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, hjavadi@ee.iust.ac.ir

** Department of Electrical Engineering, Iran University of Science and Technology, shoulaie@iust.ac.ir

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.

References

[1]. G.Seguier ,F.Labrique ,“Power Electronic Converters” ,North Oxford Academic Publishers Ltd, volume 4, 1989.

[2]. Isobe, T.; Usuki, K.; Arai, N.; Kitahara, T.; Fukutani,

K.;Shimada,R.; "Variable frequency induction heating using magnetic energy recovery switch (MERS)"

Power Electronics Specialists Conference, Page(s):2139 -

2145.IEEE 15-19 June 2008.

[3]. B. Davson and P. Jain, “A comparison of load commutated inverter system inductor heating and melting applications, ” IEEE

Trans. PowerElectron.,vol. 6, pp. 430–441, 1991.

[4]. Jordan, J.; Esteve, V.; Dede, E.; Magraner, J.M.; Sanchis, E.; Maset, E.; Ferreres, J.E.; "Automated Calculation Method to

Determine the Output Power Based on the Frequency for Induction

Heating IGBT Parallel Inverte", Power Electronics Specialists Conference, IEEE Page(s):713 – 718, 17-21 June 2007.

[5]. John Davies , Peter Simpson , "Induction Heating Handbook", McGRAW-HILL 1979.

[6]. M. K. Kazimierczuk and R. C. Cravens II, “Current-source

parallel resonant DC/AC inverter with transformer,” IEEE Trans.

Power Electron., vol. 11, pp. 275–284, Mar. 1996.

[7]. E. J. Dede, J. V. González, J. A. Linares, J. Jordán, D. Ramírez, P. Rueda; “25kW/50kHz Generator for induction heating,” IEEE

Trans. On Industrial Electronics, Vol. 38, No. 3, pp. 203-209june 1991.

[8]. M. Matsuo, T. Suetsugu,, S. Mori, I. Sasase; " Class DE Current-Source Parallel Resonant Inverter",IEEE TRANS. ON

INDUSTRIAL ELECTRONICS, VOL. 46, NO. 2, APRIL 1999.

440