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SILICON CONTROLLED RECTIFIER (SCR) 1. INTRODUCTION The word Thyristor means a family of similar controlled rectifiers of different types. The most popular type is called the Silicon Controlled Rectifier (SCR), Figure 1. Only this type will be covered in this course. The SCR is a 4-layer P-N-P-N device used as a latching electronic switch to control voltage, current, and power in circuits with an AC supply. (They are not used much where circuits have a DC supply because modern switching transistors are much better for this use.)

Figure 1: SCR

They have the advantage that they can convert and control large amounts of electrical power while only needing small amounts of power for control. SCR’s are widely used for applications such as regulated power supplies, heaters, lighting and DC motor control. 2. DESCRIPTION OF THYRISTOR The SCR symbol is shown in Figure 2. Under normal conditions, it can only conduct if: •

the SCR is forward biased, (anode A is more positive than the Cathode K), AND

•

a pulse of current, i G flows into the gate.

Figure 2 SCR Circuit Symbol

The SCR then latches ON, which means it will not switch OFF until the anode current I A drops in value until I A ≈ 0 A. This happens naturally every cycle when the supply is AC. But not with DC, so extra complicated circuit components are required to get switch OFF when V S is DC.

(a) No Gate Current, SCR is OFF

(b) Gate current applied. SCR latches ON

(c) Gate current removed, SCR continues to conduct

Figure 3: How an SCR is triggered to latch ON

Look at Figure 3 (a). The SCR is forward biased, but no current will flow in the SCR, because no gate current is applied. An SCR is triggered as follows: In Figure 3 (b) the push button switch is closed and gate current flows. The anode current grows and the SCR latches ON. Next in Figure 3 (c) the gate current is removed. The SCR has latched ON so it will continue conducting until the anode current I A drops to zero. With AC supply this will happen naturally, and the SCR will return to the nonconductive state. •

The gate current cannot be used to switch OFF the SCR. Only the drop of anode current I A to zero can cause the SCR to return to the non-conductive state.

•

The SCR will block current in the reverse direction like a diode.

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3. TWO TRANSISTOR ANALOGY OF A THYRISTOR The PNP structure is shown in Figure 4 (a). Imagine that it can be split into two pieces, one piece having PNP and the other having NPN as shown in Figure 4 (b). These now become equivalent to PNP and NPN connected in a connection as shown in Figure 4 (c).

(a) PNPN Structure

(b) Splitting Structure into (c) One PNP, and one Two Transistors NPN Transistor Figure 4: Two Transistor Analogy of SCR

The SCR triggers ON as follows (Figure 4 (c)): •

I G flows as the base current for the NPN transistor.

•

The NPN transistor switches ON and takes base current from the PNP transistor

•

The PNP transistor now switches ON.

•

Collector current from the PNP transistor now feeds the NPN transistor and it remains ON when the gate current is removed.

BOTH transistors will continue to conduct until the anode current I A ≈ 0. 4. THYRISTOR V-I CHARACTERISTIC CURVES The V-I characteristic of a SCR and its primary parameters are in Figure 5. The reverse blocking state is similar to a diode, as is the forward blocking state, and each is characterized by a leakage current and a breakdown voltage. With an AC supply, v S goes positive and negative and the V-I trace will follow A, O, C, O, A, etc. Only very small leakage currents will flow, so the SCR is OFF. When v S is between O and C, the SCR is forward biased. If a pulse of gate current is applied at B, the characteristic will follow O, B, D, the SCR is conducting, it is ON. Following v S , the characteristic will return back along D, B, O, A as the SCR returns to non-conducting state-OFF.

Figure 5: SCR Characteristic Curves

(Remember the gate cannot be used to switch OFF the SCR.) To summarise: •

No Gate Trigger: A, O, C, O, A, O etc. – SCR is OFF.

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•

With Gate trigger: A, O, B, D, B, O, and then sequence (A, O, C, O, A, …) repeats until…there is another gate pulse between O and C.

So, the SCR conducts until the SCR anode current drops to a low value ≈ 0 A and then stays OFF until it is retriggered, (gate pulse applied). 5. HOW A THYRISTOR CONTROLS THE LOAD VOLTAGE AND CURRENT The circuit in Figure 6 shows a half-controlled bridge rectifier with resistive load. The supply voltage and load voltage and current waveforms are shown in Figure 7. The half-controlled rectifier is triggered at a delay angle of α. Note the following: •

Supply voltage v S is the normal sinusoidal shape.

•

Load voltage v O and load current i O are rectified, but only begin when θ ≥ α because the SCR is only triggered ON at the angle .

•

A section of v O and i O has been left out (omitted) from the waveform between 0 and α, and between π and (α + π).

•

Since a part of the waveform has been left out, it must be that the average value and the rms value of the load waveforms v O and i O drop.

Figure 6: Half-Controlled Bridge Rectifier

Figure 7: Supply Voltage and Load Voltage and Current Waveforms

The average value and the rms value of the load waveforms v O and i O are derived using Figure 7 as follows: 6. AVERAGE LOAD VOLTAGE FOR A RESISTIVE LOAD

VO =

1

π

⋅∫

π

2 ⋅VS ⋅ sinθ ⋅ dθ

α

= Be able to do this derivation in full 2 ⋅ VS

VO =

π

IO =

(1)

⋅ ( 1+ cosα )

VO R

(2)

2 VS IO = ⋅ ⋅ (1+ cos α ) π R PETR301

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7. RMS LOAD VOLTAGE DERIVATION rms VO2 =

1

π

( =

π

⋅ ∫α

(

2 ⋅V S

2 ⋅VS ⋅ sin θ

)

)

2

⋅ dθ

2

1 π ⋅ ⋅ ∫α (1 − cos 2θ ) ⋅ dθ 2 π = Be able to do this derivation in full rms VO =VS ⋅ 1 −

(3)

α sin 2α + π 2π

8. RIPPLE VOLTAGE VALUE As for diodes, the ripple voltage is calculated using: VRI2 = VO2 − VO2

(4)

9. EXAMPLE BT148W-600R surface mount SCR’s are to be used in a half-controlled bridge rectifier. The rectifier is supplied from an AC source of 24 V, 50Hz. The load is purely resistive, being 17 Ω. The SCR is triggered at a delay angle of α = 60 °. Assume a temperature of T J = 25 °C. 1. Draw the circuit diagram of the piecewise model of the SCR. [See Figure 9] 2. Use Figure 8 to write the mathematical version of the piecewise model of the SCR.

2 ⋅V S R 2 ⋅ 24 = 17 2A = 

Im =

Figure 8

∆ VAK ∆I A

R AK =

( 1.5 − 1.2) (2 − 0 )

=

= 150 mΩ  Figure 9: SCR Piecewise Model

v AK = E0 + i A ⋅ R AK ( Note the notation for SCR's ) 1.5 = E0 + 2 ⋅ 0.15 E0 = 1.2 V  So it follows that: v AK = 1.2 + i A ⋅ 0.15

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3. Calculate the losses in each SCR. [NOTE: SCR losses are calculated exactly the same as for Diodes, because we always assume that α = 0 ° when calculating losses. Only notation in the formula is changed.]

Im2 ⋅ RAK π 4 2 1.2 ⋅ 2 2 ⋅ 0.15 = + π 4 914 mW = 

PSCR =

E0 ⋅ Im

+

4. Calculate the average load voltage and current. VO = =

2 ⋅ VS

π 2 ⋅ 24

π

⋅( 1 + cosα ) ⋅ ( 1+ cos 60)

V = 16.2  VO R 16.2 = 17 953 mA = 

IO =

5. Calculate the rms load voltage and current.

rms VO =VS ⋅ 1− =24 ⋅ 1−

α sin 2α + π 2π 1.047

π

+

sin ( 2 ⋅1.047 ) 2π

 ( 60° = 1.047 radians )

V = 21.53  VO R 21.53 = 17 A = 1.27 

rms IO =

6. Calculate the ripple voltage in the load.

V RI2 = VO2 −V O2 2 2 V RI = VO −VO

= 21.532 − 16.22 V = 14.2 

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7. Calculate the power in the load. 2

VO R 21.532 = 17 = 27.26 W 

P=

10. DRIVING THE THYRISTOR GATE The SCR is obviously different to the transistors and some attention needs to be given to the gate drive circuits. The main problem is: •

Gate drive (or gate control) circuits are often computer based and may work at low voltages like 5 V or 12 V. Industry instrumentation often uses 24 V.

•

SCR’s are used to drive machinery at much higher voltages like 220 V, 380 V. In South Africa the highest voltage that SCR’s are used for is 500 kV!

So, there must be some kind of electrical isolation between the control circuit and the power circuit. Common methods of isolation are: o Pulse transformers as shown in Figure 10 (a) (They can give voltage isolation to 30 kV or more). When a pulse of current is given from the gate supply v g the gate current i g is induced into the SCR gate current. Diode D is to prevent negative voltage on the SCR gate. •

Opto-isolators as shown in Figure 10 (b) can give isolation up to about 500 V. When a pulse of current is given from the gate supply v g , the light emitting diode in the opto-isolator shines onto the base of the opto-transistor. This acts the same as a base current and the transistor switches ON. This supplies gate current i g to the SCR gate. o Resistor R a limits the current through the transistor and the SCR gate. o Resistor R gk makes sure the voltage v ak on the SCR gate is zero when the transistor is switched off.

(a) Pulse Transformer

(b) Opto-isolator Figure 10: To Isolate SCR from Control Circuit

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11. EXAMPLE The BT148W-600R surface mount SCR used in a previous example requires the following to ensure successful SCR triggering. Assume the gate voltage V g = 12 V. The SCR needs a gate trigger current, I g = 200 A and gate trigger voltage, V gk = 1.5 V. 1. Calculate gate resistance, R g in Figure 11.

Using Kirchhoff's voltage law: Vak + Ig ⋅ Rg = Vg

Figure 11

Re-arrangement gives: Rg =

Vg − Vak Ig

12 −1.5 200 ⋅10 −6 kΩ = 52.5  =

2. Assume the transformer has a voltage ratio of 3:1. Calculate gate resistance, R g in Figure 12. To answer this, we must transfer the gate values on the secondary of the transformer to the primary side to “get rid of” the transformer. [If you do not understand the following calculation, you must revise transformer ratios. We are assuming that the pulse transformer is ideal.] Vak 1 = Vak ⋅

V1 V2 Figure 12

3 1 V = 4.5  = 1.5 ⋅

Ig 1 = I g ⋅

V2 V1

= 200 ⋅10 −6 ⋅

1 3

= 66.7 µA 

Rg =

Vg1 − Vak I g1

12 − 4.5 66.7 ⋅ 10−6 = 112 kΩ  =

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3. This is the same SCR as Q 2 above. The SCR gate is to be triggered using an opto-isolator as shown in Figure 13. You are given the led operates with a forward current of I led = 20 mA and has a forward voltage of V led = 1.2 V. The transistor saturation voltage is v ON = 0.4 V. a) Calculate the led resistance, R led . Assume V gg = 12 V. Rled =

Vgg −Vled Iled

12 − 1.2 20 ⋅10 −3 540 Ω = 

Figure 13

=

b) Calculate the gate resistance, R g . You are given V S = 120 V. Rg =

VS − vON Ig

120 − 0.4 −6 200 ⋅ 10 598 kΩ =  =

[Note the following: (i) 120 ≫ 0.4, so v ON may be omitted.] c) Calculate the resistance, R gk . Assume R gk ≫ R g . Let R gk = 10 ⋅ Rg Rgk = 5980 k Ω MΩ = 5.98 

[We make R gk ≫ R g . Let R gk be 10× larger than R g so that it does not take current away from the gate.]

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12. SOME QUESTIONS 1. Why are SCR’s mainly used in circuits with AC supply? 2. What is the difference between a Thyristor and an SCR. Name and draw the circuit symbols of at least three Thyristors that are not SCR’s? Explain how they are different. 3. What is meant by the term “two transistor analogy of an SCR”? Use diagrams to explain. 4. It is said that an SCR “Latches ON”. What is meant by this term? Explain using diagrams. 5. Draw the SCR characteristic curve. Use the curve to explain: a) What happens when there is no gate current pulse, and b) When a gate pulse is applied. 6. What two conditions must be met for an SCR to be successfully triggered ON? [Answer: Must be forward biased, AND gate pulse must be applied.] 7. How is an SCR represented using a piecewise model? Use the SCR characteristic in your answer. 8. What is the piecewise representation of an SCR model used for? 9. Why is it important to be able to isolate the SCR power circuit from the gate control circuit? 10. Give examples of circuit components that may be used to isolate an SCR from the gate control circuit. 11. Draw a circuit diagram to show how a computer control circuit may be isolated from an SCR controlled power circuit. 12. A C106M sensitive gate SCR is shown in Figure 14. This SCR is being used in a halfcontrolled bridge rectifier to control the voltage to a resistive load of 99 Ω. The supply voltage is 220 V, 50 Hz. The SCR’s are triggered at an angle of 60 °. a) Sketch the circuit diagram. b) Sketch the load voltage and current waveforms.

Figure 14

c) Calculate the average load voltage and current. [148.5 V; 1.5 A] d) Calculate the rms load voltage and current. [197 V; 2 A] e) Use Figure 15 to find the maximum SCR ON-state power dissipation. [≈ 1 W] f) Calculate the Total Power to the load. [393 W] g) Calculate the DC Power to the load. [223 W] h) Calculate the Ripple Power to the load. [171 W] i) Use your answer to the Ripple Power to calculate the rms value of the ripple voltage. [129 V]

Figure 15 ON-State Power Dissipation

j) What are the current and voltage ratings of the SCR? i) I m = 3.14 A ii) DC current of I SCR = 0.75 A iii) Rms current of I SCR = 1.41 A PETR301

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iv) PIV = 311.1 V k) The gate trigger current is 100 A and the gate voltage typically needs to be 0.6 V. The gate drive voltage from the microprocessor is 5 V. i) Draw the gate drive circuit diagram where the gate drive is directly applied to the SCR gate. Calculate the value of the gate drive resistor. [44 k] ii) Draw the gate drive circuit diagram where a pulse transformer with a ratio of 2:1 is used to apply current to the SCR gate. Calculate the value of the gate drive resistor. [76 k] iii) Draw the gate drive circuit diagram where an opto-isolator is used to apply current to the SCR gate. The led in the opto-isolator needs a current of 25 mA and a voltage of at least 1.35 to switch ON. Calculate the values of the led resistor and the gate drive resistor. [183 ; 1.48 M, (Use the average value of the load voltage here.)]

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