Working Principle of the Circuit :
The impulse capacitor is cgarged via a high charging resistance (R0) to the
direct voltage U0, and then discharged by ignition of the switching gap F. During this
charging and discharging period, the desired impulse voltage U(t) appears across the
load capacitor Cy.
If Rd >> Ra Cd is charged rapidly (short time to front), and discharged slowly
(long time to tail).
U
U m ax
1,0
0,9
fro
nt
t a il
0,5
0,3
O O′ T /
t
T
TC
TS
Waveform of the impulse voltage is determined by circuit components.
T = time between (%30 - %90)
Tc = rise time
(Tc = 1,67 . T )
Ts = decay time (tail)
( T’ = 0,3 . Tc = 0,5 . T )
( Tc / Ts = 1,2 / 50 μ s Standard Impulse Voltage )
The smaller the Ra.Cy, the faster the time to Um.
As the Rd get higher, decay time will be longer.
According to charging polarity of Cd Impulse voltage can be either positive or
negative
To increase the efficiency, C d ⟩⟩ C y .
Impulse Energy:
The impulse energy transformed during a discharge:
W=
1
. Cd . U02 [ Joule=Wh ]
2
‘ dan hesaplanır.
Multiple Stage Impulse Generators:
The multiplier circuit can be obtained by cascading single stage impulse
generators. Workig principle is based on the charging of several idendical impulse
capacitor in parallel and then discharged in series.
Cd '
Cd '
Cd '
Cd '
Cd '
Cd '
U0'
3U 0 '
Charged in parallel by U0
Discharged in series
To prevent the discharging of charged
capacitors through Ra. Since its value is
high, current flow though Ra and F spheres.
R'
o
U0'
R'
C '+
d
F
C'
d
−
Rd' R a'
Discharging of Cy (Decay
Time)
R'
+
F
F
C' +
d
−
Rd'
R a'
−
Rd'
R a'
Cy
U(t)
R'
3 Stage b – type impulse generator
When all the
switching gaps
are broken
down(Cds are in
series)
The impulse capacitors of stages Cd’ are charged to the stage charging
voltage U0’ via the high charging resistors R’. When all the switching gaps F break
down, Cd’ will be connected in series, so that Cy is charged via the series
connection of all the damping resistors Ra’. Finally, all Cd’ and Cy will
discharge again via the resistors.
The n-stage circuit can be reduced to a single stage equivalent circuit as below:
Ra = n . Ra’
Rd = n . Rd’
Cd =
Cd '
n
U0 = n . U0’
Output voltage of Impulse generator can be given in terms of circuit
elements as below:
t
t
−
⎞
T1 . T2 ⎛⎜ − T1
U0
T2 ⎟
u (t ) =
.
e −e
⎟
R a . Cy
T1 − T2 ⎜
⎝
⎠
u (t )
+ U0
U0 . e
Um
−
t
T1
μ (t )
Tcr
t
T2
−U0 . e
−U0
−
t
T2
T1
For circuit a
T1 ≅ (Ra + Rd ). (C d + C y )
⎛ R .R
T2 ≅ ⎜⎜ a d
⎝ Ra + Rd
⎞ ⎛ C y .C d
⎟⎟ . ⎜
⎜
⎠ ⎝ Cd + C y
⎞
⎟
⎟
⎠
η=
Rd
Cd
.
R a + Rd C d + C y
For circuit b
T1 ≅ Rd . (C d + C y )
⎛ Cd . C y
T2 ≅ Ra . ⎜
⎜C +C
y
⎝ d
Cd
η=
Cd + C y
⎞
⎟
⎟
⎠
( T1 >> T2)
To calculate the output voltage, Laplace transform can be used.:
Ra
1
+ s . Cd
Rd
U0
s
Ι2
Ι1
1
s . Cy
U (s )
Ra
1
+ s . Cd
U0
s
1
s . Cy
Rd
Ι1
Ι2
U (s)
Problem:
The component values of each single stage in a 5-stage a-type impulse generator are given as
below:
Cd’=100 nF
Ra’=80 Ω
Rd’=1 kΩ
Cy=1 nF
R’=1 MΩ ’dur.
This impulse generator is supplied by a 100 kV DC voltage source. Calculate the following:
a) Output Voltage [U(t)=?]
b) Efficiency
c) Tc (rise time) ve Ts (decay time)?
Solution:
a) The 5 stage circuit can be reduced to a single stage equivalent circuit:
Ra = 5 . Ra’ = 5.80 = 400 Ω
Rd = 5.1 = 5 kΩ
Cd =
Cd ' 100
=
= 20nF
5
5
Cy = 1nF
U0 = 5.100 = 500 kV
By the ignition of the switching gap, following circuit in s-domain can be drawn:
By writing the 1st and 2nd loop equations;
( Ra + Rd +
U
1
) ⋅ I 1 − Rd ⋅ I 2 = 0
s ⋅ Cd
s
− R d ⋅ I 1 + ( Rd +
U (s) = I 2 ⋅
1
) ⋅ I2 = 0
s ⋅Cy
1
s ⋅Cy
1st equation
2nd equation
3rd equation
Take the I1 from equation 2 and insert into (1):
I 1 = (1 +
1
s ⋅ Rd ⋅ C y
( Ra + Rd +
) ⋅ I2
1
1
U
) ⋅ (1 +
) ⋅ I 2 − Rd ⋅ I 2 = 0
s ⋅ Cd
s ⋅ Rd ⋅ C y
s
⎡
⎢
U
1
I2 = 0 ⎢
Ra
1
1
1
s ⎢
⎢ ( Ra + s ⋅ C + s ⋅ R ⋅ C + s ⋅ C + s 2 ⋅ R ⋅ C ⋅ C
d
d
y
y
d
d
y
⎣
⎤
⎥
⎥
⎥
⎥
⎦
By rearranging the equation I2:
⎡
⎢
U0 ⎢
1
I2 =
1
1
1
1
s ⋅ Ra ⎢ 1
⎢1 + s ⋅ ( R ⋅ C + R ⋅ C + R ⋅ C ) + s 2 ⋅ R ⋅ C ⋅ C ⋅ R
a
d
d
y
a
y
d
d
y
a
⎣
⎤
⎥
⎥
⎥
⎥
⎦
place the equation I2 into (3);
⎡
U 0 ⎢⎢
1
U ( s) =
1
1
1
1
Ra ⋅ C y ⎢ 2
⎢s + s ⋅(R ⋅C + R ⋅C + R ⋅C ) + R ⋅C ⋅C ⋅ R
a
d
d
y
a
y
d
d
y
a
⎣
⎤
⎥
⎥
⎥
⎥
⎦
Assuming A =
U0
1
1
1
1
; b=
; c=
;
+
+
Ra ⋅ C y
Ra ⋅ C d Rd ⋅ C y R a ⋅ C y
Rd ⋅ C d ⋅ C y ⋅ Ra
U(s) is reduced to:
⎡ 1
1
A
A
1 ⎤
=
⋅⎢
−
U (s) = A ⋅ 2
=
⎥
s + s ⋅ b + c ( s − s1 ) ⋅ ( s − s 2 ) s1 − s 2 ⎣ s − s1 s − s 2 ⎦
s1, 2 =
− b m b2 − 4 ⋅ a ⋅ c
2⋅a
u (t ) = K ⋅ ⎡⎢e
⎣
s1⋅t
−e
K=
s2⋅t ⎤
A
s1 − s 2
⎥
⎦
By writing the numerical values:
b = 2,825 ⋅ 10 6
c = 0,025 ⋅ 1012
K = 445,3 ⋅ 10 3
s1 = −8,877 ⋅10 3
s 2 = −2,8161225 ⋅ 10 6
3
6 ⎤
⎡
u (t ) = 445,3 ⋅103 ⋅ ⎢e − 8,877.10 ⋅ t − e − 2,8161225.10 ⋅ t ⎥
⎢
⎣
⎥
⎦
Um
.
To find the Um the derivative of dU(t)/dt should be taken, and
U0
should be equalized to zero (For this curve, we have only max. value. Hence the value we will
find is the peak value);
b) Efficiency is η =
⎛ s1⋅t
du (t )
= K ⋅ ⎜e m
⎜
dt
⎝
⇒ s1 ⋅ e
s1⋅tm
s1 s1⋅tm
⋅e
s2
=e
−e
= s2e
s2⋅tm
s2⋅tm ⎞
s2⋅tm
⎛
⎟=0
⎟
⎠
⎞
⇒ ln⎜⎜ s1 ⎟⎟ + s1 ⋅ t m = s2 ⋅ tm
⎝ s2 ⎠
⎛s ⎞
ln⎜⎜ 1 ⎟⎟
s
tm = ⎝ 2 ⎠ = 2,05μs ⇒ U m = u (t m ) = 436 ⋅ 103[V ]
s2 − s1
η=
436
= %87,2
500
c)
s ⋅t
u (t A ) = 0,9 ⋅ U m ⇒ 392,4 ⋅ 103 = 445,3 ⋅ 103 ⋅ (e 1 A
s ⋅t
− e 2 A ) ⇒ find t A
s ⋅t
s ⋅t
u (t B ) = 0,3 ⋅ U m ⇒ 130,8 ⋅ 103 = 445,3 ⋅ 103 ⋅ (e 1 B − e 2 B ) ⇒ find t B
s ⋅t
s ⋅t
u (t E ) = 0,5 ⋅ U m ⇒ 218 ⋅ 103 = 445,3 ⋅ 103 ⋅ (e 1 E − e 2 E ) ⇒ find t E
TC = (t A − t B ) ⋅ 1,67
TS ≅ T E
How we can find the tA?
By following the simple iteration technique, tA can be found easily. Let us take the initial
value of tA as 0,5 µs.
⎛ s ⋅t
0,88 = ⎜ e 1 A
⎝
tA =
−e
⎛ s ⋅t
1
⋅ ln⎜ e 1 A
s2 ⎝
s2 ⋅t A ⎞
s ⋅t
⎟⇒e 2 A
⎠
=e
s1⋅t A
− 0,88
− 0,88⎞⎟
⎠
t A (0) = 0,5μs
t A (1) = 0,76625 ⋅10 −6
t A (2) = 0,77355 ⋅10 −6
t A (3) = 0,773752 ⋅ 10 −6
t A = 0,773μs .
tB and tE can also be calculated by the same way.