Firing Circuit

Firing Circuit of PE Converter

1. Power Electronics
TRIGGERING CIRCUITS
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM
1
2018 Dr.Francis M. Fernandez

2. Gate Characteristics
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 2
OA and OB represent the
spread of characteristics for
the thyristor of same rating
V
g
I
g
V
gmax
V
gmin
I
gmin I
gmax
O
B
A
E
S
1
S
2

3. R -Triggering Circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 3
1
max
R
V
I
m
g
max
1
g
m
I
V
R
R
1is the gate current limiting
resistance
R
2is used to vary the gate current
and hence firing angle
max
1max
gm
g
VV
RV
R


R limits the voltage at Gate terminal
Diode D prevents build-up of negative
voltage at Gate terminal

4. R-Trig Waveforms
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 4
V
i
V
L
V
T
t
α
The phase angle at which
the SCR starts
conducting is called
firing angle, α

5. Features of R-Trig Circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 5
V
i
V
L
V
T
t
α
Simple circuit
Disadvantages:
Performance depends on
temperature and SCR
characteristics
Minimum phase angle is typically
2-4 degrees only (not zero degree)
Maximum phase angle is only 90
degrees

6. Problem
Design an R-triggering circuit for a half wave controlled rectifier circuit
for 24 V ac supply. The SCR to be used has the following data.
I
gmin= 0.1 mA, I
gmax= 12 mA, V
gmin= 0.6V, V
gmax= 1.5 V
6DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM
P
o=242 1 3
max
24 2
2.8 3.3
12 10
m
g
V
R k k
I

    


≤
P
lctn
i
P
o−P
lctn
=
1.5×3.3×10
U
242−1.5
=145.8 h≃120  h
P
lcgu
P
o−P
lcgu
=


i+
v

v=
P
o−P
lcgu
P
lcgu
−
i=
242−0.6×120
0.6
−3.3×10
U
=3.37  
For finding R2
Select 4.7 k Potentiometer for R
2
(Drop across diode D is neglected)
Solution:

7. Problem
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 7
In a resistance firing circuit for SCR, the following parameters are
applicable
I
gt(min) = 0.5 mA,
V
gt(min) = 0.7V,
Supply voltage = 48 V, 50Hz
Resistance in gate current path = 70 kΩ.
a)Find the firing angle for the above
condition
b)What is the resistance for which
firing angle is 90 degrees

8. Solution
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 8
 
V3.36
7.06.01070105.0
33
21




gDgt
VVRRIe
1
sin
36.3 2 48sin
36.3
sin 0.535
2 48
sin 32.3
t m
e V t
t
t
t t



 


 
 

  
Firing angle = 32.3 degrees
a)

9. Solution
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 9
 
kΩ133
105.0
6.66
105.06.66
7.06.0105.0482
3
3
3
21









R
R
R
VVRRIe
gDgt
b)

10. RC Triggering Circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 10
Capacitor charges during the negative
half cycle through D
2
When SCR is turned on, capacitor C is
suddenly discharged through D
2
D
1 protects the SCR during negative
half cycle
2
3.1T
RC
1
min mins g g D
v R I V V  
To ensure minimum gate current
1
min
min
s g D
g
v V V
R
I
 

Advantage over R-triggering Circuit:
Controls upto180 degrees

11. RC Trig Waveforms
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 11
2
3.1T
RC

12. RC Full wave trigger circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 12
2
50T
RC
Initial Capacitor voltage in each
half cycle is almost zero
min
min
s g
g
v V
R
I



13. Unijunction Transistor (UJT)
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 13
Has a lightly doped n-type silicon
layer to which a heavily doped
p-type emitter is embedded
The inter-base resistance is in the
range of 5 –10 kΩ
This device cannot ‘amplify’

14. UJT Equivalent Circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 14
BBBB
BB
B
AB
VV
RR
R
V 


21
1
1
ηis called intrinsic standoff ratio
Value of ηvaries from 0.5 –0.8
When V
eis more than V
1+V
D, then the
diode is forward biased and a current
flows through R
B1
Number of carriers in R
B1 increases
and the resistance reduces
V
edecreases with increase in I
eand the
therefore the device is said to exhibit
negative resistance

15. UJT Characteristics
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 15
At peak point, V
e= V
1+V
D,
At Valley point, R
B1is minimum

16. UJT parameters
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 16
Maximum emitter reverse voltage
◦Maximum reverse bias which the emitter –base2 junction can tolerate without
breakdown. Typ: 30V
Maximum inter-base voltage
◦Maximum voltage possible between base1 and base2. Decided by the power
dissipation. Typ: 35 V
Interbaseresistance
◦Typ: 4.7 k –9.1 k
Intrinsic stand off ratio
◦Typ: 0.56 –0.75
Maximum peak emitter current
◦Typ: 2A
Emitter leakage current
◦The emitter current when V
eis less than V
pand the UJT is off.
◦Typ12 μA
Typical values are of 2N2646

17. UJT Oscillator
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 17
•R
1and R
2are much less than the inter-base resistance
•The output pulses can be used to trigger an SCR

18. Design
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 18








RC
t
BBC
eVV 1
Time required for C to charge from
V
vto V
pis obtained as follows








RC
t
BBvDBBp
eVVVVV 1
Assuming
D v
V V 







RC
t
e1










1
1
ln
1
RC
f
T
Tt,case For this
R
1is selected based on
voltage level required to
trigger the SCR
R
2is selected using the
empirical formula:
BB
V
R

4
2
10


19. UJT firing circuit for Half Wave Controller
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 19

20. Waveforms for Half Wave Controller
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 20
V
i
V
P
V
O
t
V
C
V
V

21. Full wave UJT trigger Circuit
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 21

22. Problem
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 22
Design a UJT relaxation oscillator using UJT2646 for triggering an SCR. The
UJT has the following parameters
η= 0.63, V
BB= 20 V, V
P= 13.2 V, I
P= 50 μA
V
V= 2 V, I
V= 6 mA, R
BB= 7 kΩ, leakage current = 2.5 mA
Also find the minimum and maximum time period of oscillation.
Solution:
Assume C = 0.1 μF

ctn=
P
−
(
J
=
20−13.2
50×10
)q
=136  

cgu=
P
−P
p
(
p
=
20−2
6×10
)U
=3  

23. DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 23
Approximate value of
2=
10
(
z D

=
10
(
0.63×20
=794 Ω

I=
D
FLse
9wVDV.w  TbFFw7k
=
0.7
2.5×10
Rr
=280 Ω
I
LCR= ln
1
1−z
=136×10
r
×0.1×10
RM
×ln
1
1−0.63
=13.5  
I
Lse=3×10
r
×0.1×10
RM
×ln
1
1−0.63
=0.3  

24. Commutation
Commutation us the process by which a thyristor is turned off or
current diverted to another path.
There are two types of commutation
Natural Commutation
Forced Commutation
24DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM
•In natural commutation, the
reversing nature of alternating
voltages turn off the thyristor
•Suitable for AC circuits only
•Current passes through a zero in
every half cycle
•No external circuit is required for
natural commutation
•In DC circuits, external circuits are
necessary for turn off of thyristors
•Turn off with external circuits is called
forced commutation

25. Pulse Transformer
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 25
Used to trigger SCR, TRIAC etc
Provides Electrical isolation between power circuit and control circuit

26. Optical Isolation
DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 26
Used to trigger SCR, TRIAC etc
Provides Electrical isolation between power circuit and control circuit