Gatedriver

The mosfet is to be controlled by a microcontroller with a pwm. The output pin of the microcontroller has a maximum voltage of $3.2V$, but in order to be able to switch the power mosfet completely, a voltage of at least $10V$ is required and thus the mosfet cannot be operated directly with the microcontroller. For this application, so-called logic-level mosfets exist that switch through at a gate-source voltage of less than $V_{GS}=5V$. In this work, however, another way is to be shown, since these logic-level mosfets buy their properties through a high input capacitance and other disadvantages. Too high a capacitance at the input would slow down the switching process. The easiest way to drive this mosfet with the required voltage is to use a gate driver. This is compatible with the logic voltage level of the microcontroller and can switch a higher voltage. A gate driver also makes it possible to provide a very high current to the gate of the mosfet in a very short time. This allows the mosfet to be switched much faster. The following questions are asked for the selection of a gate driver:

• How many inputs/outputs are required?

• What is the voltage rating?

• How much gate current does the mosfet need?

• Gate voltage of $12V$ possible?

Only one Mosfet is to be switched, so only one input and one output is needed at the gate driver. The required dielectric strength should be at least three times higher than the normal operating voltage of the application, in this case ${V_{max}=3\cdot 12V=36V}$.

The required current of at least ${I_{gate} = 360mA}$ and the associated switching times were calculated in the last chapter. The half-bridge driver IR2104S from international Rectifier is selected, its characteristics meet the above requirements. 

The dimensioning of the gate resistor $R_{gate}$ is an important aspect of this circuit. If it is chosen too small, unwanted oscillations can occur at the gate. On the other hand, the switching process would be greatly slowed down if the resistor is chosen too large, because too little current flows into the gate. According to \eqref{eq:Rgate}, the gate resistor from the E24 series is chosen with $R_{gate}=37\Omega$.

\[\label{eq:Rgate} R_{gate} = \frac{V_{12V}}{I_{driver}} = \frac{12V}{360mA} = 33.33\Omega\]

The gate resistor should only limit the current for the rise time, during the descent time and the discharging process of the gate capacitance, it should not act. For this reason, a diode $D$ is connected in parallel to the gate resistor so that the current is not limited during the descent time. In figure the complete control of the gate is shown. The IR2104S is a half-bridge driver where the high-side connections are not used in this application. These are connected to ground with an $R_{3}=10k\Omega$ resistor so that there are no open pins in the circuit.

\[\label{eq:Rpwm} R_{1} = R_{2} = \frac{V_{vcc}}{I} = \frac{3.2V}{1mA} = 3.2k\Omega\]

The PWM and SD signals come from the microcontroller with a $3.3V$ level. The series resistors are calculated according to \eqref{eq:Rpwm}. Here the current is limited to $1mA$ to avoid losses.

Another component is the bypass capacitor $C_{bypass}=1\mu F$, its task is to decouple the chip from the voltage supply, especially in driver circuits this is an important element. Because higher currents flow very quickly here and the power supply is more heavily loaded as a result.

Gate driver schematic