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I have some doubts about this so-called transconductance which can be either calculated for BJT or for FET transistors. As far as I know, since BJTs are current controled devices, its transconductance (gm) differ from the FETs.

BJT's gm=Ic/Vt (Vt -> thermal voltage ~= 25mV at room temperature)

FET's gm= Id/Vgs (Vgs -> Gate-Source Voltage)

My question is: When you design a MOSTFET common-source amplifier and you want to know its gain, do you always need to calculate it's gm first? And if so, how do you start your calculations?

When I look at the datasheet of a MOSFET, the only thing is given related to transconductance is:

enter image description here

Is this helpful for our calculation?

Here's the schematic of the amplifier:

enter image description here

They say the fallowing: "The gain of this amplifier is determined partly the transconductance of the MOSFET. This depends on the bias point of the circuit, here it averages about 9mA/V. This means that a change in the gate voltage causes a change in the drain current that is 9mA/V times the gate voltage change"

Does this mean that the MOSFET's transconductance is 9mA/V And if so, how did they found out this ?

They even say that " A change in input of 50mV causes a change in drain of voltage of 9mA/V * 50mV * 4000 ohm which is equal to 1.8V". And dividing this 1.8V with the input voltage 50mV will result the gain of this amplifier which is 36. But how did they reach to this calculation?

On the other hand my book says that the voltage gain can be calculated with this formula:

Gain = vd/vgs = (-Rd*id)/Vgs and we can rewrite this as: Gain = -gm * Rd

So if I compare this formula to the one that they used above, with gm * Vgs * Rd there will be obviously a difference which makes me confused.

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  • \$\begingroup\$ When I design a common source or common emitter amplifier I define the voltage gain by leaving an unadulterated resistor in the source so that voltage gain is Rd / Rs. This improves distortion dramatically. \$\endgroup\$ – Andy aka Jun 2 '18 at 12:46
  • \$\begingroup\$ I know, maybe I shouldn't bother myself doing in this way, but still, I'm very curious about this transconductance thing.. \$\endgroup\$ – Simon Maghiar Jun 2 '18 at 14:39
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    \$\begingroup\$ @SimonMaghiar gm is a "gain" for any transconductance amplifier. gm = Iout/Vin. So if gm is 9mA/V any change in the input voltage by 1V will change the output current by 9mA \$\endgroup\$ – G36 Jun 3 '18 at 11:08
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\$g_{m}\$ is indeed an important value when you are designing an amplifier, not only does it dictate the gain but also the bandwidth, noise performance, and linearity of the circuit. When we place a resistor from source to ground, we decouple ourselves from the intrinsic \$g_{m}\$ of the FET so while still important, it becomes little a less relevant because we no longer have a 1 to 1 dependency on it.

I'm going to try to explain \$g_{m}\$ in terms of MOSFET characteristics first and then I'll move to a small design example see (Design example below)

gm in terms of MOSFET parameters
For a N-channel MOSFET in saturation region, the drain current can be defined as:

\$I_{D} = \frac{1}{2} \mu_{n}C_{ox}\frac{W}{L}(V_{GS} - V_{th})^2(1 + \lambda V_{DS})\$

For simplicity let's assume that the channel length modulation \$\lambda = 0\$ thus

\$I_{D} = \frac{1}{2} \mu_{n}C_{ox}\frac{W}{L}(V_{GS} - V_{th})^2\$

Now, the transconductance \$g_{m}\$ for a MOSFET is defined as the change in drain current \$I_{D}\$ with respect to input voltage \$V_{GS}\$:

\$g_{m}=\frac{\delta I_{D}}{\delta V_{GS}} = \mu_{n}C_{ox}\frac{W}{L}(V_{GS} - V_{th})\$

After some algebraic manipulation, \$g_{m}\$ can also be written as:

\$g_{m}=\sqrt{2\mu_{n}C_{ox}\frac{W}{L}I_{D}}=\frac{2I_{D}}{V_{GS} - V_{th}}\$

Where:

  • W: Width of transistor (you can control this if you are designing the IC)
  • L: Length of transitor (you can control this if you are designing the IC)
  • \$\mu_{n}C_{ox}\$: Are the mobility and the oxide capacitance respectively (assume constant)
  • \$V_{GS}\$: the gate to source voltage
  • \$V_{th}\$: the threshold voltage (assume constant)

So unless you are designing the IC and have control over \$\frac{W}{L}\$, the only "knobs" you can control are the DC operation point \$I_{D}\$ & \$V_{GS}\$. Once these two parameters are set, the \$g_{m}\$ is pretty much set for that \$I_{D}\$, \$V_{GS}\$ combination.

Now, as far as the small signal gain of the circuit is concerned , the output voltage \$v_{out}\$ can be defined as: (using lower case variables to refer to small signal)

Assuming the source resistance is zero to simplify the analysis

\$v_{out}\$ = \$i_{D}R_{eq}\$

Where \$R_{eq}\$ is the total equivalent resistance at the output:

\$R_{eq} = R_{Drain} || R_{Load} || r_{o}\$

where \$r_{o}\$ is the output resistance of the transistor in saturation region which you can assume to be large. Assuming \$R_{Drain} << R_{Load} << r_{o}\$

\$R_{eq} \approx R_{Drain}\$

thus

\$v_{out}\$ = \$i_{D}R_{Drain}\$

defining the small signal transconductance as

\$g_{m}=\frac{i_{D}}{v_{GS}}\$

rearranging

\$i_{D}=-g_{m}v_{GS}\$, where \$v_{GS}\$ is your input AC voltage

bringing everything together, the voltage gain can be defined as:

\$\frac{v_{out}}{v_{in}}=A_{v}=-g_{m}R_{Drain}\$

What happens when we have a resistor connected from source to ground

When we put a resistor at the source terminal we introduce local feedback which reduces the \$g_{m}\$ but increases the linearity of the circuit in a process called "source degeneration". By analyzing the small circuit equivalent we end up with:

\$g_{m} \approx -\frac{1}{R_{Source}}\$

So the gain \$A_{v}\$ can be expressed as:

\$\frac{v_{out}}{v_{in}}=A_{v}=-\frac{R_{Drain}}{R_{Source}}\$

Now you have more control of the gain and linearity of the circuit.

Design example
Say, you were to design a stand alone amplifier, and you have a target gain of 10 V/V and this gain is dictated by amplification needed, noise level, etc. Thus:

\$A_{v}=-g_{m}R_{Drain} = 10\$

for the MOSFET to be in saturation, we need \$V_{DS} > V_{GS} - V_{th}\$ and the higher \$V_{DS}\$ the better linearity we get from our circuit. However, there's never free lunch, the higher the \$V_{DS}\$, the smaller the \$v_{out}\$ can swing in the positive direction without hitting the supply. So to be conservative lets say you set your \$V_{DS} = 200mV\$. Now what happens if \$V_{out}\$ swings \$100mV\$ in the negative direction? That will reduce our \$V_{DS} = 200mV\$ to \$100mV\$ and we risk leaving the saturation region. Since we know the swing requirement for \$V_{out}\$, we can add a \$100mV\$ margin to provide some headroom. Now:

\$V_{DS} = 300mV\$

\$R_{Drain} = \frac{V_{DD} - V_{DS}}{I_{D}}\$

As a rule of thumb and in the interest of linearity, the DC bias current \$I_{D}\$ can be set as

\$I_{D} = 10\times i_{D}\$ where \$i_{D}\$ is the ac output current we are trying to deliver.

Now you have all the values you need to start your design. Notice though that our design heavily depends on \$g_{m}\$ which varies across temperature and unless we are designing the IC, we have little control over.

Having precise control over \$g_{m}\$ might or might not be an issue but that's why people use the source resistance \$R_{Source}\$ to rely on the ratio of two resistors rather than the internals of the FET. So when we take this approach we don't really need to worry much about the value of \$g_{m}\$.

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  • \$\begingroup\$ Thank you for the great reply! Even though my questions wasn't answered, you pointed out some good facts that helped me to analyse my circuit :) \$\endgroup\$ – Simon Maghiar Jun 3 '18 at 9:01
  • \$\begingroup\$ @SimonMaghiar I realized that I kinda got sidetracked in the intrinsics of the FET and forgot to answer your question =]. I've added an intro and a design example section to my answer to try to fix that. \$\endgroup\$ – StaticBeagle Jun 4 '18 at 15:42

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