Choosing of CT For the ADE7758

Okay
1. I will check out the voltage across the Burden resistance which should be AC and max input to the ADE which also will be AC since I am not making use of any rectifier ckt to convert into DC.
2. & the peak value of the AC is approximately 1.414 times the r.m.s value. Practically what does it mean.

SANATH EMBEDDED

firstly, am sorry to have missed ur chat box greeting in the morning. was just glimpsing through the queries and did not have enough time to reply. :-)
1. check page 18, fig 37 of the datasheet for ADE, u will know what i am talking about.
2. Vp = Vrms x sqrt(2) where Vp is peak voltage and Vrms is rms value.( for a symmetrical sine wave)

sqrt (2) = 1.414.
I could not find a way to represent the square root sign so i thought i would write the resultant value.
sorry if i confused u.. .

Thankz PDI for the valuable Reply :-)

I have again gone through the data sheet details. So the voltage across ( Vrms) should be around Vpeak/ (sqrt(2)) i.e. 500 mV / 1.414 = 353.60 mV. . So what can be standard Burden resistance value I can go for inorder to choose the precise CT for the design.






But I haven't understood what is that Common-Mode Vcm +- 25mV MAX.

:bye

Vcm is the voltage which appears equally on both the signal lines wrt the circuit ground.
It is generally noise induced in both the lines in an equal amount.

Suppose you have a pair of lines carrying a differential signal of +- 500mV.So the p-p differential voltage between them is 1V.
Now suppose a noise voltage of +25 mV appears on both lines.The signal rides on this noise voltage so the signal will now be +525mV and -475 mV.

The differential voltage is still 525 - (-475) = 1V p-p ,but this is now riding on a
(525 + (-475))/2 = 25mV common mode voltage.

Thanks Sashijoseph for the valuable Reply :-)

:-) Currently I am rigging up the ckt design for this ADE7758. So initially I want to be very friendly with this ADE7758 chip by understanding all its registers & hidden features of it.
So Can I do some basic testing without connecting CT to the circuit mean to say manually giving the required input to the each current& voltage channel and den check for the o/p.

right thing to do bro! initially, u need not even give a current source to the required input. just temporarily remove the burden resistance and give an equivalent a.c. voltage instead and check it out.
:-)

Ya thankz PDI,

But how Can I generate such a Small Varying Voltage of milli volts in AC.

use a simple voltage divider network from the secondary of any A.C. step down transformer except the one u are using for the power supply to the circuit.

Thankz PDI, :-)

Actually I was searching for the supply board which will give the precise AC varying o/p in Milli volts in the range of 0 to 500mV. But I think it is not there. :-s

:bye

use a simple voltage divider network from the secondary of any A.C. step down transformer except the one u are using for the power supply to the circuit.

pdi33

Ya PDI , :-)

As u told I will make use of voltage divider n/w from the Step down Transformer & as of now 5V power supply to the Circuit I am giving out the from my DC Eliminator

check out this below ckt of voltage divider.So i will be taking the o/p across 300 Ohms to get that maximum of 330 miili Volts AC. As I checked in Simulator Since it is 9V AC input to the Voltage divider from the secondary of Transformer, the maximum voltage drop across 300ohm resistor is 262.05mV , Maximum Current 870uA, and the maximum power dissipated is 225uW.
Is it sufficient or I need to look for some thing else.




:bye