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[SOLVED]Drain-current calculation in High-side N-channel Mosfet switch.

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chandresha1

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I am working on H-bridge circuit with PWM. It is basically class-D amplifier with four N-channel Mosfet. However, i am stuck at one-point. How to calculate the drain current for high side Mosfet? The configuration is shown in figure.

I am applying 30V Gate voltage(Respect to Ground). Drain is connected to the positive power supply rail of 20V. Source is connected with load.

What i have understood is in such a configuration, when i apply the Gate voltage higher than Vds, Mosfet works in triode mode. In other words it will act as a switch with very small ON-resistance of Rds(on). Because of this, the potential difference between drain and source becomes almost zero or around 200-300 mV.(Vs = Vd-id*Rds(ON).) At this point i am getting confused. According to Mosfet characteristic curve, for Vds=0, the drain current is zero. Than how can we say mosfet acts as a switch? Because according to curve, Mosfet will conduct very small current.

Curves on a datasheet are for a "typical" device that you cannot buy. You might get a better one or get a "minimum" one which has spec's PRINTED in the datasheet. All are guaranteed to have a maximum on-resistance at the temperature and gate-source voltage printed in the datasheet.

The on-resistance follows Ohm's Law. With an ID higher current then the VDS voltage drop across it is higher as shown on the graph. With no current then its voltage drop is nothing.
Sure it is a switch. When it is turned on then it is a very small resistance between the power supply voltage and the load.

I have drawn this curve by myself just for understanding.

I actually understood that Mosfet will act as a switch with very low on-resistance. But i m getting confused at below equation of drain current in triode mode.
Id=K(Vgs-Vth)Vds, for Vds<<Vgs-Vth.

According to this equation for very small voltage drop(Vds), the drain current id should also be very small. Than how can we say it will act as a switch and the drain current will follow the ohm's law?

When the Mosfet is turned on then it is a low value resistor with a maximum resistance printed on the datasheet. An old IRF540 has a maximum on-resistance of 0.044 ohms (44milli ohms) when its chip is at 25 degrees C and its VGS gate-source voltage is 10V.
Then with no current it is a piece of wire (wire has resistance too you know) with no voltage drop, with 1A its voltage drop is 44mV and with 30A its voltage drop is 30A x 44m= 1.32V or less.
Its resistance rises when it heats then the voltage drop also rises.

Not sure where you got this old chart, but isn't Saturation on the Left side of the curve where Vds approaches zero?

Here is a modern N-ch MOSFET that costs \$0.10 in production volume with 54mOhms Max RdsOn with 4.5V "Logic Level Vgs control.

Over the last few decades Vgs thresholds have dropped from 4 to 2.5 typ and now production types are as low as 400mV for Vgs

chandresha1

chandresha1

Points: 2
Not sure where you got this old chart, but isn't Saturation on the Left side of the curve where Vds approaches zero?
That is for a bipolar transistor. A Mosfet is the opposite. Look at all the different names for the regions in Wikipedia.

That is for a bipolar transistor. A Mosfet is the opposite. Look at all the different names for the regions in Wikipedia.

It looked like that was a J-Fet Graph but I see now, they refer to the Gate saturation as opposed to BJT Vce saturation.

I have read these terminology in plenty of document. The left part of characteristic graph termed as Triode region or ohmic region or liner region for the condition of Vgs-Vth>> Vds. The right side termed as saturation region(Vgs-Vth<Vds) as drain current becomes constant regardless of Vds. You can see this terminology in fairchild document.
https://www.fairchildsemi.com/application-notes/AN/AN-9010.pdf
As sunnySkyguy has mentioned the drain current for different value of Vgs it is helpful to understand drain current when Mosfet is fully turned-on or in triode region. But still i am not able to understand mathematical relationship between drain current and Vds.
According to drain current equation id = k(Vgs-Vth)Vds, for very small Vds in my case, the drain current should also be very small and in other words it is in controlled manner than how can we say the mosfet acts as a switch? Because a switch will not impose any limits on current flowing through it. In worst case, if the switch is not able to handle excess current than it will burn out but it won't resist the current to flow through it. But here small drain current introduces a limit to maximum current. Am i right or missing something?

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Through your posts, you're comparing a real MOSFET with an ideal (zero ohms) switch and wondering why they are not the same. To bring the discussion down to real world, we should state firstly state that any real real switch has a resistance and isn't able to conduct infinite currents with no voltage drop. This is also true for mechanical contacts which are offen postulated as ideal switches in simplified equivalent circuits.

Secondly, did you look at the characteristic of real MOSFETs, e.g. the Si2356 output characteristic shown above? I think the diagram illustrates well why the transistor can be described as resistive switch for Id < 20 A. There's still a slight curvature, but V/I is roughly a line, as expected for a resistor. What's the problem?

for very small Vds in my case, the drain current should also be very small
You understand it backwards. Since a turned on Mosfet has an extremely low resistance then for a very small drain current the Vds is also very small. Its Vds is also fairly small when its drain current is fairly high which is the same for a mechanical switch.

chandresha1

chandresha1

Points: 2
Yes it seems i understood the whole concept in a different way. Yes the current depends on potential difference only. So if Vds is small than definitely the current will be small. Thanks everyone for clearing my doubt.

If Vds is in loop with V+ and low impedance in series,
the current will be according to Ohm's Law
with all the loop resistance include RdsOn and supply & load series resistance.

Usually one chooses a supply ESR< 1% of load and MOSFET RdsON of 1~5% of load unless you want to reduce temp rise at very high currents, then <<1% of load R

Choices of RdsOn range wide from 300 Ohms for CD4xxx CMOS to < 1mOhm for very low loss.

If Vds is in loop with V+ and low impedance in series,
the current will be according to Ohm's Law
with all the loop resistance include RdsOn and supply & load series resistance.

Actually i am designing Class-D amplifier and for the high-side MOSFET, yes Vds is in loop with V+. I have attached diagram. But still i have one doubt. As shown in my block diagram, when high-side MOSFET Q1 turns-on the Vds becomes very small due to the small on-resistance. It is around 100mohm. I am using PMV130ENEA MOSFET. Gate drive voltage Vgs is around 11 volt higher than the input voltage at drain. But in MOSFET datasheet output characteristic curve, it is shown that for small Vds near to zero the drain current is around 1A only. For a moment assume my load is only 1 ohm than according to ohm's law the current should be 25A for 25V supply voltage at drain terminal.(little bit less because of MOSFET on-resistance). But according to datasheet it is not possible to draw more than 1A for such small Vds. I know something i am not understanding correctly. I would really appreciate if you can explain me this concept.

I wonder why you're permanently switching between different transistor with rather different data. PMV130ENEA has about 0.1 ohm rdson, which shows in the output characteristic.

But according to datasheet it is not possible to draw more than 1A for such small Vds.
According to the datasheet, minimal Vds for 1 A is about 100mV. Just Ohms law.

The curves in a datasheet are for a "typical" device that you cannot buy. Some are maximum and some are minimum. You get whatever they have.
The text in a datasheet states the guaranteed values for a minimum device so the on-resistance at a certain current is maximum at a cool temperature. Your Mosfet might have a typical on-resistance of 0.1 ohms at 1A then at 5A its Vds will be 0.1 ohms x 5A= 0.5V which is shown on the curves on the datasheet. You might end up with Mosfets with a much higher on-resistance so design the cooling for them.

With a 25V supply and a 1 ohm load the typical current will be 25V/ (0.1 ohms + 1 0hms)= 22.727A, not 25A.

Please try to understand that a turned on Mosfet is a resistor. A high current in a resistor produces a higher voltage drop across it than a lower current.

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chandresha1

chandresha1

Points: 2
Thanks a lot to everyone....... I felt i am such a stupid, i couldn't understand even a ohm's law properly.:-( As the current will increase, voltage drop will also increase across the device. Its as simple as its sound. Now i am satisfied with explanation.

Goody, now you understand.:smile:
You are not a stupid dimwit, you simply misunderstood and now you know how it works.

Just for the record. The Vgs should never exceed the Vgsmax voltage, normally +/-20V. You can add a zener diode between gate and source, and a series resistor for the gate input, this will limit the Vgs to the zener voltage, like 12V.

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