EE101 Lecture 11 October 26 (F), 2012 A brief review of materials covered thus far: (Chapters 1-5) Basic concepts in electronic circuits- Circuit elements, Ohm’s Law (V=RI), Power P=VI, Series and Parallel Connections of Rs. Basic Laws- KVL, KCL, and finding Req, Voltage Division, Current Division. Nodal Analysis, Loop/Mesh Analysis to find 𝑉 𝑛 , 𝐼 𝑘 . Solving simultaneous equations using Cramer’s rule to find node voltages, loop/mesh currents. Thevenin’s and Norton’s equivalent circuits using open circuit voltage, short circuit current and equivalent resistance. Source transformation to simplify circuit analysis. Maximum power transfer to 𝑅 𝐿 . Circuit analysis including BJTs, Op Amps (ideal and non-ideal cases). Op Amp Circuits- summing, difference, DAC (digital to analog converter).

EE101 Fall 2012 Lect - Kang

q= C v, When C is time-invariant , 𝑑𝑞 𝑑𝑡 = i = C 𝑑𝑣 𝑑𝑡 Up to now, circuit analysis has been limited to Resistive Circuits which are static (v= Ri), no dynamics! From Chapter 6 on, we will move onto RC, RL (first-order) circuits and RLC (second-order) circuits for which we need to solve ordinary differential equations (O.D.E.s) to find voltages, currents, etc. C and L are energy storing elements and thus of much more practical use for electronic circuits and systems. q= C v, When C is time-invariant , 𝑑𝑞 𝑑𝑡 = i = C 𝑑𝑣 𝑑𝑡 Ø = Li, When L is time-invariant, 𝑑ø 𝑑𝑡 = v = L 𝑑𝑖 𝑑𝑡

D (𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦) = ɛ E, and E = 𝑉 𝑑 olume olume Where D (𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦) = ɛ E, and E = 𝑉 𝑑

CCD Camera (Photon to Electron to Voltage Conversions)

Capacitor in a DRAM (Dynamic Random Access Memory) Cell Type: JPG

ε0 = 8.854187817.. × 10−12 F/m is the vacuum permittivity. Example. If A = 1 𝒄𝒎 𝟐 and d= 1 mm, and air is between two plates, then then C= 8.854x 10 −12 𝐹 𝑚 10 −4 𝑚 2 10 −3 𝑚 =0. 8854 𝑝𝐹

i= 𝐶 𝑒𝑞 𝑑𝑣 𝑑𝑡 = ( 𝑘 𝐶 𝑘 ) 𝑑𝑣 𝑑𝑡

V= 𝑘 𝑣 𝑘 = 𝑘 1 𝐶 𝑘 𝑖 𝑑𝑡= 1 𝐶 𝑒𝑞 𝑖 𝑑𝑡

- 1 𝑅𝐶 dt = 𝑣 𝑑𝑣 Assumed 𝑡 0 =0, A= 𝑉 0

Find 𝑣 𝑥 (t)

How to solve 𝒅𝒗 𝒅𝒕 = - 𝒗− 𝑽 𝒔 𝑹𝑪 ? We first note that dv=d(v- 𝑉 𝑠 ) and rewrite the equation above: 𝑑(𝑣− 𝑉 𝑠 ) 𝑑𝑡 = - 𝑣− 𝑉 𝑠 𝑅𝐶 𝑑(𝑣− 𝑉 𝑠 ) 𝑣− 𝑉 𝑠 = - 1 𝑅𝐶 dt Integrating both sides, we get ln (v- 𝑉 𝑠 )|( 𝑣 0 − 𝑉 𝑠 ) 𝑡𝑜 (v− 𝑉 𝑠 ) = - 1 𝑅𝐶 (t- 𝑡 0 ) V- 𝑉 𝑠 = ( 𝑣 0 − 𝑉 𝑠 ) 𝑒 − 𝑡− 𝑡 0 𝑅𝐶 Finally, v(t) = 𝑽 𝒔 + ( 𝒗 𝟎 −𝑽 𝒔 ) 𝒆 − 𝒕− 𝒕 𝟎 𝑹𝑪 V( 𝒕 𝟎 )= 𝒗 𝟎 , and v(∞)= 𝑽 𝒔