Mosfet

Selecting the right switching element for the boost converter is not an easy task; various bipolar transistors, IGBTs or mosfets can be used for this purpose. In this work, a suitable mosfet is to be selected, as it causes very low switching losses, can be switched at a higher frequency and is very often used for this task. For the second switching element, the Schottky diode MBRB1060 is used. The requirements for the boost converter to the mosfet are as follows:

Maximum load current: $I_{l a s t} = 10 A$
Voltage of the load: $U_{D S} = 40 V D C$
Gate voltage from driver: $12 V D C$
PWM switching frequency: $50 k H z$
No forced air cooling

The switching element of the boost converter is switched in a low-side-switch configuration because the load is switched via drain and the source is connected directly to ground. This configuration requires an N-channel mosfet so that it can be easily switched by a PWM with ground reference.

The first criterion for choosing the mosfet is its $R_{D S (o n)}$ resistance. During the switch-on phase, losses and thus heat are generated. These should amount to a maximum of $1 W$ and can thus be absorbed by most enclosure types without active cooling.

\begin{matrix} (1) & R_{D S (o n) m a x} = \frac{P}{I^{2}} = \frac{1 W}{(10 A)^{2}} = 10 m Ω \end{matrix}

As calculated in equation $(1)$ , the $R_{D S (o n)}$ of the sought mosfet must not exceed $10 m Ω$ . The design with the continuous drain current $I_{D}$ is avoided here, as this specification can be misleading. In most cases, a mosfet is limited by the heat generated by $R_{D S (o n)}$ or the switching losses and not by the actual continuous current. Of course, you should still not choose an mosfet that has an $I_{D}$ less than $10 A$ .

The second criterion is the maximum voltage $V_{D S S}$ across drain and source, this should be dimensioned significantly larger than is normally present at the load, a maximum drain-source voltage of at least $V_{D S S_{m a x}} = 80 V$ is selected.

The third criterion is the maximum gate-source voltage $U_{G S_{m a x}}$ . The mosfet is to be operated with a gate driver, which is selected in the next section. The operating voltage of the gate driver is $V_{v s} = 12 V$ and thus the maximum gate-source voltage should be at least $V_{G S_{m a x}} = 12 V$ .

The last criterion is the maximum gate charge $Q_{g s_{m a x}}$ , it describes how much charge flows into the gate when fully switched. Smaller charges allow faster switching, so the charge is chosen to allow the switching frequency of $f_{s} = 50 k H z$ .

Mosfet characteristics

Parameter	Wert
$V_{D S S}$	$100 V$
$R_{D S (o n)} a t V_{G S} = 10 V$	$5.9 m Ω$
$I_{D}$	$75 A$
$V_{G S_{m a x}}$	$\pm 20 V$
$Q_{G S} a t U_{G S} = 10 V$	$14 n C$
$R_{t h J A}$	$40^{\circ} C / W$
$R_{J}$	$+ 175^{\circ} C$

The N-channel mosfet SQM70060EL from Vishay Siliconix is selected, using the properties from table the actual power dissipation can be calculated according to $(2)$ .

\begin{matrix} (2) & P = I^{2} \cdot R_{D S (o n)} = (10 A)^{2} \cdot 5.9 m Ω = 0.59 W \end{matrix}

The question of whether the mosfet can be operated without a heat sink is answered by calculating the expected temperature rise $(3)$ , where $R_{t h J A}$ corresponds to the thermal resistance to ambient air.

\begin{matrix} (3) & Δ T = P \cdot R_{t h J A} = 0.59 W \cdot 40^{\circ} C / W = {23.6}^{\circ} C \end{matrix}

At room temperature of $25^{\circ} C$ , this gives an operating temperature of $25^{\circ} C + {23.6}^{\circ} C = {48.6}^{\circ} C$ which is far below the maximum operating temperature of $R_{J} = + 175^{\circ} C$ and thus no heat sink is required.

Another question is whether the mosfet can be switched at the frequency of $f_{s} = 50 k H z$ without the gate capacitance slowing down the switching too much. At $V_{G S} = 12 V$ , the maximum gate charge $Q_{G} = 100 n C$ is assumed so that the rise time can be calculated according to $(4)$ . The maximum current of the gate driver $I_{g a t e} = 360 m A$ is included in the equation.

\begin{matrix} (4) & t_{a u f} = \frac{Q_{G}}{I_{g a t e}} = \frac{100 n C}{360 m A} = 277.8 n s \end{matrix}

If it is assumed that the descent time also corresponds to $t_{a b} = 277.8 n s$ , then the total switching time is approximately $t_{s w i t c h} = 2 \cdot t = 555.6 n s$ for the mosfet, which corresponds to a frequency of $f_{s w i t c h} = 1 / t = 1.6 M h z$ .

The switching frequency $f_{s}$ has a period of $t_{s} = 20 μ s$ . If the switching time is subtracted from this $t_{s} - t_{s w i t c h} = 19.45 μ s$ , this results in a percentage of the pure switching time of $1 %$ . This means that switching accounts for only a fraction of the time, while keeping the switching losses small and the switching frequency $f_{s}$ can be used with the mosfet.