What is the value of resistor?

Thread Starter

onlyvinod56

Joined Oct 14, 2008
363
Hi
I want to measure the current in line. I want the sample of the current. I used a low power rating transformer as current transformer. As the secondary is HV side, my digital circuit may damage due to high voltage. I attached a circuit here. Can i directly connect the transformer to my electronic circuit? Or should i use any isolators?

And what is the value of resistor R. I just need the range of the resistor.
My digital circuit operating voltage is 5v. can any body help me?
 

Attachments

mik3

Joined Feb 4, 2008
4,843
The maximum current through the secondary is 25mA so if you want to have a voltage drop of 3 volts across the resistor:

R=3/25mA=120 ohms

It would be safer to use an optocoupler.
 

Ron H

Joined Apr 14, 2005
7,014
The maximum current through the secondary is 25mA so if you want to have a voltage drop of 3 volts across the resistor:

R=3/25mA=120 ohms

It would be safer to use an optocoupler.
12/230*5A=260mA.
How did you get 25mA?
 

mik3

Joined Feb 4, 2008
4,843
Oh yes, you are right Ron H. I calculated 0.25 and i though it was mA but it was Amps. Thus, the correct answer is:

R=3/0.25=12 ohms or to be accurate is 11.5 ohms

I apologize onlyvinod56 for my mistake. It was a stupid one :confused:
 

Thread Starter

onlyvinod56

Joined Oct 14, 2008
363
Hi mik3,
Yesterday itself i got the correct answer for the Resistor. I understud that u did a small mistake.
Anyways i got the idea from u. U are warning me again regarding CT.
Can u help in detail about the precautions in using CT.
 

mik3

Joined Feb 4, 2008
4,843
Here is another article from wikipedia:

Safety precautions

Care must be taken that the secondary of a current transformer is not disconnected from its load while current is flowing in the primary, as the transformer secondary will attempt to continue driving current across the effectively infinite impedance. This will produce a high voltage across the open secondary (into the range of several kilovolts in some cases), which may cause arcing. The high voltage produced will compromise operator and equipment safety and permanently affect the accuracy of the transformer.
 

mik3

Joined Feb 4, 2008
4,843
If you assume an ideal transformer and use it as a current transformer then the voltage on the output voltage will be:

V1={[μ*A*2*pi*f*cos(2*pi*f*t)]/L}*(I2*N2-I1*N1)

where

V1= output voltage
μ=permeability of the core
A=cross sectional area of the core
f=supply frequency
t=time
L=effective length of the core
I1=current in the output winding
I2=current in the input winding
N1=turns of the output winding
N2=turns of the input winding

Here you see that if I1 (output current) is zero (open circuit) the value of the output voltage becomes very high. However, if you have a load (resistor) on it a current flows and the voltage is proportional to I1*R.

If you solve for I1 by setting V1=0V to get the maximum output current, which means short circuit the output, you get this expression:

I1=I2*N2/N1

Thus, it is good practise to short circuit the output while is nothing connected on it for safety reasons.
 
Last edited:
Top