ELEC 353 Practice Problem Set #2

ELEC 353

Practice Problem Set #2

1. A voltage source has internal resistance 10 ohms. It produces a pulse of 0.5 ns duration starting att=0. The open-circuit voltage is 5 volts. The generator is connected to a transmission line with characteristic resistance 50 ohms, and speed of propagation 20 cm/ns. The length of the line is 1 cm. The line is terminated with a resistorR_L.

(a) Find the voltage across the load for R_L=50 ohms from t=0 until t=0.65 ns. (b) If the termination is R_L=1000 ohms, then use a bounce diagram to find the voltage at the load for t=0 until t=1.1 ns. This is a lot of reflections back and forth, but is easy to calculate.

(c) The logic gate which is the load for this circuit samples the voltage at 0.18 ns to determine if it is a “logical 0” or a “logical 1”. The input resistance to the chip is

L

R =1000 ohms. A “logical 0” is a voltage less than 1.2 volts; a “logical 1” is a voltage greater than 3.8 volts. Is the sampled voltage a “0”, or a “1”, or

indeterminate?

(d) Use BOUNCE to verify your solution.

2. A logic gate has an output resistance of 10 ohms and produces a step function voltage of 10 volts (open-circuit) starting at t=0. The gate is connected in series with two transmission lines, both having length 5 cm and speed of propagation 20 cm/ns. The first line has characteristic resistance 50 ohms, and the second line has characteristic resistance 73 ohms. The load at the end of the second line is a 1000 ohm resistor. Plot the voltage at the junction and the voltage across the load for 0<t<1.1 ns. Use BOUNCE to verify your solution.

3. A logic gate has internal resistance 10 ohms and produces a 10 volt step function (open-circuit) starting at t=0. It is connected to transmission line #1, having length 0.4 cm. The end of line #1 branches into line #2 in parallel with line #3. Line #2 has length 1.5 cm and is terminated with a 50 ohm load. Line #3 has length 0.7 cm, and is terminated with a high-impedance logic gate of input resistance 1000 ohms. The speed of propagation on all three lines is 20 cm/ns, and the characteristic resistance of all three lines is 50 ohms. Find the voltage across the 50 ohm load for 0<t<0.180 ns. Use BOUNCE to verify your solution.

ELEC 353

Solution to Practice Set #2

A voltage source has internal resistance 10 ohms. It produces a pulse of 0.5 ns duration starting at t=0. The open-circuit voltage is 5 volts. The generator is connected to a transmission line with characteristic resistance 50 ohms, and speed of propagation 20 cm/ns. The length of the line is 1 cm. The line is terminated with a resistor R_L. (a) Find the voltage across the load for R_L=50 ohms from t=0 until t=0.65 ns. (b) If the termination is 1000 ohms, then use a bounce diagram to find the voltage at the load for t =0 until t=1.1 ns. This is a lot of reflections back and forth, but is easy to calculate.

(c) The logic gate which is the load for this circuit samples the voltage at 0.18 ns to determine if it is a “logical 0” or a “logical 1”. The input resistance to the chip is

L

R =1000 ohms. A “logical 0” is a voltage less than 1.2 volts; a “logical 1” is a voltage greater than 3.8 volts. Is the sampled voltage a “0” or a “1”, or indeterminate?

Solution

(a) Draw the circuit and find the one-way transit time:

Draw a bounce diagram. Show the leading edge and the trailing edge separately:

Since the load is matched to the transmission line, there is no reflection. The voltage at the load is a copy of the voltage at the input, delayed by the one-way transit time of the transmission line.

(b) Now terminate the line with a 1000 ohm resistor. The reflection coefficients are: 905 . 0 50 1000 50 1000 ₌ + − = + − = Γ c L c L L R R R R 667 . 0 50 10 50 10 ₌₋ + − = + − = Γ c s c s s R R R R

Remark: the values in this bounce diagram would be more accurate if I had used four significant figures for all the calculations.

Read off the step changes in the output from the bounce diagram:

At t=0.18 ns, the voltage across the load is 3.14 volts. This is too low for a “logical 1” but too high for a “logical 0” so the input state of the gate is indeterminate.

Using BOUNCE

Enter the values for the generator, line and load into BOUNCE then run the simulation.

With a matched load, the output is simply a delayed version of the input. The delay time is equal to the one-way transmit time of 0.05 ns. Since the minimum rise time in

BOUNCE is one time step, the voltage at the load reaches its maximum value in one transmit time plus one time step, or 0.0501 ns.

Change the load to 1000 ohms. Run BOUNCE until say 1.3 ns. The voltage at 0.18 ns is 3.149 volts as stated above. You can verify by clicking the mouse on the waveform that each step found by hand-calculation is approximately equal to the values found with the program.

Problem 2

2. A logic gate has an output resistance of 10 ohms and produces a step function voltage of 10 volts (open-circuit) starting at t=0. The gate is connected in series with two

transmission lines, both having length 5 cm and speed of propagation 20 cm/ns. The first line has characteristic resistance 50 ohms, and the second line has characteristic

resistance 73 ohms. The load at the end of the second line is a 1000 resistor. Plot the voltage at the junction and the voltage across the load for 0<t<1.1 ns.

Solution Fig. 2-2a

Draw the circuit and then work out the reflection coefficients:

Source: 0.6667 50 10 50 10 1 1 ₌₋ + − = + − = Γ c s c s s R R R R Load: 0.8693 73 1000 73 1000 2 2 ₌ + − = + − = Γ c L c L L R R R R

Junction from the left:

1870 . 0 50 73 50 73 1 2 1 2 1 ₊ = − = + − = Γ c c c c R R R R 187 . 1 73 50 73 2 2 2 1 2 12= ₊ = ₊ = x R R R T c c c

Junction from the right:

1870 . 0 50 73 73 50 2 1 2 1 2 ₊ =− − = + − = Γ c c c c R R R R

8130 . 0 73 50 50 2 2 2 1 1 21= ₊ = ₊ = x R R R T c c c

Initial voltage at the generator:

( )

It’s handy to write the reflection coefficients on the circuit diagram because then it is easier to keep track of which one to use for each interaction in the bounce diagram. Fig. 2-2c

The bounce diagram is straightforward until 0.75 ns when two steps arrive simultaneously.

From the left, a step of -1.039 volts arrives:

Reflected back onto line #1: -1.039Γ₁=−1.039x0.1870=−0.194 Transmitted to line #2: −1.039T₁₂ =−1.039x1.187=−1.233 From the right, a step of 8.545 volts arrives:

Reflected back onto line #2:8.545Γ₂ =8.545x−0.1870=−1.598 Transmitted to line #1: 8.545T₂₁=8.545x0.8130=6.947

Add the two voltages on line #1 to get a step of 6.753 volts travelling back towards the source, and add the voltages on line #2 to get a step of -2.831 volts travelling towards the load. The load voltage is reflected at the load to give a reflected step of -2.446 volts. Fig. 2-2d

By inspection of the bounce diagram we can find the voltages at the junction and at the load as a function of time.

For the load, at 0.50 ns, the incident step is 9.891 volts and the reflected step is 8.545 volts, and these add up to a net step of 18.436 volts. The next voltage step to arrive is at 1.00 ns, and is a step of -2.831 volts, with a reflection of -2.446 volts. The net step is (incident + reflected)=-2.831-2.446=-5.277 volts, and this adds to the voltage of 18.436 volts already present at the load, so the net voltage is 18.436-5.277=13.159 volts.

Solution with BOUNCE

The initial step from the generator has amplitude 8.333 volts.

The initial step is reflected from the junction and the voltage on the transmission line steps up by 1.558 to 8.333+1.558=9.891 volts.

Also the step that is transmitted towards the load has amplitude 9.892 volts.

The 1.558 volt step is reflected from the generator, and the reflected step has amplitude -1.039 volts. The voltage on line #1 changes from 9.892 to 9.892-1.039=8.853 volts.

The 9.891 volt step incident on the 1000 ohm load is reflected, and the reflection is a step up of 8.545 volts. So the voltage on line #2 rises to (9.891+8.545)=18.436 volts.

These two steps in voltage are incident on the junction at the same time.

From the left, the -1.039 volt step gives a reflection of -0.1943 volts and a transmission onto line #2 of -1.233 volts.

From the right, the 8.545 volt step gives a reflection of -1.598 volts and a transmission onto line #1 of 6.947 volts.

Adding up the two contributions: Line #1: -0.1943+6.947=6.753 volts. Line #2: -1.233-1.598=-2.831 volts.

So the voltage on line #1 should step up from 8.853 volts to (8.853+6.753)=15.606 volts. And the voltage on line #2 should step down from 18.436 volts to (18.436-2.831)=15.605 volts.

And the voltage on Line #2 is also 15.61 volts.

At 1 ns, the -2.831 volt step on line #2 is reflected from the load and becomes a step of -2.446 volts. So the voltage at the load changes from 18.436 to

18.436+(-2.831-2.446)=13.159 volts.

Click “Plot v(t)” to make a good-quality graph of the voltage waveforms at the bottom of the BOUNCE simulation menu:

This is in agreement with the sketch we made in the pen-and-paper solution above.

This is the voltage at the junction and load up to 1.1 ns. Problem 3

3.A logic gate has internal resistance 10 ohms and produces a 10 volt step function (open-circuit) starting at t=0. It is connected to transmission line #1, having length 0.4 cm. The end of line #1 branches into line #2 in parallel with line #3. Line #2 has length 1.5 cm and is terminated with a 50 ohm load. Line #3 has length 0.7 cm, and is terminated with a high-impedance logic gate of input resistance 1000 ohms. The speed of propagation on all three lines is 20 cm/ns, and the characteristic resistance of all three lines is 50 ohms. Find the voltage across the 50 ohms load for 0<t<0.180 ns. Use BOUNCE to verify your solution.

Draw the circuit. Find the transit times: 02 . 0 20 4 . 0 1 1= = = u L T ns 075 . 0 20 5 . 1 2 2 = = = u L T ns 035 . 0 20 7 . 0 3 3 = = = u L T ns

Work out the reflection and transmission coefficients.

Source: 0.6667 50 10 50 10 ₌₋ + − = + − = Γ c s c s s R R R R Loads:

The 50 ohm load on line #2: 0.0 50 50 50 50 2 2 2 ₊ = − = + − = Γ c L c L L R R R R

The 1000 ohm load on line #3: 0.9048 50 1000 50 1000 3 3 3 ₊ = − = + − = Γ c L c L L R R R R Junction: 25 50 50 50 50 ₌ + = x R_p ohms 3333 . 0 50 25 50 25 − = + − = + − = Γ c p c p J R R R R 6667 . 0 50 25 25 2 2 = + = + = x R R R T c p p J

Initial voltage at the generator:

( )

Draw a bounce diagram:

Verify the Solution with BOUNCE

The initial step from the generator has height 8.333 volts.

The first step transmitted towards R1 has amplitude 5.556 volts. The step on the branch (line #3) also has amplitude 5.556 volts. The reflected step on line #1 changes the voltage from 8.333 to 5.556 volts, so is a step of -2.777 volts.

The first reflection from the generator steps the voltage on line #1 from 5.556 to 7.407 volts, so is a step of +1.851 volts.

The first reflection from the 1000 ohm load steps the voltage on line #3 from 5.556 up to 10.58 volts, so is a step of 5.024 volts.

Also the step up of 1.851 volts on line #1 is reflected as a step down from 7.407 to 6.790 volts of -0.617 volts. The transmitted step onto line #2 and 3 is +1.234 volts.

At the source there is a small reflection of 0.412 volts and the voltage steps up to 7.202 volts.

There is a reflection from line #3 back onto line #3 and the voltage on line #3 steps up to 10.14 volts.

A small step up is transmitted from line #1 to line #2 of (10.42-10.12)=0.28 volts. Note that at this time instant the voltage at the load has stepped up to 5.556 volts.

If we allow time to advance to 0.190 ns, only the three steps up that are already seen on line #2 will reach the load:

This is the second step at the load, increasing the voltage to 6.79 volts.

This is the 4th step at the load, increasing the voltage to 10.42 volts.