SiC MOSFETs: Method for Determining Losses from Switching Waveforms

In this article, a method is explained for using switching waveforms to determine the losses of SiC MOSFETs in a switching circuit that uses SiC MOSFETs. A switching waveform is divided within a range in which linear approximation is possible, and an approximating equation is used to calculate power losses. The exposition is planned as follows.

• Method of measurement of switching waveforms
• Method of loss calculation by division of a linear approximation of a waveform
• Example of calculation of power loss from a measured waveform
• Example of switching loss calculations for individual waveforms
• Example of conduction loss calculations for individual waveforms

Method of Measurement of Switching Waveforms

First, the method of measurement of a switching waveform is explained. Modern oscilloscopes include models that can automatically calculate and display power losses for an observed waveform; if the oscilloscope to be used does not have such a function, the waveform must be measured and the losses calculated. To do so, an understanding of the measurement method and of the measured waveform are needed.

Figure 1 shows a switching circuit and probes used to monitor and measure a waveform. The MOSFET drain-source voltage is measured using differential voltage probes. The drain current is measured using a current probe.

Figure 2 is a schematic diagram of different waveforms and of power losses (shaded portions).

Here t_on is the turn-on time and t_off is the turn-off time; switching losses occur in the portions of these intervals in which V_DS and I_D overlap. Because this circuit has an inductive load, at turn-on I_D begins changing first, and after the change in current has completed, V_DS begins to change. At turn-off the changes are in opposite order: first V_DS begins to change, and after the voltage change has completed, I_D begins to change.

t_on is the MOSFET on-interval; in this interval, conduction losses occur due to I_D and the MOSFET on-resistance.

There are some matters to be aware of during measurements. The first is the number of oscilloscope samples. If the number of samples is small, detailed parts of a waveform are omitted, and errors occur in the measurement results. The sampling points must be displayed to ensure that the waveform is being accurately traced.

The second matter is that, because there are differences in the characteristics of delay times for voltage probes and for current probes, a measured waveform will include errors caused by these delay differences. If no corrections are performed, shifts occur between voltages and currents in the time-axis direction, so that the area of the shaded part in Fig. 2 becomes inaccurate, and errors occur in the calculated loss. In order to eliminate delay differences in the measurement systems, de-skewing must be performed. For details on de-skewing, please refer to user’s manuals for the measurement equipment being used or technical materials provided by the measurement equipment manufacturer.

Apart from the above matters, please conform to basic procedures for observing waveforms in MOSFETs in which high voltages and large currents are switched rapidly, such as issues surrounding measurement points and the handling of probes.

Method of Loss Calculation by Division of a Linear Approximation of a Waveform

a method of calculating power losses from the above measured switching waveform using linear approximations is explained. The measured waveform is divided into regions that can be linearly approximated to calculate the power loss.

Switching Losses During Turn-on and Turn-off Intervals

To begin with, the power losses Pt_on and Pt_off that are consumed during turn-on and turn-off are calculated. The example in Figure 3 is used as the waveform. The power loss is calculated using the approximating equations of Table 1. The equations differ depending on the shape of the waveform, and so the equations for a shape close to the measured waveform are selected.

In the waveform example of Figure 3, the waveform at turn-on is divided in two, and case 2 in Table 1 is used for the first half (t_on1). As a condition, the equation I_D1≔0 is used. For the second half (t_on2), the equation for case 3 with V_DS2≔0 is used.

Power Losses During Conduction

Total Losses

As indicated in equation (4), the total power losses during MOSFET switching operation are the sum total of the switching losses and conduction losses calculated above.

\(P_D = P_{ton} + P_{toff} + P_{ON} \, [W]\)　(4)

There are notes referring to appendices in each of the cases in Table 1 and Table 2; the appendices give more detailed calculation examples for each case. The calculation examples are presented in the subsequent “Example of Switching Loss Calculations for Individual Waveforms”, as well as in “Example of Conduction Loss Calculation for Individual Waveforms”.

Example of Calculation of Power Loss from a Measured Waveform

Next, an example is presented in which power loss is actually calculated from a measured switching waveform. Figure 5 is an overall image of an actually measured switching waveform that repeats on and off operations. The upper waveform is I_D, and the lower is the V_DS waveform. Based on these waveforms, we proceed to determine three losses, which are those during turn-on (switching to the on state), during conduction (the on state), and during turn-off (switching to the off state).

Calculation of Loss During Turn-on

Figure 6 is an enlargement of the turn-on waveform of Figure 5; to determine losses, I_D (above) and V_DS (below) are used. Because the inclination changes midway through the waveform, division is into sections with the same inclination; but because the waveform is complex, division into sections is subjective. The starting voltage and current, ending voltage and current, and time can be read off.

Power losses are found by substituting values into the equations (A) in Table 1 shown in “Method of Loss Calculation by Division of a Linear Approximation of a Waveform”. In what follows, calculations are performed in order according to the equations given in Table 1; please refer to Table 1 as necessary. On the right in Figure 6 is shown a calculation example for turn-on. The divided sections are here labeled t1 to t5.

As indicated by the final equation P_ton, the losses at turn-on are the sum of the losses for each of the divided sections t1 to t5.

Calculation of Losses during Conduction

Figure 7 is an enlarged conduction waveform. The losses during conduction are, as before, obtained by substituting values into the equations (E) in Table 2 appearing in “Method of Loss Calculation by Division of a Linear Approximation of a Waveform”. As the SiC MOSFET on-resistance, the largest value appearing on the data sheet is used.

Calculation of Losses during Turn-off

Figure 8 shows the enlarged turn-off waveform. The losses at turn-off are, similarly to the case for turn-on, found by substituting values into the equations (A) in Table 1. Here, the divided sections are labeled t1 to t8. Below are shown loss calculation results for each of the turn-off sections, and the losses at turn-off (P_toff =total of all section losses).

Calculation of Total Power Loss

The total power loss can be found using the following equation. As the equation indicates, it is the sum of the turn-on loss, loss during conduction, and turn-off loss, calculated above.

\(P = P_{ton} + P_{ON} + P_{toff} \\
= 114.8 + 16.7 + 63.8 \\
= 195.3 \, [W]\)

Example of Switching Loss Calculations for Individual Waveforms

“Example of Calculation of Power Loss from a Measured Waveform”, an example was introduced in which an actually measured waveform was divided into regions for which linear approximation was possible, and a relevant equation from among equations presented in “Method of Loss Calculation by Division of a Linear Approximation of a Waveform” was used to calculate the switching loss and conduction loss in order to calculate the total loss.

From this point, calculation examples will be presented for cases 1-9 (Appendixes A to I) of all switching loss calculation equations using linear approximation that are shown in Table 1 of ” Method of Loss Calculation by Division of a Linear Approximation of a Waveform “. Please see these for switching loss calculations using equations that did not appear in ” Example of Calculation of Power Loss from a Measured Waveform “. Calculation examples employing conduction loss calculation equations using linear approximations, also shown in Table 2, are presented separately, and should be consulted together with these results.

Example of switching loss calculations for individual waveforms

• Case 1: Waveform with I_D rising, V_DS rising (Appendix A)
• Case 2: Waveform with I_D rising, V_DS constant (Appendix B)
• Case 3: Waveform with I_D rising, V_DS falling (Appendix C)
• Case 4: Waveform with I_D constant, V_DS rising (Appendix D)
• Case 5: Waveform with I_D constant, V_DS constant (Appendix E)
• Case 6: Waveform with I_D constant, V_DS falling (Appendix F)
• Case 7: Waveform with I_D falling, V_DS rising (Appendix G)
• Case 8: Waveform with I_D falling, V_DS constant (Appendix H)
• Case 9: Waveform with I_D falling, V_DS falling (Appendix I)

Example of Switching Loss Calculations for Individual Waveforms: Case 1: Waveform with ID rising, VDS rising (Appendix A)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure A-1.

The power loss P in the interval 0-t1 shown in Figure A-1 can generally be calculated by integrating the product of the current and voltage, as in equation (A-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（A-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure A-1, I_D(t) and V_DS(t) can be represented as in equations (A-2) and (A-3).

\(I_D (t) = I_{D1} + \frac{I_{D2} – I_{D1}}{t_1} t = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（A-2）

\(V_{DS} (t) = V_{DS1} + \frac{V_{DS2} – V_{DS1}}{t_1} t = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（A-3）

Equations (A-2) and (A-3) are substituted into equation (A-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（A-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t + \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^2 \right) dt\)　（A-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^3 \right]_{0}^{t_1}\)　（A-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t_1^3 \right]\)　（A-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} (I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})) t_1 + \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) t_1 \right]\)　（A-8）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2})(I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f \text{ [W]}\)　（A-9）

Here f is the switching frequency［Hz］

Equation (A-10) is substituted into equation (A-9):

・\(\underline{V_{DS1} := 0}\)　（A-10）

Equation (A-10) is substituted into equation (A-9):

\(P = \left[ \frac{1}{3} (0 – V_{DS2}) (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (0 – V_{DS2}) – \frac{1}{2} \times 0 (I_{D1} – I_{D2}) + 0 \times I_{D1} \right] t_1 f\)　（A-11）

\(\quad = \left[ -\frac{1}{3} V_{DS2} (I_{D1} – I_{D2}) + \frac{1}{2} I_{D1} V_{DS2} \right] t_1 f\)　（A-12）

\(\quad = \frac{1}{6} V_{DS2} (I_{D1} + 2I_{D2}) t_1 f \, [W]\)　（A-13）

・\(\underline{I_{D1} := 0}\)　（A-14）

Equation (A-14) is substituted into equation (A-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (0 – I_{D2}) – \frac{1}{2} \times 0 (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (0 – I_{D2}) + V_{DS1} \times 0 \right] t_1 f\)　（A-15）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (-I_{D2}) + \frac{1}{2} V_{DS1} I_{D2} \right] t_1 f\)　（A-16）

\(\quad = \frac{1}{6} (V_{DS1} + 2V_{DS2}) I_{D2} t_1 f \, [W]\)　（A-17）

Example of Switching Loss Calculations for Individual Waveforms: Case 2: Waveform with ID rising, VDS constant (Appendix B)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure B-1.

The power loss P in the interval 0-t1 shown in Figure B-1 can generally be calculated by integrating the product of the current and voltage, as in equation (B-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（B-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure B-1, I_D(t) and V_DS(t) can be represented as in equations (B-2) and (B-3).

\(I_D (t) = I_{D1} + \frac{I_{D2} – I_{D1}}{t_1} t = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（B-2）

\(V_{DS}(t) = V_{DS1}\)　（B-3）

Equations (B-2) and (B-3) are substituted into equation (B-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) (V_{DS1}) \, dt\)　（B-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t \right) dt\)　（B-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 \right]_{0}^{t_1}\)　（B-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 \right]\)　（B-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \left( V_{DS1} (I_{D1} – I_{D2}) \right) t_1 \right]\)　（B-8）

\(\quad = \frac{1}{2} V_{DS1} (I_{D1} + I_{D2}) t_1 f \, [W]\)　（B-9）

Find the power losses under the following conditions.

・\(\underline{I_{D1} := 0}\)　（B-10）

Equation (B-10) is substituted into equation (B-9):

\(P = \frac{1}{2} V_{DS1} I_{D2} t_1 f \, [W]\)　（B-11）

Example of Switching Loss Calculations for Individual Waveforms: Case 3: Waveform with ID rising, VDS falling (Appendix C)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure C-1.

The power loss P in the interval 0-t1 shown in Figure C-1 can generally be calculated by integrating the product of the current and voltage, as in equation (C-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（C-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure C-1, I_D(t) and V_DS(t) can be represented as in equations (C-2) and (C-3).

\(I_D (t) = I_{D1} + \frac{I_{D2} – I_{D1}}{t_1} t = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（C-2）

\(V_{DS} (t) = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（C-3）

Equations (C-2) and (C-3) are substituted into equation (C-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（C-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t + \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^2 \right) dt\)　（C-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^3 \right]_{0}^{t_1}\)　（C-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t_1^3 \right]\)　（C-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \left( I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2}) \right) t_1 + \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) t_1 \right]\)　（C-8）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f \, [W]\)　（C-9）

Find the power losses under the following conditions.

・\(\underline{I_{D1} := 0}\)　（C-10）

Equation (C-10) is substituted into equation (C-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (0 – I_{D2}) – \frac{1}{2} \times 0 (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (0 – I_{D2}) + V_{DS1} \times 0 \right] t_1 f\)　（C-11）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (-I_{D2}) + \frac{1}{2} V_{DS1} I_{D2} \right] t_1 f\)　（C-12）

\(\quad = \frac{1}{6} (V_{DS1} + 2V_{DS2}) I_{D2} t_1 f \, [W]\)　（C-13）

・\(\underline{V_{DS2} := 0}\)　（C-14）

Equation (C-14) is substituted into equation (C-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – 0)(I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – 0) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f\)　（C-15）

\(\quad = \left[ \frac{1}{3} V_{DS1} (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} V_{DS1} – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f\)　（C-16）

\(\quad = \frac{1}{6} V_{DS1} (2I_{D1} + I_{D2}) t_1 f \, [W]\)　（C-17）

・\(\underline{I_{D1} := 0, \, V_{DS2} := 0}\)　（C-18）

Equation (C-18) is substituted into equation (C-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – 0)(0 – I_{D2}) – \frac{1}{2} \times 0 (V_{DS1} – 0) – \frac{1}{2} V_{DS1} (0 – I_{D2}) + V_{DS1} \times 0 \right] t_1 f\)　（C-19）

\(\quad = \left[ \frac{1}{3} V_{DS1} (-I_{D2}) – \frac{1}{2} V_{DS1} (-I_{D2}) \right] t_1 f\)　（C-20）

\(\quad = \frac{1}{6} V_{DS1} I_{D2} t_1 f \, [W]\)　（C-21）

Example of Switching Loss Calculations for Individual Waveforms: Case 4: Waveform with ID constant, VDS rising (Appendix D)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure D-1.

The power loss P in the interval 0-t1 shown in Figure D-1 can generally be calculated by integrating the product of the current and voltage, as in equation (D-1)

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（D-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure D-1, I_D(t) and V_DS(t) can be represented as in equations (D-2) and (D-3).

\(I_D (t) = I_{D1}\)　（D-2）

\(V_{DS} (t) = V_{DS1} + \frac{V_{DS2} – V_{DS1}}{t_1} t = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（D-3）

Equations (D-2) and (D-3) are substituted into equation (D-1):

\(P = f \int_{0}^{t_1} I_{D1} \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（D-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t \right) dt\)　（D-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t^2 \right]_{0}^{t_1}\)　（D-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t_1^2 \right]\)　（D-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \left( I_{D1} (V_{DS1} – V_{DS2}) \right) t_1 \right]\)　（D-8）

\(\quad = \frac{1}{2} (V_{DS1} + V_{DS2}) I_{D1} t_1 f \, [W]\)　（D-9）

Find the power losses under the following conditions.

\(\underline{V_{DS1} := 0}\)　（D-10）

Equation (D-10) is substituted into equation (D-9):

\(P = \frac{1}{2} (0 + V_{DS2}) I_{D1} t_1 f\)　（D-11）

\(\quad = \frac{1}{2} V_{DS2} I_{D1} t_1 f \, [W]\)　（D-12）

Example of Switching Loss Calculations for Individual Waveforms: Case 5: Waveform with ID constant, VDS constant (Appendix E)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure E-1.

The power loss P in the interval 0-t1 shown in Figure E-1 can generally be calculated by integrating the product of the current and voltage, as in equation (E-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（E-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure E-1, I_D(t) and V_DS(t) can be represented as in equations (E-2) and (E-3).

\(I_D (t) = I_{D1}\)　（E-2）

\(V_{DS} (t) = V_{DS1}\)　（E-3）

Equations (E-2) and (E-3) are substituted into equation (E-1):

\(P = f \int_{0}^{t_1} I_{D1} V_{DS1} \, dt\)　（E-4）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t \right]_{0}^{t_1}\)　（E-5）

\(\quad = V_{DS1} I_{D1} t_1 f \, [W]\)　（E-6）

Example of Switching Loss Calculations for Individual Waveforms: Case 6: Waveform with ID constant, VDS falling (Appendix F)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure F-1.

The power loss P in the interval 0-t1 shown in Figure F-1 can generally be calculated by integrating the product of the current and voltage, as in equation (F-1)

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（F-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure F-1, I_D(t) and V_DS(t) can be represented as in equations (F-2) and (F-3).

\(I_D (t) = I_{D1}\)　（F-2）

\(V_{DS} (t) = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（F-3）

Equations (F-2) and (F-3) are substituted into equation (F-1):

\(P = f \int_{0}^{t_1} I_{D1} \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（F-4）

\(= f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t \right) dt\)　（F-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t^2 \right]_{0}^{t_1}\)　（F-6）

\(\quad = f \left( V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t_1^2 \right)\)　（F-7）

\(\quad = f \left( V_{DS1} I_{D1} t_1 – \frac{1}{2} \left( I_{D1} (V_{DS1} – V_{DS2}) \right) t_1 \right)\)　（F-8）

\(\quad = \frac{1}{2} (V_{DS1} + V_{DS2}) I_{D1} t_1 f \, [W]\)　（F-9）

Find the power losses under the following conditions.

\(\underline{V_{DS2} := 0}\)　（F-10）

Equation (F-10) is substituted into equation (F-9):

\(P = \frac{1}{2} (V_{DS1} + 0) I_{D1} t_1 f\)　（F-11）

\(\quad = \frac{1}{2} V_{DS1} I_{D1} t_1 f \, [W]\)　（F-12）

Example of Switching Loss Calculations for Individual Waveforms: Case 7: Waveform with ID falling, VDS rising (Appendix G)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure G-1.

The power loss P in the interval 0-t1 shown in Figure G-1 can generally be calculated by integrating the product of the current and voltage, as in equation (G-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（G-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure G-1, I_D(t) and V_DS(t) can be represented as in equations (G-2) and (G-3).

\(I_D (t) = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1}\)　（G-2）

\(V_{DS} (t) = V_{DS1} + \frac{V_{DS2} – V_{DS1}}{t_1} t = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（G-3）

Equations (G-2) and (G-3) are substituted into equation (G-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（G-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2})}{t_1} t – \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t + \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^2 \right) dt\)　（G-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^3 \right]_{0}^{t_1}\)　（G-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t_1^3 \right]\)　（G-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} (I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})) t_1 + \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) t_1 \right]\)　（G-8）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f \, [W]\)　（G-9）

Find the power losses under the following conditions.

\(\underline{I_{DS2} := 0}\)　（G-10）

Equation (G-10) is substituted into equation (G-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – V_{DS2})(I_{D1} – 0) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – 0) + V_{DS1} I_{D1} \right] t_1 f\)　（G-11）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) I_{D1} – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} I_{D1} + V_{DS1} I_{D1} \right] t_1 f\)　（G-12）

\(\quad = \frac{1}{6} (2V_{DS1} + V_{DS2}) I_{D1} t_1 f \text{ [W]}\)　（G-13）

\(\underline{V_{DS1} := 0}\)　（G-14）

Equation (G-14) is substituted into equation (G-9):

\(P = \left[ \frac{1}{3} (0 – V_{DS2}) (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (0 – V_{DS2}) – \frac{1}{2} \times 0 (I_{D1} – I_{D2}) + 0 \times I_{D1} \right] t_1 f\)　（G-15）

\(\quad = \left[ -\frac{1}{3} V_{DS2} (I_{D1} – I_{D2}) + \frac{1}{2} I_{D1} V_{DS2} \right] t_1 f\)　（G-16）

\(\quad = \frac{1}{6} V_{DS2} (I_{D1} + 2I_{D2}) t_1 f \text{ [W]}\)　（G-17）

\(\underline{I_{D2} := 0, V_{DS1} := 0}\)　（G-18）

Equation (G-18) is substituted into equation (G-9):

\(P = \left[ \frac{1}{3} (0 – V_{DS2}) (I_{D1} – 0) – \frac{1}{2} I_{D1} (0 – V_{DS2}) – \frac{1}{2} \times 0 (I_{D1} – 0) + 0 \times I_{D1} \right] t_1 f\)　（G-19）

\(\quad = \left( -\frac{1}{3} V_{DS2} I_{D1} + \frac{1}{2} I_{D1} V_{DS2} \right) t_1 f\)　（G-20）

\(\quad = \frac{1}{6} V_{DS2} I_{D1} t_1 f \text{ [W]}\)　（G-21）

Example of Switching Loss Calculations for Individual Waveforms: Case 8: Waveform with ID falling, VDS constant (Appendix H)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure H-1.

The power loss P in the interval 0-t1 shown in Figure H-1 can generally be calculated by integrating the product of the current and voltage, as in equation (H-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（H-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure H-1, I_D(t) and V_DS(t) can be represented as in equations (H-2) and (H-3).

\(I_D (t) = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（H-2）

\(V_{DS} (t) = V_{DS1}\)　（H-3）

Equations (H-2) and (H-3) are substituted into equation (H-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) (V_{DS1}) \, dt\)　（H-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t \right) dt\)　（H-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 \right]_{0}^{t_1}\)　（H-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 \right]\)　（H-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \left(V_{DS1} (I_{D1} – I_{D2})\right) t_1 \right]\)　（H-8）

\(\quad = \frac{1}{2} V_{DS1} (I_{D1} + I_{D2}) t_1 f \text{ [W]}\)　（H-9）

Find the power losses under the following conditions.

\(\underline{I_{D2} := 0}\)　（H-10）

Equation (H-10) is substituted into equation (H-9):

\(P = \frac{1}{2} V_{DS1} I_{D1} t_1 f \text{ [W]}\)　（H-11）

Example of Switching Loss Calculations for Individual Waveforms: Case 9: Waveform with ID falling, VDS falling (Appendix I)

The switching waveform drain-source voltage V_DS and drain current I_D are used to determine power losses at turn-on and turn-off (switching losses), through linear approximations. The waveforms used in loss calculations are shown in Figure I-1.

The power loss P in the interval 0-t1 shown in Figure I-1 can generally be calculated by integrating the product of the current and voltage, as in equation (I-1).

\(P = f \int_{0}^{t_1} I_D (t) V_{DS} (t) \, dt\)　（I-1）

Here f is the switching frequency［Hz］

Moreover, using the slopes in Figure I-1, I_D(t) and V_DS(t) can be represented as in equations (I-2) and (I-3).

\(I_D (t) = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（I-2）

\(V_{DS} (t) = V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t\)　（I-3）

Equations (I-2) and (I-3) are substituted into equation (I-1):

\(P = f \int_{0}^{t_1} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right) \left( V_{DS1} – \frac{V_{DS1} – V_{DS2}}{t_1} t \right) dt\)　（I-4）

\(\quad = f \int_{0}^{t_1} \left( V_{DS1} I_{D1} – \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t + \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^2 \right) dt\)　（I-5）

The integration is performed according to the formula:

\(P = f \left[ V_{DS1} I_{D1} t – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t^3 \right]_{0}^{t_1}\)　（I-6）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} \frac{I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})}{t_1} t_1^2 + \frac{1}{3} \frac{(V_{DS1} – V_{DS2}) (I_{D1} – I_{D2})}{t_1^2} t_1^3 \right]\)　（I-7）

\(\quad = f \left[ V_{DS1} I_{D1} t_1 – \frac{1}{2} (I_{D1} (V_{DS1} – V_{DS2}) + V_{DS1} (I_{D1} – I_{D2})) t_1 + \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) t_1 \right]\)　（I-8）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f \, [W]\)　（I-9）

Find the power losses under the following conditions.

\(\underline{V_{DS2} := 0}\)　（I-10）

Equation (I-10) is substituted into equation (I-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – 0)(I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} (V_{DS1} – 0) – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f\)　（I-11）

\(\quad = \left[ \frac{1}{3} V_{DS1} (I_{D1} – I_{D2}) – \frac{1}{2} I_{D1} V_{DS1} – \frac{1}{2} V_{DS1} (I_{D1} – I_{D2}) + V_{DS1} I_{D1} \right] t_1 f\)　（I-12）

\(\quad = \frac{1}{6} V_{DS1} (2I_{D1} + I_{D2}) t_1 f \text{ [W]}\)　（I-13）

\(\underline{I_{D2} := 0}\)　（I-14）

Equation (I-14) is substituted into equation (I-9):

\(P = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) (I_{D1} – 0) – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} (I_{D1} – 0) + V_{DS1} I_{D1} \right] t_1 f\)　（I-15）

\(\quad = \left[ \frac{1}{3} (V_{DS1} – V_{DS2}) I_{D1} – \frac{1}{2} I_{D1} (V_{DS1} – V_{DS2}) – \frac{1}{2} V_{DS1} I_{D1} + V_{DS1} I_{D1} \right] t_1 f\)　（I-16）

\(\quad = \frac{1}{6} (2V_{DS1} + V_{DS2}) I_{D1} t_1 f \text{ [W]}\)　（I-17）

Example of Conduction Loss Calculations for Individual Waveforms

“Example of Switching Loss Calculations for Individual Waveforms”, we present examples of conduction loss calculation for each type of waveform. Calculation examples are presented for each of cases 1 to 3 (appendices J to L) for conduction loss calculation formulas using linear approximation appearing in Table 2 of “Method of Loss Calculation by Division of a Linear Approximation of a Waveform “.

Example of conduction loss calculations for individual waveforms

• Case 1: Waveform with ID rising （Appendix J）
• Case 2: Waveform with ID constant （Appendix K）
• Case 3: Waveform with ID falling （Appendix L）

Example of Conduction Loss Calculations for Individual Waveforms: Case 1: Waveform with ID Rising (Appendix J)

The power loss during conduction (0 to t1) is determined, using linear approximations, from the MOSFET on-resistance R_ON and the drain current I_D of the switching waveform. The waveform used in loss calculations is shown in Figure J-1

In Figure J-1, the MOSFET is conducting over the interval 0 to t1, so that V_DS is equal to the product of the MOSFET on-resistance R_ON and I_D.

The power loss P in the interval 0 to t1 can generally be calculated by integrating the product of the resistance and the square of the current, as indicated in equation (J-1).

\(P = f \int_{0}^{t_1} R_{ON} I_D (t)^2 \, dt\)　（J-1）

where:
R_ON is the MOSFET on-resistance［Ω］

f is the switching frequency［Hz］

Moreover, using the slopes in Figure J-1, ID(t) can be represented as in equations (J-2).

\(I_D (t) = I_{D1} + \frac{I_{D2} – I_{D1}}{t_1} t = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（J-2）

Equations (J-2) is substituted into equation (J-1):

\(P = f \int_{0}^{t_1} R_{ON} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t \right)^2 dt\)　（J-3）

\(\quad = f \int_{0}^{t_1} R_{ON} \left( I_{D1}^2 – 2I_{D1} \frac{(I_{D1} – I_{D2})}{t_1} t + \frac{(I_{D1} – I_{D2})^2}{t_1^2} t^2 \right) dt\)　（J-4）

The integration is performed according to the formula:

\(P = f R_{ON} \left[ I_{D1}^2 t – 2I_{D1} \frac{(I_{D1} – I_{D2})}{2t_1} t^2 + \frac{(I_{D1} – I_{D2})^2}{3t_1^2} t^3 \right]_{0}^{t_1}\)　（J-5）

\(\quad = f R_{ON} \left( I_{D1}^2 t_1 – 2I_{D1} \frac{(I_{D1} – I_{D2})}{2t_1} t_1^2 + \frac{(I_{D1} – I_{D2})^2}{3t_1^2} t_1^3 \right)\)　（J-6）

\(\quad = f R_{ON} \left( I_{D1}^2 t_1 – I_{D1} (I_{D1} – I_{D2}) t_1 + \frac{(I_{D1} – I_{D2})^2}{3} t_1 \right)\)　（J-7）

\(\quad = f R_{ON} \left( I_{D1}^2 – I_{D1} (I_{D1} – I_{D2}) + \frac{(I_{D1} – I_{D2})^2}{3} \right) t_1\)　（J-8）

\(\quad = f R_{ON} \left( I_{D1} I_{D2} + \frac{I_{D1}^2 – 2I_{D1} I_{D2} + I_{D2}^2}{3} \right) t_1\)　（J-9）

\(\quad = f R_{ON} \left( \frac{I_{D1}^2 – 2I_{D1} I_{D2} + I_{D2}^2 – 3I_{D1}^2 + 3I_{D1} I_{D2} + 3I_{D1}^2}{3} \right) t_1\)　（J-10）

\(\quad = \frac{1}{3} R_{ON} (I_{D1}^2 + I_{D1} I_{D2} + I_{D2}^2) t_1 f \text{ [W]}\)　（J-11）

Example of Conduction Loss Calculations for Individual Waveforms: Case 2: Waveform with ID Constant (Appendix K)

The power loss during conduction (0 to t1) is determined, using linear approximations, from the MOSFET on-resistance R_ON and the drain current I_D of the switching waveform. The waveform used in loss calculations is shown in Figure K-1.

In Figure K-1, the MOSFET is conducting over the interval 0 to t1, so that V_DS is equal to the product of the MOSFET on-resistance R_ON and I_D.

The power loss P in the interval 0 to t1 can generally be calculated by integrating the product of the resistance and the square of the current, as indicated in equation (K-1).

\(P = f \int_{0}^{t_1} R_{ON} I_D (t)^2 \, dt\)　（K-1）

where:
R_ON is the MOSFET on-resistance［Ω］

f is the switching frequency［Hz］

Moreover, using the slopes in Figure K-1, I_D(t) can be represented as in equations (K-2).

\(I_D (t) = I_{D1}\)　（K-2）

Equations (K-2) is substituted into equation (K-1):

\(P = f \int_{0}^{t_1} R_{ON} I_{D1}^2 \, dt\)　（K-3）

The integration is performed according to the formula:

\(P = f \left[ R_{ON} I_{D1}^2 \right]_0^{t_1}\)　（K-4）

\(P = R_{ON} I_{D1}^2 t_1 f \text{ [W]}\)　（K-5）

Example of Conduction Loss Calculations for Individual Waveforms: Case 3: Waveform with ID Falling (Appendix L)

The power loss during conduction (0 to t1) is determined, using linear approximations, from the MOSFET on-resistance R_ON and the drain current I_D of the switching waveform. The waveform used in loss calculations is shown in Figure L-1.

In Figure L-1, the MOSFET is conducting over the interval 0 to t1, so that V_DS is equal to the product of the MOSFET on-resistance R_ON and I_D.

The power loss P in the interval 0 to t1 can generally be calculated by integrating the product of the resistance and the square of the current, as indicated in equation (L-1).

\(P = f \int_{0}^{t_1} R_{ON} I_D (t)^2 \, dt\)　（L-1）

where:
R_ON is the MOSFET on-resistance［Ω］

f is the switching frequency［Hz］

Moreover, using the slopes in Figure L-1, ID(t) can be represented as in equations (L-2).

\(I_D (t) = I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} t\)　（L-2）

Equations (L-2) is substituted into equation (L-1):

\(P = f \int_{0}^{t_1} R_{ON} \left( I_{D1} – \frac{I_{D1} – I_{D2}}{t_1} \right)^2 dt\)　（L-3）

\(\quad = f \int_{0}^{t_1} R_{ON} \left( I_{D1}^{\hspace{0.5em} 2} – 2I_{D1} \frac{I_{D1} – I_{D2}}{t_1} t + \frac{(I_{D1} – I_{D2})^2}{t_1^2} t^2 \right) dt\)　（L-4）

The integration is performed according to the formula:

\(P = f R_{ON} \left[ I_{D1}^{\hspace{0.5em} 2} t – 2I_{D1} \frac{I_{D1} – I_{D2}}{2t_1} t^2 + \frac{(I_{D1} – I_{D2})^2}{3t_1^2} t^3 \right]_{0}^{t_1}\)　（L-5）

\(\quad = f R_{ON} \left( I_{D1}^{\hspace{0.5em} 2} t_1 – 2I_{D1} \frac{I_{D1} – I_{D2}}{2t_1} t_1^2 + \frac{(I_{D1} – I_{D2})^2}{3t_1^2} t_1^3 \right)\)　（L-6）

\(\quad = f R_{ON} \left( I_{D1}^{\hspace{0.5em} 2} t_1 – I_{D1} (I_{D1} – I_{D2}) t_1 + \frac{(I_{D1} – I_{D2})^2}{3} t_1 \right)\)　（L-7）

\(\quad = f R_{ON} \left( I_{D1}^{\hspace{0.5em} 2} – I_{D1} (I_{D1} – I_{D2}) + \frac{(I_{D1} – I_{D2})^2}{3} \right) t_1\)　（L-8）

\(\quad = f R_{ON} \left( I_{D1} I_{D2} + \frac{I_{D1}^{\hspace{0.5em} 2} – 2I_{D1} I_{D2} + I_{D2}^{\hspace{0.5em} 2}}{3} \right) t_1\)　（L-9）

\(\quad = f R_{ON} \left( \frac{I_{D1}^{\hspace{0.5em} 2} – 2I_{D1} I_{D2} + I_{D2}^{\hspace{0.5em} 2} – 3I_{D1}^2 + 3I_{D1} I_{D2} + 3I_{D1}^2}{3} \right) t_1\)　（L-10）

\(\quad = \frac{1}{3} R_{ON} (I_{D1}^{\hspace{0.5em} 2} + I_{D1} I_{D2} + I_{D2}^{\hspace{0.5em} 2}) t_1 f \text{ [W]}\)　（L-11）