Loss Analysis and Air-Cooled Design for a Cascaded Electrical Source Transmitter

Abstract

Air-cooling method is adopted on the basis of the requirements for the thermal stability and convenient field use of an electrical source transmitter. The power losses of the transmitter are determined after calculating the losses of the alternating current (AC)-direct current (DC) power supply, the constant-current circuit, and the output circuit. According to the analysis of the characteristics of a heat sink with striped fins and a fan, the engineering calculation expression of the Nusselt number and the design process for air-cooled dissipation are proposed. Experimental results verify that the error between calculated and measured values of the transmitter losses is 12.2%, which meets the error design requirements of less than 25%. Steady-state average temperature rise of the heat sink of the AC-DC power supply is $22^{\circ}C$, which meets the design requirements of a temperature rise between $20^{\circ}C$ and $40^{\circ}C$. The transmitter has favorable thermal stability with 40 kW output power.

NOMENCLATURE

 

I. INTRODUCTION

Electromagnetic prospecting methods directly obtain underground information by using electromagnetic theory, which is widely used in water, mineral, and hydrocarbon resource explorations, as well as in other fields [1]-[3]. In controlled source audio-frequency magnetotelluric method, direct current resistivity method, and induced polarization method, the electrical source transmitter is the artificial excitation source for electromagnetic prospecting. The load of the electrical source transmitter is two electrode holes away from each other for 1–3 km on the ground, with an equivalent resistance of approximately 10–80 Ω. The transmitter must provide an exciting current to the earth load in practical applications, and the exciting current is approximately 10–40 A. Maximum output voltage usually needs to reach 1000 V, and transmitter power usually needs to reach tens of kW. For transmitters with a power of tens of kW, thermal losses can reach several kW.

Several transmitters have been developed in the field of geophysical instruments[4]-[8]. A transmitter developed by the Central South University of China and used for wide-area electromagnetic sounding has 200 kW output power [5]. The Institute of Geophysical and Geochemical Exploration under the Chinese Academy of Geological Sciences developed the DEM-T70 transmitter, which has 70 kW output power [7], [8]. An American company, Zonge International, developed the GGT-30 transmitter, which has 30 kW output power. These power devices of transmitters, used to acquire direct current (DC) bus voltage, work below 400 Hz. These transmitters have high efficiency and output power. However, the system response speed is low, and the effect of suppressing transient earth load fluctuation should be improved. Therefore, these transmitters achieve high thermal stability at the cost of control precision. The Canadian Phoenix Company developed the TXU-30 transmitter. The power devices of TXU-30 work at 20 kHz, which has high control precision. However, China is the main area where electrical source transmitters are used, and TXU-30 uses a three-phase 220 V generator instead of the three-phase 380 V mainstream generator used in China. Therefore, the thermal design of TXU-30 is not suitable for transmitters with 380 V input. In addition, the power of TXU-30 is 20 kW, and its thermal losses are relatively low.

In the field of power electronics, research is focused on the loss calculation of power devices, whereas thermal dissipation analysis is comparatively scarce. The analysis method of thermal dissipation proposed in Ref. [9] involves several coefficients, but Ref. [9] does not show how to determine the coefficients, which limits its use for engineering applications. Ref. [10] disregards the effect of the efficiency of heat sink fin, and the thermal dissipation analysis is relatively rough. In Refs. [11]-[17], the analysis of heat sink is comprehensive, and several calculation expressions are introduced. However, the expressions proposed in these studies are generally complex, adding to the complexity of the engineering design. These studies also focus on a heat sink with smooth fins rather than the widely used one with striped fins. In the heat transfer field [18], thermal dissipation theories are systematically illustrated. However, these theories are generally macroscopic, and few studies show the corresponding analysis of air-cooled design on the basis of power circuit.

A heat sink with striped fins is analyzed in the present study on the basis of the characteristics of an electrical source transmitter, the classical theory of heat transfer, and the research results for heat sinks. Fan characteristics are studied based on the research results for hydromechanics. An engineering design process for air-cooled thermal dissipation is proposed. Experimental results verify that the transmitter has favorable thermal stability when the outputs are 1000 V and 40 A. The calculation error of thermal losses of the transmitter and the temperature rise of the heat sink are reasonable. The proposed air-cooled method can be applied to other power electronic equipment below 100 kW to a certain extent.

This paper includes several symbols that are not easy to remember. Hence, the definitions of the main symbols are listed after the Abstract.

 

II. DESIGN PURPOSES

The main purpose of the thermal design is to ensure the safety of devices in the transmitter, especially the security of power semiconductor devices. When the main purpose is met, another secondary purpose of the thermal design is considered. The secondary purpose is to ensure that the size and weight of the heat sink are as small as possible, making it convenient for field application.

For power semiconductor devices, the maximum allowable junction temperature of the tube core is generally 125 ℃ –150 ℃. The maximum junction temperature Tjmax can be obtained from device specifications, and the junction temperature is represented as Tj or Tvj in device specifications. The temperature loss, ΔT1, between the tube core and the case of the device is 10 ℃–25 ℃. ΔT1 = P·RthJC, where P represents the power loss of the device, and RthJC represents the thermal resistance between the tube core and the case. RthJC can be obtained from device specifications, and some specifications also adopt Rth(j-c). Insulation pad, contact gap, and so on will induce a loss, ΔT2, of 10 ℃–25 ℃ between the device and the heat sink. For power modules, the case is generally insulated against device pins. ΔT2 = P·RthCK, where RthCH represents the contact thermal resistance between the case and the heat sink. RthCK can be obtained from device specifications, and some specifications also adopt Rth(c-k). For discrete power devices, the case is generally connected to one pin of a device. Thus, an insulation pad should be added between the case and the heat sink. ΔT2 = P·(RthCK + Ric), where Ric represents the thermal resistance of the insulation pad. The calculation of Ric will be shown in Section IV.A. The local temperature of the heat sink where the power device is located is higher than the average temperature of the heat sink, and the temperature loss ΔT3 is approximately 10 ℃. ΔT3 = P·Rcb. The calculation of Rcb will be shown in Section IV.A, which is from Ref. [17]. The maximum temperature Tamax in the field in summer may be 35 ℃–38 ℃. Therefore, the average temperature rise ΔThs of the heat sink should be less than 40 ℃, where ΔThs = Tjmax − ΔT1 − ΔT2 − ΔT3 − Tamax. Although a low temperature rise will increase the reliability of the devices, it will also increase the size and weight of the heat sink. A significantly low temperature rise does more harm than good. Therefore, the temperature rise should be kept between 20 ℃ and 40 ℃.

To control the temperature rise of the heat sink, the device losses should be calculated. When loss error is too high, the selected heat sink will be obviously small or large. The small size of the heat sink will cause a high temperature rise, which endangers the safety of the devices. The large size of the heat sink will cause the transmitter to become very heavy. Given that the power dissipation of the heat sink is approximately proportional to the size and weight of the heat sink with the same structure and with the size difference below several tens of percent (Figs. 2(d) and 5(a) in Ref. [19]), the loss error between 0% and +25% does not significantly increase the transmitter weight. Volumetric flow rate Qm and static pressure pm are also different for same-sized fans, as shown in Table I [20]. Q ∝ P/ΔTa, where P represents the power loss, Q represents the volumetric flow rate of the fan, and ΔTa represents the temperature difference between inlet and outlet air. The error can be feasibly and easily compensated by replacing the fan with the same size when the loss error is between 0% and −25%. A low loss error needs a highly accurate mathematical model and must consider numerous factors [21]-[23]. Some stray losses are difficult to calculate, and the actual conditions are not exactly the same as the testing conditions in the specification. These conditions include voltage [21], current [21], junction temperature [21], device driver [22], pulse width [23], and circuit topologies. Therefore, selecting 25% as the accepted loss error is a relatively reasonable trade-off between application value and calculation complexity.

TABLE IFAN PARAMETERS OF DELTA COMPANY

 

III. TRANSMITTER LOSSES

A. Cascaded Topology of the Transmitter

The transmitter adopts the cascaded topology shown in Fig. 1, where ZL represents the earth load of the transmitter. When VT11, VT14 and VT21, VT24 turn on, the load voltage is vL = E1 + E2 = 2E, and the output is a positive pulse. When VT12, VT13 and VT22, VT23 turn on, the load voltage is vL = −(E1 + E2) = −2E, and the output is a negative pulse. The output level of the transmitter is only +2E and −2E without +E, −E, and 0, which is different from the conventional cascaded multilevel inverters [24]-[26]. E1 and E2 are in a floating state that refers to the load ZL. Therefore, the rated voltage of the insulated gate bipolar transistors (IGBTs) can be half of the integrated transmitter, which can reduce the design difficulty and the switching losses of the transmitter.

Fig. 1.Principle of the cascaded topology.

Table II shows the relationship between maximum rated voltage and switching losses of different IGBTs. In Table II, the switching loss parameters of 600, 1200, and 3300 V IGBTs are from the IGBT modules marked FF200R06KE3, FF200R12KT4, and FF200R33KF2C, respectively. Eon, Eoff, and Err represent turn-on, turn-off, and diode reverse recovery energy losses per pulse, respectively. Qrate is the energy losses under the rated power. For 600, 1200, and 3300 V IGBTs, the rated powers are 300 V/100 A, 600 V/100 A, and 1800 V/100 A, respectively. For comparison, the energy losses are normalized to per kW, which is represented as Q/kW in Table II. Table II shows that the switching losses of 600 V IGBT are approximately half of those of 1200 V IGBT. Meanwhile, the switching losses of 3300 V IGBT are approximately five times those of 1200 V IGBT. Therefore, cascaded topology can reduce switching losses.

TABLE IISWITCHING LOSSES OF DIFFERENT IGBTS

The circuit of the single-stage transmitter involved in the cascaded technology is shown in Fig. 2. The AC–DC power supply converts the 380 V AC provided by the generator to 550 V DC, and the high-frequency transformer isolates the input and the output. The constant-current circuit makes the average current in the chopper inductance L 2 the set value by adjusting the chopper-circuit switch VT2 ; that is, it maintains the output current of the chopper circuit at the set value. The output full-bridge Z2 of the output circuit with another output full bridge in another single-stage transmitter comprises the cascaded topology shown in Fig. 1.

Fig. 2.Circuit diagram of a single-stage transmitter.

B. Single-Stage Transmitter Losses

Considering that the entire electrical source transmitter consists of two single-stage transmitters in series, only the thermal losses of the single-stage transmitter must be analyzed. 1) Switching Losses: When any IGBT of the four IGBTs involved in the full-bridge inverter U2 is represented as VT1, any diode of the high-frequency rectifier U4 is represented as D1, the chopper switch IGBT is represented as VT2, and the chopper diode is represented as D2. The voltage and current waveforms of IGBTs and diodes during the turn-on and turn-off are shown in Fig. 3. In Fig. 3, iVT1 and vVT1 represent the current and voltage of VT1 when VT1 turns on and turns off respectively. iD1 and vD1 represent the current and voltage of D1 during the reverse recovery respectively. iVT2 and vVT2 represent the current and voltage of VT2 when VT2 turns on and turns off respectively. iD2 and vD2 represent the current and voltage of D2 during the reverse recovery respectively.

Fig. 3.Switching loss analysis. (a) AC–DC power supply. (b) The constant-current circuit.

In Fig. 3(a), the overlapping area between iVT1 and vVT1 is small when VT1 turns on because the leakage inductance of the transformer suppresses the changing rate of the current. The slope of the diode current iD1 is relatively low when iD1 drops from the positive steady-state current IF to the maximum reverse value −IRM. The overlapping area between iD1 and vD1 is small. This situation is not in accordance with the test conditions for the turn-on loss of the IGBT in the specifications. Therefore, the turn-on loss of VT1 and the reverse recovery loss of D1 are lower than the turn-off energy loss Eon and the reverse recovery energy loss Err in the specification. When VT1 turns off, no other circulation loop exists, except VT1 for the current in the primary side of the transformer before the voltage vVT1 rises from 0 to bus voltage. Therefore, iVT1 remains the initial current value, and the overlapping area between iVT1 and vVT1 is large. The turn-off is in accordance with the test conditions in the specifications. Therefore, the turn-off energy loss Eoff can be used to calculate the turn-off loss.

In Fig. 3(b), the overlapping area between iVT2 and vVT2 is large when VT2 turns on because vVT2 is clamped to the input voltage before iVT2 rises from 0 to the steady-state value. The changing rate that the current iD2 in diode D2 decreases from the positive steady-state current IF to 0 is determined by the inherent rate of the IGBT. The changing rate of the IGBT inherent current is fast when it turns on. Therefore, the peak value IRM of the reverse recovery current of the diode D2 is large. The overlapping area between iD2 and vD2 is also large. When VT2 turns off, the current of chopper inductance L2 can only flow through VT2 before the voltage vVT2 increases from 0 to the input voltage. Therefore, the overlapping area between iVT2 and vVT2 is large. The turn-on and turn-off of VT2 and the reverse recovery of the diode D2 are in accordance with the specifications of IGBT test conditions. Eon, Eoff, and Err can thus be used to calculate VT2 and D2 losses.

For the IGBTs involved in the output full-bridge Z2 in the output circuit, the turn-off process is consistent with that of VT1. The turn-off energy loss Eoff can be used to calculate the turn-off loss. The turn-on process is also consistent with that of VT1. However, the load inductance is larger than the leakage inductance of the transformer. Turn-on loss is lower than that of VT1, which can thus be ignored.

Eq. (1) can be used to calculate the corresponding switching losses when the situation is equivalent to the test conditions in the specification. It is expressed as follows:

where x can be on, off, or rr, which respectively represents the turn-on of the IGBT, turn-off of the IGBT, or reverse recovery of the diode. Ex represents the energy losses with the testing voltage VREF and the testing current IREF in the specification. V and I represent the actual working voltage and current of the power device respectively. Px represents the power losses. fx represents the switching frequency.

For some diodes, the specification does not give Err . Eq. (2) can be used to calculate Err of the diode, that is,

where VRM is the reverse steady-state voltage of the diode. Qrr is the reverse recovery charge, which can be found in the specification.

2) Total Losses: The output of the AC–DC power supply is designed to be 550 V and 45 A, and the power is 25 kW. The 50 Hz rectifier U1 of the AC–DC power supply adopts VUO110-16NO7, the input contactor adopts CJX2-6511, and the full-bridge inverter U2 adopts the IGBT module FF200R12KT4, which operates in 18 kHz. The high-frequency rectifier U4 adopts a full-bridge rectifier, and the single-bridge leg consists of four diodes in series, which adopt the diode module MEE250-012DA. The full-bridge inverter U2 adopts the resistor–capacitor (RC) snubber circuit to eliminate resonance. Each IGBT parallels an RC snubber, with R of 100 Ω and C of 3.3 nF. The high-frequency transformer U3 adopts 3 EE110 cores in parallel, 7 turns in primary, and 10 turns in secondary. Each turn in primary and secondary uses double Liz wires, Φ 0.1 mm × 800, in parallel. The high-frequency rectifier U4 also adopts an RC snubber, with R of 50 Ω and C of 6.8 nF. The RC snubber parallels the secondary of the high-frequency transformer U3, and the output inductance L1 adopts four EE110 cores in parallel. The air gap is 12.8 mm, and the number of turns is 27. Each turn uses double Liz wires, Φ 0.1 mm × 800, in parallel. Table III shows the calculated value of the AC–DC power supply losses.

TABLE IIIAC–DC POWER SUPPLY LOSSES

In Table III, P represents the losses, the subscript U1 represents the 50 Hz rectifier, U2 represents the full-bridge inverter, U2R represents the snubber resistance of the full-bridge inverter U2, U3 represents the high-frequency transformer, U4 represents the high-frequency rectifier, U4R represents the RC snubber resistance of the high-frequency rectifier, L1 represents the output inductor, U6 represents the fan and its power supply, U7 represents the resistance used to balance the voltage of the series capacitor C1, U8 represents the input contactor, and U9 represents other stray losses. The losses of U1, U2, U2R, U4, and U4R are 1258 W. These devices are fixed on a heat sink. The total losses of the AC–DC power supply are 1688 W. The efficiency of the AC–DC power supply is 93.2%.

The constant-current circuit output is designed to be 500 V and 50 A, and the power is 25 kW. The working frequency of the chopper circuit is 18 kHz. The chopper switch VT2 and the chopper diode D2 adopt the internal integration devices of the IGBT module FF200R12KT4. The constant-current inductance L2 is in accordance with the output inductance L1 of the AC–DC power supply. The diode of the short-circuit protection circuit Z1 adopts MEK250-12DA, which has two parallel internal diodes. The inductance of the short-circuit protection circuit Z1 adopts nine EE55 cores in parallel, the air gap is 2.25 mm, and the number of turns is 7.5. Each turn uses double Liz wires, Φ 0.1 mm × 800, in parallel. The full-bridge Z2 adopts the IGBT module 2MBI200U4H-170. Each IGBT parallels an RC snubber, with R of 100 Ω and C of 3.3 nF. The working frequency is 1–10 kHz. Given that the load is inductive, the current generally cannot be more than 20 A when the transmitting frequency of the output full-bridge Z2 is higher than 1 kHz. Therefore, 2 kHz and 20 A are adopted to calculate the losses of Z1 and the output full-bridge Z2. Table IV shows the calculated losses of the constant-current circuit and the output circuit.

TABLE VILOSSES OF CONSTANT-CURRENT CIRCUIT AND OUTPUT CIRCUIT

In Table IV, P represents the losses, the subscript VT2 represents the chopper switch, D2 represents the chopper diode, L2 represents the inductance of the constant-current circuit, Z1D represents the diode of the short-circuit protection circuit, Z1L represents the short-circuit protection inductance, Z2 represents the output full bridge, Z2R represents the RC snubber of the full-bridge Z2, Z3 represents the resistance used to balance the voltage of series capacitor C2, Z4 represents the fan and its power supply, and Z5 represents other stray losses. The total losses of the constant-current circuit and the output circuit are 944 W. The losses of VT2, D2, Z1D, Z1L, Z2, and Z2R are 619 W. These devices are fixed on a heat sink. Therefore, the comprehensive efficiency of the constant-current circuit and the output circuit is 96.2%.

The preceding analysis indicates that the design losses of the single-stage transmitter are 2632 W, and the efficiency is 89.5%.

 

IV. AIR -COOLED DESIGN

A. Thermal Resistance Model

Fig. 4 shows the thermal resistance model. The length and width of the heat source are Ls and Ws respectively. The heat source is attached to the heat sink with the insulation pad. The length and width of the heat sinkare L and W respectively. The height of the fin and the thickness in the root are H and ta respectively. The fin spacing is s. Ai is the area of the outside of the baseplate, and Ao is the total surface area of the fins. The thickness of the baseplate is tb. The equivalent circuit of thermal resistance is shown at the bottom of Fig. 4. Ti, Tc Tb, Tf, and Ta represent the temperature of the heat source, the baseplate side of the insulation pad, the outer surface of the baseplate, the inside of the baseplate, and the air, respectively. Ric, Rcb, Rbf, Rfa, and Rba represent the thermal resistance of the insulation pad, the insulation pad to the outer surface of the baseplate, the baseplate, the fins to the air, and the outer surface of the baseplate to the air [17].

where tc and kc represent the thickness and the thermal conductivity of the insulation pad respectively.

where

h is the heat transfer coefficient of the heat sink to air. ηf is the efficiency of the heat sink, δm = 2mπ/L, λm = 2nπ/W, and kf is the thermal conductivity of the heat sink.

Fig. 4.Thermal resistance model.

B. Efficiency of Striped Fins

The shapes of common heat sinkfins, including rectangular, trapezoidal, and triangular fins, are shown in Fig. 5. Stripes are usually engraved onfins to increase the surface area of thermal dissipation in engineering applications. Refs. [11]-[18] provided some calculation methods for smooth fins. However, few reports have discussed the effect of striped fins. The efficiency of striped fins will be analyzed in the following section on the basis of the known analysis of smooth fin.

Fig. 5.Heat sink fins. (a) Straight fin. (b) Trapezoidal fin. (c) Triangular fin. (d) Straight fin with stripes. (e) Trapezoidal fin with stripes. (f) Triangular fin with stripes.

For the heat sink of thermal dissipation power above 100 W, air velocity Va in the fin channel is lower than 5 m/s, and fin spacing s is lower than 7 × 10−3 m in general. Air kinematic viscosity ν is approximately 17 × 10−6 m2 /sand around 40 ℃. Therefore, Reynolds number Re = Va ·s/ν < 2200, and the air flow in the fin channel is laminar flow [18]. Fig. 6 shows the laminar flow in the striped fin channel. When the air flows in the fin channel along y direction, given that the direction of the stripe groove coincides with the direction of the air flow, the striped fin cannot change the air flow direction. The air only flows in the xOy plane, as shown in Fig. 6(b). The air velocity is uniform at x direction in inlet. When air from the developing flow section enters the fully developed section, the air velocity is larger in the middle and lower in both sides. The nature of air flow in striped fins and smooth fins is the same.

Fig. 6.Laminar flow in striped fin channel.

The main difference between striped and smooth fins is the surface area of the fins. The stripe coefficient c1 in Eq. (8) can be used to express the difference, that is,

where Ao represents the total surface area of the fins, and nf represents the fin number of the heat sink.

Fig. 7 shows the 1D analysis model of the efficiency of striped fins. According to the thermal conduction differential equation,

where T is the temperature variable, and ρp is the intensity of the internal thermal source. Af (x) represents the cross-sectional area of the fin along the x direction. Af (x) = taL for straight fin, and Af (x) = taLx/H for triangular fin. The boundary conditions for straight fin are x = 0, T = Tf and x = H, dT/dx = 0. The boundary conditions for straight fin are x = 0, T ≠ ∞ and x = H, T = Tf. According to Fig. 7, the thermal power losses on dx are as follows:

where C is the perimeter of the fin, which approximately equals 2L; ta(x) is the function of the fin thickness along x direction; ta(x) = ta for straight fin; and ta(x) = ta(x)/H for triangular fin. According to Eq. (10),

Fig. 7.Analysis diagram of the efficiency of striped fins. (a) Straight fin analysis. (b) Triangular fin analysis.

The function of the fin temperature T along the x direction can be obtained by uniting Eq. (9), Eq. (11), and ta(x), and it will be denoted by t(x). According to t(x) and the expression of the Fourier’s law shown in Eq. (12), the efficiency expressions of the rectangular striped fin and the triangular striped fin can be obtained, as shown respectively in Eqs. (12) and (13).

In Eqs. (13) and (14), P represents the actual thermal dissipation of the fin, and Po represents the thermal dissipation when the fin keeps the temperature in the root from the root to the terminal. I1 and I0 represent the 1-order and 0-order first-kind-modified Bessel functions respectively.

According to Eqs. (13) and (14), the efficiency curves of the striped fins shown in Fig. 8 can be obtained. Therefore, the efficiencies of the striped rectangular striped fin and the triangular striped fin can be obtained by evaluating the efficiency curves shown in Fig. 8. Calculating accurate results for the striped trapezoidal fin is difficult. The efficiency can be replaced by the average value of the triangular striped fin efficiency and the rectangular striped fin efficiency when AV = (ta + tm)H/2 is used. tm represents the thickness of the terminal of the trapezoidal fin. When c1 = 1, Eq. (13), Eq. (14), and Fig. 8 can also be used to calculate the efficiencies of smooth fins.

Fig. 8.Efficiency curves of striped fins.

C. Heat Transfer Coefficient

Heat transfer coefficient h is the key parameter in calculating Eqs. (4), (6), and (7), as well as the function of the Nusselt umber, Nu. Unless otherwise stated, h = Nuka/s is considered in the following sections, and ka is the air thermal conductivity. This section focuses on the calculation of Nu.

Air flows in the heat sink. Air flow can be divided into developing flow, transition, and fully developed sections. Except for special specifications in the following sections, Re = Vas/v, where Va is the air velocity, and v is the kinematic viscosity. Before analysis, x+ = Re·s/L and x* = Re·Pr·s/L are defined, where Pr is the Prandtl number, and Pr is approximately equal to 0.7 for air.

1) Developing Flow Section: For the developing flow section, Refs. [11] and [12] adopted the expression shown in Eq. (15), Ref. [13] adopted the expression shown in Eq. (16), Ref. [18] adopted the expression shown in Eq. (17), and Refs. [14] and [15] adopted the expressions shown in Eq. (18).

When only one channel of fins is analyzed, Eq. (18a) can be simplified as follows:

In Eqs. (15)–(18), d is the equivalent hydraulic diameter of the conduit, NTU is the number of transfer units, and Aa is the total surface area of the heat sink. ηa is the viscosity of the air under the inlet and the outlet average air temperatures, and ηh is the viscosity of air under the fin surface temperature. This study considers that ηa = ηh. ρa is the air density. Vamax is the maximum velocity of the air flow between fins, and is approximately twice the average velocity of Va.

2) Fully Developed Section: For the fully developed section, Ref. [11] adopted the expression shown in Eq. (19), and Ref. [13] adopted the expression shown in Eq. (20).

3) Composite Mode: To facilitate the calculation, Refs. [11] and [13] proposed Nu expressions shown as Eqs. (21) and (22) respectively; such expressions can be applied in the developing flow, transition, and fully developed sections.

4) Engineering Expression of Nu: Nu expressions are complex in Eqs. (15)-(22). Therefore, the engineering expression shown in Eq. (23) is introduced by mathematically fitting Nu expressions proposed in Refs. [11]-[18]. This expression is suitable for the process from the developing flow section to the fully developed section, and it is concise and easy to calculate.

The heat transfer coefficient h can be obtained by using Eq. (23), x+ = Re·s/L, and h = Nuka/s.

To compare the error of Eq. (23), the curves of Nu shown in Fig. 9 are drawn according to Eqs. (15)-(18) and Eqs. (21)-(23) of the developing flow section; the curves of Nu shown in Fig. 10 are drawn according to Eqs. (19)-(23) of the fully developed section. For comparison, Pr is 0.7 when drawing Figs. 9 and 10, and d = s is approximately taken. The curve of Eq. (18) represents only one channel of fins. In Fig. 9, given that the upper limit range of the curve of Eq. (18) is Re·s/L = 125, no reference value is used when crossing the border [14], [15]. The expressions of the developing flow section shown in Eqs. (15)-(18) are valid only when it takes approximately Re·s/L > 20. The expressions of the fully developed section shown in Eqs. (19) and (20) are valid only when they utilize approximately Re·s/L < 3.

Fig. 9.Nu of the developing flow section.

Fig. 10.Nu of the fully developed section.

As shown in Figs. 9 and 10, the curves of Eq. (23) are located among the curves of Eqs. (15)-(18), (21), and (22) in the developing flow section, that is, when Re·s/L > 20. The curve of Eq. (23) is slightly lower than those of Eqs. (19)-(22) in the fully developed section, that is, when Re·s/L < 3. The curve of Eq. (23) is insignificantly lower than those of Eqs. (21) and (22) in the transition section, that is, when 3 < Re·s/L < 20.

According to the preceding analysis, the complexity of Eq. (23) is significantly lower than that of existing expressions. Meanwhile, the accuracy is only slightly lower than that of existing expressions.

D. Pressure Drop of Heat Sink

The frictional air pressure drop in a heat sink can directly influence the volumetric flow rate of the fan, which is related to the effect of the air-cooled design. This section investigates the frictional air pressure drop of heat sinks.

Heat sink causes the total pressure drop, as follows:

where Δp1 represents the sum of the inlet and outlet pressure drop, and Δpc represents the core pressure drop. Δp1 is

Δpc is

where f app is the apparent frictionfactor. The calculation of 2f app Re in Refs. [14] and [15] adopted Eq. (27), and Ref. [16] adopted Eq. (28). For high-power cooling fins, s/H < 1/6 generally, and Eqs. (27) and (28) are similar. Therefore, Δpc can be calculated with either of the following equations:

where f is the friction factor.

E. Fan Characteristics

For fans, the relationship between volumetric flow rate and static pressure is shown in Fig. 11 [27]. This figure is generally given in the fan specification. When the total pressure drop of the heat sink is Δpa, the corresponding actual volumetric flow rate is shown as Q in Fig. 11. Qm and pm in Fig. 11 represent the rated volumetric flow rate and the rated static pressure of the fan respectively.

Fig. 11.Relationship between volumetric flow rate and static pressure.

According to Ref. [28], the following equation is reasonable to some extent:

where n, p, and N represent the rotational speed, static pressure, and shaft power of the fan, respectively. The subscript m represents the rated value. Therefore, Eq. (29) and the given parameters in the fan specification can be used to estimate the fan performance.

The relationship between the actual volumetric flow rate Q and the thermal dissipation power P of the fan is shown in Eq. (30). This equation can be used to estimate the volumetric flow rate of the fan. The unit of Q in Eq. (30) is m3/min, and the other variables use System International units.

where ca is the specific heat capacity of air with a value of 1000 J/(kg·℃). ρa = 1.20 kg/m3 at 20 ℃. ΔTa is the temperature difference between inlet and outlet air. The outlet temperature should not be higher than 71 ℃.

When the power is relatively high, a single fan generally cannot provide thermal dissipation. Several fans running together are needed. Fans in series mainly change the air pressure, and fans in parallel mainly change the volumetric flow rate. Therefore, if the total pressure drop of the heat sink is relatively high, the series operation will be efficient. If the volumetric flow rate is insufficient, the parallel operation will be efficient.

For fans with the same size, their volumetric flow rate and air pressure differ [20], and the effect of thermal dissipation has a great difference. Therefore, fans with medium performance should be adopted for an air-cooled design. Replacing medium-performance fans with high-performance fans that have the same size is simple and efficient if the air-cooled design cannot meet the requirements.

F. Air-cooled Design Process

 

V. MEASURED RESULTS

A. Experimental Conditions

The transmitter runs 8 h a day at 40 kW and operates normally after working for three days. Fig. 12 shows a photo of the working transmitter at 40 kW. Fig. 13 shows the output voltage and current waveforms after thermal stabilization. Fig. 13 shows that the output voltage is more than 1000 V, the current is more than 40 A, and the total output power is more than 40 kW. During the test, waveforms are obtained by using a TDS2012C oscilloscope. The current waveform in Fig. 13 is obtained by using a current probe A622, and the current waveforms in Figs. 14 and 15 are obtained by measuring the voltage on 0.1 Ω resistance. Current values are measured with the oscilloscope. Voltage is obtained by using a UT39A multimeter. Temperature is obtained by using an ST380 infrared thermometer, and the resolution of the infrared thermometer is 1 ℃.

Fig. 12.Photo of a 40 kW transmitter.

Fig. 13.Output voltage and current waveforms.

Fig. 14.Transition waveforms in AC–DC power supply. (a) VT1 turn-on waveforms. (b) VT1 turn-off waveforms. (c) D1 reverse recovery waveforms.

Fig. 15.Transition waveforms in constant-current circuit. (a) VT2 turn-on waveforms. (b) VT2 turn-off waveforms. (c) D2 reverse recovery waveforms.

B. Transmitter Losses

Fig. 14 shows the turn-on and turn-off waveforms of VT1, and the reverse recovery waveforms of D1 in AC–DC power supply. The overlapping area of iVT1 and vVT1 in Fig. 14(a) is small, that of iVT1 and vVT1 in Fig. 14(b) is large, and that of iD1 and vD1 in Fig. 14(c) is small, which are the same as in Fig. 3(a). The overlapping area in Figs. 14(a) and 14(c) are small because the leakage inductance of the transformer in AC–DC power supply limits the change rate of the current. The overlapping area in Fig. 14(b) is large because the current in the leakage inductance of the transformer can only flow through VT1 before vVT1 up to the bus voltage.

Fig. 15 shows the turn-on and turn-off waveforms of VT2, and the reverse recovery waveforms of D2 in constant-current circuit. The overlapping areas in Fig. 15(a)-15(c) are large, the same as in Fig. 3(b). The overlapping area in Fig. 15(a) is large because the voltage is clamped to the bus voltage before the current up to the inductance current. The overlapping area in Fig. 15(b) is large because the current is clamped to the inductance current before the voltage up to the bus voltage. Given that the change rate of iD2 from 40 A to 0 A is fast, which is determined by the change rate of iVT2 from 0 A to 40 A, the overlapping area in Fig. 15(c) is large.

The turn-off loss of VT1, the turn-on loss and turn-off loss of VT2, and the reverse recovery loss of D2 can be calculated using Eq. (1). However, the turn-on loss of VT1 and the reverse recovery loss of D1 are only small parts of Eq. (1) and cannot be calculated using the equation. Therefore, when switching losses are calculated, the circuit topologies should be considered.

The bus voltage and current value after 50 Hz rectifier U1 are measured to approximately calculate the input power of the AC–DC power supply, which is represented as P1 in Table V. The output voltage and current of the AC–DC power supply are measured to approximately calculate the output power of the AC–DC power supply, which is represented as P2. P2 also represents the input power of the constant-current circuit. The bus voltage and current values between short-circuit protection circuit Z1 and output full-bridge Z2 are measured to approximately calculate the output power of the constant-current circuit, which is represented as P3. The test results are shown in Tables V and VI. To compare the error of loss calculation, the calculated value of the efficiency of the AC–DC power supply, represented as η1, in Table VI excludes the losses of 50 Hz rectifier bridge U1 and the losses of the input contactor U8. The calculated value of the constant-current circuit efficiency, represented as η2, excludes the losses of the output full-bridge Z2 and its RC snubber Z2R from Table IV. The efficiency of the transmitter, represented as η, is determined by the product of η1 and η2.

TABLE VMEASURED LOSS DATA

TABLE VIEFFICIENCY OF CALCULATED AND MEASURED VALUES

The results in Table VI show that the deviation between calculated value of the transmitter and measured value is 1.2%; the calculation deviation of losses is 1.2/(100 − 90.2) × 100% = 12.2%. The error is reasonable. The proposed thermal loss calculation method in this study is effective and feasible.

C. Temperature Rise of Heat Sink

According to this analysis, a heat sink, SRX-YTJ, with striped rectangular fins is adopted as the heat sink of the AC–DC power supply with parameters of s = 4.8 mm, ta = 2.5 mm, tb = 13 mm, c1 = 2, H = 62 mm, L = 350 mm, W = 360 mm, nf = 49, and kf = 160 W/(m·℃)．Three parallel fans are adopted for thermal dissipation. The fan model is FFC1212DE, which has the parameters Qm = 5.38 m3/min, pm = 170 Pa, and Nm = 29 W. The temperature of the five test points are measured on the heat sink of the AC–DC power supply shown in Fig. 16, and the results are shown in Table VII. The representation of the devices U1, U2, U2R, U4, U4R, and U6 is indicated in Table III. The ambient temperature is 21 ℃. Table VII shows that the average temperature rise is 22 ℃, the temperatures of the five points are relatively even, and the design of the thermal dissipation is reasonable.

Fig. 16.Distribution of measured points.

Table VIITEMPERATURE OF THE HEAT SINK

 

VI. CONCLUSIONS