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Power Electronics

Dr. Imtiaz HussainAssistant Professor

email: imtiaz.hussain@faculty.muet.edu.pkURL :http://imtiazhussainkalwar.weebly.com/

Lecture-6Thyristor Gate Control Circuits

Outline

• Introduction• Voltage Divider Triggering• RC Triggering• Double RC Triggering

Introduction• The popular terms used to describe how SCR is operating are

conduction angle and firing delay angle.

– Conduction angle is the number of degrees of an ac cycle during which the SCR is turned ON.

– Firing delay angle is the number of degrees of an ac cycle that elapses before the SCR is turned ON.

• Of course, these terms are based on the notion of total cycle time (3600)

Introduction• An SCR is fired by a short burst of current into the gate (IG).

• The amount of gate current needed to a fire particular SCR is symbolized as IGT.

• Most SCRs require current between 0.1 and 50mA.

• Since there is a standard pn-junction between gate and cathode, voltage between these two terminals (VGK) must be slightly greater than 0.7 volt.

Example-1• For the circuit shown in figure below, what voltage is required

at point X to fire the SCR? The gate current needed to fire 2N3669 is 20mA under normal conditions.

Solution

• The voltage between point X and cathode must be sufficient to forward bias the junction between X and K (0.7V).

• And also at least cause 20mA to flow from 150Ω resistor.

• For 20mA current to flow in XG branch we need

• Therefore,

𝑉 𝑋𝐺=𝐼𝐺𝑇×150 𝑉 𝑋𝐺=20𝑚×150=3𝑉

𝑉 𝐺=𝑉 𝑋𝐺+0.7=3.7V

6

Gate Control Circuits• Gate Control Circuit Design• Consideration must be given to the following points when designing gate

control circuits.

• The gate signal should be removed after the thyristor has been turned on. A continuous gate signal will increase the power loss in the gate junction.

• No gate signal should be applied when the thyristor is reversed biased. If a gate signal is applied under these conditions, the thyristor may fail due to an increased leakage current.

• The width of the gate pulse must be greater than the time required for the anode current to rise to the holding current. In practice, the gate pulse width is made wider than the turn-on time of the thyristor.

• A simple type of gate control circuit (triggering circuit) is shown in following figure.

Gate Control Circuits

• When SW is closed, there will be current into the gate when supply voltage goes positive.

• Firing delay angle is determined by setting of R2.

• One disadvantage of this simple triggering circuit is that the firing delay angle is adjustable is only from about 00 to 900.

Gate Control Circuits

• This can be understood by referring to following figure.

Example-2• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 20o. To what value should R2 be adjusted?Solution

• At 20o instantaneous supply voltage is

34

𝑉 𝑖𝑛𝑠=55.62𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-2• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=55.62−0.6−0.7

15𝑚

𝑅𝑇=54.3215𝑚

=3.6𝐾 Ω

𝑅2=3.6𝐾 −3𝐾=600Ω

Example-3• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 30o. To what value should R2 be adjusted?Solution

• At 30o instantaneous supply voltage is

𝑉 𝑖𝑛𝑠=𝑉 𝑝sin 30 °=¿√2𝑉 𝑟𝑚𝑠×0.5¿𝑉 𝑖𝑛𝑠=81.3𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-3• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=81.3−0.6−0.7

15𝑚

𝑅𝑇=8015𝑚

=5.3𝐾Ω

𝑅2=5.3𝐾−3𝐾=2.3𝐾 Ω

Example-4• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 60o. To what value should R2 be adjusted?Solution

• At 30o instantaneous supply voltage is

𝑉 𝑖𝑛𝑠=𝑉 𝑝sin 60 °=¿√2𝑉 𝑟𝑚𝑠×0.866 ¿𝑉 𝑖𝑛𝑠=140.84𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-4• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=140.84−0.6−0.7

15𝑚

𝑅𝑇=139.5415𝑚

=9.3𝐾 Ω

𝑅2=9.3𝐾 −3𝐾=6.3𝐾 Ω

Example-5• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 90o. To what value should R2 be adjusted?Solution

• At 90o instantaneous supply voltage is

𝑉 𝑖𝑛𝑠=𝑉 𝑝=√2𝑉 𝑟𝑚𝑠

𝑉 𝑖𝑛𝑠=√2×115=162𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-5• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=162−0.6−0.7

15𝑚

𝑅𝑇=160.715𝑚

=10.7𝐾 Ω

𝑅2=10.7𝐾−3𝐾=7.7𝐾 Ω

Example-6• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 150o. To what value should R2 be adjusted?Solution

3KΩ

40Ω

• At 150o instantaneous supply voltage is

𝑉 𝑖𝑛𝑠=𝑉 𝑝sin 150 °=¿√2𝑉 𝑟𝑚𝑠×0.5¿𝑉 𝑖𝑛𝑠=81.3𝑉

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-6• Total resistance in the gate

lead is given by

3KΩ

5Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=81.3−0.6−0.7

15𝑚

𝑅𝑇=8015𝑚

=5.3𝐾Ω

𝑅2=5.3𝐾−3𝐾=2.3𝐾 Ω

• R2 is same as it was for firing angle of 30o. Therefore with this circuit arrangement it is not possible to fire SCR beyond 90o.

Example-7• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 10o. To what value should R2 be adjusted?Solution

• At 10o instantaneous supply voltage is

3

𝑉 𝑖𝑛𝑠=28.24𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-7• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=28.24−0.6−0.7

15𝑚

𝑅𝑇=26.915𝑚

=1.79𝐾 Ω

𝑅2=1.79𝐾−3𝐾=−1.21𝐾 Ω

• Cannot have firing angle of 10o. For extended firing angle R3 can be made smaller.

Example-8• For following figure assume that the supply is 115V rms,

IGT=15mA, and R1=3KΩ. The firing delay is desired to be 18o. To what value should R2 be adjusted?Solution

• At 15o instantaneous supply voltage is

𝑉 𝑖𝑛𝑠=𝑉 𝑝sin 15 °=¿√2𝑉 𝑟𝑚𝑠×0.3¿𝑉 𝑖𝑛𝑠=48.79𝑉

3KΩ

40Ω

• Voltage drop across Load

𝑉 𝐿=𝐼𝐺𝑇×𝑅𝐿𝑜𝑎𝑑

𝑉 𝐿=15𝑚×40=0.6𝑉

Example-8• Total resistance in the gate

lead is given by

3KΩ

40Ω

• Therefore, R2 is

𝑅2=𝑅𝑇−𝑅1

𝑅𝑇=𝑉 𝑖𝑛𝑠−𝑉 𝐿𝑜𝑎𝑑−𝑉 𝐺𝐾

𝐼𝐺𝑇=48.79−0.6−0.7

15𝑚

𝑅𝑇=47.4915𝑚

=3.16 𝐾Ω

𝑅2=3.16𝐾 −3𝐾=160Ω

Conclusion

• The value of resistor R2 is increasing as firing angle is further delayed.

S. No Firing Angle R2

1 10o -1.21KΩ

2 18 160Ω

3 20o 600Ω

4 30o 2.3KΩ

5 60o 9.3KΩ

6 90o 7.7KΩ

7 150o 2.37.7KΩ

Range of Firing

Angles

RC Triggering Circuits

• The simplest method of improving gate control is to add a capacitor at the bottom of the gate lead resistance as shown in following figure.

• Advantage of this circuit is that the firing delay angle can be adjusted past 90o.

RC Triggering Circuits

• This can be understood by focusing on the voltage across Capacitor C.

• When the ac supply is –ve, the reverse voltage across SCR is applied to RC triggering circuit, charging the capacitor –ve on top plate and +ve on bottom plate.

• When the supply enters its positive half cycle, the forward voltage drop across SCR tends to charge C in opposite direction.

• However, voltage buildup in new direction is delayed until the –ve charge is removed.

RC Triggering Circuits• The idea can be extended to achieve even extended firing

angles by modifying the circuit slightly.

• A resistor has been inserted into the gate lead, requiring the capacitor to charge higher than 0.7 V to trigger the SCR.

• With the resistor in place, capacitor voltage must reach a value large enough to force sufficient current (IGT) through the resistor.

RC Triggering Circuits• The firing delay angle can further be extended by the use of

double RC network as shown in following figure.

• The delayed voltage across C1 is used to charge C2 resulting in even further delay in building up the gate voltage.

Triggering • 50Hz sine wave takes 1/50 seconds to complete one

cycle.

𝑇=150

=20𝑚𝑠

𝟐𝟎𝒎𝒔

𝟏𝟎𝒎𝒔

𝟏𝟎𝒎𝒔

RC Triggering Circuits• Capacitors in RC triggering circuits usually fall in the range

from 0.01µF to 1µF.

• For the given capacitor sizes minimum firing delay angle (maximum load current) is set by fixed resistors R1 and R3.

• The maximum firing angle (minimum load current) is set mostly by variable resistor R2.

• When these gate control circuits are used with 50Hz AC supply, the time constant of the RC circuit should fall in the range of 1-20ms.

RC Triggering Circuits• For single RC circuit of fig (a) the product (R1+R2)C1 should fall in the

range 1ms to 20ms.

• For double RC circuit of fig(b) (R1+R2)C1 should fall in that range and R3C2 should also fall in that range.

𝑇=(𝑅1+𝑅2)𝐶

𝑇 1=(𝑅1+𝑅2 )𝐶1

𝑇 2=𝑅3𝐶2

Fig(a)

Fig(b)

Example-9

• For the circuit shown in following figure approximate the R1, R2 and R3 to give wide range of firing adjustment.

𝑪𝟏=𝟎 .𝟎𝟔𝟖𝝁𝑭

𝑪𝟐=𝟎 .𝟎𝟑𝟑𝝁𝑭

Example-9

• The total time constant must fall in the range of 1ms-20ms.

• Let us set and . RC ntwork-1 must provide 1ms-18ms of firing delay and RC netwrork-2 2ms of delay.

𝐶1=0.068𝜇𝐹

𝐶2=0.033𝜇𝐹

• Minimum time constant occurs in RC network-1 when R2 is set to minimum.

𝑇 1=(𝑅1+𝑅2 )𝐶1

1𝑚= (𝑅1+0 )0.068𝜇

𝑅1=1𝑚

0.068𝜇  

𝑅1=1𝑚

0.068𝜇  14.7𝐾 Ω

𝑇=𝑇 1+𝑇 2

Example-9

𝐶1=0.068𝜇𝐹

𝐶2=0.033𝜇𝐹

• Maximum time constant occurs in RC netwrok-1 when R2 is set to maximum.

𝑇 1=(𝑅1+𝑅2 )𝐶1

18𝑚=(14.7 𝐾+𝑅2 )0.068𝜇

𝑅2=18𝑚0.068𝜇  

−14.7𝐾

𝑅2=250𝐾 Ω

Example-9

𝐶1=0.068𝜇𝐹

𝐶2=0.033𝜇𝐹

• Time constant of RC netwrok-2 is 2ms.

𝑇 2=𝑅3𝐶2

2𝑚=𝑅30.033𝜇

𝑅3=2𝑚

0.033𝜇  

𝑅3=60.6𝐾 Ω

Example-9

𝐶1=0.068𝜇𝐹

𝐶2=0.033𝜇𝐹

• Minimum and maximum firing angles are (18o=1ms)

𝑇𝑚𝑖𝑛=1𝑚𝑠+2𝑚𝑠=3𝑚𝑠

𝑇𝑚𝑎𝑥=18𝑚𝑠+2𝑚𝑠=20𝑚𝑠

𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑓𝑖𝑟𝑖𝑛𝑔𝑎𝑛𝑔𝑙𝑒=3×18=54𝑜

𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑖𝑟𝑖𝑛𝑔𝑎𝑛𝑔𝑙𝑒=20×18=360𝑜

𝟑𝟎𝟔°

Example-10

• For the circuit shown in following figure, to what value the potentiometer be set to obtain a firing delay angle of 120o.

𝑪𝟏=𝟎 .𝟎𝟔𝟖𝝁𝑭

𝑪𝟐=𝟎 .𝟎𝟑𝟑𝝁𝑭

𝑹𝟏=𝟏𝟒 .𝟕𝑲𝛀

𝑹𝟑=𝟔𝟎 .𝟔𝑲𝛀

Use of Break Over Devices

• The firing circuits discussed so far share two disadvantages.1. Temperature dependence2. Inconsistent firing behaviour between SCRs of

same type• These problems can be eliminated by

introducing a break over device at gate terminal

Use of Break Over Devices• Four layer diode (Shockley Diode) has certain

break over voltage.

Shockley Diode

END OF LECTURE-6

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