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How to determine mosfet gate driver current requirement?

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seyyah

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how to determine mosfet

How can we determine that a mosfet gate draws how much current and choose a proper mosfet driver?
 

jdhar

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mosfet driver application note

Not entirely sure, but it should draw no static power. If you are asking about power drawn upon switching, look at the capacitance of the gate and you can get an estimate with I = Cdv/dt, based on the rise-time of your pulse.
 

batdin

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determine mosfet gate voltage

If you talk about the average current,in TC4422 datasheet I found a graph helping me calculate it.For instance,if you drive 4700pF gate capacitance,at 100kHz the average current will be about 18mA.
The max.current is limited by the output resistance of the driver.TC4422 is capable of providing 9A peak current.
 

Iouri

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fet gate driver+rg

MOSFET it is voltage device, and it is drive by voltage source not curret like a regular BJT, data sheet for MOSFET ussually shows what level should be at the gate i.e. 10V or logic drive. Also data sheet call for the input capacitance Ciss. which is ussually makes to match trouble in the switching power supplies, depends how fast you want to drive your MOSFET, the you input circuit for the MOSFET will be different. There are some nice app notes avalible from the Texas INstrument website www.ti.com for TL5001 PWM controller

Have fun
 

SkyHigh

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ig current gate current

seyyah,

How can we determine that a mosfet gate draws how much current and choose a proper mosfet driver?

To choose a proper MOSFET driver, check with datasheets to find out the Vth, Vds, Ids, max Vgs, max Vds, max Ids, max freq ft, etc.

I am surprised how would you think MOSFET draws power at the Gate. It doesn't.

MOSFET as a driver passes current Source-Drain (PMOS) or Drain-Source (NMOS) because of the load connected either at the Source for NMOS or Drain for PMOS. In other words, MOS used as a driver is a Voltage-Controlled Current Source by varying Vgs.

Increase in load impedance causes higher power consumption, because you will need to increase the Vgs, hence increase the Ids, i.e the current to feed your load.

From Norton equivalent of an ideal current source or even an ideal controlled current source, the greater the load impedance, the lower the power efficiency because by means of KCL, the current divider is such that the current distribution between internal source impedance and load impedance is narrowing the magnitude gap or difference, resulting more current flowing into the internal impedance of the current source.
 

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how to calculate gate driver power consumption

The gate current requirement is ESTABLISHED by the designer, based on the required switching speed, since the switching action happens during the "plateau" region of the gate voltage. To shorten this time, you must drive the gate with higher current. The gate charge associated with this region is Qgd, due to the MIller effect. You supply this charge by injecting a current into the gate. The switching time is then tsw=Qgd/Ig. Thus, you choose Ig based on the desired switching time tsw and the MOSFET's Qgd.
The switching time is selected based on the the switching frequency and the allowable switching losses.

Now the gate of the MOSFET behaves as a (variable) capacitor. Therefore, the current Ig can only be limited by the driver's resistance or an external resistance. You select that resistor (if external) such that
Ig=(Vgmax-Vplateau)/Rg
where Vgmax is the maximum gate drive voltage you apply and Vplateau is calculated from the MOSFET datasheet for the actual drain current.

Based on the info above, you select a suitable gate driver (with a peak current at least equal to Ig) and an external resistor Rg to limit that current (if needed).

For more info, see application note U-118 from Unitrode (now TI). https://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=slua054

The average gate drive requirement (yes, you will need power to drive the MOSFET) is calculated based on the total gate charge of the MOSFET and the maximum applied gate voltage, as well as the switching frequency. That is because the energy you deliver to the gate capacitance when you turn on the MOSFET is actually lost when you turn it off. The net effect is a power loss, commonly referred to as gate drive loss.


If you need more info, please post.
 

gilbertomaldito

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leomecma

Hi, do you have other good references about this?

proper gate drivers for POWER MOS?
I have some problems with my gate MOS signal. I have a very long stepping at 6v during ton of my gate, and after sometime, it will reach 10v. This stepping is really unwanted and abnormal. I still dont know how to remove it.

_ please help
 

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the step in the gate voltage is due to the miller-effect talked about earlier in this post.

This is a common mistake in driving MOSfets, you do not calculate with the Cgs capacitance to determine the switching time (and so current needed). The Cgd contributs more to the actual switching time than the Cgs.

Switch on cycle :

1. When the gate driver switched on, you start charging the Cgs capacitance, nothing happens yet to the Vds until the Vth of the FET is reached.

2. When the Vds of the MOSfet starts to change (the beginning of the switching), you need to change the voltage on the Cgd capacitance from about the Vds to -Vth. This because Vds = Vdg + Vgs (the voltage on Cgs + the voltage on Cgd = Vds)
The rate Vds going down ONLY depends on how fast you can discharge the Cgd-capacitor! This energy (current) comes entirely from your gate drive circuit, so the more current you supply to the gate, the faster Cgd discharges, the faster Vds goes down, and the faster the switching takes place.

3. When the switching is over (Vds ~ 0V), you only charge Cgs to a higher voltage.

That's why the manufacturers of MOSFets give curves with 'total gate charge', and this on different Vds levels, because the importand parameters are Vds and Cgd. Since Cgd is voltage depended on its own, it is generally not enough to use only a single capacitance value.

So you can compare the switching time with the duration of the plateau on the gate voltage. Doubling the current, will double the switching speed.

88_1264283191.jpg
 

gilbertomaldito

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2. When the Vds of the MOSfet starts to change (the beginning of the switching), you need to change the voltage on the Cgd capacitance from about the Vds to -Vth. This because Vds = Vdg + Vgs (the voltage on Cgs + the voltage on Cgd = Vds)
The rate Vds going down ONLY depends on how fast you can discharge the Cgd-capacitor! This energy (current) comes entirely from your gate drive circuit, so the more current you supply to the gate, the faster Cgd discharges, the faster Vds goes down, and the faster the switching takes place.


For number 1. Yes you are right, charging Cgs, is not really a problem.

For number 2.
I just know that I can increase the current by adding a larger taper buffer. But it seems it still couldnt charge Cgd that fast. Is there something wrong with by taper buffer? I still couldnt solve my stepping. The problem worsens especially during high frequency switching.At high freq, stepping at 6v stays longer than my gate input of 10v.

Please check my gate signals in the attached screen shots. My stepping problem is worse during high freq switching of Power MOS.
 

girish1989

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Hi everyone,

I av a CAM(Content Addressable Memory) circuit, and my task is to find the W/L values of each and every transistor how do I find those...
I am using Tanner n if der's any procedure pls do let me know...
Genearally speaking how do I calculate those values coz at eacy transistor the cureent flow is different????
pls do the needful
 

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Re: how to calculate gate driver power consumption

The gate current requirement is ESTABLISHED by the designer, based on the required switching speed, since the switching action happens during the "plateau" region of the gate voltage. To shorten this time, you must drive the gate with higher current. The gate charge associated with this region is Qgd, due to the MIller effect. You supply this charge by injecting a current into the gate. The switching time is then tsw=Qgd/Ig. Thus, you choose Ig based on the desired switching time tsw and the MOSFET's Qgd.
The switching time is selected based on the the switching frequency and the allowable switching losses.

Now the gate of the MOSFET behaves as a (variable) capacitor. Therefore, the current Ig can only be limited by the driver's resistance or an external resistance. You select that resistor (if external) such that
Ig=(Vgmax-Vplateau)/Rg
where Vgmax is the maximum gate drive voltage you apply and Vplateau is calculated from the MOSFET datasheet for the actual drain current.

Based on the info above, you select a suitable gate driver (with a peak current at least equal to Ig) and an external resistor Rg to limit that current (if needed).

For more info, see application note U-118 from Unitrode (now TI). https://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=slua054

The average gate drive requirement (yes, you will need power to drive the MOSFET) is calculated based on the total gate charge of the MOSFET and the maximum applied gate voltage, as well as the switching frequency. That is because the energy you deliver to the gate capacitance when you turn on the MOSFET is actually lost when you turn it off. The net effect is a power loss, commonly referred to as gate drive loss.


If you need more info, please post.


please tell me how to calculate the gate drive current..
 

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