DESIGN METHOD OF AN OPTIMAL INDUCTION HEATER FOR MAXIMUM POWER DISSIPATION AND MINIMUM POWER LOSS CAUSED BY ESR

Jung-gi Lee, Sun-kyoung Lim, Kwang-hee Nam ∗ and Dong-ik Choi ∗∗

∗ Department of Electrical Engineering, POSTECH University, Hyoja San-31, Pohang, 790-784 Republic of Korea. Tel:(82)54-279-2218, Fax:(82)54-279-5699, E-mail:[email protected]

∗∗ POSCO Gwangyang Works, 700, Gumho-dong, Gwangyang-si, Jeonnam, 541-711, Korea.

Abstract: In the design of a parallel resonant system, choosing a proper capacitance for the resonant circuit is quite important. The capacitance affects the resonant , output power, Q-factor, heating efficiency and power factor. In this paper, the important role of equivalent series resistance (ESR) in the choice of capacitance is recognized. Without the effort of reducing temperature rise of the , the life time of capacitor tends to decrease rapidly. This paper, therefore, presents a method of finding an optimal value of the capacitor under voltage constraint for maximizing the output power of an induction heater, while minimizing the power loss of the capacitor at the same time. Based on the equivalent circuit model of an induction heating system, the output power, and the capacitor losses are calculated. The voltage constraint comes from the voltage ratings of the capacitor bank and the switching devices of the inverter. The effectiveness of the proposed method is verified by simulations and experiments.

Keywords: Induction heating, Equivalent series resistance, Design

1. INTRODUCTION four and a parallel resonant circuit com- prised capacitor bank and heating coil. Thyris- tors are naturally commutated by the ac current Induction heating is widely used in metal industry flowing through the resonant circuit. Therefore, for melting or heating of thin slab in a contin- this type of inverter called as a load commutated uous casting plant because of good heating effi- inverter (Y. kwon, 1999; F.P. Dawson, 1991; R. ciency, high production rate, and clean working Bonert 1994). environments. A typical parallel resonant inverter circuit for induction heater is shown in Fig. 1. The The induction heater was kept up the metal slab of o phase controlled rectifier provides a constant DC 1100 C for the next milling process. In the parallel current source. The H-bridge inverter consists of resonant inverter, if the switching frequency is 3-phase H-bridge Heating Idc Inverter is iL Coil

DC link vC

Capacitor Heated Bank Slab (a1, a2, a3, a4, a5, a6 )

Power VoltageLimiter Reference Firing angle a, P* Vmax vC S DC link current Idc control + -V - max Output Power P Output Power vC Calulator 1 i P = V I cosa s out 2 C s

Fig. 1. Block diagram of induction heating system. closer to the resonant frequency for high power, elled as an plus a series resistance, and higher voltage is generated at capacitor bank the capacitor bank is modelled as a pure capac- (F.P. Dawson, 1991). However, due to the limit in itance with an equivalent series resistance(ESR) the voltage tolerance of the capacitor bank, the (Dec.2002 issue of Practical Wireless). Utilizing inverter output voltage vC needs to be limited this model, the output power of the induction below the rated voltage Vmax. One method of heater and the power loss in the capacitor are limiting vC is to reduce the DC-link current Idc derived as functions of capacitance. An optimal by increasing the firing angle of the rectifier. value of the capacitance under the voltage con- However, it results in the decrease of Idc, thereby straint is found with the help of Lagrange multi- the decrease of the output power. plier (D.G. Luenberger, 1989). At the mini-mill in a POSCO steel plant, eight This procedure looks very useful in the design induction heaters of load commutated type were of the induction heating system, specifically, in installed to heat thin slabs for next milling pro- determining the size of capacitor bank. All the cess in the continuous casting plant. The rated parameters used in this work come from the output power of each induction heater is 1.5 MW. induction heater operating in POSCO. However, there were two problems: One was the insufficiency in the output power, and the other was the frequent damages of the capacitor bank. 2. EQUIVALENT CIRCUIT AND POWER Insufficiency in output power was caused by a EQUATIONS poor power factor of the inverter. On the other hand, the damage to the capacitor bank was due In general, the heating coil and the load are to a little voltage margin between v and V , C max modelled as a with a single turn sec- and it resulted in a large power dissipation in the ondary winding as shown in Fig. 3(a). Almost capacitor causing a high temperature rise. all magnetic flux generated by the induction coil The capacitance of the capacitor bank affects the (primary winding) penetrates into the slab (sec- overall operating factors of induction heater such ondary winding). Hence, in the secondary circuit as resonant frequency, Q-factor, efficiency, and no leakage inductance appears and the coupling power factor (P. Jain,1988; E. J. Davis,1979; E. coefficient is equal to one. The secondary circuit J. Davies, 1990). Hence, in this work, we pro- can be moved to the primary part as shown in pose a method of choosing an optimal capacitance Fig. 3(b) (E. J. Davis, 1979). Denoting the mutual value Copt, which maximizes the output power, inductance by LM , it follows that and at the same time, minimizes the capacitor loss. Firstly, an equivalent model of the induction heater is developed based on previous works (Y. kwon, 1999; T. Imai, 1997; E. J. Davis, 1979; E. J. L1 = Ll + NLM , (1) L Davies, 1990). The heating coil and slab is mod- L = M , (2) 2 N 1.2 10−6 Ω F to 1.5 10−6 Ω F (Dec.2002 issue of Practical× Wireless).· × ·

Fig. 2. Model of slab for one turn coil. where L1 and Ll denote the inductance of the heating coil and the leakage inductance, re- spectively. Using Wheeler’s formula (Harold A. Wheeler, 1942), the inductance of the heating coil is calculated as follows: r2 N 2 L (µH) = × , (3) 1 9r + 10h where r is the radius, N is the turn number and h is the height of the coil. The slab resistance RL for one turn coil is given by L 2(w + 2b) R = ρ = ρ , (4) L A lδ where L and A are length and area of eddy cur- rent, l is the effective length of the slab occupied by one turn coil and b and w are defined in Fig. 2 (S. Zinn, 1988), δ and ρ are skin depth almost distributed over the surface of slab and electrical resistivity of the material. Simplified equivalent Fig. 3. (a) Based circuit of the induction heater. model for a transformer can be represented in (b) Equivalent circuit of the induction heater Fig. 3(c) by an equivalent inductance Leq and based on a transformer concept. (c) Simpli- resistance Req (Y. kwon, 1999; E. J. Davis, 1979). fied equivalent circuit of the induction heater. These equivalent parameters depend on several variables including the shape of the heating coil, A useful variable to calculate the power is a total the spacing between the coil and slab, the elec- impedance seen from the impedance of equivalent trical conductivity and magnetic permeability of circuit in Fig. 3(c) (H. Kojima, 1990). The total the slab, and the angular frequency of the varying impedance is given by current ωs (E. J. Davis, 1979).

(ZC + ZESR) (ZL + ZR) Zt = · 2 Z + Z + Z + Z Leq = L1 A L2, (5) L R C ESR − 2 2 (ωskReq + ωsLeq) + j(ωs Leqk Req) Req = R1 + A RL, (6) = − ,(7) ωs(CReq + k) + j(ωsLeqC 1) − where R1 denotes the resistance of the heating Z Z Z coil, RL denotes the resistance of the heated slab, where L = jωsLeq, C = 1 jωsC, R = Req, 2 2 2 and ZESR = RESR = k/C. and A = ωsLM ωs L2 + RL . It is noted that ± the inductance of heating coil L1 is not affected by The rectifier and H-bridge inverter of the induc- ±p the existence of the slab in the heating coil, since tion heater are represented by a square waved at about 1100oC temperature the permeability of current source whose magnitude is equal to the the iron slab is equal to that of air, i.e., µ = 4π DC-link current I . Therefore, the current source × dc 10−7 (H/m) (E. J. Davis, 1979). expanded in a fourier series is described as follows: To represent the power dissipation in the capaci- ∞ 4I tor bank, it is modelled by a pure capacitance C i (t) = dc sin nω t n = 1, 3, 5, . . . .(8) s nπ s and an equivalent series resistance (ESR) R . n−1 ESR X It is noted that RESR is inversely proportional to the capacitance, hence, it is modelled as RESR = The fundamental component of the square waved k C, where k is a coefficient of ESR ranged from current source is given by ± 4 where VC is the peak of vC , Vmax is rated voltage is(t) = Idc sin ωst = Is sin ωst, (9) π of the capacitor bank, and Zt is total impedance of capacitor bank and heating parts. One can see where Is = 4Idc/π is a peak of is. The current that Zt(j1.1ω0) is also a function of capacitance, | | through Req and RESR are represented by iL and since ω0 = 1 LeqC. iC , respectively. The phasor expression of iL and The aim is to find an optimal capacitance value iC are described as follows: ±p that maximize the output power of the induction heater and minimize the capacitor losses simulta- neously under the voltage constraint (14). If the VC ZC + ZESR IL = = Is,(10) cost function is selected as Pout Ploss, maximiz- ZL + ZR ZL + ZR + ZC + ZESR − ing Pout Ploss is equivalent to maximizing Pout VC ZL + ZR − IC = = (11)Is, and at the same time minimizing Ploss. ZC + ZESR ZL + ZR + ZC + ZESR where V and I are phasors of v and i and C s C s maximizing Pout Ploss VC = Zt Is. − π · subject to Zt(j1.1ω0) ( Idc) Vmax. In Fig. 3(b), the power consumption generate | | · 4 ≤ equivalent resistor Req and ESR RESR of the capacitor bank. therefore, the output power of the It leads to apply Kuhn-Tucker theorem (D.G. induction heater Pout and the capacitor loss Ploss Luenberger, 1989). Firstly, the cost function is are given by defined by

2 IL Pout(C) = Zeq √2 J(C) = Pout(C) Ploss(C) µ ¶ − 2 π 1 ZC + ZESR + λ Vmax Zt(j1.1ω0) Idc ,(15) = I2Z (12) · − | | · 4 2 Z + Z + Z + Z s eq ¯ L R C ESR ¯ ³ ´ ¯ ¯ 2 8I¯2 ω k j ¯ where λ 0 is the Lagrange multiplier. The = ¯dc s ¯ R , ≥ 2 k − 2 eq maximum point is found under the following con- π ¯ωsC Req + + j(ωs CLeq 1)¯ ¯ C − ¯ ditions, ¯ 2 ¯ ¯ ¡IC ¢ ¯ Ploss(C¯) = ZESR ¯ √2 µ ¶ 1 Z + Z 2 = L R I2Z (13) 2 Z + Z + Z + Z s ESR ¯ L R C ESR ¯ ¯ ¯ 2 8I¯2 ω C(R + jω¯L ) k = ¯dc s eq s¯ eq , 2 k 2 π ¯ωsC Req + + j(ωs CLeq 1)¯ C ¯ C − ¯ ¯ ¯ where I ¯ and¡I denote¢ the peak of i and¯ i , L¯ C L ¯ C respectively (S. Dieckerhoff, 1999). It is noted that Pout and Ploss are function of capacitance C, since all the parameters except capacitance are known values in (12) and (13).

3. OPTIMAL CAPACITANCE FOR MAXIMIZING OUTPUT POWER AND MINIMIZING CAPACITOR LOSS

In the load commutated inverter, the switching frequency of the inverter must be higher than the resonant frequency of the L-C load to guarantee commutation of the thyristors (R. Bonert, 1994). Hence, for more suitable value for the inverter while working close to the resonant frequency, we let ωs = 1.1ω0, then the voltage constraint is given by Fig. 4. The output power Pout, the capacitor loss π Ploss, and the power difference (cost function) VC = Zt(jωs) Is = Zt(j1.1ω0) Idc Vmax,(14) Pout Ploss versus capacitance value. | | · | | · 4 ≤ − ∂J = 0, (16) ∂C π λ Vmax Zt(j1.1ω0) Idc = 0. (17) · − | | · 4 ³ ´

The optimal capacitance value was found by using MATLAB. But, the resulting equation is quite long, hence it is described in Appendix.

Fig. 6. The source current is and the capacitor voltage vC when (a) C = 115 µF and (b) C = 126 µF, respectively. 4. SIMULATION AND EXPERIMENTAL RESULTS

Simulation was performed with SIMPLORER circuit simulator. The parameters of POSCO induction heater were utilized in this simula- tion: Leq=8.3 µH, Req=0.053 Ω, Vmax =1700 V, −6 Is=1300 A, and k = 1.35 10 Ω F . With the help of MATLAB, the optimal× capacitance· value is found to be Copt = 126µF and λ = 0 by (19) and (20) in Appendix. Fig. 4 shows the output power Pout, the capacitor loss Ploss, and the power difference Pout Ploss versus capacitance value. MATLAB is used− for deriving various powers in functional forms, and SIMPLORER is used for simulating powers versus time. One can see that the cost function is maximized at the optimal capacitance C = 126µF. Fig. 5 shows circuit simulation results of the source current is, the capacitor voltage vC , the output power Pout, and the capacitor loss Ploss when C = 90 µF, C = 126 µF, and C = 170 µF, respectively.

Fig. 5. The source current is, the capacitor voltage With C = 90 µF, the largest output power vC , the output power Pout, and the capacitor (1.38 MW) is achieved. However, the capacitor loss Ploss when C = 90 µF, C = 126 µF, and loss (0.62 MW) is impracticably large. With the C = 170 µF, respectively. optimal value C = 126 µF, the capacitor loss is reduced to a reasonable value, 0.4 MW, while frequency induction heating”, Industry Applica- the output power is 1.27 MW. Fig. 6 shows tions Conference, Conference Record of the 1999 experimental results of the source current is and IEEE, vol.3, pp. 2039 -2045, 1999. capacitor voltage v when C = 115 µF and C H. Kojima, T. Shimizu, M. Shioya, G. Kimura, C = 126 µF, respectively. Note that the source “High frequency power transmission in plural in- current i is constant at 1300A and the peak of s duction heating load circuits”, Industrial Elec- capacitor voltage v falls to 1250V at the optimal C tronics Society, 16th Annual Conference of IEEE, operating point(C = 126 µF). Note also that vol.2 pp. 990 -995, 1990. the experimental results are identical with the simulation results. Harold A. Wheeler, “Formulas for the Skin Ef- fect”, Proceedings of the I.R.E., pp. 412-424, September 1942. 5. CONCLUSION S. Zinn and S. L. Semiatin, Elements of Induction Heating. Ohio: Electric Power Research Institute, This paper suggests a new method in the choice Inc. 1988. of capacitance and operating frequency with the considerations on voltage tolerance and ESR of E. J. Davis and P. G. Simpson, Iduction Heating the capacitor, maximum output power, and high Handbook. New York: MaGraw-Hill, 1979. power factor. The optimal solution is found with E. J. Davies, Conduction and Iduction Heating. Lagrange multiplier. In the derivation of the de- Londonn: Peter Peregrinus Ltd. 1990. sired solution, MATLAB is utilized. This optimal choice is thought to contribute to increasing the D.G. Luenberger, Linear and Nonlinaer Program- life time of the capacitor bank and generating a ming. Stanford University: Addison-Wesley Pub- maximum output power. lishing Company. 1989. Dec.2002 issue of Practical Wireless in World REFERENCES Wide Web, “http://www.valveandvintage.co.uk /pw/esr.htm” Y. Kwon, S. Yoo, and D. Hyun, “Half-bridge series resonant inverter for induction heating applica- tions with load- adaptive PFM control strategy”, IEEE-APEC Conf. Rec., pp. 575-581, 1999. F.P. Dawson, and P. Jain, “A comparison of load commutated inverter systems for induction heating and melting applications”, IEEE Trans. on Power Electronics, Vol. 6, No. 3, pp. 430 -441, July, 1991. R. Bonert, and J.D. Lavers, “Simple starting scheme for a parallel resonance inverter for induc- tion heating”, IEEE Trans. on Power Electronics, Vol. 9, No. 3, pp. 281 -287, May, 1994. T. Imai, K. Sakiyama, I. Hirota, and H. Omori, “A study of impedance analysis for an induction heating device by applying a new interpolation method”, IEEE Trans. on Magnetics, Vol. 33, No. 2, pp. 2143-2146, March, 1997. Y. Kim, S. Okuma, and K. Iwata, “Characteristics and starting method of a cycloconverter with a tank circuit for induction heating”, IEEE Trans. on Power Electronics, Vol. 3, No. 2, pp. 236 -244, April, 1988. P. Jain, and S.B. Dewan, “Designing a com- plete inverter system for a series tuned induc- tion heating/melting load”, Power Electronics and Variable-Speed Drives, Third International Conference on, pp. 456-461, 1988. S. Dieckerhoff, M. J. Ruan, De Doncker, “Design of an IGBT-based LCL-resonant inverter for high- 6. APPENDIX Based on these parameters, the cost function and the condition equations are calculated as follows:

2 2 Idc 10000 LeqC + 12100 k Req J(C) = 8 2 2 2 2 π 12100 Req C¡ + 24200 CkReq + 12100¢ k + 441 LeqC 2 2 Idc 121 Leq + 100 Req C k 968 ¡ ¢ 2 2 2 2 − π 12100 Req C +¡ 24200 CkReq + 12100¢ k + 441 LeqC 2 2 V ¡ 121 Leq + 100 Req C (100 LeqC + 121¢ k ) + λ max , (18) 2 2 2 Ã Idc − sC 12100¡ Req C + 24200 Ck¢Req + 12100 k + 441 LeqC ! ∂J(C) ¡ I 2L R ¢ = 80000 dc eq eq 2 2 2 2 ∂C π 12100 Req C + 24200 CkReq + 12100 k + 441 LeqC 2 2 2 Idc 10000¡ LeqC + 12100 k Req 24200 Req C + 24200 kReq¢+ 441 Leq 8 2 − 2 2 2 2 ¡ π 12100 Req C + 24200¢ C¡ kReq + 12100 k + 441 LeqC ¢ I 2R 2k 96800 ¡ dc eq ¢ 2 2 2 2 − π 12100 Req C + 24200 CkReq + 12100 k + 441 LeqC 2 2 2 Idc ¡121 Leq + 100 Req C k 24200 Req C + 24200 kReq +¢441 Leq + 968 2 2 2 2 2 ¡π 12100 Req C + 24200¢ ¡ CkReq + 12100 k + 441 LeqC ¢ 2 2 2 2 2 L C¡ Req 100 LeqC + 121 k 12100 Req C +¢ 24200 CkReq + 12100 k + 441 LeqC + λ eq 100 2 3 − 2 × ¡ ¢ ¡ Leq C ¢ { { 2 2 2 2 121 Leq + 100 Req C 12100 Req C + 24200 CkReq + 12100 k + 441 LeqC + 100 3 ¡ ¢ ¡ LeqC ¢ 2 2 2 121 Leq + 100 Req C 100 LeqC + 121 k 24200 Req C + 24200 kReq + 441 Leq + 2 3 ¡ ¢ ¡ Leq C¢ ¡ ¢ 2 2 2 2 2 121 Leq + 100 Req C 100 LeqC + 121 k 12100 Req C + 24200 CkReq + 12100 k + 441 LeqC 3 2 4 − ¡ ¢ ¡ ¢ ¡ Leq C ¢} 2 3 Leq C 3 × v 2 2 2 2 2 u 121 Leq + 100 Req C (100 LeqC + 121 k ) 12100 Req C + 24200 CkReq + 12100 k + 441 LeqC u t 2 2 ¡ 121 Leq + 100¢Req C (100 LeqC + 121¡ k ) ¢ 3 2 2 2 − sC 12100¡ Req C + 24200 Ck¢Req + 12100 k + 441 LeqC 2 ¡ 2 2 2 ¢ 121 Leq + 100 Req C (100 LeqC + 121 k ) 24200 Req C + 24200 kReq + 441 Leq + 3 = 0, (19) v 2 2 2 u¡ C 12100¢Req C + 24200 CkR¡eq + 12100 k + 441 LeqC ¢ u t } ¡ ¢ V 1.1 λ( max Z j ) Idc − ¯ Ã LeqC !¯ ¯ ¯ ¯ ¯ 2 2 ¯ V p ¯ 121 Leq + 100 Req C (100 LeqC + 121 k ) = λ¯ max ¯ = 0. (20) 2 2 2 Ã Idc − sC 12100¡ Req C + 24200 Ck¢Req + 12100 k + 441 LeqC ! ¡ ¢