Question

Work the following problems.

1)Express the speed of light…

a. In ps/cm

b. In ns/foot

2)Consider a signal with a rise time of 200 ps.

a. What is the knee frequency of the signal?

b.Consider the signal on a trace over an insulator with ε_r = 4. How long is the rising edge, in inches?

c.How long does a trace have to be in order to be considered “distributed”?

3)The same signal is routed on a circuit board. The output capacitance is 15 pF and the resistance of the gate that drives the signal is 20 ohms.

a. What is the risetime of the signal, taking into account the R and C of the output driver?

b. What is the new knee frequency, after RC filtering?

4) A circuit board is being designed with ε_r = 4.2. A trace is laid out with a length of 2.5 cm.

a.What risetime is needed for the trace to be considered distributed and not lumped?

b.What knee frequency corresponds to the risetime you calculated in a?

c.The signal has a risetime of 50 ps. What RC (that is, an RC time constant of what value) is needed to slow the signal down to the risetime you calculated in a?

d.The load capacitance of the signal is 20 pF. What value of resistor should be inserted onto the line to obtain the value of RC you calculated in c?

Answer 1

2. Consider a signal with a rise time of 200 ps.

(a) What is the knee frequency of a signal?

using a formula, we have

f_knee = 1 / T_rise 1 / (200 x 10^-12 s)

f_knee = 5 x 10⁹ Hz

3. The same signal is routed on a circuit board.

(a) What is the risetime of a signal, taking into account the R and C of an output driver?

using a formula, we have

_rise = RC

where, R = resistance of the gate = 20

C = output capacitance = 15 x 10^-12 F

then, we get

_rise = [(20 ) (15 x 10^-12 F)]

_rise = 3 x 10^-10 s

_rise = 0.3 ns

(b) What is the new knee frequency, after RC filtering?

we know that, f'_knee = 1 / _rise

f'_knee = 1 / (3 x 10^-10 s)

f'_knee = 3.33 x 10⁹ Hz

4. A circuit board is being designed with ε_r = 4.2. A trace is laid out with a length of 2.5 cm.

(c) The signal has a risetime of 50 ps. What RC (that is, an RC time constant of what value) is needed to slow the signal down to the risetime you calculated in part (a)?

we know that, _rise = RC

(50 x 10^-12 s) = RC

RC = 50 x 10^-12 s

(d) The load capacitance of the signal is 20 pF. What value of resistor should be inserted onto the line to obtain the value of RC we calculated in part (c)?

using a formula, we have

= R C (50 x 10^-12 s) = R (20 x 10^-12 F)

R = [(50 x 10^-12 s) / (20 x 10^-12 F)]

R = 2.5