Current Transformers and Burdens
Current Transformers and Burdens
(OP)
Hi
I am going around in circles with CT burden calculations… Can someone assist
Say we have the following
Ip= 100A, Is=5A, Type=5P, ALF=10, VA=15 and say Rct=0.2 ohm, Rl=0.45 Ohm (50 metres of 4mm2) and Relay Resistance Rr= 0.004 ohm and the fault level is 900 A and the maximum running current 80A.
We know that the impedance at rated burden and current is 15/(5*5) = 0.6 Ohm and hence we know that the CT must be capable of developing an internal voltage of (5 x 10 x 0.6) = 30V to deliver Is x ALF at the rated burden. This is as I understand it is the minimum or roughly appox knee voltage.
We know that at the fault level the voltage at the CT terminals to drive the secondary fault current through the cable and protection burden is 900*5*(0.45+0.004)/100 = 20.4V
I have read that the saturation voltage must be greater than ALF*Is*(Rct+Rl+Rr) = 10*5*(0.2+0.454) = 32 V
How is the actual saturation voltage calculated or assessed? How does the internal knee voltage of the CT calculated above relate to the required terminal voltage or the saturation voltage ? Is the saturation voltage calculation an approximation as it does not include reactive impedance of the CT or is it negligible enough to neglect? What is the pass/fail criteria for these calculations? Obviously AFL to fault level is one and Ip to load is another. Will this CT do the job?
Toddlers quide anyone!
I am going around in circles with CT burden calculations… Can someone assist
Say we have the following
Ip= 100A, Is=5A, Type=5P, ALF=10, VA=15 and say Rct=0.2 ohm, Rl=0.45 Ohm (50 metres of 4mm2) and Relay Resistance Rr= 0.004 ohm and the fault level is 900 A and the maximum running current 80A.
We know that the impedance at rated burden and current is 15/(5*5) = 0.6 Ohm and hence we know that the CT must be capable of developing an internal voltage of (5 x 10 x 0.6) = 30V to deliver Is x ALF at the rated burden. This is as I understand it is the minimum or roughly appox knee voltage.
We know that at the fault level the voltage at the CT terminals to drive the secondary fault current through the cable and protection burden is 900*5*(0.45+0.004)/100 = 20.4V
I have read that the saturation voltage must be greater than ALF*Is*(Rct+Rl+Rr) = 10*5*(0.2+0.454) = 32 V
How is the actual saturation voltage calculated or assessed? How does the internal knee voltage of the CT calculated above relate to the required terminal voltage or the saturation voltage ? Is the saturation voltage calculation an approximation as it does not include reactive impedance of the CT or is it negligible enough to neglect? What is the pass/fail criteria for these calculations? Obviously AFL to fault level is one and Ip to load is another. Will this CT do the job?
Toddlers quide anyone!






RE: Current Transformers and Burdens
I've never had do those types of calculations so I can't really help. I read these types of questions here periodically and wonder why the effort. Use a standard that works most anywhere on the system and don't worry about optimizing every CT for just the specific location they'd go today. What happens when the system changes a bit and that precisely engineered CT needs to be replaced because the available fault current went up a smidgen? "Overkill" can be good engineering at times.
RE: Current Transformers and Burdens
The example is 100/1 5P10 15VA
I must admit the Cxxx designations seem easier to understand and specify..
RE: Current Transformers and Burdens
RE: Current Transformers and Burdens
Very nice video series. Thanks for sharing.
RE: Current Transformers and Burdens
Thanks
RE: Current Transformers and Burdens
Absolutely wonderful material, very well done. Thank you for sharing.
EEJaime
RE: Current Transformers and Burdens
Absolutely useful and wonderful. Keep it up.
RE: Current Transformers and Burdens
To calculate voltage available at saturation, multiply 10 (ALF) * 5 (Is) * 0.2+0.6 (Rct + Rb) = 40V
The CT is rated to have a 0.6 ohm burden connected to it. You have 0.454 ohm. You still have 40V available, but you have a smaller than rated burden, so the ALF increases to 40V / 5*(0.2+0.454) = 12.2. Your maximum secondary current is now 61A and primary is 1220A which is below your fault current.
This excludes any DC component to the fault current which can saturate CTs very quickly. OTOH, modern relays can deal with saturation and act in the short time period prior to CT saturation.
RE: Current Transformers and Burdens
@surefire01 - I'm using a Wacom Intuos tablet, Inkscape graphic editor, Camtasia video editor, and Blue Yeti microphone. I spent countless hours experimenting with different hardware, software, and audio equipment until I found the perfect match for me.