Connecting a resistor to Mosfet's gate

[EngIntoHW]

Hello,

I talked to an electric engineer today which said that when designing an H-bridge which comprises Mosfets, a resistor should be connected to each gate.

He said that the resistor's value is determined by: C*V = I * t.
Where:
C - Gate's capacitance.
t - Mosfet's switching time.
I - Current through gate.
(I assume that V is the capacitor's final voltage).

Could you explain please why R is needed and how is its value calculated?

Thanks.

 

[MikeMl]

This calculation is to calculate how much current it will take to charge the gate in a specified time, or knowing the available charging current, how long it will take the gate to charge.

The answer is that NO resistor should be connected between the gate driver and the gate. If you want to connect a resistor somewhere, then connect a high value (>10K) from gate to source of each FET. This will guarantee that the FET is OFF in case the wire to the gate driver gets disconnected.

 

[crutschow]

Your electric engineer friend is misinformed. Adding a resistor will slow down the transistors switching time, which will increase its dissipation, and is not what you want. As Mike ML said, you want no resistor in series with the gate drive.

 

[EngIntoHW]

Hey, perhaps he meant a resistor from Gate to Source.

All he said was connecting a resistor to the Gate.

I'm still not sure how does a Gate-Source resistor help.

 

[MikeMl]

...

I'm still not sure how does a Gate-Source resistor help.

Quoting myself: "If you want to connect a resistor somewhere, then connect a high value (>10K) from gate to source of each FET. This will guarantee that the FET is OFF in case the wire to the gate driver gets disconnected."

 

[crutschow]

Hey, perhaps he meant a resistor from Gate to Source.

All he said was connecting a resistor to the Gate.

He did not. The formula he gave you is for a resistor in series with the gate to its driver, not a resistor from gate to source.

 

[shortbus=]

International Rectifiers gate drivers use a low value- 4 - 20 Ohm gate resistor to prevent "ringing" in the circuit. Go to IRF website and check out the Application Notes on gate drivers. International Rectifier - Application Notes

 

[MikeMl]

International Rectifiers gate drivers use a low value- 4 - 20 Ohm gate resistor to prevent "ringing" in the circuit. Go to IRF website and check out the Application Notes on gate drivers. International Rectifier - Application Notes

That is a requirement to guarantee stability of the gate driver IC (because it is driving a capacitive load); it is not something required by the FET!!!

 

[ronv]

I always add a gate resistor when I am using a breadboard for power fets as the wiring tends to add to the inductance causing ringing. The downside to adding resistance is that it slows down turn on/off time due to the gate capacitance. Data sheets now quote total gate charge (nC). A fet with a nC of 20 will turn on in 20 usec with 1 ma or 20 ns with 1 amp of gate drive. A lot depends on your application and your driver. Some drivers don't like the ringing. Some also spec a peak current. Below is a bit from IR on the subject.

The gate resistor in series with the output impedance of the circuit used to drive the gate in a given application, controls the switching on and off times.

MOSFET gate charge in Coulombs = I t. So if a gate driver having say 0.2A output current capability switches a particular MOSFET in 100ns and by increasing the series gate resistance it reduces the current into the gate to 0.1A, then the MOSFET will switch in 200ns.

The optimum value for Rg is very application dependant. You want the MOSFET to switch as quickly as possible to minimise switching losses, but not so fast that parasitic inductances and capacitances associated with the pcb layout and any wiring to a load, will cause high di/dt voltage spiking or ringing.If you find that an optimised value of Rg controls switch on OK but slows the turn off too much, then a fix is to place a diode across Rg with its cathode towards the gate drive circuit. This will bypass Rg during turn off thereby speeding up the turn off. Placing a resistor in series with the diode will enable you to control turn off time independantly of turn on.

 

[EngIntoHW]

I always add a gate resistor when I am using a breadboard for power fets as the wiring tends to add to the inductance causing ringing. The downside to adding resistance is that it slows down turn on/off time due to the gate capacitance. Data sheets now quote total gate charge (nC). A fet with a nC of 20 will turn on in 20 usec with 1 ma or 20 ns with 1 amp of gate drive. A lot depends on your application and your driver. Some drivers don't like the ringing. Some also spec a peak current. Below is a bit from IR on the subject.

The gate resistor in series with the output impedance of the circuit used to drive the gate in a given application, controls the switching on and off times.

MOSFET gate charge in Coulombs = I t. So if a gate driver having say 0.2A output current capability switches a particular MOSFET in 100ns and by increasing the series gate resistance it reduces the current into the gate to 0.1A, then the MOSFET will switch in 200ns.

The optimum value for Rg is very application dependant. You want the MOSFET to switch as quickly as possible to minimise switching losses, but not so fast that parasitic inductances and capacitances associated with the pcb layout and any wiring to a load, will cause high di/dt voltage spiking or ringing.If you find that an optimised value of Rg controls switch on OK but slows the turn off too much, then a fix is to place a diode across Rg with its cathode towards the gate drive circuit. This will bypass Rg during turn off thereby speeding up the turn off. Placing a resistor in series with the diode will enable you to control turn off time independantly of turn on.

Hey Ron, thank you so much for the detailed explanation.

Say that I want the switching time to be 1usec, and that the driver's PSU can source 0.1A, how would you calculate R in such application?
I'm not sure what is the formula for that? (which comprises R).

 

[ronv]

You first need the FET data sheet and the nC value for that FET. Let's say the nC is 100. That means that with one amp of gate drive the FET will switch in 100 ns or with .1 amp 1 usec.. You also need to see flat part of the switching curve for the gate voltage that turns the FET on. If that voltage is say 5 volts and your driver is driving from a 10 volt supply you need 5 volts/.1 amps or 50 ohms. I'm sure that is not as good an explanation as this link.
https://www.electro-tech-online.com/custompdfs/2010/08/an-944.pdf

 

[MrAl]

Hi,


I didnt read all the posts yet so maybe this has been addressed already...

Normally you want to turn the mosfet on as fast as possible and off as fast as possible. There are times however when this is not the optimal technique for the application. One example is where the output includes inductance, and that causes a high output spike. This spike can be either eliminated or partially eliminated by controlling the mosfet turn off by using a resistor in series with the gate to slow it down. Usually however this is done with an asymmetrical drive where you use a diode in one direction and a resistor in another direction, where the diode turns it on quickly and the resistor turns it off slower, but that is coupled with other timing too in for example an H bridge. Of course the down side is more power dissipation in the mosfet, so all things must be weighed in.
Another example is to help with isolating gates when driving mosfets in parallel. Each gate may have a slightly different Vth and that means it might turn on at a different time. Ultimately we want them all turning on as close as possible to the same time, and that means a gate resistor can help.
There's also the issue of the small inductance in the source circuit, which you can take a look at.

So to sum up, it really depends on the application. It's best to take a look at what is happening in the prototype and see what resistor(s) work best. Alternately, you can take a look at the spice simulations given a number of possible scenarios that you think might occur with your actual application.

 

[oingo456]

I just wanted to add that increasing gate resistance is a fairly common way of reducing EMI switching noise. Efficiency will be lowered of course, but often a good trade-off can be found. Just a few % efficiency loss may lower the harmonics quite a bit, and make it easier to pass emissions or avoid filtering.

 

[audioguru]

All Mosfet manufacturers say a low value series resistor right at the gate pin stops the Mosfet from ringing or oscillating at VHF frequencies.

 

[Mosaic]

From my calcs I can see that a a MOSFET with a gate charge max of 6 nC will switch on from a VGs of 5V with a 9.1K gate resistor in abt 10 uSec.

MOSFET = https://www.electro-tech-online.com/custompdfs/2010/10/FDN357N.pdf

I am switching relays and solenoids in an automotive environment. Max frequency is abt 50Hz.

Why do I want such a large gate resistor? Cuz the design places a small 25V transorb across the drain and gate to switch on the FET when the inductive spike comes back. I don't want any high currents making it's way back to the MCU drive. The MCU 16f886 drive has it's own built in diodes to clip the ESD at the pins to abt 5.6V.

I am trying to emulate this :
https://www.electro-tech-online.com/custompdfs/2010/10/nud3124-d.pdf

as I can't find it to buy.

I'd like some advice on calculating the avg power dissapation on the MOSFET with this slow gate drive though..

Thanks

 

[ronv]

Here is a link. **broken link removed**
Your swithcing rate is so low (50HZ) that the swirching losses don't matter much.

If I were you I think I would use a higher voltage fet than the one you showed the datasheet for. That way you don't have to worry about the possibility of load dump damaging it. A 2n7002 is widely available and has a 60 volt breakdown voltage. Add the back to back zeners at the input (above the gate drive voltage). Don't forget to add a diode across the load for this type of design to limit the inductive kick. I think I would switch it a little faster than the driver you referenced. Say a 2k gate resistor and 100K input to source.

 

[Mosaic]

Ronv, I need to have a logic drive and a small footprint, so I am planning to place a small transorb across the Drain/gate to active clamp the inductive spike using the FET rather than the diode.

Thus the FET would be fully on during a Load dump and the 9.1K gate resistor will allow the built in MCU ESD protection to work safely.
The reason for 9.1K is that other parts of my design call for it so i am reducing the part value counts.

I have to compensate for a load dump of up to 125V for up to 400 ms as per the SAE spec. So a 60Vds won't help me. Notably, that load dump will be limited by the 80 ohm resistance and some inductance of a 150mA, 12V relay. The max current should be 125/80 = abt 1.5A , which is below the 1.9A rated of the MOSFET.

 

[ronv]

I got it. So you are duplicating the auto protected mosfet with descretes.

 

[MikeMl]

I just went and read a bunch of app notes on the IR website. I found all kinds of circuits that have a low value resistor in the gate lead, but it is there to unload the gate driver to prevent it from oscillating, or so that the gate current can be monitored (shunt resistor). Not one of the app notes even suggested it was needed or desirable. I did read one that warned that if there was resistance in the gate, the MOSFET was more prone to spurious turn-on during dV/dt at the drain.

Where is this app note that says to add external resistance in the gate????

 

[ronv]

https://www.electro-tech-online.com/custompdfs/2010/10/slup169.pdf

This one is from TI. Page 10