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Transconductance VS Vgs in MOSFET

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Amninder

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transconductance mosfet

Hi,
I am new to this filed on analog design. I am reading the book by Behzad Razavi on "Design of Analog CMOS Integradted Circuits"

While reading on MOSFETs, I encountered an equation for transconductance which says that : gm = 2 * Id / Vgs - Vth .Its explanation states that Transcondutance decreses with overdrive (Vgs - Vth) when Id is constant. I did not undestand that how Id can be constant when we are varying the overdrive.A change in Vgs will change the drain current even in saturation region.

Can anybody explain this contradiction? It would be of great help.
 

transconductance of mosfet

Isn't transcondutance change in drain current over change in gate voltage?

gm = ΔId/Δ(Vgs - Vth)
 

mosfet transconductance

Simple first order model for MOSFET current:

Id=K*(VGS-VT)^2 (in saturation)
where K depends on W and L (transistor width and length respectively)

gm=dId/dVGS=2*K*(VGS-VT)=2*K*(VGS-VT)^2/(VGS-VT)=2*Id/(VGS-VT)

when VGS varies, Id can be held constant by adjusting W (device's width)
 

mosfet vgs

Basically when W is increased, V_overdrive will go down and hence g_m will go up and vice versa. So this gives you a feel that to increase the gain you will need a wider transistor (of course you will pay some penalty because capacitance will increase).
 

fet transconductance

Thanks to all for helping! But I still have one doubt....The width of a transisitor is a physical quantity that remains constant when the design of a chip is finished. So, Is there any way to keep the Drain current constant by varying the Vgs.
Can we do it by varying the channel length by changing the souce-substrate voltage?
 

id = gm (vgs-vt)

The length of the transistor is also a physical quantity that cannot be changed when the design is fabricated. When you change vgs, the drain current changes. Youcan suppres this change in vgs incresing the thrashold voltage via vsb, but this always requires a very large change of vsb to eliminate a small contribution of vgs.
 

vgs vs id

Amninder,
Best way to understand this phenomenon is to simulate the circuit.

You are correct that by chaning Vgs, Id will also change. But in the circuit if you don't allow the current to change, then gm will decrease by increase of Vgs over overdrive. In the circuit this can be possible if the current source or sink is Ideal Current Source.
 

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    Amninder

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mosfet gm equation

Amninder said:
Hi,
I am new to this filed on analog design. I am reading the book by Behzad Razavi on "Design of Analog CMOS Integradted Circuits"

While reading on MOSFETs, I encountered an equation for transconductance which says that : gm = 2 * Id / Vgs - Vth .Its explanation states that Transcondutance decreses with overdrive (Vgs - Vth) when Id is constant. I did not undestand that how Id can be constant when we are varying the overdrive.A change in Vgs will change the drain current even in saturation region.

Can anybody explain this contradiction? It would be of great help.
Click to expand...

it is saying that when Vgs is increasing, you have to lower gm to hold Id constant.
 

ids vs vgs

K can also change
 

In a common source amplifier with resistor as load, when vds is contant and MOSFET is completely driven to saturation by vgs, further increase in overdrive voltage will not improve the drain current. This may be because vds decreases considerable and thus MOSFET enters in triode region. In this region if Vgs-Vth increase, current remains the same. At this situation, as Idc is contant, gm decreases sharply with increase in overdrive voltage.
 

It's possible when MOS drain connected to the current source. Current source always keepin current constant.
 

    A

    Amninder

    Points: 2
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for details refer razavi page 49-51 ...
 

Thanks a lot to all for helping. @ fubsyman and freescale_bharat , I agree with your explanation and it seems to be the best possible solution. But this approach indicates that transconductance is characteristic of the Circuit used and not the transisitor itself.
In other words, say we calculated the gain of a common source amplifier using a resistive load and it comes out to be equal to (A1 = gm1 * R_load). Now if , I change the load of this circuit with current source, then say the gain is given by (A2 = gm2 * R_current_source ).
My question:- Is gm1 = gm2 ? or it would change depending upon the value of current source?
 

I see nobody except one guys has given very correct explanation of this phenomenon. I am explaining in case a new user arrive and find what exactly happening in this equation:-

Explanation assume a CS amplifier with constant current source load. This should be noted that as load is a constant current source , so its not gonna change and it is feeding to a Mosfet sized properly and given by a proper bias to get a good output common mode voltage. Now as we increase the gate voltage, we are making our mosfet to increase the current which was initially biased at Iref. Since current cannot change in the branch , so transistor come out of saturation to follow KCL at output node. Since transistor is no longer in sat mode and entered into triode region, we know gm is always less in this region.
 

I am explaining in case a new user arrive and find what exactly happening in this equation:
(...)
Since current cannot change in the branch , so transistor come out of saturation to follow KCL at output node. Since transistor is no longer in sat mode and entered into triode region, we know gm is always less in this region.
Click to expand...
Objection! Unfortunately you are misunderstanding the formula in the same way as the original poster and many contributors did. In equation 2.19 (Razavi page 21) "Id is constant" implies that transistor geometry W/L is varied. Transistor is staying in saturation.
 
" In equation 2.19 (Razavi page 21) "Id is constant" implies that transistor geometry W/L is varied. Transistor is staying in saturation."
In this equation gm = 2 * Id / Vgs - Vth, If Id is a not changing, you change Vgs - Vth , gm will change. I actually think, when you keep Id constant, you don't change size. In fact in this formula nowhere W/L is appearing. Again my explanation was based for a CS amplifier with constant current source load.

Now with what you said. " W/L should be decreased if Voverdrive is increasing to keep the current constant. I guess this is equation 2.17. But this is not the point here.

Apologies if my previous explanation was not good enough.
FvM, please explain me why my explanation is not correct. I am not an expert in analog. I want to know what I am missing. It will increase my learning.
 

The point is that you didn't understand the meaning of the equation. It doesn't describe the behaviour of a transistor circuit. It is one several equations describing the interdependence of transistor parameters. Although W/L doesn't appear in the equation, it's not assumed constant.

You can't see this intended meaning from the equation itself, you need to review the context, Razavi Design of Analog CMOS Integrated Circuits, 2.2.2 Derivation of I/V Characteristics, page 21.
 
FvM,
Thanks for correcting me. The equation of gm is dId/dVGs = gm assumes that VDs is constant. So the explanation of transistor going in triode region with output going low is plain wrong.
What I was explaining was also correct but it doesn't apply in this context ( as for as this equation is concerned )
Thanks.
 

I suggest you refer to slide 0123 from Analog Design Essentials by Willy Sansen to clarify this. The 2Id/(Vgs-Vt) equation is the one that people use while designing.

Say we have a gain of 2000 now and we want a gain of 4000, so, having the denominator fixed, If the current is doubled, then gain increases two times. However, as FvM pointed out, W (or L for that matter) is not fixed here. (gain here refers to small signal gm)

However, having w/L fixed and doubling the bias current only increases gm by around 41%. (its gm proportional to root Id. Doubling it gives 1.414 times previous gm, an increase of 41%)
 

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