# CMOS Transistor Small signal model

## What is the small-signal model?

The CMOS transistor is normally used as an amplifier when it is working in the saturation region (it has a low dependence with $V_{DS}$ and a high sensitivity to $V_{GS}$). To characterize the transistor for that purpose, it is used a model that considers that small signals are injected in the terminals of the transistor. That model linearizes the $I_D-V_{GS}$ and $I_D-V_{DS}$ curves around a point of operation, called the bias or operating point.

### The bias or operating point

Consider the following circuit:

A gate voltage $v_{GS}$ is applied to the transistor. A current will be generated that will match the current that flows through the resistance. This will in turn generate a voltage drop across the resistance and set the drain-source voltage of the transistor. This will lead to a stable condition for the transistor, for which we call bias or operating point.

The slider represents $R$. As you decrease $R$ (go left), the current sunk by the transistor has a small voltage drop at the resistance and the operating point goes to higher $v_{DS}$. If you increase $R$ (go right), the voltage drop is higher and $v_{DS}$ decreases. If the resistance was too large, it would put the transistor in triode region for this $v_{GS}$.

$R=$$k\Omega ### Linearizing the curves If a small signal v_{gs} is superimposed on a DC signal V_{GS} to produce v_{GS} = V_{GS} + v_{gs} at the gate of the transistor, the DC signal sets the bias and the small-signal will swing around that bias point. The curves around the bias point are not linear and to get a simple model for small signals that can be analyzed, we approximate the curves to lines to get a linear model. The linear model is based on taking the derivatives of the drain current with respect to the voltages at the terminals of the transistor. ### The small-signal equivalent model First let us find what components should make up this model. First we see that the transistor is a current source controlled by the gate voltage, so a voltage-controlled current source should be there. Next, we see that the current is also influenced by the drain-source voltage from the channel length modulation, which translates into an output resistance. There is no input resistance because there is no current flowing through the gate. Now it is time to determine the values of the transconductance g_m and the output resistance r_o. #### Transconductance g_m g_m is the change in current caused by a change in v_{GS}, so it is:$$g_m = \frac{\partial i_D}{\partial v_{GS}}.$$The equation for g_m depends whether the transistor is in triode or in saturation. g_m in triode g_m in saturation$$\frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(2(v_{GS} - V_{TH})v_{DS} - v_{DS}^2 ) }{\partial v_{GS}}\mu C_{ox}\frac{W}{L}v_{DS}\frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH})^2}{\partial v_{GS}}\mu C_{ox}\frac{W}{L}(V_{GS}-V_{TH})$$When in triode, g_m does not really depend on v_{GS} because i_D is linear with v_{GS} in this region. However, v_{DS} affects the transconductance linearly. When in saturation, g_m depends on V_{GS} and it is a parameter that can be tweaked to control the gain of the transistor. #### Output resistance r_o r_o is the inverse of a change in current caused by a change in the drain-source voltage, so it is:$$1/r_o = g_o = \frac{\partial i_D}{\partial v_{DS}}.$$g_o is the output conductance and is commonly used in exchange of r_o when it simplifies the equations. The equation for g_o depends whether the transistor is in triode or in saturation. g_o in triode g_o in saturation$$\frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(2(v_{GS} - V_{TH})v_{DS} - v_{DS}^2 ) }{\partial v_{DS}}\frac{1}{2}\mu C_{ox}\frac{W}{L}(2(v_{GS} - V_{TH}) - 2v_{DS} ) \mu C_{ox}\frac{W}{L}(v_{GS} - V_{TH} - v_{DS} ) \frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH})^2(1+\lambda v_{DS})}{\partial v_{GS}}\frac{1}{2}\mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH})^2 \lambda \approx I_D \lambda $$The output resistance is influenced both by v_{GS} and v_{DS} when the transistor is in triode. When in saturation, the output resistance is only defined by the bias current. From here we see that the transistor is not a very good amplifier in the triode region, because its gain is changed with the output voltage and we can only play with the transistor's size to control the gain. It is not a very good current source either, because its gain (and therefore its current) and its output resistance changes with the output voltage. The transistor is usually used in the saturation region for amplification. Also, to reduce the distortion caused by the amplification, the bias point should be somewhere in the middle of the saturation region, so that large swings of the drain voltage will not bring the transistor into the triode region or saturate to the power supply voltage. #### Include body effect The body effect changes the threshold voltage, which in turn affects the drain current. How can this be incorporated in the model? The body is like another gate (it is sometimes called a back-gate) and it is modelled as just that: Since v_{BS} is always negative, the current direction of the g_{mb} current source is actually the opposite of the g_m current source. To calculate the value of g_{mb}, we have to realize that there is a cascade of effects (v_{BS} to V_{TH} and V_{TH} to i_D):$$g_{mb} = \frac{\partial i_D}{\partial V_{TH}}\frac{\partial V_{TH}}{\partial v_{BS}}.$$\frac{\partial V_{TH}}{\partial v_{BS}} is expressed as:$$\frac{\partial V_{TH}}{\partial v_{BS}} = -\frac{\partial V_{T0_n} + \gamma \left( \sqrt{2\phi_f+V_{SB}} - \sqrt{2\phi_f} \right)}{\partial v_{SB}}\frac{\partial V_{TH}}{\partial v_{BS}} = -\frac{\gamma}{ 2\sqrt{2\phi_f+V_{SB}} }$$As in the other cases, the expression depends if the transistor is in triode or in saturation g_{mb} in triode g_{mb} in saturation$$g_{mb} = \frac{\partial i_D}{\partial V_{TH}}\frac{\partial V_{TH}}{\partial v_{BS}}g_{mb} = \frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(2(v_{GS} - V_{TH})v_{DS} - v_{DS}^2 )}{\partial V_{TH}}\frac{\partial V_{TH}}{\partial v_{BS}}\frac{1}{2}\mu C_{ox}\frac{W}{L}(-2v_{DS} ) \frac{\partial V_{TH}}{\partial v_{BS}}-\mu C_{ox}\frac{W}{L}v_{DS} \frac{\partial V_{TH}}{\partial v_{BS}}-g_m \frac{\partial V_{TH}}{\partial v_{BS}}g_m \frac{\gamma}{ 2\sqrt{2\phi_f+V_{SB}} }\frac{\partial \frac{1}{2}\mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH})^2}{\partial V_{TH}} \frac{\partial V_{TH}}{\partial v_{BS}}\frac{\partial \mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH})^2}{\partial V_{TH}} \frac{\partial V_{TH}}{\partial v_{BS}}-\mu C_{ox}\frac{W}{L}(v_{GS}-V_{TH}) \frac{\partial V_{TH}}{\partial v_{BS}} -g_m \frac{\partial V_{TH}}{\partial v_{BS}} g_m \frac{\gamma}{ 2\sqrt{2\phi_f+V_{SB}} } $$### Summary The model has three parameters: the gate-source transconductance g_m, the bulk-source transconductance g_{mb} and the output resistance r_o. The expressions for these parameters are summarized in the following table: Parameter Triode Saturation g_m$$\mu C_{ox}\frac{W}{L}v_{DS}\mu C_{ox}\frac{W}{L}(V_{GS}-V_{TH})$$r_o$$\frac{1}{\mu C_{ox}\frac{W}{L}(v_{GS} - V_{TH} - v_{DS} )}\frac{1}{I_D \lambda}$$g_{mb}$$g_m \frac{\gamma}{ 2\sqrt{2\phi_f+V_{SB}} }$\$

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