 

Use the micro to measure the IC output pulse rate, to apply the necessary scaling factor to convert that rate to instantaneous kW, and to integrate the kW value to get kWh.

 

Use the micro to measure the IC output pulse rate, to apply the necessary scaling factor to convert that rate to instantaneous kW, and to integrate the kW value to get kWh.

sir m exactly not getting ur point.can u plz brief me about scaling factor and the integration part.
m using STM32F030C6 micro,and planning to interface the output CF pin to the counter of the micro.is it correct..its just i thot this way..please correct me and guide..m very new to this domain..

 

The IC receives, for each of the 3 channels (phases), two analogue signals (voltages) Vi,Vv, proportional repectively to device-under-test (DUT) current Di and DUT voltage Dv. The proportionality constants ki,kv are unknown and depend on the sensors used to obtain Vi and Vv. The IC output pulse frequency f is proportional to Vi x Vv and hence power, but the proportionality constant is not specified. Hence the need to do a calibration by using a known DUT whose current and voltage can be measured by some means which doesn't involve the IC. If, for example, you have a DUT which when run from 240V draws 3A you know its power consumption = 240 x 3 = 720W = 0.72kW. If we assume you now use the IC to monitor that DUT and the IC produces an output frequency of 355Hz from the CF pin (I'm just using that arbitrary value for argument's sake), then the scaling factor SFcf would be 355/0.72 = 493Hz per kW (for the CF output).
Note that the 3 main outputs of the IC are frequency-divided (by some power of 2 which I can't remember) with respect to the CF output. If the division factor is R, then the overall scale factor to be applied to a main output would be SFcf/R Hz/kW.
So, for measuring energy consumed by a DUT you would need your micro to count how many pulses you get from a main output of the IC in one hour and multiply the count by SFcf/R to get kWh.

 

The IC receives, for each of the 3 channels (phases), two analogue signals (voltages) Vi,Vv, proportional repectively to device-under-test (DUT) current Di and DUT voltage Dv. The proportionality constants ki,kv are unknown and depend on the sensors used to obtain Vi and Vv. The IC output pulse frequency f is proportional to Vi x Vv and hence power, but the proportionality constant is not specified. Hence the need to do a calibration by using a known DUT whose current and voltage can be measured by some means which doesn't involve the IC. If, for example, you have a DUT which when run from 240V draws 3A you know its power consumption = 240 x 3 = 720W = 0.72kW. If we assume you now use the IC to monitor that DUT and the IC produces an output frequency of 355Hz from the CF pin (I'm just using that arbitrary value for argument's sake), then the scaling factor SFcf would be 355/0.72 = 493Hz per kW (for the CF output).
Note that the 3 main outputs of the IC are frequency-divided (by some power of 2 which I can't remember) with respect to the CF output. If the division factor is R, then the overall scale factor to be applied to a main output would be SFcf/R Hz/kW.
So, for measuring energy consumed by a DUT you would need your micro to count how many pulses you get from a main output of the IC in one hour and multiply the count by SFcf/R to get kWh.

thank u sir for d support..
1 more thing i need to ask about this IC is how to decide the neutral current??IAN,IBN,ICN.

SOME DATAS THAT WOULD HELP:m using 3 current transformer CT1010(1:1000) such that that 5 Ampere current is converted to 0.005 ampere.i have used a burden resistor of 750 ohms so that it gives 0.5V at its output..this inputS m giving to the IC WT7752B..
nw my ques is:
in the CT section the i/p current of 5A is measured with respect to ground isn't it??
n it is mentioned in the datasheet that vmax differential is 0.5 V ,so as m getting it thru the CT o/p wHich m givin to IAP,IBP AND ICP so what should be the value of IAN,IBN,ICN??should i connect those to ground??
i evn dont know how shud i ask ques...so please guide..

 

i have used a burden resistor of 750 ohms so that it gives 0.5V at its output.

According to the datasheet 0.5V is the absolute maximum for the analog inputs. It is not good practice to run things normally at that level. You should allow for current spikes exceeding 5A.

in the CT section the i/p current of 5A is measured with respect to ground isn't it?

I assume from the datasheet that the IC differential inputs of a given current channel would be applied directly from the respective CT burden resistor terminals, neither of which is directly grounded. I don't think you need to connect IAP/N,IBP/N or ICP/N to ground. The IC has an analog ground (to which bias circuitry for the opamps is presumably connected).
It's a pity the datasheet doesn't show a typical schematic using the IC.

 

Doesn't it seem that it would be better to integrate in the analog domain? It seems like if you just sample instantaneous current periodically (at the pulse output rate of the IC) that if the load being measured had certain characteristics, either accidentally or intentionally to defeat the metering, that it could result in significant measurement error.

Since a metering device like this really cares about energy, not power, it would seem better to convert sensed current to energy at the lowest level possible. I'm sure there must be chips out there that do this...

 

According to the datasheet 0.5V is the absolute maximum for the analog inputs. It is not good practice to run things normally at that level. You should allow for current spikes exceeding 5A.
I assume from the datasheet that the IC differential inputs of a given current channel would be applied directly from the respective CT burden resistor terminals, neither of which is directly grounded. I don't think you need to connect IAP/N,IBP/N or ICP/N to ground. The IC has an analog ground (to which bias circuitry for the opamps is presumably connected).
It's a pity the datasheet doesn't show a typical schematic using the IC.

yes sir for the spikes u r talking about m using a tvs diode(bidirectional)5V.as it is mentioned in datasheet that "an overvoltage of ±6 V can be sustained on these differential current inputs without risk of permanent damage"..

and the value of burden resistor i have calculated using ideal values.i wl explain .plz confirm me with ur rply.

A current of 5A to CT(1:1000),so o/p current is 0.005 A..now to get 0.5 V we calculate value of Burden resistor by formula V=IR.

 

for the spikes u r talking about m using a tvs diode(bidirectional)5V

Good idea.

we calculate value of Burden resistor by formula V=IR

So R=V/I = 0.5/0.005 = 100 Ohm. Why then are you using 750 Ohm for your burden resistor?

 

Good idea.

So R=V/I = 0.5/0.005 = 100 Ohm. Why then are you using 750 Ohm for your burden resistor?

oh by mistake i wrote 750 ohm,m using 75 ohm .
0.005 is converted to peak value.that comes as 0.00707 and
0.5/0.00707 is 70.7 ohm so m using 75 ohm...is it ok?

 

That looks ok.

 

That looks ok.

ok.thanku