Hello Islam Zead

There are three conditions you must match:

1. Vgs may not exceed the gate break down limit.

VinVgstarget*Rgs/(Rin+Rgs)

3. switching speed:

This is a bit more complex because Cgs and Cgd change with Vgs and Vds. The turn on delay is approximately

tdelay~Cgs*Rgs*Rin/(Rgs+Rin)

Slew rate dVds/dt depends on the current you can drive through Rin and Cgd.

dVds/dt=Cgd*(Vin-Vgstarget)/Rin

Since the capacities are voltage dependent optimizing the slew rate and the delay time often is done by simulation (hoping the models for Cgs and Cgd are correct) in SPICE or SPECTRE.

Hi Islam,

to complete the previous Ricardo's answer, which look same as what I would say; I would also say that there's a tradeoff to consider about the resistors values in between the speed of the switching and the consumption of the resistors bridge. Indeed, the more you reduce the resistors values, the faster (Cgs*(Rgs//Rin)) the switch goes, but the more the consumption Vin^2/(Rgs+Rin) increases at "ON" state.

best regards

Hi Islam Zead ,

as a supplement to the correct previous answers:

The value of the resistors depend on the driver circuit providing Vin, and the kind of suitable driver depends on the parameters of the application. If the MOSFET is switching low power at low frequency then the power loss in the MOSFET and hence the speed of switching isn't of much concern. In this case, the interface between a logic circuit with 3.3 V supply voltage (for example) and the Vin point might be a pnp transistor, emitter at 8 V (for example). Here Rin limits the collector current to the permissible value (Rin = 0, if IB and hFE of the BJT are low enough), and RGS, together with CIN, determines the turn-off time.

In the other extreme, switching high current through an inductive load at relatively high frequency, the power loss during turn off can well be in the range of a few kW, so fast switching is crucial. Here one would generate Vin employing a push-pull output; there are a number of integrated drivers for MOSFETs commercially available. If the maximum current provided by the driver doesn't exceed the permissible gate current (rarely given in the datasheets), then you can choose RIN = 0. Since the push-pull output of the driver pulls the gate actively to 0 V, you can omit RGS.

Because MOSFETs have a non-vanishing internal resistance in series at the gate terminal, there is a minimum time constant one cannot undercut. In order to shorten the turn-off time even further as compared to switching between VGSon and 0 V, one can supply the driver with a negative and a positive voltage, referenced to the source of the MOSFET, thus switching the gate between VGSon and -5 V, for example.

The practical answers:

• RGS is usually chosen as 10 k. Its purpose is to prevent the MOSFET from conduction when no control signal is available (e.g. driver still unpowered). You may choose a higher resistance value but 10 k is a nice compromise between current leakage and function.
• RIn is quite special: it serves to limit the peak gate current (desired effect) while at the same slowing down the transitions (due to CDG, CGS and the Miller effect). You may even find circuits where RIn is paralleled with a diode (anode towards the gate) to speed-up switchoff.

Given that RIn is usually in the 1- or 2-digit ohms, you can neglect the effect of the (formal) voltage divider RIn / RGS. This one may have to be considered once RIn / RGS exceeds 0.05 - a ratio rarely advisable for a switching MOSFET.