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Power Electronics
Dr. Imtiaz HussainAssistant Professor
email: [email protected] :http://imtiazhussainkalwar.weebly.com/
Lecture-6Thyristor Gate Control Circuits
Outline
β’ Introductionβ’ Voltage Divider Triggeringβ’ RC Triggeringβ’ Double RC Triggering
Introductionβ’ The popular terms used to describe how SCR is operating are
conduction angle and firing delay angle.
β Conduction angle is the number of degrees of an ac cycle during which the SCR is turned ON.
β Firing delay angle is the number of degrees of an ac cycle that elapses before the SCR is turned ON.
β’ Of course, these terms are based on the notion of total cycle time (3600)
Introductionβ’ An SCR is fired by a short burst of current into the gate (IG).
β’ The amount of gate current needed to a fire particular SCR is symbolized as IGT.
β’ Most SCRs require current between 0.1 and 50mA.
β’ 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-1β’ 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
β’ The voltage between point X and cathode must be sufficient to forward bias the junction between X and K (0.7V).
β’ And also at least cause 20mA to flow from 150Ξ© resistor.
β’ For 20mA current to flow in XG branch we need
β’ Therefore,
π ππΊ=πΌπΊπΓ150 π ππΊ=20πΓ150=3π
π πΊ=π ππΊ+0.7=3.7V
6
Gate Control Circuitsβ’ Gate Control Circuit Designβ’ Consideration must be given to the following points when designing gate
control circuits.
β’ 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.
β’ 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.
β’ 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.
β’ A simple type of gate control circuit (triggering circuit) is shown in following figure.
Gate Control Circuits
β’ When SW is closed, there will be current into the gate when supply voltage goes positive.
β’ Firing delay angle is determined by setting of R2.
β’ 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
β’ This can be understood by referring to following figure.
Example-2β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 20o. To what value should R2 be adjusted?Solution
β’ At 20o instantaneous supply voltage is
34
π πππ =55.62π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-2β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=55.62β0.6β0.7
15π
π π=54.3215π
=3.6πΎ Ξ©
π 2=3.6πΎ β3πΎ=600Ξ©
Example-3β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 30o. To what value should R2 be adjusted?Solution
β’ At 30o instantaneous supply voltage is
π πππ =π πsin 30 Β°=ΒΏβ2π πππ Γ0.5ΒΏπ πππ =81.3π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-3β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=81.3β0.6β0.7
15π
π π=8015π
=5.3πΎΞ©
π 2=5.3πΎβ3πΎ=2.3πΎ Ξ©
Example-4β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 60o. To what value should R2 be adjusted?Solution
β’ At 30o instantaneous supply voltage is
π πππ =π πsin 60 Β°=ΒΏβ2π πππ Γ0.866 ΒΏπ πππ =140.84π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-4β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=140.84β0.6β0.7
15π
π π=139.5415π
=9.3πΎ Ξ©
π 2=9.3πΎ β3πΎ=6.3πΎ Ξ©
Example-5β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 90o. To what value should R2 be adjusted?Solution
β’ At 90o instantaneous supply voltage is
π πππ =π π=β2π πππ
π πππ =β2Γ115=162π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-5β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=162β0.6β0.7
15π
π π=160.715π
=10.7πΎ Ξ©
π 2=10.7πΎβ3πΎ=7.7πΎ Ξ©
Example-6β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 150o. To what value should R2 be adjusted?Solution
3KΞ©
40Ξ©
β’ At 150o instantaneous supply voltage is
π πππ =π πsin 150 Β°=ΒΏβ2π πππ Γ0.5ΒΏπ πππ =81.3π
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-6β’ Total resistance in the gate
lead is given by
3KΞ©
5Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=81.3β0.6β0.7
15π
π π=8015π
=5.3πΎΞ©
π 2=5.3πΎβ3πΎ=2.3πΎ Ξ©
β’ 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-7β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 10o. To what value should R2 be adjusted?Solution
β’ At 10o instantaneous supply voltage is
3
π πππ =28.24π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-7β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=28.24β0.6β0.7
15π
π π=26.915π
=1.79πΎ Ξ©
π 2=1.79πΎβ3πΎ=β1.21πΎ Ξ©
β’ Cannot have firing angle of 10o. For extended firing angle R3 can be made smaller.
Example-8β’ For following figure assume that the supply is 115V rms,
IGT=15mA, and R1=3KΞ©. The firing delay is desired to be 18o. To what value should R2 be adjusted?Solution
β’ At 15o instantaneous supply voltage is
π πππ =π πsin 15 Β°=ΒΏβ2π πππ Γ0.3ΒΏπ πππ =48.79π
3KΞ©
40Ξ©
β’ Voltage drop across Load
π πΏ=πΌπΊπΓπ πΏπππ
π πΏ=15πΓ40=0.6π
Example-8β’ Total resistance in the gate
lead is given by
3KΞ©
40Ξ©
β’ Therefore, R2 is
π 2=π πβπ 1
π π=π πππ βπ πΏπππβπ πΊπΎ
πΌπΊπ=48.79β0.6β0.7
15π
π π=47.4915π
=3.16 πΎΞ©
π 2=3.16πΎ β3πΎ=160Ξ©
Conclusion
β’ The value of resistor R2 is increasing as firing angle is further delayed.
S. No Firing Angle R2
1 10o -1.21KΞ©
2 18 160Ξ©
3 20o 600Ξ©
4 30o 2.3KΞ©
5 60o 9.3KΞ©
6 90o 7.7KΞ©
7 150o 2.37.7KΞ©
Range of Firing
Angles
RC Triggering Circuits
β’ 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.
β’ Advantage of this circuit is that the firing delay angle can be adjusted past 90o.
RC Triggering Circuits
β’ This can be understood by focusing on the voltage across Capacitor C.
β’ When the ac supply is βve, the reverse voltage across SCR is applied to RC triggering circuit, charging the capacitor βve on top plate and +ve on bottom plate.
β’ When the supply enters its positive half cycle, the forward voltage drop across SCR tends to charge C in opposite direction.
β’ However, voltage buildup in new direction is delayed until the βve charge is removed.
RC Triggering Circuitsβ’ The idea can be extended to achieve even extended firing
angles by modifying the circuit slightly.
β’ A resistor has been inserted into the gate lead, requiring the capacitor to charge higher than 0.7 V to trigger the SCR.
β’ With the resistor in place, capacitor voltage must reach a value large enough to force sufficient current (IGT) through the resistor.
RC Triggering Circuitsβ’ The firing delay angle can further be extended by the use of
double RC network as shown in following figure.
β’ The delayed voltage across C1 is used to charge C2 resulting in even further delay in building up the gate voltage.
Triggering β’ 50Hz sine wave takes 1/50 seconds to complete one
cycle.
π=150
=20ππ
ππππ
ππππ
ππππ
RC Triggering Circuitsβ’ Capacitors in RC triggering circuits usually fall in the range
from 0.01Β΅F to 1Β΅F.
β’ For the given capacitor sizes minimum firing delay angle (maximum load current) is set by fixed resistors R1 and R3.
β’ The maximum firing angle (minimum load current) is set mostly by variable resistor R2.
β’ 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 Circuitsβ’ For single RC circuit of fig (a) the product (R1+R2)C1 should fall in the
range 1ms to 20ms.
β’ For double RC circuit of fig(b) (R1+R2)C1 should fall in that range and R3C2 should also fall in that range.
π=(π 1+π 2)πΆ
π 1=(π 1+π 2 )πΆ1
π 2=π 3πΆ2
Fig(a)
Fig(b)
Example-9
β’ For the circuit shown in following figure approximate the R1, R2 and R3 to give wide range of firing adjustment.
πͺπ=π .πππππ
πͺπ=π .πππππ
Example-9
β’ The total time constant must fall in the range of 1ms-20ms.
β’ Let us set and . RC ntwork-1 must provide 1ms-18ms of firing delay and RC netwrork-2 2ms of delay.
πΆ1=0.068ππΉ
πΆ2=0.033ππΉ
β’ Minimum time constant occurs in RC network-1 when R2 is set to minimum.
π 1=(π 1+π 2 )πΆ1
1π= (π 1+0 )0.068π
π 1=1π
0.068π
π 1=1π
0.068π 14.7πΎ Ξ©
π=π 1+π 2
Example-9
πΆ1=0.068ππΉ
πΆ2=0.033ππΉ
β’ Maximum time constant occurs in RC netwrok-1 when R2 is set to maximum.
π 1=(π 1+π 2 )πΆ1
18π=(14.7 πΎ+π 2 )0.068π
π 2=18π0.068π
β14.7πΎ
π 2=250πΎ Ξ©
Example-9
πΆ1=0.068ππΉ
πΆ2=0.033ππΉ
β’ Time constant of RC netwrok-2 is 2ms.
π 2=π 3πΆ2
2π=π 30.033π
π 3=2π
0.033π
π 3=60.6πΎ Ξ©
Example-9
πΆ1=0.068ππΉ
πΆ2=0.033ππΉ
β’ Minimum and maximum firing angles are (18o=1ms)
ππππ=1ππ +2ππ =3ππ
ππππ₯=18ππ +2ππ =20ππ
ππππππ’π πππππππππππ=3Γ18=54π
πππ₯πππ’π πππππππππππ=20Γ18=360π
πππΒ°
Example-10
β’ For the circuit shown in following figure, to what value the potentiometer be set to obtain a firing delay angle of 120o.
πͺπ=π .πππππ
πͺπ=π .πππππ
πΉπ=ππ .ππ²π
πΉπ=ππ .ππ²π
Use of Break Over Devices
β’ The firing circuits discussed so far share two disadvantages.1. Temperature dependence2. Inconsistent firing behaviour between SCRs of
same typeβ’ These problems can be eliminated by
introducing a break over device at gate terminal
Use of Break Over Devicesβ’ Four layer diode (Shockley Diode) has certain
break over voltage.
Shockley Diode
END OF LECTURE-6
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