Homework on RC Circuits (solutions)

Feel free to work on the homework in groups. The work you hand in, however, should reflect your understanding of the material and be in your own words. Students who turn in identical (or close to identical) homework assignments will be asked to explain their answers orally to the TA or prof. A student who cannot explain how he or she arrived at a given answer will be charged with academic dishonesty.

You should justify all of your answers for full credit.

RC Circuits

In order to send a signal from one transistor to another on the opposite side of a chip, current has to travel through an interconnect approximately 1.0 cm (1.0e-2 m) in length, 2 mm wide (2.0e-6 m), and 1 mm deep (1.0e-6 m). Aluminum, which is currently used in chips has a resistivity r of 2.65 x 10^-8Wm.¹ Copper, which you will hear Dr. Lu discuss in his guest lecture, has a resistivity r of 1.68 x 10^-8Wm.¹ The resistance of a wire is given by R = rL/A, where A is the cross-sectional area of the wire and L is its length.

¹Data from Physics, by Giancoli, 5th ed.

1. What is the resistance of the current aluminum interconnect? What is the resistance of the future copper interconnect?
((2.65 x 10^-8 Wm)(1 x 10^-2 m)) / ((1 x 10^-6 m)(2 x 10^-6 m)) = 130 W
((1.68 x 10^-8 Wm)(1 x 10^-2 m)) / ((1 x 10^-6 m)(2 x 10^-6 m)) = 84 W

2. If the capacitance intrinsic to the interconnect is 1.0 pF (1.0e-12 F), what is the time constant t of the aluminum interconnect? What is the rise time Dt? What are the time constant and rise time for the copper interconnect?
t = RC; (132.5 W) (1 x 10^-12 F) = 0.1325 ns; rise time is approximately double this, so 0.265 ns
For copper, (84 W) ... = 0.084 ns; 0.168 ns

3. How long will it take for a voltage signal sent down the aluminum interconnect to rise to 85% of its maximum value? This might be a threshhold we impose to determine whether the signal is on. (You must use the equation for voltage in a charging RC circuit here; you cannot just figure this out using Dt.) What is the minimum duration for a pulse if it must be at least twice as long as the time you calculated? Q(t) = Q₀(1 - e^-t/RC);
since Q = CV, we can write V(t) = V₀(1 - e^-t/RC), or
V(t)/V₀ = (1 - e^-^t/RC)
So, 0.85 = 1 - e^-t/RC, or 0.15 = e^-t/RC.
Taking a natural log, we get t/RC = 1.897, or t = 1.897RC.
t = (1.897) (132.5 W) (1 x 10^-12 F) = 0.2514 ns; 0.5028 ns for the pulse

4. A GHz processor produces one billion "clicks" per second, so each "click" is allocated one nanosecond. Assume the time per "click" is equally split between the "on" signal and the "off" spacer. How does the time you calculated above compare to the time the signal is "on" in a GHz processor with aluminum interconnects? With copper interconnects?
The "on" signal requires 0.5 ns; the time for the aluminum interconnect is greater than this. This is bad.
The time for a pulse on the copper interconnect is 0.3187 ns, which is less than the 0.5 ns figure. This is good.

5. If the speed of the circuit is determined by the rise time through this interconnect, what is the maximum speed (rate of "clicks") the chip with aluminum interconnects can attain? What is the maximum speed for the chip with copper interconnects?
Double the pulse time and take the reciprocal. For aluminuminuminum, you get 994 MHz; for copper, you get 1.57 GHz.

1.	What is the resistance of the current aluminum interconnect? What is the resistance of the future copper interconnect? ((2.65 x 10^-8 Wm)(1 x 10^-2 m)) / ((1 x 10^-6 m)(2 x 10^-6 m)) = 130 W ((1.68 x 10^-8 Wm)(1 x 10^-2 m)) / ((1 x 10^-6 m)(2 x 10^-6 m)) = 84 W
2.	If the capacitance intrinsic to the interconnect is 1.0 pF (1.0e-12 F), what is the time constant t of the aluminum interconnect? What is the rise time Dt? What are the time constant and rise time for the copper interconnect? t = RC; (132.5 W) (1 x 10^-12 F) = 0.1325 ns; rise time is approximately double this, so 0.265 ns For copper, (84 W) ... = 0.084 ns; 0.168 ns
3.	How long will it take for a voltage signal sent down the aluminum interconnect to rise to 85% of its maximum value? This might be a threshhold we impose to determine whether the signal is on. (You must use the equation for voltage in a charging RC circuit here; you cannot just figure this out using Dt.) What is the minimum duration for a pulse if it must be at least twice as long as the time you calculated? Q(t) = Q₀(1 - e^-t/RC); since Q = CV, we can write V(t) = V₀(1 - e^-t/RC), or V(t)/V₀ = (1 - e^-^t/RC) So, 0.85 = 1 - e^-t/RC, or 0.15 = e^-t/RC. Taking a natural log, we get t/RC = 1.897, or t = 1.897RC. t = (1.897) (132.5 W) (1 x 10^-12 F) = 0.2514 ns; 0.5028 ns for the pulse
4.	A GHz processor produces one billion "clicks" per second, so each "click" is allocated one nanosecond. Assume the time per "click" is equally split between the "on" signal and the "off" spacer. How does the time you calculated above compare to the time the signal is "on" in a GHz processor with aluminum interconnects? With copper interconnects? The "on" signal requires 0.5 ns; the time for the aluminum interconnect is greater than this. This is bad. The time for a pulse on the copper interconnect is 0.3187 ns, which is less than the 0.5 ns figure. This is good.
5.	If the speed of the circuit is determined by the rise time through this interconnect, what is the maximum speed (rate of "clicks") the chip with aluminum interconnects can attain? What is the maximum speed for the chip with copper interconnects? Double the pulse time and take the reciprocal. For aluminuminuminum, you get 994 MHz; for copper, you get 1.57 GHz.