1. 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).

2. EE101 Fall 2012 Lect - Kang

3. 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

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

6. CCD Camera (Photon to Electron to Voltage Conversions)

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

8. 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 −4 𝑚 −3 𝑚 = 𝑝𝐹

12. i= 𝐶 𝑒𝑞 𝑑𝑣 𝑑𝑡 = ( 𝑘 𝐶 𝑘 ) 𝑑𝑣 𝑑𝑡

13. V= 𝑘 𝑣 𝑘 = 𝑘 1 𝐶 𝑘 𝑖 𝑑𝑡= 1 𝐶 𝑒𝑞 𝑖 𝑑𝑡

17. - 1 𝑅𝐶 dt = 𝑣 𝑑𝑣
Assumed 𝑡 0 =0, A= 𝑉 0

18. Find 𝑣 𝑥 (t)

24. 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