MOSFET Amplifier

MOSFETamplifier.gif MOSFETtransfer.gif

Assume in the circuit above $V_{in}=V_{GS}>V_T$ and the transistor is in saturation region, i.e., $V_{DS}>V_{in}-V_T$, then we have

$\displaystyle \left\{ \begin{array}{l}
I_D=K(V_{in}-V_T)^2 \\
V_{out}=V_{DS}=V_{dd}-I_D R=V_{dd}-KR(V_{in}-V_T)^2 \end{array} \right.$ (156)

The second equation relates the output $V_{out}$ to the input $V_{in}$, as shown by the red segment of the curve in the plot above. As the transistor is in saturation region,

$\displaystyle V_{DS}=V_{out}=V_{dd}-KR(V_{in}-V_T)^2\ge V_{in}-V_T$ (157)

which can be solved for $V_{in}-V_T$ to get:

$\displaystyle V_{in}-V_T <\frac{-1+\sqrt{1+4KRV_{dd}}}{2KR},\;\;\;\;$i.e.$\displaystyle \;\;\;\;
V_{in}<\frac{-1+\sqrt{1+4KRV_{dd}}}{2KR}+V_T$ (158)

We see that for the transistor to be in the saturation region, $V_{in}$ needs to satisfy both $V_{in}>V_T$ and the inequality above, i.e., it needs to be in the following range:

$\displaystyle V_T<V_{in}<\frac{-1+\sqrt{1+4KRV_{dd}}}{2KR}+V_T
\;\;\;\;$i.e.$\displaystyle \;\;\;\;
0<V_{in}-V_T <\frac{-1+\sqrt{1+4KRV_{dd}}}{2KR}$ (159)

When the transistor is in saturation mode the slope of the curve (red) indicates the ratio between input $V_{out}$ and output $V_{in}$, the voltage gain of the circuit:

$\displaystyle g=\frac{d V_{out}}{d V_{in}}=\frac{d}{d V_{in}}(V_{dd}-KR(V_{in}-V_T)^2 )
=-2KR(V_{in}-V_T)$ (160)

Example: Assume $V_{dd}=10V$, $R=10 k\Omega$, $K=0.5\;mA/V^2$, $V_T=1V$. For the transistor to be in saturation region, we need

then we have

$\displaystyle V_{out}=V_{dd}-K(V_{in}-V_T)^2 R=10-5(V_{in}-1)^2$ (161)

and the voltage gain is a function of the input $V_{in}$:

$\displaystyle g(V_{in})=\frac{d}{d\,V_{in}} V_{out}(V_{in})=-10(V_{in}-1)$ (162)

This nonlinear equation can be represented by the table below:
$V_{in}$ 0 1 1.4 1.5 1.8 1.9 2.0 2.1 2.2 2.3 2.32 2.35 2.4    
$V_{out}$ 10 10 9.2 8.8 6.8 6.0 5.0 4.0 2.8 1.6 1.3 0.9 0.0    
In particular, when the input $V_{in}$ increases from $1.8V$ to $1.9V$, the output $V_{out}$ decreases from $6.8V$ to $6V$, with a gain $g(V_{in})=g(1.8)=-10(1.8-1)=-8$. Also when the input $V_{in}$ increases from $2.0V$ to $2.1V$, the output $V_{out}$ decreases from $5V$ to $4V$, with a gain $g(V_{in})=g(2)=-10(2-1)=-10$.


In summary, we see that

Next we consider a MOSFET circuit with sinusoidal input. Assume the drain resistor is $R=10\;k\Omega$, $K=1\;mA/V^2$, $V_T=1V$, $V_{dd}=5V$ and a sinusoidal input $v_{in}(t)=V_{bias}+\sin(\omega t)$. If the bias voltage is $V_{bias}=1.5V$, the input voltage $v_{in}(t)$ varies between 1.4V and 1.6V. The output voltage can be found to be:

$\displaystyle v_{out}=V_{dd}-RI_D=V_{dd}-RK(V_{in}-V_T)^2=5-10 (V_{in}-1)^2$ (163)

In particular, corresponding to $V_{in}=1.4V,\;1.5V,\;1.6V$, the output voltage $v_{out}$ and the current $i_{DS}$ are, respectively, $v_{out}=3.4V,\;2.5V,\;1.4V$, and $i_{DS}=0.16mA,\;0.25mA,\;0.36mA$, as shown in the figure below:


Biasing: In the example above, the DC offset of the input is at 1.5V, so that the transistor is working in the saturation region when the magnitude of the AC input is limited. However, if this offset is either too high or too low, the gate voltage may go beyond the saturation region to enter either the triode or the cutoff region. In either case, the output voltage will be severely distorted, as shown below:


It is therefore clear that the DC offset or biasing gate voltage has to be properly setup to make sure the dynamic range of the input signal is within the saturation region.

Method 1: One way to provide the desired DC offset is to use two resistors $R_1$ and $R_2$ that form a voltage divider, as shown in the figure below (a). As the input resistance of a MOSFET transistor is very high, therefore the gate of the transistor does not draw any current, the DC offset voltage can simply obtained as:

$\displaystyle V_{bias}=V_{dd}\frac{R_1}{R_1+R_2}$ (164)

The input AC signal through the input capacitor is then superimposed on this DC offset.


Method 2: Another way to set up the bias is the circuit shown in (b) above. Assume $R_1=84k$, $R_2=16k$, $R_d=20k$, $V_{dd}=10V$, and $K=0.5mA/V^2$. The bias voltage can be found to be $V_B=V_{bias}=1.6V$, and the voltage between gate and source is $V_{GS}=V_B-V_{in}$. The output voltage is

$\displaystyle V_{out}=V_{dd}-R_d K (V_{GS}-V_T)^2=V_{dd}-R_d K (V_B-V_{in}-V_T)^2=10-10\times
(0.6-V_{in})^2$ (165)

When $V_{in}=0$, $V_{out}=6.4V$.

To determine the dynamic range of the input $V_{in}$, recall the conditions for the transistor to be in saturation region:

Therefore the overall dynamic range for the input is

$\displaystyle -0.37V \le V_{in} \le 0.6V$ (170)

with the corresponding output range

$\displaystyle 0.6V \le V_{out} \le 10V$ (171)

and the overall voltage gain is about $g=9.7$. Note that the output voltage is in phase with the input voltage.

Source Follower: If the output is taken from the source, instead of the drain of the transistor, the circuit is called a source follower.


Assume $R_s=10 k\Omega$, $V_T=1V$ and $K=10 mA/V^2$. To find the input and output voltages and the gain of the circuit, consider the current $I_D=V_{out}/R_s$:

$\displaystyle I_D=K(V_{GS}-V_T)^2=K(V_{in}-V_{out}-V_T)^2=V_{out}/R_s$ (172)

Plugging in the given values, we get

$\displaystyle (V_{in}-V_{out}-1)^2=V_{out}$ (173)

If $V_{in}=2V$, this equation becomes:

$\displaystyle (V_{out}-1)^2=V_{out}$ (174)

which can be solved to get $V_{out}=2.6V$ or $V_{out}=0.4V$. We take the smaller voltage in order for the transistor to be outside the cutoff region:

$\displaystyle V_{GS}=V_{in}-V_{out}=2-V_{out}>V_T=1$ (175)

Similarly, if $V_{in}=3V$, the equation becomes:

$\displaystyle (V_{out}-2)^2=V_{out}$ (176)

and we get $V_{out}=1V$. The voltage gain of the source follower is

$\displaystyle g=\frac{\Delta V_{out}}{\Delta V_{in}}=\frac{1-0.4}{3-2}=0.6<1$ (177)

To maximize the dynamic range for the input AC signal, the DC operation point in terms of the DC variables $\{V_{GS}, I_D, V_{DS}\}$ needs to be set around the middle point of the saturation region. If the AC signal around the DC operation point is small enough, the behavior of the circuit can be linearized (first term of Taylor expansion of the nonlinear relationship) to simplify the analysis.


Specifically, the nonlinear relationship between $V_{GS}$ and $I_D$ can be linearized around the DC operation point for small changes:

$\displaystyle g_m=\frac{\Delta I_D}{\Delta V_{GS}}$ (178)

Here $g_m$, called incremental transconductance, is the ratio between small change in $I_D$ and the small change in $V_{GS}$.