---
title: "Thyristor V-I and opto-coupler isolation"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:thyristor-vi-opto-isolation"
source: https://www.bctengineer.com/q/semester_7/power_electronics/thyristor-vi-opto-isolation
site: BCT Engineering
---

# Thyristor V-I and opto-coupler isolation

Power Electronics · Chapter 1: Characteristics and specification of power electronic devices

- Asked in: 2076 Asoj, 8 marks
- Video: [What is Optocoupler ? How Optocoupler Works ? The Optocoupler Explained](https://www.youtube.com/watch?v=eHevtwPn27w) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/thyristor-vi-opto-isolation
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

Explain the V-I characteristics of a power thyristor. How an opto-coupler can be used to isolate the gate signal generator and power circuit.

## Short answer

A thyristor has reverse blocking, forward blocking, and forward conduction regions; reverse above V_BR and forward above V_BO cause avalanche, while gate current lowers breakover voltage. An opto-coupler separates control and power circuits: the pulse drives an LED whose light turns on a phototransistor, connecting an auxiliary supply to the gate with no electrical path.

## Answer

A **thyristor (SCR)** is a four-layer P-N-P-N, three-junction ($J_1, J_2, J_3$) semiconductor switch with three terminals: anode (A), cathode (K) and gate (G). It can be turned ON by a small gate pulse when forward biased, but once ON the gate loses control; it turns OFF only when its anode current falls below the holding current.

#### V-I characteristic

```text
              IA
               ^
               |  |<- ON state (VT = 1-2 V)
               |  |
         IL ---|--|
         IH ---|--+.
               |    `.   Ig2 > Ig1 > Ig=0
               |      `-.___      ___
     -VBR      |            `----'   |
  -----+-------+---------------------+--> VAK
       |       |  forward blocking  VBO
       |       |  (leakage current)
       | reverse blocking
       | (leakage current)
       v reverse avalanche
```

1. **Reverse blocking mode:** cathode is positive with respect to anode. Junctions $J_1$ and $J_3$ are reverse biased and $J_2$ is forward biased, so only a small reverse leakage current flows. If the reverse voltage reaches the **reverse breakdown voltage** $V_{BR}$, avalanche breakdown occurs at $J_1$ and $J_3$; the current rises sharply and the device is usually destroyed.
2. **Forward blocking mode:** anode is positive and gate is open. $J_1$ and $J_3$ are forward biased but $J_2$ is reverse biased, so only a small forward leakage current flows. The SCR is OFF and blocks the forward voltage.
3. **Forward conduction (ON) mode:** if the forward voltage is raised to the **forward breakover voltage** $V_{BO}$ (with $I_g = 0$), junction $J_2$ breaks down by avalanche and the SCR switches suddenly to the ON state (the negative-resistance jump). The voltage across it drops to about 1 to 2 V and the current is limited only by the load. With gate current, $J_2$ breaks down at a lower voltage: the larger $I_g$, the smaller the breakover voltage ($I_{g2} > I_{g1}$). In practice the SCR is always turned ON by a gate pulse at a voltage well below $V_{BO}$.

The latching current $I_L$ and holding current $I_H$ ($I_L > I_H$) are marked on the ON-state part of the curve.

#### Isolation of the gate circuit by an opto-coupler

**Opto-coupler isolation:** the gate signal generator (logic level, grounded to the control circuit) and the power circuit (hundreds of volts, cathode floating) must be electrically separated. An opto-coupler does this with light.

```text
   control side     |      power side
                    |          + Vaux
    R1              |          |
 o--/\/\--+         |          R2
 pulse    |         |          |
         LED ~~~~>  |   photo-transistor
          |  light  |   (C top, E below)
 o--------+         |          |
 GND                |          +------> G
                    |          |
                    |          R3
                    |          |
                    |  Vaux(-) +------> K
```

1. The control pulse drives a current through the LED via the current-limiting resistor $R_1$.
2. The LED emits infrared light, which falls on the phototransistor (or photo-SCR) inside the same package.
3. The phototransistor turns ON and connects the auxiliary supply $V_{aux}$ (referenced to the cathode) through $R_2$ to the gate, giving gate current. $R_3$ prevents false triggering by noise.
4. When the pulse ends, the LED is dark, the phototransistor turns OFF and gate current stops.

There is no electrical path between the two sides, so the isolation can withstand a few kV, and noise from the power circuit does not reach the logic circuit.

## Check yourself

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

1. In the reverse blocking mode, which junctions are reverse biased and which is forward biased?
   A. J1 and J3 reverse biased, J2 forward biased
   B. J2 reverse biased, J1 and J3 forward biased
   C. All three reverse biased
   D. J1 reverse, J2 and J3 forward
   Answer: A. In reverse bias J1 and J3 block while J2 is forward biased; only a small leakage current flows. (Key point: Reverse blocking junctions)

2. Which junction blocks the voltage in the forward blocking mode?
   A. J1
   B. J2
   C. J3
   D. All of them equally
   Answer: B. With anode positive, J1 and J3 are forward biased and J2 is reverse biased, so J2 blocks. (Key point: Forward blocking junction)

3. How does gate current affect the forward breakover of the SCR?
   A. Larger Ig raises the breakover voltage
   B. Larger Ig lowers the breakover voltage, so the SCR is fired by a gate pulse well below VBO
   C. Gate current has no effect
   D. Gate current makes the SCR reverse block
   Answer: B. J2 breaks down at a lower voltage as Ig increases, so in practice the SCR is turned ON by gate pulses well below VBO. (Key point: Effect of gate current)

4. How does the opto-coupler isolate the gate signal generator from the power circuit?
   A. An LED lights a phototransistor, so only light crosses and there is no electrical path
   B. A resistor divider links the two sides
   C. A common ground ties both sides
   D. A relay contact closes the gate circuit
   Answer: A. The LED light turns on the phototransistor, which connects Vaux to the gate; no electrical connection exists between the sides. (Key point: Opto-coupler isolation)

## Close variants set in other exams

- [Thyristor V-I, application and opto-isolation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athyristor-vi-application-opto) — Power Electronics, chapter 1, asked 2×

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