ELEC 353 – Solution to Assignment #5 This assignment is included on the class test.

1.A step-function voltage source with sV 10 volts and internal resistance 50sR ohms drives a

transmission line having characteristic resistance cR 50 ohms and speed-of-travel 14u cm/ns. The

line is 3 cm long. The load terminating the line has pin inductance 2L nH in series with a chip input resistance of LR =100 ohms. (a)Neglect the pin inductance by using 0L . Find the voltage across the load tvL for 5.00 t ns.

Also, find the voltage across the transmission line tvm at the middle of the line, 1.5 cm from the

generator. (b)Use a pin inductance of 2L nH and find tvm and tvL for 5.00 t ns.

(c) What is the time constant? (d) What is the value of the voltage tvL at t=0.16 ns?

(e)The voltage tvL is measured across the series connection of the inductance and resistance. The inductance represents IC’s pin and the lead inside the IC that connects the pin to the actual chip. But inside the logic chip, the voltage the logic gate “sees” is the voltage across resistor LR . Using 2L

nH, plot the voltage across LR for 5.00 t ns. (f) Use the BOUNCE program to verify that your calculations in part (a) and (b) are correct. Note that BOUNCE cannot solve part (c) because there is no way to tell the program to graph the voltage across the resistor alone. Notes: (1) Thanks to a student in the class for suggesting this problem. (2) The input of a CMOS chip is not resistive. A more realistic approach is to model the chip input as capacitive with a capacitance of a few pF. Solution to Question 1 Since the source is matched ( 50sR ohms and cR 50 ohms), so the voltage applied on the

transmission line is 0v =5v, and the reflection coefficient at source is 0s . The time delay at the

load is 2143.014

3dt ns.

(a) The time delay at the middle point is 2/dm tt . When 0L , the reflection coefficient at the

load is 3

1

50100

50100

load =0.333. Thus the voltage at the load at md tt 2 is

3/5*4)1( 0 vload =6.667 v. since there is no reflection from the source, therefore, the

voltage across the load maintains 6.667v from 2143.0dt ns, shown in Fig.1-1. And the

voltage at the middle is shown also in Fig.1-1 in dashed line.

Fig.1-1 The voltages at the load and at the middle without inductor

(b) When 2L nH, initially, when the wave arrives, the inductance functions as an open circuit, thus the wave is totally reflected. As time advances, the total resistance at the load gradually decreases, till the inductor becomes a short circuit. The net series resistance is 150 loadc RRR ohms, the time constant is

99 1001333.0150/102/ RL s=0.01333ns. Thus, the voltage is 5+5=10v (a peak) because of total reflection, at the time instance of dtt , then exponentially decays till reaches

the final value, which is 6.667 v. The voltage the middle is 5 v mtt to mmd tttt 3 . At mtt 3 , the peak arrives, the

shape of voltage is a shift of the voltage at the load. From mtt 3 , the voltage experiences the

exponential decay till the final voltage of 6.667v. detailed calculation can be found in (e).

Fig.1-2 the voltages at the load and at the middle

(c) The time constant is calculated as 0.01333 ns. (d) Since t=0.16 ns is less than 2143.0dt ns, the incident wave has not arrived the load, thus the

voltage at the load is in its initial value (assumed to be zero). (e) from dtt

The voltage across the load is

RindcLoad VVV

From basic circuit analysis we know that:

The voltage at the inductor is )()(

0 102 dd tttt

ind eevV

thus the voltage across the resistor LR is

)()(

10])3/2010(3/20[)(dttdtt

indcLoadLR eeVVtV

=

)(

1667.6 dtt

e

Which will look like follow:

The voltage in the middle is the shift (because of the distance from the load to the middle) of the voltage at the load. Note that the voltage at the load already contains the incident 5v. The graphs of the voltages are shown in Fig.1-2.

Question 2

2.A step-function generator of internal resistance 50sR ohms steps up from 0 volts to 1sV volt at

0t seconds. It drives the circuit shown above. The transmission lines both have the same speed of travel of 14 centimeters per nanosecond. They have different characteristic impedances, of 501 cR

ohms and 1002 cR ohms. The shunt load resistors are 1001 LR ohms and 502 LR ohms.

Solution to Question 2

ssc

c VRR

RV

0 = 0.5 V

14/51 u

Ltd =0.36 ns

14/32 u

Ltd =0.21 ns

cs

css RR

RR

= 0 (the generator is matched to the line)

cL

cLL RR

RR

2 = -1/3

112

11212

ceq

ceq

RR

RR

= 0 (where: 50// 1212 Lceq RRR )

1212 1 =1

221

22121

ceq

ceq

RR

RR

= -0.5 (where: 3/100// 1121 Lceq RRR )

2121 1 = 0.5 Now we are ready to draw the bounce diagram and calculate the time response.

(a)What is the voltage 1V across 1LR at t 0.36 ns? 0.5 V

(b)What is the voltage 2V across 2LR at t 0.58 ns? 0.5-0.1667=0.3333 volts

(c)What is the voltage 2V at t 1.08 ns? 0.3333+0.0833-0.0278=0.3888 volts

(d)What is the voltage 2V as t ? As t the transmission lines behave as ideal short circuits, so the equivalent circuit has RL1 and RL2 in parallel. RL1 parallel RL2 = 100x50/150= 33.33 so

𝑉 =. .

.= 0.4 volt

Verify the solution with the BOUNCE program:

The initial step from the generator is 0.5 v.

The transmission coefficient through the junction is 1.0 so the step on line #2 is 0.5 v.

The reflection coefficient at the load is -1/3 so the reflected step is -0.1667 and the load voltage drops to 0.5000-0.1667=0.3333 volts.

The reflection coefficient at the junction is -0.5 so the step of -0.1667 is reflected as a step of 0.08333 and the voltage at the junction steps up from 0.3333 to 0.4167 volts.

The reflection coefficient at the load is -1.3 so the step up of 0.08333 reflects as a step down of -0.0278 and the load voltage changes to 0.3333+0.0833-0.0278=0.3889 volts.

Question 3

3.A transmission-line circuit is driven by a pulse generator with 25sR ohms. The pulse steps up at

t=0 sec from 0 to sV 10 volts and then steps back down to zero volts at 0t .85 ns. The transmission

lines in the circuit shown above have speed of travel 20 cm per ns. Line #1 has characteristic impedance 501 cR ohms and line #2 has 2cR 200 ohms. The lengths of the lines are 101 L cm and

52 L cm. The load resistor is 200LR ohms.

Solution to Question 3

ssc

c VRR

RV

0 = 20/3 V

20/101 u

Ltd =0.50 ns

20/52 u

Ltd =0.25 ns

cs

css RR

RR

= -1/3

2

2

cL

cLL RR

RR

= 0

12

1212

cc

cc

RR

RR

= 3/5

1212 1 =8/5

21

2121

cc

cc

RR

RR

= -3/5

2121 1 = 2/5

(a)What is the junction voltage 1V at t 0.51 ns? 32/3 V (b)What is the voltage )0(V at the source terminals at t 0.9 ns? 0 (c)What is the voltage )0(V at t 1.1 ns? 2.7 V

(d)What is the voltage LV at t 1.8 ns? -32/15 V (e)What is the voltage )0(V at t 2.05 ns? -0.53 V

4.A step-function generator has amplitude sV =10 volts and turns on at t = 0 sec. Its internal resistance is

sR =50 ohms. It is connected to a transmission line of characteristic resistance cR , speed of propagation

u 20 cm/ns, and length d. The line is terminated with a load consisting of an unknown component in series with resistor LR . The graph shows the voltage 1V at the generator terminals as a function of time. (a)What is the characteristic resistance cR of the transmission line?

(b)What is the length d of the transmission line? (c)What is the value of the load resistor LR ? (d)What is the value of the unknown component?

Solution: (a) Since the initial voltage 5v applied on the transmission line is half of the source voltage 10v,

thus the characteristic resistance is 50cR ohms, the same as the internal resistance.

(b) The first peak arrives at t=1ns, thus the distance is 102/ utd cm. (c) Note that the wave form graphed is at the source end, which is 0.5ns lagged compared to the

voltage at the load, due to the distance of 10 cm. Thus at t=0.5ns, the voltage at the load is 10v, thus it is a total reflection, which means an open circuit. So he unknown must be an inductor.

The exponential voltage at the source is expressed as )1(44.45.75.2)( tetV

)2

()()(

dtt

finalinitialfinal eVVVtV

Note in the exponential expression, we used dt2 because it is at the source end.

Thus load

load

loadc

loads R

R

RR

RV

50105.2 , so 667.163/50 loadR ohms.

Since )/(44.4/1 loadc RRL , thus 1544.4/)( loadc RRL nH.

Thus, the final answers are (a) 50cR ohms, (b) d=10 cm, (c) 667.16loadR ohms, (d) the

unknown component is an inductor with 15 nH inductance.

Using the calculated parameters, below are the BOUNCE results at t=1.237 ns to show the peak occurs at t=1ns and at t=2.5ns to show the voltage is about 2.5 v.