Revision of Circuits and Electronics - Amplifiers - small signal model (Lecture 10)

Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-10

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/


To make the MOSFET operate in the saturation region,  the signal first needs to be boosted by applying a DC value, a DC voltage source, V_I , then after that apply  a signal of interest, V_A which operates within the saturation region.

The output is an inversion  of the input and there is a fair bit of distortion as the saturation region is not a straight line. There is amplification for sure  but the signal is distorted so the amplifier is non linear.

 How do we get a linear amplifier?  This can be done with the small signal trick, where we focus on a small piece of the non-linear curve. So after boosting with a DC voltage,  the input voltage needs to be shrunk to a small signal that gives a linear response.

 Let's look at the small signal method
 (1)graphically,   (2) mathematically, and (3)  from a circuit viewpoint -  in the next sequence.

(1) Graphically, recall the small Signal Model Notation  where v_I (total variable)= V_I (dc bias) + v_i (small signal)   and v_O=V_O + v_o. From  this method we will notice that a linear output can be obatined.

(2)  Mathematically,  first we need to substitute v_I=V_I+ v_i  into
v_O=V_S -( k/2(v_I - V_T)^2)*R_L.  The Bias Point Equation is  V_O=V_S - (k/2(V_I - V_T)^2)*R_L  and this is marked with an asterisk, to be used later. Then,v_O=V_O+ v_o is substituted as well. From this point we should realize that v_o  is an amplification of v_i , giving  an equation v_o= A*V_i where A is the amplification.

 Removing the terms of V_O  ( the  bias point equation), and neglecting higher  order terms of v_i, we will get v_o=- (k*R_L(V_I - V_T))*v_i.


 v_o=- g_m*R_L*v_i   where   g_m=k*(V_I - V_T)  so  v_o=- A*V_i . A is a constant w.r.t. v_i

so the circuit behaves linearly for small signals.

 Another way is to differentiate v_O=V_S -( k/2(v_I - V_T)^2)*R_L  w.r.t. v_I   taking v_I=V_I . This will be the slope at V_I.  This value will need  to be multiplied by v_i and it will be equal to v_o as v_o/v_i is  equal to the slope. (see lecture 7 for a similar case)

Revision of Circuits and Electronics - MOSFET Amplifier Large Signal Analysis (Lecture 9)

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.

Revision of Circuits and Electronics - Dependent Sources and Amplifiers (Lecture 8)

 Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-8/

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

Dependent sources can be linear or nonlinear.

A dependent source needs to have its voltage or current controlled by some other variable. A diamond symbol with a current through it  is a dependent current source. A 2 port device has a control port and an output port.

VCCS  - (Voltage Controlled Current Source) Current at the output port is a function of voltage at the input port.

First, an example with the independent source is done resulting in a result of V=IR.

The dependent source example is an abstracted view of the actual circuit. Node method is also used but probably not really necessary. The general result is still V=IR but I=K/V is used.

 The other types of dependent sources are CCVS, VCVS  and CCCS.

 In another dependent source example, v_O needs to be found as a function of v_IN.  The circuit is different from the previous one as there is an input voltage and a supply voltage V_S.

The drawing is simplified and the node method is used to find v_O.

v_O versus v_I curve can be plotted  for the dependent source example.

 When superposition theorem is used with dependent sources, all dependent sources are left in the circuit. Then, solve for one independent source at a time.

Why amplify? Signal amplification is the key to both analog and digital processing.  Amplification helps to make a signal larger but it can also aid in the process of tolerating noise during communication.

 Amplification is fundamental for the digital domain as well.  It is needed to make a digital device meet the static discipline. The minimum amplification needed is  (V_OH - V_OL)/(V_IH-V_IL)

An amplifier is a 3 ported device.  The power port is not often shown. All ports share a common reference point called ground.

How do we build one? It's already been done! It is the VCCS in the second dependent source example.

Where's the amplification? From the plot of v_O versus v_I ,  there is a region where there is gain.

But what happens in the region when v_O is more than 0 and when it is  less than zero?  . When v_O   is more than zero, the VCCS consumes power. When it is less than zero, the VCCS supplies power.

If VCCS consumes power, there is saturation at some point.

Revision of Circuits and Electronics - Incremental Analysis (Lecture 7)

Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-7/

Lecture notes : - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/ 

A small region of the transfer characteristic curve looks linear so we have to use this region for the circuit to work. The way to implement this is to boost and shrink the signal - shrink the signal of interest and add a dc offset to it. Here is where signal notation becomes important. The response i_d to small signal v_d is approximately linear.

What does this mean mathematically? The total variable i_D is a function of v_D. (Let delta v_D be equal to v_d.) Taylor's expansion  is used to expand this function. The higher order terms of delta v_D  are neglected.

Why is the small signal response linear? If we equate the DC component and the time varying component of the current with the expanded function, there will be a constant part for the time varying one. This constant part is the the slope of the transfer characteristic. The other constant part for the DC component  is the the operating point.

From the circuit view of the small signal model, where the original elements are replaced by their small signal models,  the LED behaves like a resistor.

To summarise, there are 3 steps for the small signal method:-

(1) Find the operating point using DC bias inputs from large signal circuit

(2) Develop small signal models for each of the elements around the operating point.

(3) Analyze the linearized circuit to get the small signal response.

The DC voltage source  behaves as a short to small signals.  The DC current source  behaves as an open to small signals.

For a voltage source containing both dc and small signal, the small signal model is the small signal part of the source.

For a resistor , the small signal model is the resistor itself.

For a non linear device , the small signal model is a resistor.

In the small circuit analysis example, i_d needs to be found.  For step  1,  with the large signal circuit model, nodal  analysis is  used to find V_D and I_D.  For step 2,   with the small signal circuit model,  R_D of the non-linear device needs to be found.  For step 3, i_d can be obtained from the linearized circuit.

Revision of Circuits and Electronics - Nonlinear Circuits (Lecture 6)

Non linear circuits can be analysed by the following methods

(1) Analytical method

(2) Graphical method

(3) Piecewise linear method - not a focus for the course

(4) Introduction to incremental analysis


How do we analyse them? It is possible to replace the linear part with the Thevenin equivalent.


Method 1 uses nodal analysis. From the two equations, a solution can be obtained by  by trial and error. v_D needs to be found first and this is done by substituting v_D into the arranged equation until a constant value of v_D is obtained.

Method 2 also uses nodal analysis but the equation is rearranged so that i_D becomes the main subject of the equation.  With the two equations, the graph of i_D versus  v_D is plotted.  There will be two plots representing the two equations, which intersect at a point that gives the solution.

Method 3 involves determining which of the linear region applies for the non linear component. Linear region is 1/ R_D so the  component can be replaced with a resistance value of R_D.

Method 4 in this lecture introduces incremental analysis which is also known as the small signal method. It is  a special way to make a non linear circuit,  linear. An example is  a circuit that transmit music over a light beam. The circuit consists of an input music signal, an LED, a photoresistor and an amplifier. The problem lies with the conversion at the beginning which is non linear, as the LED is  a non linear component, resulting in a distorted sound. This can be seen when i_D, the output current is plotted using the transfer characteristic and the graph of the input voltage versus time. The I_D does not look like the plot of the input voltage or v_D. So in order to work, the non linear LED needs to be turned into a linear device.


Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-6

Lecture notes : -  http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

Revision of Circuits and Electronics Part 1 - Inside The Digital Gate (lecture 5)

To build a digital gate, a switching device is needed.

The MOSFET device (Metal Oxide Semiconductor Field Effect Transistor) is a 3 terminal element that behaves like a switch with the gate being the control terminal.

If v_GS is less than a threshold voltage of say 1V then the switch is off. If v_GS is more than a threshold voltage  then the switch is on.

Looking at the simplified i_DS vs v_DS graph, it can be seen that when  i_DS flows (switch is on), v_DS=0, and when I_DS does not flow (switch is off), V_DS is recorded.

The MOSFET behaves like an inverter. The voltage transfer characteristic can be plotted - this is a graph of v_out vs v_in.

 There is a question about whether the inverter satisifies the static discipline for different thresholds. The lines for V_IL and V_IH need to be plotted first on the voltage transfer characteristic graph. The V_OL and V_OH lines need to be plotted next. Parts of the  transfer characteristic graph needs to lie within the green shaded areas to satisfy the static discipline.

The switch resistor  (SR) model of MOSFET is a more accurate MOSFET model.  When v_GS is more than a threshold voltage, the on state, there is some resistance between the drain and the source.

Under the more accurate model, the inverter satisfies the static discipline.


Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-5/


Lecture notes : - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

Revision of Circuits and Electronics Part 1 - The Digital Abstraction (lecture 4)

Node method can be used in all cases.

Analog systems lack noise immunity.

So values need to be discretized - high and low ( 1 and 0)

If we take 2.5V to be the midpoint for 0 and 1 and if the voltage sent  is 2.5V , it won't work so a forbidden region is needed.

If we set a forbidden region between 2V and 3V, there will still be problems because of a lack of noise margin. So the solution is to set tougher standards for the sender. For example, between 4V and 5V for a 1  (V_oh)  and between 0V and 1V for a 0 (V_ol).

The rest of the lecture is on digital circuits which I am not going to cover much about as I am quite familiar enough about that topic.


Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-4

Lecture notes : - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/ 

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