ELEC351 Lecture Notes Set 1

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1 ELEC351 Lecture Notes Set 1 There is a tutorial on Monday September 10! You will do the first workshop problem in the tutorial. Bring a calculator. The course web site is: Course outline Lecture notes Software: BOUNCE, TRLINE, WAVES

2 PART 1 Waves on Transmission Lines in the Time Domain Time delay Wave equation and travelling waves Solving transmission line circuits.

3 Ulaby page 53, 212, 267 Coaxial Cable 4

4 Coaxial Cable Transmission Line RG-59 A:Outer plastic sheath B:Woven copper shield C:Inner dielectric insulator D:Copper core Source: Wikipedia

5 Transmission Line Properties cc = capacitance per unit length, F/m l = inductance per unit length, H/m rr = series resistance per unit length, ohm/m gg = shunt conductance per unit length, Siemens /m

6 Coaxial Cable from ELEC251 Ulaby page 212 Ulaby page 267 You should know how to derive these formulas from ELEC251. If you don t remember this, you should review it. 7

7 Coaxial Cable - Losses Shunt Conductance The dielectric filling a coaxial cable has conductivity σσ S/m. If the center conductor has charge density ρρ l Coul/meter, then the electric field in the dielectric is EE rr = ρρ l 2ππππππ V/m The voltage between the outer and inner conductor is aa ρρ l VV = bb 2ππππππ dddd = ρρ l 2ππππ ln bb aa

Shunt Conductance, continued Since JJ = σσσσ, the current density in the dielectric is JJ rr = σσee rr = σσρρ l 2ππππππ The current flowing through the surface of a cylinder of length 1 meter and

8 Shunt Conductance, continued Since JJ = σσσσ, the current density in the dielectric is JJ rr = σσee rr = σσρρ l 2ππππππ The current flowing through the surface of a cylinder of length 1 meter and radius rr is II = 1 meter xx 0 2ππ JJ rr rrrrφφ 2ππ σσσσ l = 0 2ππππππ rrrrφφ = 2ππ σσσσ l 2ππππ = σσσσ l εε The resistance of a 1 m length of cable is RR = VV ρρ l II = 2ππππ ln bb aa σσσσ l εε = ln bb aa 2ππσσ Hence the conductance of a 1 m length of cable is gg = 1 RR = 2ππσσ ln bb aa

Series Resistance of Coaxial Cable The Surface Resistance If a conductor has conductivity σσ cc and permeability μμ cc then the ratio of the electric field to the magnetic field at the surface of the

9 Series Resistance of Coaxial Cable The Surface Resistance If a conductor has conductivity σσ cc and permeability μμ cc then the ratio of the electric field to the magnetic field at the surface of the conductor is defined as the surface resistance RR ss = EE HH It May Be Shown (Ulaby page 341) that RR ss = ππππμμ cc /σσ cc We will talk about this later in the course. If we know the magnetic field tangent to the surface HH, then we can find the electric field as EE = RR ss HH If the center conductor carries current II flowing out of the page, then the magnetic field in the coaxial cable is HH = II 2ππrr What is the series resistance of the outer conductor of a coaxial cable?

Series Resistance of Coaxial Cable, continued At the surface of the outer conductor rr = bb the magnetic field is HH = II 2ππbb The electric field is II EE = RR ss HH = RR ss 2ππππ The field is

10 Series Resistance of Coaxial Cable, continued At the surface of the outer conductor rr = bb the magnetic field is HH = II 2ππbb The electric field is II EE = RR ss HH = RR ss 2ππππ The field is oriented axially along the cable. The voltage across a 1 meter length of cable is VV = electric field xx distance II = RR ss 2ππππ xx1 meter = RR II ss 2ππππ The resistance of the 1 meter length of outer conductor is RR oooooooooo = VV II = RR II ss 2ππππ = RR ss II 2ππππ Similarly the resistance of the inner conductor is RR iiiiiiiiii = RR ss 2ππaa

conductor and back from the load to the generator on the inner conductor, the conductors are in series and the net resistance for

11 Series Resistance of Coaxial Cable, continued The resistance of the outer conductor is RR oooooooooo = RR ss 2ππππ The resistance of the inner conductor is RR iiiiiiiiii = RR ss 2ππaa Since the current flows from the generator to the load on the outer conductor and back from the load to the generator on the inner conductor, the conductors are in series and the net resistance for a 1 meter length of cable is rr = RR oooooooooo + RR iiiiiiiiii = RR ss 2ππππ + RR ss 2ππππ = RR ss 2ππ 1 aa + 1 bb rr = RR ss 2ππ 1 aa + 1 bb

12 Coaxial Cable We will show that: The speed of travel on a transmission line is u = 1 lcc m/s The characteristic impedance of a transmission line is ZZ oo = l cc ohms 13

13 Coaxial Filters: Low Pass Filter matched jjωωll 1 jjωωωω matched

Reading Assignment: Ulaby Chapter 1 Waves and Phasors Ulaby Chapter 2 Transmission Lines Transmission Lines Two-conductor transmission lines are also called TEM lines.

14 Reading Assignment: Ulaby Chapter 1 Waves and Phasors Ulaby Chapter 2 Transmission Lines Transmission Lines Two-conductor transmission lines are also called TEM lines. TEM stands for transverse electric magnetic. Both the electric field and the magnetic field vectors lie in the transverse plane, with no axial component. Hollow tubes such as rectangular waveguide do not support a TEM mode! Ulaby Fig

15 Ulaby Table 2-1 Notation: RRR = rr LL = l GG = gg CC = cc 16

16 Twin Lead Transmission Line 300-ohm twin lead Balance to unbalance transformer for connecting 300-ohm twin lead to 75 ohm coaxial cable. Source: Wikipedia

17 Stripline and Microstrip Inan, Inan and Said Figure 2.1 Stripline Microstrip

Upper dielectric (B) is typically air.[1] Microstrip 90 mitred bend.

18 Microstrip Transmission Line Cross-section of microstrip geometry. Conductor (A) is separated from ground plane (D) by dielectric substrate (C). Upper dielectric (B) is typically air.[1] Microstrip 90 mitred bend. The percentage mitre is 100x/d [1] [1] Source: Wikipedia

19 Microstrip

20 Design Formulas for Microstrip Pozar Fig Effective permittivity: Speed of travel: Characteristic impedance:

21 Hairpin Filter in Microstrip 25

22 Stripline Transmission Line Cross-section diagram of stripline geometry. Central conductor (A) is sandwiched between ground planes (B and D). Structure is supported by dielectric (C). [1] [1] Source: Wikipedia

23 Transmission Line Circuit How do we: Ulaby Fig. 2-5 Account for the capacitance-per-unit-length cc and the inductance-per-unit-length l 1 Account for the speed of travel u = c L Hence account for the time delay T = u Find the voltage at the input and the voltage at the load. 40

24 Typical Response of a Transmission Line Circuit RR ss VV ss VV 0 Transmission Line RR cc, uu RR LL VV LL Parameters: LL V s R s = = step function from 0 to 10 volts starting at t=0 internal resistance of the generator, use 1 ohm L = length of the interconnection path, say 2 cm Z 0 = R c = characteristic resistance of the interconnection path (?????????) Use 50 ohms. u = speed of travel of a voltage on the interconnection path (???????) Use 20 cm/ns R L = input resistance of the load, which is ideally matched (RR LL = RR cc )but for this example we will use a high-resistance load, RR LL = 1000 ohms.

25 Step Response The slide shows one frame from an animation of the response of the circuit, obtained from the BOUNCE program.

26 Questions: What is the physics of this behavior? How does it arise from circuit analysis? How do you solve circuits such as this one? Time domain? Frequency domain? What determines the speed of travel? What is characteristic resistance? How do you design the circuit to avoid this behavior?

Physical Structure of CMOS Integrated Circuits

Physical Structure of CMOS Integrated Circuits Dae Hyun Kim EECS Washington State University References John P. Uyemura, Introduction to VLSI Circuits and Systems, 2002. Chapter 3 Neil H. Weste and David