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Wednesday, February 25, 2015

Basic Amplifier Configurations


Q1: What are the basic amplifier configurations? \frac {g_mr_\pi R_E}{1+(\frac{R'_L}{r_o})(1+\frac{R_E}{r_\pi})}
A1: There are 8 basic amplifier configurations: 4 basic BJT amplifier configurations and 4 basic MOS amplifier configurations. Note that that taking the limit as r_\pi \to \infty of the BJT equations produces the MOS equations.




Common Base:
A_v=g_m(r_o || R'_L)
R_i=r_\pi || \frac {r_o+R'_L}{1+g_mr_o} 
Common Gate:
A_v=g_m(r_o || R'_L)
R_i= \frac {r_o+R'_L}{1+g_mr_o}  


Common Source:
A_v=-g_m(r_o || R'_L)
R_i=\infty 

Common Emitter:
A_v=-g_m(r_o || R'_L)
R_i=r_\pi 




Common Emitter with R_E:
A_v=-\frac{g_mR'_L}{1+g_mR_E+(R'_L/r_o)(1+R_E/r_\pi)}
R_i=r_\pi + \frac {g_mr_\pi R_E}{1+(R'_L/r_o)(1+R_E/r_\pi)}
Common Source with R_S:
A_v=-\frac{g_mR'_L}{1+g_mR_E+(R'_L/r_o)}
R_i=\infty










Common Collector/Emitter Follower:
A_v=\frac{g_m(r_o || R'_L)}{1+g_m(r_o || R'_L)}
R_i=r_\pi + \beta (r_o || R'_L) 

Common Drain/Source Follower:
A_v=\frac{g_m(r_o || R'_L)}{1+g_m(r_o || R'_L)}
R_i=\infty 

Introduction to Small Signal Model

Q1: How can I analyze real amplifier circuits?
A1: You can divide the amplifier circuit into its signal circuit and bias circuit. For now, we will consider the signal circuit.

Q2: How do I map real circuit elements to their signal circuit counterparts?
A2: By using their small signal models (SSM):
Resistors and capacitors behave the same in signal circuits

Independent voltage sources mapped to short circuits while
independent current sources are mapped to open circuits

Things change a bit with nonlinear elements.
The diode maps to a resistor of r_d=\frac{nV_T}{I_D}

r_\pi=\frac{V_T}{I_B}
g_m=\frac{I_C}{V_T}=\frac{\beta}{r_\pi}
r_o\approx \frac{V_A}{I_C}

Both the NPN and PNP map to the same SSM

g_m=\frac{2I_D}{V_{OV}}
r_o\approx \frac{1}{\lambda I_D}
Both the NMOS and PMO map to the same SSM

Friday, February 20, 2015

Mirror Circuit

Q1: What's a mirror circuit?
A1:
Q_{ref} and Q_1 are identical NPN transistors.
This is why current splits evenly to each of their bases.
Q2: What state is Q_{ref} in?
A2: Since i_{C,ref}>0, we know that Q_1 must be on. Hence, V_{BE,ref}=V_{D0}
Note that the base and collector share the same node in Q_{ref}.
Hence, V_{BE,ref}=V_{CE,ref}=VD_{0}
Therefore, Q_{ref} is always active.

Q3: How is I_1 related to I_{ref}?
A3: First, let's find an expression for I_{ref}:
By KCL: I_{ref}=2i_B+i_{c,ref}
Since Q_{ref} active: I_{ref}=2\frac{i_{C,ref}}{\beta}+i_{C,ref}
I_{ref}=i_{C,ref}(\frac{2}{\beta}+1)

Next, let's examine I_1:
I_1=i_C
I_1=\beta i_B
I_1=\beta \frac{i_{C,ref}}{\beta}
I_1=i_{C,ref}
I_1=\frac {I_{ref}}{\frac{2}{\beta}+1}

For \beta \gg 1:
I_1\approx I_{ref}

By KVL through Q_{ref} (either the BE or CE branch):
I_1\approx I_{ref}=\frac{V_{CC}-V{D0}+V{EE}}{R}

Q2: Why use a mirror circuit?
A2: You can use a mirror circuit to behave as a current source in order to bias a BJT.

Q3: Why not use a mirror circuit?
A3: The mirror circuit produces integer multiples of the reference current. For a more flexible option, you might want to look into current-steering circuits.

Wednesday, February 4, 2015

MOS Circuits

Q1: What states can a MOS be in?
A1: A NMOS may be in:

  1. Cutoff
    • v_{OV}\leq 0
    • i_D=0
  2. Saturation
    • v_{OV}\geq 0 and v_{DS}\geq v_{OV}
    • i_D=0.5\mu_nC_{ox}\frac{W}{L}v_{OV}^2(1+\lambda v_{DS})
  3. Triode
    • v_{OV}\geq 0 and v_{DS}\leq v_{OV}
    • i_D=0.5\mu_nC_{ox}\frac{W}{L}(2v_{OV}v_{DS}- v_{DS}^2)
A couple of notes:
  • v_{OV}=v_{GS}-v_{tn} where v_{OV} is the overdrive voltage and v_{tn} is the threshold voltage associated with the NMOS
  • The expressions for i_D in saturation and triode both contain the pre-factor 0.5\mu_nC_{ox}\frac{W}{L} 
    • \mu_n is the negative charge carrier effective mobility
    • C_{ox} is gate-oxide capacitance per unit area
    • W is the gate width
    • L is the gate length
  • The expression for i_D in saturation contains the channel-length modulation parameter \lambda
    • If channel-length modulation can be neglected, then \lambda=0
    • The channel-length modulation parameter is related to early voltage by: \lambda=\frac{1}{v_A}
    • The phenomenon of channel-length modulation in MOSFETs can be compared to Early effect in BJTs

For a PMOS:
  1. v_{OV}=v_{GS}-v_{tn}\to v_{OV}=v_{SG}-|v_{tp}|
  2. \mu_n \to \mu_p
  3. v_{DS}\to v_{SD}
  4. v_{GS}\to v_{SG}
Q2: How do I solve MOS circuits?
A2: For an NMOS circuit:
  1. Write the KVLs for v_{GS} and v_{DS}
  2. Assume cutoff by setting i_D=0. Check that you are in cutoff.
  3. Assume saturation. Check that v_{DS}\geq v_{OV}
  4. You're in triode mode. 
Note that the equations for i_D in saturation and triode mode are quadratic. Although you may find 2 roots, only keep the 1 that's valid for that particular state.

MOS Device

Q1: What is a MOSFET?
A1: A MOSFET is a metal oxide semiconductor field-effect transistor. They're often abbreviated as MOS devices. A MOSFET has 4 terminals:

  1. Gate
  2. Body
  3. Drain
  4. Source
Q2: What are some types of MOS devices?
A2: There's n-channel metal oxide semiconductor (NMOS) and p-channel metal oxide semiconductor (PMOS).
Current direction is reversed in the PMOS

Note the 4 terminals of the NMOS



In practice, the body terminal is often connected
to the source terminal of the NMOS
Same goes for the PMOS



Combining the body and source terminals
simplifies the NMOS to a 3 terminal device
Same for the PMOS














Q3: What circuit variables do I need to look for in a MOSFET?
A3: For a MOSFET in which the body and source terminals are connected, there are only 3 currents and 3 voltages to deal with. By KVL, you know that 1 voltage may be written in terms of the other 2. Same goes for current. So it looks like we need to account for 4 circuit variables, as was the case for the BJTs. However, MOSFETs differ from BJTs in that one current is already specified: i_G is always 0. Hence, you only need to keep track of 3 circuit variables.

For an NMOS you might choose to keep track of:
  1. i_D
  2. v_{GS}
  3. v_{DS}
For a PMOS, the current direction is reversed and the voltage subscripts are flipped. 

Saturday, January 31, 2015

Universal and Empty Set

Q1: What is the complement of a set A?
A1: The complement of a set A is the set that contains all elements that do not belong to A.

A^c=\{ x: x\not\in A\}

It follows that:
(A^c)^c=A

Q2: What is the universal set?
A2: The universal set \Omega is the set that contains all conceivable elements.
Hence, A \cup A^c = \Omega

Q3: What is the complement of the universal set?
A3: The empty set {} is the complement of the universal set. The empty set contains no elements.
Hence, A \cap A^c = \{\}


Union and Intersection

Q1: What is a union of sets?
A1: A union of sets is a a set whose elements are belong to at least one of the sets in the union.
Given: For 2 sets A and B:
A\cup B=\{ x: x\in A\ or\ x\in B\}
Given: For countable collection of infinite sets C_1, C_2, C_3, ...:
\bigcup\limits_{i=1}^{\infty}C_i=C_1 \cup C_2 \cup C_3 \cup ... =\{ x: x\in C_i\ for\ some\ i\}

Q2: What is an intersection of sets?
A2: An intersection of sets is a set that contains only those elements that are shared by all sets that form the intersection.
Given: For 2 sets A and B:
A\cap B=\{ x: x\in A\ and\ x\in B\}
Given: For countable collection of infinite sets C_1, C_2, C_3, ...:
\bigcap\limits_{i=1}^{\infty}C_i=C_1 \cap C_2 \cap C_3 \cap ... =\{ x: x\in C_i\ for\ all\ i\}

Q3: What is a disjoint collection of infinite sets C_1, C_2, C_3, ...?
A3: A disjoint collection of infinite sets is a collection of sets whose intersection is the empty set.
\bigcap\limits_{i=1}^{\infty}C_i=\{\}

Q4: What does it mean if a family of sets P composed of cells P_1, P_2, P_3, ... partition a set A?
A5: If P partitions a set A, the following 3 conditions hold:

  1. The partition does not contain the empty set
    • \{\}\not\in P
  2. The union of the cells is the set A. The partition covers A.
    • \bigcup\limits_{i=1}^{\infty}P_i=A
  3. No 2 cells share any elements. The partition is pairwise disjoint.
    • P_i\cap P_j=\{\}\ for\ all\ i\not =j