Analysis of H-bridge Current Source Parallel Resonant Inverter
for Induction Heating
Abstract: This paper gives the theory and experimental
results for a current-source parallel-resonant inverter
employed for induction heating. The analysis is performed
in the frequency domain using Fourier series techniques to
predict output power, efficiency, dc-to-ac voltage transfer
function, and component voltage and current stresses. The
inverter consists of four switches, a large choke inductor,
and a parallel-resonant circuit. Each switch consists of a
MOSFET in series with a fast diode. An inverter was
designed and constructed. The dc input voltage was 70 V
and the output voltage was a sine wave with a peak value of
100 V at an operating frequency of 83 kHz. The output
power at full load was 400 W.
Index Terms: induction heating, parallel resonant,
current source inverter.
1. Introduction
Voltage source inverters suffer from a pulsating
input current [1,3]. An alternate topology that draws a
constant current from the dc supply is the current
source parallel-resonant inverter. The constant current
drawing property of the current-source inverter is an
advantage over the voltage-driven inverters because of
the lack of harmonics introduced to the line[7].
Another advantage of the current-source parallel-
resonant inverter topology is that the switches only
carry the active power of the resonant circuit [2]. The
objectives of this paper are to present an analysis of a
full bridge current source parallel-resonant inverter for
induction heating, give a design example, and present
the experimental results.
2. Principle of Operation
Fig. l(a) shows the circuit diagram of the full
bridge current source parallel-resonant inverter. It
consists of a large choke inductor DCL , four switches
S1-S4, and an LLCR , parallel-resonant circuit. The dc
input source IV and the filtering choke DCL , form a
dc input current source II . LR is the calculated
work-piece resistance reflected to the heating coil
terminals[5]. Each switch is made up of a MOSFET
in series with a fast recovery diode. This series diode
disables the internal body-drain diode of the
MOSFET that would normally allow the negative
current to flow when the switch is off [4]. The
MOSFET’s are driven by rectangular gate-to-source
voltages GSV at the operating frequency f = 1/T and
an on-duty cycle of approximately 70%. The overlap
period ( sµ1 ) when all switches are on is necessary to
prevent the filtering choke from being open circuited.
Fig. l(b) shows an equivalent circuit of the inverter
with parasitic resistances and offset voltage sources,
where DCLR is the equivalent series resistance
(ESR) of the input inductor DCL , DSr is the
MOSFET on-resistance, FR is the diode forward
resistance, FV is the diode offset voltage, LpR is the
equivalent parallel resistance (EPR) of the resonant
inductor (work coil without work-piece) L, and CpR
is the EPR of the resonant capacitor C.
When switch S4 and S3 are OFF and switch S1
and S2 are ON, the current through the resonant
circuit is IIi = . When S1 and S2 are OFF and S3
and S4 are ON, the current through the resonant
circuit is IIi −= . Assuming that the input inductor
LDC is large enough, the circuit composed of IV ,
DCL , and S1-S4 can be modeled by a square-wave
current source IIi ±= . Shown in Fig. l(c).
The resonant frequency of the parallel resonant
H.javadi*,A.Shoulaie**
*Iran University of Science and Technology, [email protected]
** Department of Electrical Engineering, Iran University of Science and Technology, [email protected]
435
circuit is LCf /10 = . When 0ff < , the parallel-
resonant circuit represents an inductive load. This
operation is rarely used for two main reasons. 1- The
semiconductor devices have to ensure forced turn-off.
However, in the current source inverters, the
inductances limiting dtdi / bring about high voltage
spikes across the semiconductor devices at the end of
turn-off. 2- Operation occurs in the unfavorable side
of the frequency characteristics of the inverter. Where
0/ ωω is less than 1 and the output voltage is not
sinusoidal any more [1].
S1
S3 S2
S4
LDC
L
C
RL
D2D3
D4D1
(a)
(b)
(c)
Fig.1- a) current source parallel resonant inverter b) Equivalent parasitic resistance and offset voltage sources. c)simplified model
3. Analysis of the H-Bridge Current Source
Parallel- Resonant Inverter
In this section the fundamental analysis of the H-
bridge current source inverter with parallel resonant load such as induction heating systems is presented.
3.1 Analysis of the Parallel-Resonant Circuit
Considering to the model of the current source
parallel-resonant circuit shown in Fig. l (b), it can be seen that the parallel-resonant circuit is made up of a
capacitor C , an inductor L , and a resistor R [6]. For
this circuit, we can write.
dL
dL
RR
RR
GR
+==
1 (1)
Where LR is the ac load resistance and dR is the
equivalent parasitic resistance of the resonant tank,
given by
CpLp
CpLpd
RR
RRR
+= (2)
The parallel-resonant circuit can be characterized by
the resonant Frequency LC/10 =ω , the
characteristic impedance LCLZ 00 / ω== , and the
loaded quality factor RCLRQ 00 )/( ωω == . The
unloaded quality factor is: .)/( 000 dd CRLRQ ωω ==
The input admittance of the parallel-resonant circuit
can be written as
−+=++= )(1
1 0
0 ω
ω
ω
ω
ωω LjQG
LjCjGY (3)
One may therefore write the magnitude of the
admittance as
20
0
20
11
−+
==
ω
ω
ω
ωL
out
rms
Q
ZV
IY
(4)
Where rmsI is the rms value of the fundamental
component of the current through the resonant circuit
defined in equation (9) and outV is the rms value of
the ac output voltage of the inverter. The input power
of the resonant circuit is RVP outR /2= , the output
power of the resonant circuit is LoutR RVPL
/2=
and the power dissipated in the resonant circuit
is doutRd RVP /2= . Hence, the efficiency of the
resonant tank is
dL
L
RR
d
R
Rrc
R
P
P
+==η (5)
3.2 Voltage Transfer Function of H-bridge
Current Source Inverter
The input current of the resonant circuit i is a square
436
wave of magnitude II
<<−
<<=
πωπ
πω
2
0
tforI
forIi
I
tI (6)
Fourier analysis of the current i gives its fundamental
component
tIi mi ωsin1 = (7)
where
Im IIπ
4= (8)
The rms value of 1i is
πIm
rmsII
I22
2== (9)
The dc to ac current transfer function from the dc
input current II to the fundamental component rmsI
at the input of the resonant circuit is
π
221 =≡
I
rmsI
I
IM (10)
Equations (4) and (10) can be combined to obtain the
magnitude of the transfer function of the dc current II
to the rms output voltage Vout
20
02
0
12
1
22
−+
=
=×=≡
ω
ω
ω
ωπ
L
I
I
rms
rms
R
I
RI
Q
Z
Y
M
I
I
I
V
I
VM LL
(11)
The dc input power of the current-source inverter is
given by
III IVP = (12)
and the ac output power of the inverter is
LoutR RVPL
/2= (13)
Dividing (13) by (12) gives the efficiency Iη of the
inverter
L
I
I
out
I
RI
R
M
V
V
P
PL 2
==η (14)
Substituting (11) into (14) yields the magnitude of the
voltage transfer function
0
20
02
2 22
1
Z
QR
M
R
V
VM
L
LI
I
LI
I
outVI
−+
==≡ω
ω
ω
ωπη
η
(15)
Assuming the inverter efficiency %100=Iη , the
range of VIM is from 4/2π to ∞ . In Fig. 2, VIM
is plotted as a function of 0/ ff for various values
of LQ .
It can be seen that the output voltage of the inverter
increase as switching frequency increase, as a result
the output power of the inverter can be controlled by
switching frequency variation. The limitation of this
control method is the switches rated voltage that
should be consider.
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.50
2
4
6
8
10
12
f/f0
MV
I
QL=2
QL=3
QL=7
Fig.2- voltage transfer function of the inverter versus 0/ ff .
3.3 Output Power of the H-bridge Current-
Source Inverter
Solving (15) for outV and substituting into (13) gives
the ac output power of the inverter to be
R
QVR
R
VP
LIIL
L
outRL 8
1
20
0
222
2
−+
==
ω
ω
ω
ωηπ
(16)
Fig. 3 shows the ac output power LRP as a function
of LR .
The peak value of the switch current is ISM II = .
10 10.5 11 11.5 12 12.5 13 13.5 14518
520
522
524
526
528
530
532
534
536
538
PR
L(W
)
RL
Fig.3- output power versus LR .
The maximum values of the amplitude of the currents
through the resonant inductor L and the resonant
capacitor C occur at the resonant frequency and are
given by
πI
LCmLmI
QII4
== (17)
4. Efficiency of the Inverter
4.1 Power Loss in the filtering choke DCL
Fig. l(b) shows an equivalent circuit of the inverter
with parasitic resistances. Substituting (11) into (13)
gives
LR
LL
I PR
QR
I2
20
0
22
2
8
1
−+
=
ω
ω
ω
ωπ
(18)
437
Neglecting the ripple current in the inductor DCL , the
input current of the inverter contains only a dc
component II . The power loss in the dc ESR DCLr of
DCL is
2ILL IrP
DCDC= (19)
4.2 Conduction Power Loss in the MOSFET's
The MOSFET's are modeled by switches with on-
resistance DSr . The rms value of the switch current
is2
II. The conduction power loss in each switch can
be expressed by (20)
2
22 IDS
SrmsDSrIr
IrPDS
== (20)
4.3 Turn-on switching Loss in the MOSFET In reality, however, there is a problem associated with
the discharging of the transistor output
capacitance ossC [8]. When the switch voltage
increases, the MOSFET output capacitor ossC is
charged via diode 1D to the peak value of the switch
voltage max.outSM VV = and then remains at that
voltage until the transistor turns on. At this time, the
capacitor ossC is discharged through the transistor,
resulting in a turn-on switching loss in the MOSFET
according (21).
2/2SMosston VfCP = (21)
Where SMV is the peak value of the switch voltage
[1]. Total power in each power MOSFET can be
calculated by adding (21) and (20).
4.4 Conduction Power Loss in the series Diodes
The diode that is conducting is modeled by a voltage
source FV and a forward resistance FR . A diode that
is not conducting is modeled by an open switch.
Assuming that the forward resistances of the diodes
are identical and equal to FR , the power loss due to
FR can be found by using of (18) and (22)
2
22 IF
SrmsFrFIR
IRP == (22)
The average current through the diode is 2
II and the
power loss associated with FV is
L
L
F RI
F
II
RF
I
IFIFV P
V
V
V
PV
V
PVIVP
2222≈===
η (23)
Adding (22) and (23) gives the total conduction power
loss in each diode.
4.4 Power Loss in the resonant tank
The resistance dR is the parallel combination of the
equivalent parallel resistance LPR of work coil L and
the equivalent parallel resistance CPR of resonant
capacitor bank C . The total conduction power loss in
the resonant tank can be obtained as
Ld Rd
L
d
outR P
R
R
R
VP ==
2
(24)
The total conduction power loss rP in the current
source inverter is simply the sum of the power losses
in the filtering choke, MOSFET's, diodes, parallel
resonant circuit and turn-on switching losses.
tonRDrLr PPPPPPdDSDC
444 ++++= (25)
The dc input power of the inverter is
rRI PPPL
+= (26)
Using (16), (25), and (26), the efficiency of the
inverter can be found as
L
L
LL
R
rrR
R
I
RI
P
PPP
P
P
P
+
=+
==
1
1η (27)
The inverter efficiency Iη is shown in Fig. 4 as a
function of RL for 1/ 0 <ff and 1/ 0 >ff . Notice
that the efficiency decreases as f moves away from
0f and the inverter is more efficient above resonance
than below resonance.
5 10 15 20 25 300.6
0.62
0.64
0.66
0.68
0.7
0.72
RL
Effic
iency
X: 13
Y: 0.7083f/f
0=0.96
f/f0=1.04
Fig.4 -efficiency versus LR .
5. Experimental Results
A current source parallel-resonant inverter was
implemented, using four IRFP460 MOSFET's
(International Rectifier), four ultrafast STTA2006PI
diodes in series with the MOSFET's and twenty five
KP-6 capacitors (Alcon Electronics) as a capacitor
bank. The capacitors EPR can be calculated with the
δtan versus frequency curve provided in the
capacitor datasheet. Circuit parameters are shown in
Table 1.
Next, the inverter efficiency can be calculated as (25),
(27)
W
PPPPPP tonRDrLr dDSDC
1942.*452)3.4(*411*480
444
=++++
=++++=
The input power of the inverter,
Wp I 72010*72 ==
Hence, the efficiency of the inverter Iη is
438
%731 =−=I
rI
P
Pη .
Fig.5 shows the output voltage of the inverter. The
inverter DC input current is 10A, switching frequency
is slightly above resonant frequency ( KHzf r 5.82= )
and the work-coil current peak value is 100A.
Fig.5 -inverter output voltage f=83300 Hz . peak value=
100V.;Horizontal: 5us/div.
Fig.6 -inverter output voltage f=71000 Hz ( 86.0/ 0 =ff ).
peak value= 68V.;Horizontal: 5us/div.
Fig.6 shows the output voltage of the inverter when
the output frequency is less than the load resonant
frequency. In this case the resonant tank has an
inductive behavior. Hence, high voltage spikes will
appear on the load, which is undesirable.
TABLE 1. System parameters
DCL 27 mH DCLr 0.8 ohm
CPR 6.1 k ohm C 1 Fµ
FR 0.017 ohm DSr 0.22 ohm
L 3.7 uH LPR 190 ohm
f 83300 Hz IV 72 V
LR 13 ohm ossC 500pF
Fig.7 shows the voltage transfer function of the
implemented inverter versus 0/ ff that is in good
agreement with the experimental results.
0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 1.250.8
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8
f/f0
MV
I
Calculation
Experimental
Fig.7 –voltage transfer function of the implemented inverter versus
0/ ff ( %75=Iη ).
Fig.8 shows voltage across switches. No voltage
overshoot is present hence; the need for any snubber
circuit was eliminated, which reduce the number of
components needed for the inverter stage.
Fig.8 -switch voltage f=83300 Hz . peak value= 90V.;Horizontal:
5us/div.
The heating load consists of a 10-turn copper coil
made from 6mm hollow tube. The cast iron work-
piece is placed inside a crucible which is made of a
alumina, capable of withstanding temperature up to
12000C. the maximum recorded temperature was
8000C which is suitable for induction hardening
application.
Fig.9 -work-coil; alumina crucible and work piece
439
6. Conclusion In this article the topology of the current source
parallel-resonant inverter was investigated and the loss
equation for every circuit component has been
calculated. These equations can be used for
minimizing loss in design step and proper heat sink
selection in cooling system. The equations provide
easy-to-use design tools and good insight into inverter
operation. Experimental results prove the proposed
equations.
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