Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-9-part-1/
Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/
There needs to be a device that has a control port and output port which is a VCCS. It turns out that the MOSFET under certain conditions will do what is needed to be done.
The MOSFET has a gate , drain and a source terminals but it does not look like the VCCS which is a four terminal device. In order to make it look like the VCCS, the source terminal needs to be shared with the input and output port. To work, some voltage needs to be applied between the gate and source terminals and that is treated as the input port. Then there will be a current that goes from the drain to the source terminal.
The MOSFET has a high input gate resistance so a very small current is needed to turn it on. For the course, the input current is taken as zero. The on-state behaviour of the MOSFET is a bit more complex than the ideal switch or switch resistor model. When v_GS is more than V_T , the MOSFET is turned on.
What really happens as can be seen graphically (i_DS vs v_DS), is that the MOSFET behaves like a resistor in the beginning (small v_DS) but later saturates and behaves like a current source. Saturation occurs when v_DS is more than or equal to v_GS minus V_T. The triode region is where the MOSFET behaves like a resistor. The saturation region is where it behaves like a current source.
For different values of v_GS, the current through the MOSFET changes. When v_GS is lower, the current will also be lower.
In the saturation region, i_DS is some function of v_GS.
i_DS=k/2 ( v_GS - V_T)^2
For digital designs, the SR model is used. For analog models, the SCS model is used. This is the model where the MOSFET behaves like a dependent current source (saturation region).
In more advanced courses, a combination of SR and SCS models can be used.
To ensure the MOSFET operates as a VCCS , it must be operated in the saturation region only so it needs to adhere to the saturation discipline. This is the case where v_GS is more than or equal to V_T AND v_DS is more than or equal to (v_GS - V_T).
Let's analyze the circuit. First, replace the MOSFET with its SCS model.
i_DS=k/2 ( v_GS - V_T)^2 for the saturation region where v_GS is more than or equal to V_T AND v_DS is more than or equal to (v_GS - V_T).
There are two ways of analyzing the circuit- (1) the analytical method (2) the graphical method.
How does the output change with respect to the input (v_O versus v_I) ?
Node analysis is used. The equation for I_DS is substituted into the equation obtained from nodal analysis. A result v_O=V_S - (k/2(v_I - V_T)^2)*R_L is obtained for the saturation region. v_O=V_S when the MOSFET is off.
For the graphical method, the equations for v_DS (marked as A) and v_O (marked as B) are used.
Graphs of i__DS versus v_DS for the rearranged equations of A and B will be plotted.
Saturation begins at a point where v_O = v_I - V_T . This equation is substituted into the equation marked A giving an equation i_DS=(k*v_O^2)/2 to find where saturation begins.
Equation B is rearranged to get i_DS=(V_S - v_O)/R_L .
The plot of the rearranged equation of B known as the load line is then superimposed on the plot of A. Given a value of V_I, I_DS and V_O should lie on the load line.
Large signal analysis of amplifier has two steps -
(1) v_O versus v_I
(2) Find valid input operating range and valid output operating range (to guarantee saturation region operation.)
The graph of v_O versus v_I is plotted. First take note of V_T. When v_I is less than V_T, the MOSFET is in the cut-off region and v_O =V_S. When v_I is more than or equal to V_T AND v_O is more than or equal to (v_I - V_T), the MOSFET is in the saturation region and v_O=V_S - (k/2(v_I - V_T)^2)*R_L. When v_O is less than (v_I - V_T), the MOSFET is in the triode region. For example when v_I=3V and V_T=1V, then the v_O point where it gets into the triode
region = 3-l = 2V
The saturation region is the valid region for analog systems. The cutoff region and the triode region are for digital systems.
The valid operating ranges under the saturation discipline is the loadline that meets the boundary of the cut off region and the boundary of the triode region. The point at cutoff occurs when i_DS=0 (v_I=V_T ) and v_O=V_S. The point at the triode region is solved from the two equations of i_DS :-
i_DS=(V_S - v_O)/R_L and i_DS=(k*v_O^2)/2
Equating these two equations, it is possible to get a quadratic equation. The solution of this quadratic equation is v_O=(-1+((1+2*K*R_L*V_S)^0.5))/(k*R_L)
With that solution, it is possible to find v_I and i_DS. v_O=v_I - V_T at the boundary of the triode region so v_I=v_O + V_T .
Another way of obtaining the valid operating range is using the v_O versus v_I curve. The valid input operating range lies between v_I=V_T and the boundary point of the triode region where v_I=v_O + V_T. The valid output operating range lies between v_O=V_S and v_O=v_I - V_T.
To get the point at the triode region graphically, recall that v_O=v_I - v_T. If you draw the line v_O=v_I , you will find that v_O=v_I - V_T is a translation of v_O=v_I. From the translation, the point can be obtained graphically. The other point at the cutoff region boundary is at (V_T, V_S).
To get the point at the triode region mathematically, substitute v_O=v_I - V_T into
v_O=V_S - (k/2(v_I - V_T)^2)*R_L and solve for v_O. Once v_O is obtained, v_I= v_O + V_T.
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