1. Reduce Arcing Time – Bus Differential
Bus Differential (87B)
MV or HV
IM1 = IF1 + IF2 + IF3
IM1 = IF1 + IF2 + IF3 + IF4
87B
IM1
IF2
IF3
IF1
IM1  IF1 + IF2 + IF3 X
IF4
Instantaneous trip!
CT issues
CT balancing
CT accuracy / linearity
CT “burden” capacity
Load impedance - saturation
Zone Selective Interlocking is a way of communicating between circuit breakers to let upstream breakers know when they should delay tripping and when they should trip instantaneously. Normally, the upstream breaker must delay since you want only the downstream breaker to clear the fault and the upstream breaker to remain closed on the fault. But if the fault occurs upstream from the downstream breaker, the downstream breaker can’t clear the fault. The Zone Selective Interlocking works is that a restraining signal is transmitted from any downstream breaker that sees a fault up to the upstream breaker feeding that downstream breaker. The upstream breaker sees the fault current, but also sees the restraining signal and delays tripping, allowing the downstream breaker to clear the fault. On the other hand, if the downstream breaker doesn’t see the fault, it won’t send the restraining signal and that is a green flag for the upstream breaker to trip instantaneously, as it is closest breaker that clear the fault. The result is that the fault is cleared in 80 milliseconds instead of 1/2 second and the incident energy released at the point of the fault drops from 43.7 calories (which exceeds the 40 calorie limit of PPE 4), down to only 7 calories, safely below the 8 calorie PPE 2 limit which is only a fire resistant shirt and pants.

2. CT Error 𝐼 𝑒 = 𝐸′ 𝑋 𝑒 𝐸′=𝐼′∙ 𝑍 ′ + 𝑍 𝐵 𝐶𝑇 𝑒𝑟𝑟 = 𝐼 𝑒 𝐼′+ 𝐼 𝑒
Primary Current
Secondary Current
CT excitation
current
𝐼 𝑒 = 𝐸′ 𝑋 𝑒
𝐸′=𝐼′∙ 𝑍 ′ + 𝑍 𝐵
𝐶𝑇 𝑒𝑟𝑟 = 𝐼 𝑒 𝐼′+ 𝐼 𝑒

3. CT Saturation #14 AWG (0.002525 /ft) 100’ total wiring (50’ x 2)
22.5V
ISEC or I’
I PRI
600:5
Z’/2 (conductor impedance)
Z’/2 + - E’
ZB (relay burden)
Emax
~5V
𝐸′=𝐼′∙ 2 𝑍 ′ /2 + 𝑍 𝐵
E’ < EMax
𝐸′=𝐼′∙ 𝑍 𝑇𝑜𝑡𝑎𝑙
#14 AWG ( /ft)
100’ total wiring (50’ x 2)
ZB = 0.2  load
Z’ = 100 * = 0.25 
ISEC = I’ = 5A * 10 = 50A (10x fault)
E’ = I’*ZTotal = 50*( ) = 22.5V
22.5V
0.04A
0.5% (or better) accuracy?

4.  CT Error 𝐶𝑇 𝑒𝑟𝑟 = 𝐼 𝑒 𝐼′+ 𝐼 𝑒 𝐶𝑇 𝑒𝑟𝑟 = 0.04 50+0.04 =0.08% I’ = 50 A
I’ Secondary Current
I Primary Current
𝐶𝑇 𝑒𝑟𝑟 = 𝐼 𝑒 𝐼′+ 𝐼 𝑒
I’ = 50 A
Ie = 0.04
𝐶𝑇 𝑒𝑟𝑟 = =0.08% 
Better than 0.5%