Power Electronics

# Power MOSFET switching characteristics It is very important to understand switching characteristics of power MOSFET. To understand the turn-on and turn–off process in power MOSFETs, we have to consider the simplified equivalent circuits of the power MOSFET in turn-on and turn-off states. When the power MOSFET is off, ${V}_{source}=0$, ${V}_{DS}={V}_{DD}$ and ${I}_{D}={I}_{G}$. Let’s first consider turn-on processes among power MOSFET switching characteristics.

1. $t={t}_{0}$, when the voltage ${V}_{GG}$ is applied, the gate source voltage starts to control the drain-source current, and the capacitor ${C}_{GS}$ charges through the resistor $R$.
2. ${t}_{0}, when the ${V}_{GS}<{V}_{Th}$, the transistor  is in the cut-off mode and ${i}_{D}=0.$ This time $∆t={t}_{1}–{t}_{0}$ is needed to charge capacitor ${C}_{GS}$, and this means delay in time before transistor will turn on. Capacitor ${C}_{GS}$ charges to the voltage level  ${V}_{Th}$. The gate current${i}_{G}\left(t\right)=\frac{{V}_{source}–{V}_{GS}}{R}={i}_{GS}+{i}_{GD}={C}_{GS}\frac{d{i}_{GS}}{dt}–{C}_{GD}\frac{d\left({v}_{G}–{v}_{D}\right)}{dt}$, here ${v}_{G}$ is gate-to-ground voltage,  and ${v}_{D}$ is a drain-to-ground voltage. And gate current  ${i}_{G}\left(t\right)=\left({C}_{GS}+{C}_{GD}\right)\frac{d{v}_{GS}}{dt}.$Resolving these exponential equations we can show that ${v}_{GS}\left(t\right)={V}_{source}\left(1–{e}^{\frac{t–{t}_{0}}{\tau }}\right)$, where $\tau =R\left({C}_{GS}+{C}_{GD}\right)$, and gate current , ${i}_{D}=0$. Resistance ${r}_{ds}$ is characterising conducting state of power MOSFET.
3. $t={t}_{1}$, then ${V}_{GS}\left({t}_{1}\right)={V}_{Th}$ MOSFET start to conduct current. Delay time $∆{t}_{10}={t}_{1}–{t}_{0}=–\tau \mathrm{ln}\left(1–\frac{{v}_{Th}}{{v}_{GG}}\right)$.
4. and ${V}_{GS}>{V}_{Th}$, so the ${i}_{D}$ becomes a function of ${v}_{GS}$ and ${V}_{Th}$.
5. and ${V}_{GS}>{V}_{Th}$, raises exponentially  and is characterised by function ${i}_{D}\left(t\right)=k\left({v}_{GS}–{v}_{Th}\right)$, here is a coefficient, and ${i}_{D}<{I}_{o}$, ${v}_{DS}={V}_{DD}$.
6.  the drain turns off.  Resolving exponential equation we can obtain .
7. $t>{t}_{2}$, the drain is closed, .  From the equations above ${v}_{GS}=\frac{{I}_{max}}{k}+{V}_{Th}$.
8. ${t}_{2}, MOSFET turns off ,${i}_{D}={I}_{max}$, capacitance ${C}_{DS}$ is discharging, ${v}_{G}=const$, current flows through ${C}_{GD}$.
9.   flows through capacitance ${C}_{GD}$, ${v}_{GS}\left(t\right)={V}_{source}\left(1–{e}^{–\frac{t–{t}_{2}}{\tau }}\right)$, gate voltage raises until moment of time $t={t}_{3}$, when gate current ${i}_{G}=0$ and MOSFET is completely turned off. Time interval $∆{t}_{32}=R{C}_{GD}\frac{{V}_{DD}–{I}_{D}{r}_{DS}}{{v}_{source}–{V}_{Th}}$.
10. Total delay when power MOSFET is on-state is ${t}_{on}=∆{t}_{32}+∆{t}_{21}+∆{t}_{10}$, there is a high current and voltage goes through the device during periods of time $∆{t}_{21}$ and $∆{t}_{32}$, that provokes high power losses in MOSFET. Smaller resistance $R$ will decrease power losses.

Now we know turn-on part of power MOSFET switching characteristics, so we can consider the turn-off part in the power MOSFET as well. We can assume that the device is on for $t>{t}_{0}$.

1. When The equivalent circuit is depicted on figures 2-4. When ${v}_{DS}=const$${C}_{GS}$ and ${C}_{GD}$ are discharging, gate-to-source voltage is ${v}_{GS}\left(t\right)={v}_{GS}\left({t}_{0}\right){e}^{–\frac{t–{t}_{0}}{\tau }}$. Current through the capacitor ${C}_{GD}$ reaches the constant value ${i}_{DS}\left({t}_{0}\right)={I}_{max}$. So ${v}_{GS}\left({t}_{0}\right)=\frac{{I}_{max}}{k}+{V}_{Th}$.
2. When , so the current goes through the ${C}_{GD}$.
3. When ${t}_{2}, the drain current ${i}_{D}\left(t\right)$ becomes 0, and the service is turned off, ${v}_{GS}\left({t}_{3}\right)={V}_{Th}$.
4. When the gate voltage continues to fall to 0, and the voltage function is exponential. The gate-to-drain capacitance ${C}_{GD}$ charges to the ${V}_{DD}$ value.
5. When ${t}_{3} drain current ${I}_{D}$ decreases to 0. 