elec353 lecture notes set 5 - concordia universitytrueman/elec353/elec353_2017_5.pdf · elec353...

ELEC353 Lecture Notes Set 5

1

The homework assignments are posted on the course web site.http://users.encs.concordia.ca/~trueman/web_page_353.htm

Homework #3: Do homework #3 by February 3, 2017. Homework #4: Do homework #4 by February 10, 2017.Homework #5: Do homework #4 by February 17, 2017.

Mid-term test: Thursday February 16, 2017• The mid-term test will include a question on the “life long learning”

topic of A.C. circuit analysis. • See the course web site for sample mid-term tests, including question 3

on A.C. circuit analysis.• Study tip:

• Download the question paper for a mid-term from a previous year.• Spend one hour 15 minutes solving the test with your calculator

and the formula sheet, but no textbook or notes.• Grade your answer against the solution to the test!

Tentative Final Exam DateMonday May 1, 2017, from 9:00 to 12:00.

http://users.encs.concordia.ca/%7Etrueman/web_page_353.htm

Junctions between Transmission-Lines

( ) ( ) ( )tzVtzVtzV ,,, 111−+ +=

( ) ( ) ( )1

1

1

11

,,,cc R

tzVR

tzVtzI−+

−=

( ) ( ) ( )tzVtzVtzV ,,, 222−+ +=

( ) ( ) ( )2

2

2

22

,,,cc R

tzVR

tzVtzI−+

−=

𝐿𝐿1 𝐿𝐿2𝑧𝑧 = 𝐿𝐿1

𝑧𝑧 = 𝐿𝐿1 + 𝐿𝐿2𝑧𝑧 = 0

𝑅𝑅𝑐𝑐1,𝑢𝑢1 𝑅𝑅𝑐𝑐2,𝑢𝑢21

𝑉𝑉1+

𝑉𝑉1−𝑉𝑉2+

𝑉𝑉2−

Incidence from the Left

𝐿𝐿1 𝐿𝐿2𝑧𝑧 = 𝐿𝐿1 𝑧𝑧 = 𝐿𝐿1 + 𝐿𝐿2𝑧𝑧 = 0

𝑉𝑉1+

𝑉𝑉1−𝑉𝑉2+𝑇𝑇12


Γ1

For incidence from the left: (𝑉𝑉2− = 0)

+

−

=Γ1

11 V

V

+

+

=1

212 V

VT

Find the Reflection and Transmission Coefficientsfor incidence from the left, with 𝑉𝑉2− = 0:

Enforce KVL:

At the junction, z= 1L

Incidence from the left, given with +1V 02 =−V


𝑉𝑉1+



For incidence from the left:

At the junction:

Enforce KCL:


𝑉𝑉1+



KVL:

KCL:

Solve the equations:

Reflection Coefficient:

Transmission Coefficient:21

212

2

cc

c

RRRT+

=

+

−

=Γ1

11 V

V

+

+

=1

212 V

VT

+− Γ= 111 VV

++ = 1122 VTV


𝑉𝑉1+



For incidence from the right:

Homework: Enforce KVL and KCL and show that

21

212

cc

cc

RRRR

+−

=Γ −+ Γ= 222 VV

−− = 2211 VTV21

121

2

cc

c

RRRT+

=

Incidence from the right, given with −2V 01 =+V



𝑉𝑉2−Γ2

𝑇𝑇21


Example

A voltage source has internal resistance 𝑅𝑅𝑠𝑠 = 50 ohms and generates a 10 volt step function that starts at 𝑡𝑡 = 0. It is connected to a 2 cm transmission line with 𝑅𝑅𝑐𝑐1 = 50ohms, which is in turn connected to a 3 cm line with 𝑅𝑅𝑐𝑐2 = 100 ohms. Both lines have speed of propagation 𝑢𝑢 = 20 cm/ns. The load at the end of line #2 is 𝑅𝑅𝐿𝐿 = 50 ohms.

1.Graph the voltage across the load and across the generator as a funcation of tine.

2.Find the final value of the load voltage as as 𝑡𝑡 → ∞.

Delay times:

Find 𝑉𝑉1+: So 𝑉𝑉1+ = 5 volts

Draw a bounce diagram:

At the junction:

Final values:

5=sV

5=LV

volts

volts


25


Homework #3: Do homework #3 by February 3, 2017. Homework #4: Do homework #4 by February 10, 2017.Homework #5: Do homework #4 by February 17, 2017.








Transmission Line with Shunt Load

Incidence from the left:



𝑅𝑅𝑠𝑠

𝑉𝑉𝑠𝑠 𝑅𝑅𝐿𝐿𝑅𝑅𝑀𝑀

𝑉𝑉1+


Γ1

𝑇𝑇12

𝑧𝑧 = 𝐿𝐿1

𝑅𝑅𝑀𝑀

𝑉𝑉1− = Γ1𝑉𝑉1+

𝑉𝑉2+ = 𝑇𝑇12𝑉𝑉1+

Use KCL and KVL to find the reflection coefficient and the transmission coefficient:

At the junction at 𝑧𝑧 = 𝐿𝐿1:

𝑉𝑉1+



𝑅𝑅𝑀𝑀

KVL:

KCL: ( ) ( ) ( ) ( )Mccc R

tLVR

tLVR

tLVR

tLV ,,,, 12

2

12

1

11

1

11++−+

+=−

( ) ( ) ( )tLVtLVtLV ,,, 121111+−+ =+ +−+ =+ 211 VVV

Mccc RV

RV

RV

RV ++−+

+=− 2

2

2

1

1

1

1

Solve the equations:KVL:

KCL:

+−+ =+ 211 VVV

Mccc RV

RV

RV

RV ++−+

+=− 2

2

2

1

1

1

1

Homework: solve the equations to show that

Reflection and transmission coefficients:

𝑉𝑉1+


Γ1

𝑇𝑇12


𝑅𝑅𝑀𝑀

Example

A logic chip with a 10 ohm output resistance produces a step function voltage. It drives two other logic chips over a short interconnect on a circuit board, as shown in the figure. The circuit board paths behave as transmission lines with characteristic resistance 50 ohms and speed-of-propagation 20 cm/ns. At a distance of 1 cm from the logic chip there is a memory chip with input resistance 100 ohms, and at a further 1 cm distance there is another memory chip of input resistance 100 ohms.

Find the voltage across the load 2 cm from the driver chip.Find the final value as 𝑡𝑡 → ∞.

Solution

Γ1 = −0.2000𝑇𝑇12 = 0.8000

Γ𝐿𝐿 = 0.3333Γ𝑠𝑠 = −0.6667

Decompose into two simpler problems:

Γ1 = −0.2000𝑇𝑇12 = 0.8000

Γ2 = −0.2000𝑇𝑇21 = 0.8000

Γ𝐿𝐿 = 0.3333Γ𝑠𝑠 = −0.6667

Carry on for one more reflection:

Find the load voltage:

0.7777 0.2223

-0.51850.07409

Verify the solution with BOUNCE:


43


Homework #4: Do homework #4 by February 10, 2017.Homework #5: Do homework #4 by February 17, 2017.








Done!Done!

Branching Transmission Lines

++ = 1133 VTV

++ = 1122 VTV

+− Γ= 111 VV

Reflection and transmission coefficients:

Enforce KVL

Enforce KCL

Solving the equations:

Example

Ω= 50cR5=sV Ω=100sR 14=u cm/nsvolts

Draw a bounce diagram for each of the three transmission lines:

Γ𝑠𝑠 = +0.3333

Γ𝐿𝐿 = +1

Γ𝐿𝐿 = +1

Put all the information onto one bounce diagram:

Solution by Computation with BOUNCE

Find the time for the load voltage to reach 4.5 volts and remain above 4.5 volts.

Time required to rise above 4.5 volts and remain above 4.5 volts:

The load voltage remains above 4.5 volts for t>4.706 ns

Let the rise time of the source voltage be 0.1 ns

Model the input of the CMOS chip as a 1 pF capacitor.

Open-circuit loads compared to capacitive loads:

elec353 lecture notes set 5 - concordia universitytrueman/elec353/elec353_2017_5.pdf · elec353...

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