Comparative Study between Series Resonant

Converter and Parallel Resonant Converter

S.M.Ferdous
Design Specification:
Input voltage: 400 V

Output voltage (for all the range);

200V
Maximum frequency: 100 KHz

Output power: from 1kW to

5kW
So,

V n=

200
=0.5
400

Series Resonance Circuit:

A Series resonant Circuit is given below-

Fig. 1: Series Resonance Circuit

The input and output voltage relation for Series Resonance Circuit can be written as
follows-

V out
=
V

For a set of values of

1
1

j
n

jn

and normalized frequency

n the voltage and power

plot for series resonance circuit is shown in following figure.

Page 1 of 23

Fig. 2. Normalized Frequency ( n ) vs. voltage curve for different values of

in

SRC circuit

Fig. 3. Normalized Frequency ( n ) vs. Power curve for different values of

in

SRC circuit
As ratio between maximum power and minimum power is 5, so from figure 2,

=0.6

and

=3 .0

has been selected.

Page 2 of 23

V n=0.5

line is also plotted in figure 2. So from figure 2, it can be seen that at the

=0.6

intersecting point of

n=1.646 .

frequency,

n=1.646

V n=0.5

curve and

the corresponding normalized

From figure 3, it is found that for

=0.6

and

corresponding Pn=0. 4171 .

Similarly, for

=3 .0

corresponding

Pn

and

are 5.382 and 0.0833

respectively.
The obtained values are listed in the following table.

Pn

0.6
3.0

1.646
5.382

0.4171
0.0833

Table 1. Showing the value of

n and Pn

for chosen value of

Now, as maximum frequency corresponds to maximum normalized frequency, so it

can be written,

n=

>1.646=

> B =

2 (1 00 kHz)
B

1
=381.725 10 3
LC s

LC s=2.62 106
Again

if

minimum

power

(=1kW)

is

considered

Pn ( 0.0833) has to be taken to calculate the value of

P n=

But,

then

L
Cs

P
PB

V B2
PB =
ZB

Page 3 of 23

minimum

value

of

P n=
=>

P
2
VB
ZB

=> Z B=

Pn V B 0.0833 (400)2
=
3
P
1 10

L
=13.333
Cs
C s =0.642 F

So the value of L=114.12 H and

V out

(200)2
=8
3
5 10

For maximum power the value of Load Resistance,

R Lmin =

For minimum power the value of Load Resistance,

V out 2
(200)2
R Lmax =
=
=40
1 103 1 103

5 10

Frequency range can be calculated as follows.

nmin=

f min=

2 f min 2 f min
=
B
1
LC s

1.646
2 8.56 106

f min=30.606 kH z
Similarly,

nmax=

f max=

2 f max 2 f max
=
B
1
LC s

5.382
6
2 8.56 10

f max=100 kH z

Page 4 of 23

=
3

From theory, it is known that maximum frequency

( f max=100 kHz )

corresponds to

minimum power (=1kW) and maximum load resistance ( R Lmax =40 ) .

Alternatively, minimum frequency

( f min =30.606 kHz )

corresponds to maximum

power (=5kW) and minimum load resistance ( R Lmin =8 ) .

Now, simulate the SRC circuit in PSIM without parasitic values and with minimum
value of load resistance (=8 ) and frequency

( f min=30.606 kHz )

in order to find

out the maximum voltage across capacitor.

Fig.4. Finding Maximum voltage across capacitor

Fig.5. Maximum Voltage across Capacitor waveform

So it is seen that the maximum voltage across capacitor C s is 351.26 volt

Page 5 of 23

Now, we have to choose a capacitor with value of 0.68 F and voltage rating around
350 volt (ac).
From the manufacturers website, suitable capacitor for this specific requirement is
0.68 F and voltage is 460 volt (ac). Model number is 5MP2_K474K.
So, chosen value of inductance and new value of capacitance is listed below.
Inductor, L
114.12 H
Capacitor, Cs
0.68 F
Table 2. Chosen value of Inductor and Capacitor for SRC
As the value of capacitance has been changed so it is required to re-calculate the
value of

and frequency range.

Z B=

New Base Impedance,

Now, for new value of

L
114.12 H
=
=12.9546
Cs
0.68 F

R Load
8
=
=0.6175
ZB
18.99

New

min =

New

max =

R Load
40
=
=3.087
ZB
18.99

voltage vs. normalized frequency and power vs.

normalized frequency is plotted in order to find out new value of

Page 6 of 23

n and Pn .

Fig.6. Normalized frequency vs. Voltage plot for

Fig.7. Normalized frequency vs. Power plot for

=0.61753.087

=0.61753.087

The newly obtained values are listed in the following table.

Pn

0.6175
3.087

1.684
5.557

0.3934
0.08046

Table 3. Showing the value of

n and Pn

for new value of

Page 7 of 23

For the above calculated value, Frequency range can be calculated as follows.

nmin=

2 f min 2 f min
=
B
1
LC s

1
1
1
=
=
6
6
LC s 114 .944 10 0.68 10
8.841 106

f min=

1.684
2 8.841 106

f min=30.425 kH z
Similarly,

nmax=

f max=

2 f max 2 f max
=
B
1
LC s

5.55
2 8.841106

f max=100 kH z
From theory, it is known that maximum frequency

( f max=100 kHz )

corresponds to

minimum power (=1kW) and maximum load resistance ( R Lmax =40 ) .

Alternatively, minimum frequency

( f min=30.425 kHz )

corresponds to maximum

Lm=8
R
power (=5kW) and minimum load resistance
.

Calculation of Parasitic Resistance:

From the Data sheet of the selected capacitor, the parasitic resistance value can be
found.
So, for the selected model of the capacitor, the value of Equivalent Series
Resistance (ESR) is 0.008.
For Inductors parasitic, thumb rule is equivalent series resistor around 10 m /
30H.
So for 118.944 H inductor the value of equivalent series resistance is 38.04 m .
The final circuit with all the parasitic value is shown below.
Page 8 of 23

Fig. 8. SRC circuit with parasitic value

The waveforms of output voltage and current drawn by the load resistance for R Load=
8 and input frequency is 30.425 kHz ( f min

is given below.

(a) Voltage across the Load Resistor

Page 9 of 23

(b) Current drawn by load Resistor

(b) Time vs. Power

Fig. 9. SRC circuits (a) output voltage and current drawn by the load resistance and
(b) power for minimum input frequency and minimum output load resistance
The power drawn by the load resistance should be maximum (around 5028 W)
during minimum frequency (30.425 kHz) and minimum load resistance (8). So the
obtained result matches with theory.
The current and voltage at the switching instants for maximum power is shown
below-

Page 10 of 23

Figure 9: Source Voltage and Switching Current

Similarly, the waveforms of output voltage and current drawn by the load resistance
for RLoad= 40 and input frequency is 100 kHz ( f max

is given below.

Fig. 10. SRC circuits output voltage and current drawn by the load resistance for
maximum input frequency and maximum output load resistance
The power drawn by the load resistance should be minimum (1050 W) during
maximum frequency (100 kHz) and maximum load resistance (40). So again the
obtained result matches with theory.
The obtained values can be summarized in the table below.
L

Vmax at
the

Imax of
the

Frequen
cy

Maximu
m

Page 11 of 23

SR
C

114.12
H

0.6
8
F

reacti
ve
eleme
nt
785 V
across
Induct
or

reacti
ve
eleme
nt

range

Output
Power

33 A

30.425
kHz to
100 kHz

5.028kW

Table 4. Calculated parameters for SRC circuit

Parallel Resonance Circuit:
Using First Harmonic Approach the simplified circuit will be as follows.

Fig. 11: Parallel Resonance Circuit

For parallel Resonance Circuit, we know

V out
=
V

For a set of value of

1
j
1+ n n2

and normalized frequency

n the voltage and power plot

for the circuit is shown in following figure.

Page 12 of 23

Fig. 12. Normalized Frequency ( n ) vs. voltage curve for different values of

in PRC circuit

Fig. 13. Normalized Frequency ( n ) vs. Power curve for different values of

in

PRC circuit
As ratio between maximum power and minimum power is 5, so from figure 2,

=0.6

and

=3.0

has been selected.

Page 13 of 23

V n=0.5

line is also plotted in figure 14. So from figure 14, it can be seen that at

=0.6

the intersecting point of

normalized frequency,
and

n=1.177

Similarly, for

V n=0.5

curve and

the corresponding

n=1.177 . From figure 15, it is found that for

=0.6

corresponding Pn=0.4166 .

=3.0

corresponding

Pn

and

are 1.708 and 0.08334

respectively.
The obtained values are listed in the following table.

Pn

0.6

1.188

0.4075

3.0

1.708

0.08333

Table 5. Showing the value of

n and Pn

for chosen value of

Now, as maximum frequency corresponds to maximum normalized frequency, so it

can be written,

n=

>1.708=

> B =

2 (1 ookHz)
B

1
=367.88 103
LC
P

LC P =2.7184 106
Again

if

minimum

power

(=1kW)

is

considered

Pn ( 0.08333) has to be taken to calculate the value of

P n=

But,

then

L
CP

P
PB

V B2
PB =
ZB

Page 14 of 23

minimum

value

of

P n=
=>

P
2
VB
ZB

=> Z B=

Pn V B 0.08334 (400)2
=
3
P
1 10

So the value of L=36.25 H and

L
=13.3344
CP
C P =0.21 F
V out

(200)2
=8
3
5 10

For maximum power the value of Load Resistance,

R Lmin =

For minimum power the value of Load Resistance,

V out 2
(200)2
R Lmax =
=
=40
1 103 1 103

5 10

Frequency range can be calculated as follows.

nmin=

f min=

2 f min 2 f min
=
B
1
LC P

1.247
2 2.7184 106

f min=70.8806 kH z
Similarly,

nmax=

f max=

2 f max 2 f max
=
B
1
LC s

1.708
6
2 2.7184 10

f max=97.216 kH z

Page 15 of 23

=
3

From theory, it is known that maximum frequency

( f max=97.216 kHz )

corresponds to

minimum power (=1kW) and maximum load resistance ( R Lmax =40 ) .

Alternatively, minimum frequency

( f min =70.8806 kHz )

corresponds to maximum

power (=5kW) and minimum load resistance ( R Lmin =8 ) .

As CP is parallel to output load resistance R Load so the voltage across C p will be equal
to the output voltage i.e. 200 volt.
Now, we have to choose a capacitor with value of 0.21 F and voltage rating around
327 volt (ac).
From the manufacturers website it is seen that the nearest value is 0.47 F. So we
can add two 0.47 F in series so that the equivalent capacitance will be 0.23 F.
So we are taking two capacitance of value 0.47 F and voltage is 400 volt (ac).
Model number is 5MP22J474F. And we will connect the two capacitor in series so
that the equivalent capacitance will be 0.23 F.
So, chosen value of inductance and new value of capacitance is listed below.
Inductor, L
36.25 H
Capacitor, Cs
0.23 F
Table 6. Selected value of Inductor and Capacitor for PRC
As the value of capacitance has been changed so it is required to re-calculate the
value of

and frequency range.

New Base Impedance,

Z B=

New

min =

R Load
8
=
=0.64
ZB
12.55

New

max =

R Load
40
=
=3.2
ZB
12.55

Now, for new value of

L
36.25 H
=
=12.55
CP
0.23 F

voltage vs. normalized frequency and power vs.

normalized frequency is plotted in order to find out new value of

Page 16 of 23

n and Pn .

Fig.14. Normalized frequency vs. Voltage plot for

Fig.15. Normalized frequency vs. Power plot for

The newly obtained values are listed in the following table.

Page 17 of 23

=0.643.2

=0.643.2

Pn

0.64
3.2

1.247
1.711

0.3807
0.0781

n and Pn

Table 7. Showing the value of

for new value of

For the above calculated value, Frequency range can be calculated as follows.

nmin=

2 f min 2 f min
=
B
1
LC P

1
1
1
=
=
6
6
LC P 36.25 10 0.23 10 2.8 106

f min=

1.247
2 2.8 106

f min=70.8806 kH z
Similarly,

nmax=

f max=

2 f max 2 f max
=
B
1
L CP

1.711
2 2.8 106

f max=97.216 kH z
From theory, it is known that maximum frequency

( f max=97.216 kHz )

corresponds to

minimum power (=1kW) and maximum load resistance ( R Lmax =40 ) .

Alternatively, minimum frequency

( f min=70.2kHz )

corresponds to maximum power

(=5kW) and minimum load resistance ( R Lmin =8 ) .

Calculation of Parasitic Resistance:
From the Data sheet of the selected capacitor, the parasitic resistance value can be
found.
So, for the selected model of the capacitor, the value of Equivalent Series
Resistance (ESR) is 0.016. As we are connecting two capacitor in series, so the
total ESR will be 0.032.

Page 18 of 23

For Inductors parasitic, thumb rule is equivalent series resistor around 10 m /

30H.
So for 36.25 H inductor the value of equivalent series resistance is 12.08 m .
The final circuit with all the parasitic value is shown below.

Fig. 16. PRC circuit with parasitic value

The waveforms of output voltage and current drawn by the load resistance for R Load=
8 and input frequency is

70.8806 kHz ( f min

is given below.

(a) Voltage across the Load Resistor

Page 19 of 23

(b) Time vs. Power

Fig. 17. PRC circuits (a) output voltage and current drawn by the load resistance
and (b) power for minimum input frequency and minimum output load resistance
The output power as obtained from the simulation is 4.58 kW.
The power drawn by the load resistance should be maximum (around 5 kW) during
minimum frequency (70.8802 kHz) and minimum load resistance (8). So the
obtained result matches with theory.

Figure 18 : Source Current and Switching waveform of the circuit

Similarly, the waveforms of output voltage and current drawn by the load resistance
for RLoad= 40 and input frequency is 100 kHz ( f max

is given below.

Page 20 of 23

Fig.19. PRC circuits output voltage across the load resistance for maximum input
frequency and maximum output load resistance

Figure 20: Minimum power vs time

Output Power = 811.4W.
The power drawn by the load resistance should be minimum (around 1 kW) during
maximum frequency (100 kHz) and maximum load resistance (40). So again the
obtained result matches with theory.

Page 21 of 23

The obtained values can be summarized in the table below.

PR
C

Vmax at
the
reactiv
e
elemen
t

36.25
H

0.2
3
F

650 V
across
Inductor

Imax of
the
reacti
ve
eleme
nt
30 A
throug
h
Inducto
r

Frequen
cy
range

Maximu
m
Output
Power

70.8802
kHz to
97.216
kHz

4.56kW

Table 8. Calculated parameters for PRC circuit

The current and switching waveform are shown below-

Figure 21 : Switching waveform of the H-Bridge and the current

Comparison between PRC and SRC:

Vmax at
the
reacti
ve
eleme
nt

PR
C

36.25
H

0.2
3
F

650 V
across
Induct
or

SR

118.94

0.3

785 V

Imax of
the
reacti
ve
eleme
nt
40 A
throug
h
Induct
or
33A

Frequen
cy
range

Maximu
m
Output
Power

70.8802
kHz to
100 kHz

4.56kW

30.415

5.02kW

Page 22 of 23

4 H

3
F

across
Induct
or

kHz to
100 kHz

Table 9. Calculated parameters for PRC and SRC circuit

From the above table it is seen that, for SRC the frequency range is large so SRC has
smaller quality factor thus it has poor selectivity.
On the other hand, for PRC the frequency range is small so PRC has larger quality
factor thus PRC has good selectivity. Small frequency range makes easier switching
and control for the PRC circuit.
For both PRC and SRC the required value of L and C are available in market.
The size of inductor and capacitor for PRC will be lesser than SRC. At the same time
cost of the component for PRC will be relatively less than SRC.
In SRC circuit maximum voltage across Inductor is higher than the PRC circuit. So
the rating of the inductor has to be high. The same goes for capacitor too. The
rating of the capacitor is high for SRC where as for PRC it is less.
But in terms of power delivered to the load, the SRC is capable of delivering a larger
power than PRC. In case of SRC is very close to rated power whereas for PRC it is
less than SRC and lower than the rated power to be delivered. The same is true for
minimum power delivery too.
So, depending on the designing requirement, the choice or selection of SRC and PRC
should vary. PRC is better in terms of cost and size of the components where for SRC
the delivered power is more accurate and close than SRC. So SRC can be used for
better power quality whereas PRC can be used where cost of the converter is of
more concern than power quality.
A drawback of the series converter is that the output cannot be regulated for the
no-load condition. As RL goes to infinity, Q goes to zero. The output voltage is then
independent of frequency. However, the parallel converter is able to regulate the
output at no load. For the parallel converter Q becomes larger as the load resistor
increases, and the output remains dependent on the switching frequency.
A drawback of the parallel converter is that the current in the resonant components
is relatively independent of load. The conduction losses are fixed, and the efficiency
of the converter is relatively poor for light loads.
The series-parallel converter combines the advantages of the series and parallel
converters. The output is controllable for no load or light load, and the light load
efficiency is relatively high.

Page 23 of 23