---
title: "Design IDMT grading for relays A and B"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:radial-feeder-idmt-design-ab"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/radial-feeder-idmt-design-ab
site: BCT Engineering
---

# Design IDMT grading for relays A and B

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2070 Bhadra, 8 marks
- Video: [Relay setting calculation|IDMT relay|Protection|Electrical Technology and Industrial Practice](https://www.youtube.com/watch?v=y2C_8CJkVhw) — ELECTRICAL TECHNOLOGY and INDUSTRIAL PRACTICE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/radial-feeder-idmt-design-ab
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

Design a time-current graded protection scheme based on two identical IDMT relays at points A and B for the following radial feeder. Select appropriate current tap setting and time dial settings for the relays. Find out actual time of operation of relay at A for fault in its zone and as back up. Relay at A should also back up for faults at protection zone of B.
[Figure: Source – relay point A – feeder section – relay point B – feeder – far end C.]

| Relay point | CT ratio | Fault current, A |
|---|---|---|
| A | 300/5 | 4000 |
| B | 200/5 | 3000 |
| Far end of feeder at point C | – | 2000 |

IDMT characteristics (for TDS = 1):

| PSM | 2 | 3.6 | 5 | 8 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|---|
| T.op. sec | 10 | 6 | 3.9 | 3.15 | 2.8 | 2.2 | 2.2 |

## Short answer

Both relays set PS=100%. Relay B (200/5 CT) set TDS=0.1, giving 0.22s at its 3000A fault. Relay A graded at 3000A: PSM=10, required 0.72s (0.22+0.5), giving TDS=0.72/2.8=0.26. Actual times: Relay A operates at 0.624s for own 4000A fault and 0.728s for backup at B (margin 0.508s).

## Answer

In a time-graded scheme the relay farthest from the source (B) is set fastest, and relay A behind it is set slower by a discrimination margin for every fault in B's zone, so that A acts as back-up.

**Assumptions:** relay rated current 5 A; load current is below the CT primary rating, so PS = 100% is used on both relays (pick-up = CT primary current); discrimination margin = 0.5 s (breaker time + relay overshoot + safety); TDS steps of 0.01; the curve between table points is interpolated linearly.

```text
Source ─[A]──── AB ────[B]──── BC ──── C
       300/5           200/5
      4000 A          3000 A        2000 A
```

#### Relay B (last relay, set fastest)
```text
PS = 100% → pick-up = 200 A primary
Fault at B (3000 A): PSM = 15 → t = 2.2 s (TDS 1)
Fault at C (2000 A): PSM = 10 → t = 2.8 s (TDS 1)
Choose minimum TDS = 0.1
t_B (fault at B) = 0.1 × 2.2 = 0.22 s
t_B (fault at C) = 0.1 × 2.8 = 0.28 s
```

#### Relay A
```text
PS = 100% → pick-up = 300 A primary
Grading point: fault at B, 3000 A
(maximum fault current in B's zone)
PSM_A = 3000/300 = 10 → t = 2.8 s (TDS 1)
Required t_A = 0.22 + 0.5 = 0.72 s
TDS_A = 0.72 / 2.8 = 0.257 → choose 0.26
```

#### Actual operating times of relay A (TDS = 0.26)

| Fault at | Current | PSM_A | t (TDS 1) | t_A | Relay B time | Margin |
|---|---|---|---|---|---|---|
| A (own zone) | 4000 A | 13.33 | 2.40 s | 0.624 s | – | – |
| B (back-up) | 3000 A | 10.0 | 2.80 s | 0.728 s | 0.22 s | 0.508 s |
| C (back-up) | 2000 A | 6.67 | 3.48 s | 0.906 s | 0.28 s | 0.626 s |

Interpolations used: PSM 13.33 → 2.8 − (3.33/5)(0.6) = 2.40 s; PSM 6.67 → 3.9 − (1.667/3)(0.75) = 3.483 s.

#### Settings

| Relay | CT | PS | TDS |
|---|---|---|---|
| A | 300/5 | 100% (5 A) | 0.26 |
| B | 200/5 | 100% (5 A) | 0.10 |

**Answer:** relay B: PS 100%, TDS 0.1; relay A: PS 100%, TDS 0.26. Relay A operates in **0.624 s** for a fault at A in its own zone; as back-up it operates in **0.728 s** for a fault at B and **0.906 s** for a fault at C, always at least 0.5 s after relay B.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In a radial time-graded scheme, which relay is set fastest and why?
   A. The relay nearest the source, as it carries most current
   B. The relay farthest from the source (B), with the upstream relay slower by a discrimination margin to act as back-up
   C. Both are set identical
   D. The relay with the larger CT ratio
   Answer: B. The downstream relay must clear its own zone first; the upstream relay is delayed by a margin for back-up. (Key point: Grading principle)

2. For relay A (CT 300/5, PS 100%) with a fault at B of 3000 A, what is the PSM?
   A. 15
   B. 6.67
   C. 10
   D. 13.33
   Answer: C. Pick-up is 300 A, so PSM = 3000/300 = 10. (Key point: Relay A PSM at grading point)

3. Relay B operates in 0.22 s for the fault at B; with a 0.5 s margin, relay A at PSM 10 (2.8 s at TDS 1) needs TDS of about:
   A. 0.26
   B. 0.72
   C. 0.10
   D. 2.0
   Answer: A. Required t_A = 0.22 + 0.5 = 0.72 s; TDS = 0.72/2.8 = 0.257, chosen as 0.26. (Key point: TDS calculation)

4. For a fault at A (4000 A), PSM = 13.33 giving 2.40 s at TDS 1. With TDS 0.26, relay A operates in about:
   A. 0.624 s
   B. 2.40 s
   C. 0.728 s
   D. 0.906 s
   Answer: A. 0.26 x 2.40 = 0.624 s. (Key point: Relay A own-zone time)

## Related in Switchgear and Protection

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- [Time-current grading for relays X, Y, Z](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-grading-xyz) — Switchgear and Protection, chapter 7
- [IDMT settings for 20 MVA transformer and feeder](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Atransformer-feeder-idmt-20mva) — Switchgear and Protection, chapter 7, asked 2×
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