10.8. Mount Package Types of Semiconductor Devices and Types of Heat Sinks

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10.8. Mount Package Types of Semiconductor Devices and Types of Heat Sinks

Fig. 10.64 shows different types of mount packages of power semiconductor devices. As can be seen from Fig. 10.64(a), the mount packages TO-220, TO-268, and TO-264 are for low power semiconductor devices. The mount packages shown in Fig. 10.64(b) and (c) are for higher power applications.

The high power semiconductor switches analyzed in the previous sections, due to the high density of current that flows through their junctions, they exhibit power losses and, consequently, generate internally high temperature levels. Since the high temperature levels are affecting negatively the operation and the life time of the semiconductor devices, there is a need to design a proper heat sink system for temperature dissipation to keep the device under manufacturer's temperature specifications. The aim of the heat sinks is to increase the temperature conductivity between the device module case and the ambient temperature in order for the heat dissipation to become more effective. Fig. 10.65 shows different types of heat sinks that can be used to cool and keep the semiconductor devices to manufacturer's temperature levels. The heat sink is considered to be a natural cooling system.

Figure 10.64 Different types of mount packages of power semiconductor switches mount packages.
(a) Single semiconductor switches with different types of mount packages; (b) two-semiconductor switches in one mount package to implement one phase leg; (c) six-semiconductor switches in one mount package to implement three-phase legs.

Fig. 10.66 shows two different heat sinks with natural air cooling for power semiconductor devices.

Besides the natural heat sink air cooling systems there are the so-called heat forced air or liquid cooling techniques which are using air fans or liquid pumps for temperature dissipation. These techniques are applied in applications where the power electronics converter needs high power density. Fig. 10.67 shows two different forced cooling techniques.

Figure 10.65 Different types of heat sinks with different thermal resistance.

Figure 10.66 Heat sinks with natural air cooling.

Figure 10.67 Heat sinks with forced cooling.
(a) Forced air cooling with fans; (b) forced cooling with liquid.

10.8.1. Heat Sink Computations

When the power semiconductor devices are operating, they exhibit power switching and conduction losses which are converted to heat. If the junction temperature of the power semiconductor exceeds the manufacturer's specifications the device will be destroyed. Therefore, this high value temperature has to be reduced. The usual way to increase the transfer of excess heat from the crystal of a semiconductor device to the ambient air is to attach the semiconductor device to a heat sink and many times an air fan is also include for faster air flow. The heat sink is made of metal and most of the times from aluminum and is constructed with fins to allow maximum contact with ambient air. The heat flow power is equal to the power losses. The heat flows from the highest temperature region, which is the junction of the crystal, to the lowest, which is the ambient air. The heat flows from the junction to the ambient air through a thermal path consisted of the case of the device, a thermal grease or insulator and a heat sink. Fig. 10.68 shows a heat sink which is used for natural air cooling of four power semiconductor switches.

Figure 10.68 Heat sink with four power semiconductor switches.

Fig. 10.69(a) shows the direction of the heat flow and the steady-state thermal resistances between different points. Also, Fig. 10.69(b) shows the steady-state thermal resistance equivalent model where the thermal resistance of the thermal grease and the thermal resistance are added to a single thermal resistance case-sink. The power losses of a semiconductor device are equal to the heat transfer power.

To understand the thermal analysis of semiconductor switch mounted on a heat sink Table 10.4 presents how the electrical parameters are associated with the thermal parameters.

To estimate and select the right heat sink for the right application, it is necessary to understand the heat transfer power between two points which is given by:

$P_{heattrasf} = \frac{Δ T}{R_{th}}$ $P_{heattrasf} = \frac{Δ T}{R_{th}}$

(10.15)

where Rth = thermal resistance between the two points in °C/W or °K/W (since Rth is based on a temperature difference and not on the absolute temperature, both units are perfectly equivalent (for example, 5°C/W = 5°K/W)); ΔT = Temperature difference (rise).

In case of semiconductor switches that are mounted on a heat sink, the total thermal resistance consists of the following three main components:

1) Thermal resistance, R_jc: It is the thermal resistance between the silicon junction and the case of the semiconductor device. This resistance is given in the manufacturer's specifications.

2) Thermal resistance, R_cs: It is the thermal resistance between the case of the device and the heat sink. This resistance also depends on the thermal grease or insulator that there is between the case and the heat sink. If the thermal grease is applied correctly between the case and the heat sink, then its thermal resistance value is negligible and can be ignored. The thermal resistance of an insulator is provided by the manufacturer.

Figure 10.69 Heat sink thermal resistances for single semiconductor device.
(a) Heat flow through a heat sink and thermal resistances between different points; (b) steady-state thermal resistances equivalent model.

Table 10.4

Association between electrical and thermal parameters

Electrical parameters

R_el = electrical resistance (Ω)

V = potential difference (V)

I = electrical current

V = I × R_el

Thermal parameters

R_th = thermal resistance (°C/W or °K/W)

ΔΤ = temperature difference (°C or °K)

Ρ_losses = power losses of the semiconductor device (W)

ΔT = Ρ_losses × R_th (°C)

3) Thermal resistance, R_sa: It is the thermal resistance between the heat sink and the ambient temperature. This resistance depends on the characteristics of the heat sink (i.e., material, color, shape, and dimensions).

Therefore, the total steady-state (i.e., constant power losses) thermal resistance between the junction and the ambient air is given by:

$R_{ja} = R_{jc} + R_{cs} + R_{sa} = \frac{T_{jmax} - T_{a}}{P_{losses (av)}}$ $R_{ja} = R_{jc} + R_{cs} + R_{sa} = \frac{T_{jmax} - T_{a}}{P_{losses (av)}}$

(10.16)

where T_jmax = maximum junction temperature; T_a = ambient air temperature; P_losses(av) = maximum permissible power average losses of the semiconductor device.

Also, from Fig. 10.69(b), the following equations can be obtained:

$T_{j} = T_{a} + T_{c} + T_{s} = T_{a} + R_{jc} P_{losses (av)} + P_{losses (av)} R_{cs} + P_{losses (av)} R_{sa} = T_{a} + P_{losses (av)} (R_{jc} + R_{cs} + R_{sa}) = T_{a} + P_{losses (av)} \times R_{ja}$ $T_{j} = T_{a} + T_{c} + T_{s} = T_{a} + R_{jc} P_{losses (av)} + P_{losses (av)} R_{cs} + P_{losses (av)} R_{sa} = T_{a} + P_{losses (av)} (R_{jc} + R_{cs} + R_{sa}) = T_{a} + P_{losses (av)} \times R_{ja}$

(10.17)

$Δ T_{ja} = T_{j} - T_{a} = P_{losses (av)} \times R_{ja}$ $Δ T_{ja} = T_{j} - T_{a} = P_{losses (av)} \times R_{ja}$

(10.18)

$T_{c} = T_{a} + P_{losses (av)} \times (R_{cs} + R_{sa})$ $T_{c} = T_{a} + P_{losses (av)} \times (R_{cs} + R_{sa})$

(10.19)

$T_{s} = T_{a} + P_{losses (av)} \times R_{sa}$ $T_{s} = T_{a} + P_{losses (av)} \times R_{sa}$

(10.20)

$R_{sa} = heat sink thermal resistance = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{js} ° C / W$ $R_{sa} = heat sink thermal resistance = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{js} ° C / W$

(10.21)

Example 10.2

Calculate the maximum average power that an IGBT can dissipate (a) without a heat sink and (b) with a heat sink. The following data are given: R_jc = 1.8°C/W, R_cs = 0.8°C/W, R_sa = 4°C/W, R_ja = 50°C/W, T_a = 35°C, and T_jmax = 130°C.

Solution

$P_{losses (av) without - sink} = \frac{T_{jmax} - T_{a}}{R_{ja}} = \frac{130 - 35}{50} = 1.9 W$ $P_{losses (av) without - sink} = \frac{T_{jmax} - T_{a}}{R_{ja}} = \frac{130 - 35}{50} = 1.9 W$

$P_{losses (av) with - sink} = \frac{T_{jmax} - T_{a}}{R_{jc} + R_{cs} + R_{sa}} = \frac{130 - 35}{1.8 + 0.8 + 4} = 14.4 W$ $P_{losses (av) with - sink} = \frac{T_{jmax} - T_{a}}{R_{jc} + R_{cs} + R_{sa}} = \frac{130 - 35}{1.8 + 0.8 + 4} = 14.4 W$

Example 10.3

An IGBT is operating with average conduction and switching losses of 10 W. Find its operating temperatures, given the following data: R_jc = 3.8°C/W, R_cs = 1°C/W, R_sa = 6°C/W, and T_a = 35°C.

Solution

$\begin{matrix} T_{j} = T_{a} + P_{losses (av)} (R_{jc} + R_{cs} + R_{sa}) = 35 + 10 (3.8 + 1 + 6) = 143 ° C \\ T_{c} = T_{a} + P_{losses (av)} (R_{cs} + R_{sa}) = 35 + 10 (1 + 6) = 105 ° C \\ T_{s} = T_{a} + P_{losses (av)} (R_{sa}) = 35 + 10 (6) = 95 ° C \end{matrix}$ $\begin{matrix} T_{j} = T_{a} + P_{losses (av)} (R_{jc} + R_{cs} + R_{sa}) = 35 + 10 (3.8 + 1 + 6) = 143 ° C \\ T_{c} = T_{a} + P_{losses (av)} (R_{cs} + R_{sa}) = 35 + 10 (1 + 6) = 105 ° C \\ T_{s} = T_{a} + P_{losses (av)} (R_{sa}) = 35 + 10 (6) = 95 ° C \end{matrix}$

Example 10.4

An IGBT is operating with average conduction and switching losses of 25 W for junction temperature of 150°C. Find the thermal resistance of the heat sink that is needed to keep the maximum junction temperature to 150°C when the ambient temperature is 50°C. The following data are given: R_jc = 1.5°C/W (given by the manufacturer) and R_cs = 0.6°C/W (insulator and thermal grease).

Solution

Using the above given specifications the following steady-state thermal resistances equivalent model is obtained:

$R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{150 - 50}{25} - 1.5 - 0.6 = 1.9 ° C / W$ $R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{150 - 50}{25} - 1.5 - 0.6 = 1.9 ° C / W$

Therefore to keep the device junction to 150°C the device should be mounted on a heat sink with thermal resistance of 1.9°C/W. From the heat sinks of Fig. 10.70 and according to their specifications, which are given in Table 10.5, the best choice is the heat sink no. 1.

Table 10.5

Specifications of the aluminum type heat sinks shown in Figure 10.71

Heat sink no	R_sa (^oC/W)	Height mm	Width mm	Length mm
1	1.9	27	65	300
2	0.95	40	96	300
3	0.8	50	125	300
4	0.71	50	140	300
5	1.05	25	100	300
6	0.9	40	100	300
7	0.6	50	125	300
8	0.45	73	145	300

Example 10.5

A BJT has an average power loss of 25 W at a junction temperature of 120°C. The transistor manufacturer specifies a thermal resistance R_jc = 0.9°C/W. A 75-μm thick mica insulator is used with thermal grease, and the thermal resistance of the combination is 0.5°C/W. Find the thermal resistance of the heat sink that is needed to keep the maximum junction temperature to 120°C when the worst case ambient temperature is 60°C.

Solution

$R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{120 - 60}{25} - 0.9 - 0.5 = 1.0 ° C / W$ $R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - R_{jc} - R_{cs} = \frac{120 - 60}{25} - 0.9 - 0.5 = 1.0 ° C / W$

Therefore, from the heat sinks of Fig. 10.70 the best choice is the heat sink no. 5.

When multiple power semiconductor devices are mounted on a common heat sink, their thermal resistance model can be one of the models shown in Figs. 10.71 and 10.72. Fig. 10.71 presents the thermal resistance steady-state model when the power semiconductor devices have individual mounting case and Fig. 10.72 presents the thermal resistance steady-state model when the power semiconductor devices have common mounting case.

Figure 10.71 Thermal resistance model for multiple semiconductors mounted on the same heat sink with individual mounting cases.

Figure 10.72 Thermal resistance models for multiple semiconductors mounted on the same heat sink with common mounting case.

According to Fig. 10.71, when power semiconductors have individual mounting case and are mounted on a heat sink then the total thermal resistance of the heat sink is given by:

$R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - \frac{R_{jc} + R_{cs}}{n} ° C / W$ $R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - \frac{R_{jc} + R_{cs}}{n} ° C / W$

(10.22)

Also, according to Fig. 10.72, when power semiconductors have common case the total thermal resistance of the heat sink is given by:

$R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - \frac{R_{jc}}{n} - R_{cs} ° C / W$ $R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (av)}} - \frac{R_{jc}}{n} - R_{cs} ° C / W$

(10.23)

where n is the number of semiconductor devices.

Example 10.6

Find the required heat sink to mount eight power semiconductor devices with individual cases. The following specifications are given:

T_j = power semiconductors junction temperature (by manufacturer) ≤ 150°C

T_a = ambient temperature = 50°C

R_jc = power semiconductors thermal resistance between junction and case (by manufacturer) = 1.5°C/W

R_cs = thermal grease and isolator thermal resistance (by manufacturer) = 0.6°C/W

P_losses(av) = average value of a single semiconductor power losses = 24.23 W

Solution

Using the thermal model of Fig. 10.71, the following thermal resistance model is obtained.

Using Eq. (10.22), the following results are obtained:

$R_{sa} = \frac{150 - 50}{194.4} - \frac{1.5 + 0.6}{8} = 0.5144 - 0.26 = 0.255 ° C / W$ $R_{sa} = \frac{150 - 50}{194.4} - \frac{1.5 + 0.6}{8} = 0.5144 - 0.26 = 0.255 ° C / W$

$T_{s} = T_{a} + P_{losses (av)} (R_{sa}) = 50 + 194.4 (0.255) = 99.57 ° C$ $T_{s} = T_{a} + P_{losses (av)} (R_{sa}) = 50 + 194.4 (0.255) = 99.57 ° C$

Therefore to maintain the junction temperature of the semiconductor devices T_j ≤ 150°C they have to be mounted on a heat sink that will have thermal resistance of R_sa ≤ 0.255°C/W. Fig. 10.73 presents a proposed heat sink and the respective air flow speed characteristics.

Company: Birmingham Aluminum

Series: 1500 series

Length: 300 mm

Thermal resistance: 0.255°C/W

The steady-state thermal resistance is not enough for finding peak junction temperatures for pulsed current applications. Using the peak power value, results in overestimating the actual junction temperature, while using the average power value underestimates the peak junction temperature at the end of the power pulse. The reason for the discrepancy lies in the thermal capacity of the semiconductor and its housing (i.e., its ability to store heat and to cool down before the next current pulse). Fig. 10.74 presents the modified thermal equivalent circuit when transient thermal analysis is needed.

Figure 10.73 Proposed heat sink that can be used in this example.
(a) Type and dimensions of the heat sink; (b) heating–length characteristics of the heat sink.

Figure 10.74 Transient thermal resistance equivalent model.

The normally distributed thermal capacitances have been lumped into single capacitors labeled C_j, C_c, and C_s. This simplification assumes current is evenly distributed across the silicon crystal and that the only significant power losses occur in the junction. When a step of heating power, P_losses(max), is introduced at the junction of the semiconductor, T_j will rise exponentially to some steady-state value dependent upon the response of the thermal network. When in a time instant the current pulse becomes zero and, consequently, the power losses become zero, T_j will decrease exponentially until the next current pulse. The transient thermal impedance between junction and case at time t is given by:

$Z_{jc} (t) = \frac{Δ T_{jc} (t)}{P_{losses (\max)}}$ $Z_{jc} (t) = \frac{Δ T_{jc} (t)}{P_{losses (\max)}}$

(10.24)

where ΔT_jc(t) = difference of temperature between junction and case at a given time t; P_losses(max) = amplitude of the step power losses.

Figure 10.75 Variation of junction and case temperature when the pulse time is small compared to thermal time constant of the heat sink.

The junction to case temperature is always assumed to be constant under transient operation. This is valid in practice when the power pulse width is less than about 1 s. As can be seen from Fig. 10.75, T_jc does not change significantly under this condition. This is because heat sinks have a high thermal capacity and, consequently, a high thermal time constant. Thus, Eq. (10.24) is valid for transient operation when the pulse time is less than 1 s. Therefore, using Eq. (10.24) the value of T_j can be calculated as follows:

$T_{j} = P_{losses (\max)} Z_{jc} + T_{c}$ $T_{j} = P_{losses (\max)} Z_{jc} + T_{c}$

(10.25)

If the power pulse width t_p exceeds 1 s (see Fig. 10.76), the semiconductor device is temporarily in thermal equilibrium since such a pulse width is significantly greater than the thermal time constant of the semiconductor. Therefore, in this case Eq. (10.24) becomes:

$T_{j} = P_{losses (\max)} R_{jc} + T_{c}$ $T_{j} = P_{losses (\max)} R_{jc} + T_{c}$

(10.26)

The following expression describes how to select the heat sink that keeps the junction at/or below a given temperature:

$R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (\max)}} - Z_{jc} - R_{cs}$ $R_{sa} = \frac{T_{j} - T_{a}}{P_{losses (\max)}} - Z_{jc} - R_{cs}$

(10.27)

Figure 10.76 Variation of junction and case temperature when the pulse time is not small compared to thermal time constant of the heat sink.

Example 10.7

Calculate the junction temperature of the IGBT type SGP20N60 under the following operating conditions:

D = duty cycle = 0.5

f_s = switching frequency = 75 kHz

T_a = ambient temperature = 40°C

T_c = case temperature = 80°C

P_losses(max) = 45 W

R_cs = thermal resistance between case and heat sink = 0.45

\frac{° K}{W}

$\frac{° K}{W}$

\frac{° C}{W} is equivalent to \frac{° K}{W}

$\frac{° C}{W} is equivalent to \frac{° K}{W}$

Fig. 10.77 presents the thermal impedance of the IGBT type SGP20N60 for different duty cycles given by the manufacturer.

Solution

Since the operating duty cycle is 0.5 and the switching frequency 75 kHz, then the pulse width of the power losses is given by:

$D = \frac{t_{on}}{T_{s}} = \frac{t_{p}}{\frac{1}{75 \times 10^{3}}} = 0.5 or t_{p} = 6.67 μ s$ $D = \frac{t_{on}}{T_{s}} = \frac{t_{p}}{\frac{1}{75 \times 10^{3}}} = 0.5 or t_{p} = 6.67 μ s$

Figure 10.77 Thermal impedance of IGBT type SGP20N60 for different duty cycles (provided by the manufacturer).

From Fig. 10.77, for t_p = 6.67 μs and D = 0.5 the thermal impedance is

Z_{jc} = 0.35 \frac{° K}{W}

$Z_{jc} = 0.35 \frac{° K}{W}$

. At this point it should be mentioned that

\frac{° K}{W} is equivalent to \frac{° C}{W}

$\frac{° K}{W} is equivalent to \frac{° C}{W}$

. Therefore using Eqs. (10.26) and (10.27), the following results are obtained:

$T_{j} = (45 W) (0.35 \frac{° C}{W}) + 80 ° C = 95.75 ° C$ $T_{j} = (45 W) (0.35 \frac{° C}{W}) + 80 ° C = 95.75 ° C$

$R_{sa} = \frac{95.75 ° C - 40 ° C}{45 W} - 0.35 \frac{° C}{W} - 0.45 \frac{° C}{W} = 0.44 \frac{° C}{W}$ $R_{sa} = \frac{95.75 ° C - 40 ° C}{45 W} - 0.35 \frac{° C}{W} - 0.45 \frac{° C}{W} = 0.44 \frac{° C}{W}$

Therefore, from the heat sinks of Fig. 10.70 according to their specifications, which are given in Table 10.5, the best choice is a heat sink with higher dimensions of no. 8.

Example 10.8

For the step-down converter shown below, the following parameters are given:

IGBT

Conduction voltage = 2 V

Conduction resistance = 3 mΩ

W_on = 20 mJ (measured at 400 A with 600 V dc supply)

W_off = 40 mJ (measured at 400 A with 600 V dc supply)

Thermal resistance junction case 0.05°K/W

Diode

Conduction voltage = 2.5 V

Conduction resistance = 2 mΩ

Thermal resistance junction case = 0.1°K/W

a) Calculate the power dissipated in the IGBT and diode.

b) Estimate the junction temperature of the IGBT and diode for a case temperature of 60°C.

Solution

$1 J = 1 W s$ $1 J = 1 W s$

$D = \frac{{\bar{V}}_{o}}{V_{in}} = \frac{198}{600} = 0.33$ $D = \frac{{\bar{V}}_{o}}{V_{in}} = \frac{198}{600} = 0.33$

Assuming that the semiconductor switch operates with a duty cycle D = 0.33 and switching frequency 13.266 kHz, then the power losses of the IGBT are:

$P_{turn - on} = W_{turn - on} \times f_{s} = 20 mJ \times 13266 Hz = 265.32 W$ $P_{turn - on} = W_{turn - on} \times f_{s} = 20 mJ \times 13266 Hz = 265.32 W$

$P_{turn - off} = W_{turn - off} \times f_{s} = 40 mJ \times 13266 Hz = 530.64 W$ $P_{turn - off} = W_{turn - off} \times f_{s} = 40 mJ \times 13266 Hz = 530.64 W$

$P_{conduction} = W_{conduction} \times f_{s} = I_{s} \times V_{conduction} \times t_{on} \times f_{s} = I_{s} \times V_{conduction} \times t_{on} \times \frac{1}{T_{s}} = I_{s} \times V_{conduction} \times D = 400 A \times 2 V \times 0.33 = 264 W$ $P_{conduction} = W_{conduction} \times f_{s} = I_{s} \times V_{conduction} \times t_{on} \times f_{s} = I_{s} \times V_{conduction} \times t_{on} \times \frac{1}{T_{s}} = I_{s} \times V_{conduction} \times D = 400 A \times 2 V \times 0.33 = 264 W$

Therefore, the total power losses of the IGBT are:

$P_{total (IGBT)} = P_{turn - on transition} + P_{turn - off transition} + P_{on - state} = 265.32 W + 530.64 W + 264 W = 1060 W$ $P_{total (IGBT)} = P_{turn - on transition} + P_{turn - off transition} + P_{on - state} = 265.32 W + 530.64 W + 264 W = 1060 W$

Moreover, the power losses of the diode are calculated as follows:

$V_{F} = Total forward voltage drop across the diode = {diode}^{,} s resistance voltage drop + voltage drop across drift region = V_{R} + V_{J} = 2 \times 10^{- 3} \times 400 + 2.5 V = 3.3 V$ $V_{F} = Total forward voltage drop across the diode = {diode}^{,} s resistance voltage drop + voltage drop across drift region = V_{R} + V_{J} = 2 \times 10^{- 3} \times 400 + 2.5 V = 3.3 V$

The conduction losses of the diode are given by:

$P_{conduction (diode)} = I_{D} \times V_{F} \times D_{D} = I_{D} \times V_{F} \times (1 - D) = 200 \times 3.3 (1 - 0.33) = 224.4 W$ $P_{conduction (diode)} = I_{D} \times V_{F} \times D_{D} = I_{D} \times V_{F} \times (1 - D) = 200 \times 3.3 (1 - 0.33) = 224.4 W$

where D_D = diode duty cycle.

Also, it is known that:

$P_{D} = dissipated semiconductor losses = \frac{maximum junction temperature - case temperature}{thermal resistance between junction and case} = \frac{T_{j (\max)} - T_{c}}{R_{jc}}$ $P_{D} = dissipated semiconductor losses = \frac{maximum junction temperature - case temperature}{thermal resistance between junction and case} = \frac{T_{j (\max)} - T_{c}}{R_{jc}}$

Therefore, the junction temperature of a semiconductor device is given by:

$T_{j (\max)} = P_{D} R_{jc} + T_{c}$ $T_{j (\max)} = P_{D} R_{jc} + T_{c}$

Therefore, the junction temperature of the IGBT is:

$T_{j (\max) IGBT} = P_{totalIGBT} R_{jc} + T_{c} = (1060 W) (0.05 K / W) + 60 ° C = 113 ° C$ $T_{j (\max) IGBT} = P_{totalIGBT} R_{jc} + T_{c} = (1060 W) (0.05 K / W) + 60 ° C = 113 ° C$

and the junction temperature of the diode is:

$T_{j (\max) (diode)} = P_{conduction (diode)} R_{jc} + T_{c} = (224.4 W) (0.1 K / W) + 60 ° C = 82.44 ° C$ $T_{j (\max) (diode)} = P_{conduction (diode)} R_{jc} + T_{c} = (224.4 W) (0.1 K / W) + 60 ° C = 82.44 ° C$

The total power that a power device can dissipate is based on the maximum junction temperature and the thermal resistance R_θjc at a case temperature of 25°C.

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Table of Contents for 10.8. Mount Package Types of Semiconductor Devices and Types of Heat Sinks

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10.8. Mount Package Types of Semiconductor Devices and Types of Heat Sinks

10.8.1. Heat Sink Computations

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Table of Contents for
10.8. Mount Package Types of Semiconductor Devices and Types of Heat Sinks