---
title: "Static Ward-Leonard four quadrant control"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:static-ward-leonard-four-quadrant"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/static-ward-leonard-four-quadrant
site: BCT Engineering
---

# Static Ward-Leonard four quadrant control

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2073 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/static-ward-leonard-four-quadrant
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

Explain static Ward-Leonard control scheme for four quadrant control of separately excited motor. Discuss non circulating and circulating current schemes. Mention its merits over conventional Ward-Leonard control.

## Short answer

Static Ward-Leonard uses a dual converter back-to-back for four-quadrant control; the non-circulating scheme blocks one converter with dead time, while the circulating scheme fires both with alpha sum 180 degrees and uses a reactor. It has no rotating machines, higher efficiency, faster response, and lower maintenance than conventional Ward-Leonard control.

## Answer

The **static Ward-Leonard drive** replaces the M–G set of the conventional Ward-Leonard system by a thyristor phase-controlled converter. For **four-quadrant** control, a **dual converter** (two fully controlled converters connected back to back) feeds the armature of a separately excited dc motor, whose field is kept constant.

#### Scheme
```text
          +--[Converter P (+I)]--+
 ac ------|                      |---[ M ]  field const
          +--[Converter N (-I)]--+
   P: alpha_P,  N: alpha_N = 180 - alpha_P
```
- Converter P carries positive armature current; converter N carries negative current.
- Each converter can give positive or negative average voltage ($V = V_{do}\cos\alpha$).

| Quadrant | Converter | $\alpha$ | Mode |
|---|---|---|---|
| I | P | < 90° | Forward motoring |
| II | N | > 90° (inverting) | Forward regenerative braking |
| III | N | < 90° | Reverse motoring |
| IV | P | > 90° (inverting) | Reverse regenerative braking |

To brake from forward motoring, P is blocked, N is fired at $\alpha_N > 90^\circ$ so that its voltage is just below $E_b$; current reverses and energy returns to the supply.

#### Non-circulating current scheme
- Only one converter conducts at a time; the other is blocked (no firing pulses).
- On reversal, current in the working converter is brought to zero, a **dead time** (about 2–10 ms) is allowed for thyristors to recover, then the other converter is fired.
- No inter-group reactor needed; smaller converters, higher efficiency.
- Slower response and possible discontinuous current near zero.

#### Circulating current scheme
- Both converters are fired continuously with $\alpha_P + \alpha_N = 180^\circ$, so their average voltages are equal and opposite in sign around the loop.
- Instantaneous voltages differ, so a **reactor** is connected between them to limit the circulating current.
- Current can pass smoothly from one converter to the other without dead time: fast, smooth reversal and continuous conduction.
- Costlier: reactor, converters rated for load plus circulating current, extra losses.

#### Merits over conventional Ward-Leonard
- No rotating machines: less space, weight, foundation, noise and maintenance.
- Higher efficiency (above 95 %) and fast response (milliseconds).
- Lower cost for most ratings; easy closed-loop control.
- Regeneration still possible.

Drawbacks: harmonics and poor power factor at large $\alpha$ (low speed).

## Check yourself

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

1. How is four-quadrant control achieved in static Ward-Leonard?
   A. One converter with a rheostat
   B. Two fully controlled converters back to back (dual converter)
   C. A single diode bridge
   D. A motor-generator set
   Answer: B. Converter P carries positive current, N negative current, so voltage and current both reverse. (Key point: Dual converter)

2. In the circulating current scheme, firing angles satisfy:
   A. alpha_P = alpha_N
   B. alpha_P + alpha_N = 180°
   C. alpha_P + alpha_N = 90°
   D. alpha_P = 0
   Answer: B. Average voltages are then equal and opposite; a reactor limits the circulating current from the instantaneous difference. (Key point: Circulating relation)

3. A feature of the non-circulating scheme is:
   A. Both converters fire all the time
   B. Only one converter conducts, with a dead time at reversal
   C. It needs an inter-group reactor
   D. It has the fastest reversal
   Answer: B. The other converter is blocked; dead time of 2-10 ms is needed for thyristor recovery. (Key point: Non-circulating)

4. A merit over conventional Ward-Leonard is:
   A. Lower power factor
   B. No rotating machines, efficiency above 95 % and fast response
   C. Cheaper at all ratings
   D. Less harmonics
   Answer: B. Static converters avoid the M-G set and respond within milliseconds. (Key point: Merit)

## Related in Utilization of Electrical Energy

- [Ward-Leonard system of speed control](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Award-leonard-speed-control) — Utilization of Electrical Energy, chapter 3
- [Four quadrant operation in DC motor control](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afour-quadrant-dc-motor-control) — Utilization of Electrical Energy, chapter 3
- [Speed of separately excited DC motor (given J, B)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-motor-speed-load-100nm) — Utilization of Electrical Energy, chapter 3
- [Plugging of 200 V, 1000 rpm separately excited motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-plugging-200v-1000rpm) — Utilization of Electrical Energy, chapter 3
- [Plugging of 220 V separately excited motor (100 A)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-plugging-220v-100a) — Utilization of Electrical Energy, chapter 3
- [Chopper ON time for separately excited DC motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Achopper-on-time-separately-excited-motor) — Utilization of Electrical Energy, chapter 3

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- [25 kW separately excited DC machine as motor or generator](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Asep-excited-25kw-125v-terminal-voltage) — Electrical Machines I, chapter 3
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