1. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 1 CIRCUITS, NODES, AND BRANCHES Last time: Looked at circuit elements, used to model the insides of logic gates Saw how the resistor and capacitor in the gate circuit model make the change in output voltage gradual Looked at the mathematical equations that describe the output voltage Today: Derive the mathematical equation for the change in gate output voltage Practice an easy method used to graph and write the equation for the changing output voltage Look at how R and C affect the rate of change in output voltage Look at how R and C affect how quickly we can put new signals through Review ideas we will need to understand Kirchoff’s laws

2. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 RC RESPONSE V in R V out C Internal Model of Logic Gate time V out 0 0 0 V in time 0 V out (t=0) V in V out (t=0) Behavior of V out after change in V in

3. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 RC RESPONSE We said that V out (t) has the general exponential form V out (t) = A + Be -t/  Initial value V out (t=0) = A + B Final value V out (t→∞) = A But we know V out (t→∞) = V in so rewrite V out (t) = V in + [V out (t=0)-V in ]e -t/  We will now prove that time constant  is  RC Remember, V out (t) has gone 63% of the way from initial to final value after 1 time constant.

4. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 RC RESPONSE: Proof of V out equation V in R V out C i C i R

5. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 RC RESPONSE: Proof V out (t) = V in + [V out (t=0)-V in ]e -t/RC We know Isa solution? Substitute: This is the solution since it has the correct initial condition!

6. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 Example: Draw graph and write equation Input nodeOutput node ground R C Vin Vout + - Assume Vin has been zero for a long time, and steps up to 10 V at t=0. Assume R = 1KΩ, C = 1pF. Ingredients for graph: time Vin 0 0 10 Vout 1nsec 6.3V 1)Initial value of V out is 0 2)Final value of V out is V in =10V 3)Time constant RC is 10 -9 sec 4)V out reaches 0.63 X 10 in 10 -9 sec V out (t) = 10 - 10e -t/1nsec V out (t) = V in + [V out (t=0)-V in ]e -t/RC => Ingredients for equation:

7. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 Charging and discharging in RC Circuits (The official EE40 Easy Method ) Input nodeOutput node ground R C Vin Vout + - 4) Find the asymptotic value V out (t→∞), in this case, equals V in after step 5) Sketch the transient period. After one time constant, the graph has passed 63% of the way from initial to final value. 6) Write the equation by inspection. 1) Simplify the circuit so it looks like one resistor, a source, and a capacitor (it will take another two weeks to learn all the tricks to do this.) But then the circuit looks like this: Method of solving for any node voltage in a single capacitor circuit. 2) The time constant is  = RC. 3) Find the capacitor voltage before the input voltage changes. This is the initial value for the output, V out (t=0).

8. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 PULSE DISTORTION What if I want to step up the input, wait for the output to respond, then bring the input back down for a different response? time V in 0 0 time Vin 0 0 Vout time Vin 0 0 Vout

9. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 Vin + - O The pulse width (PW) must be greater than RC to avoid severe pulse distortion. 0 1 2 3 4 5 6 012345 Time Vout PW = 0.1RC 0 1 2 3 4 5 6 012345 Time Vout PW = RC 0 1 2 3 4 5 6 0510152025 Time Vout PW = 10RC PULSE DISTORTION

10. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 EXAMPLE V in R V out C Suppose a voltage pulse of width 5  s and height 4 V is applied to the input of the circuit at the right. Sketch the output voltage. R = 2.5 KΩ C = 1 nF First, the output voltage will increase to approach the 4 V input, following the exponential form. When the input goes back down, the output voltage will decrease back to zero, again following exponential form. How far will it increase? Time constant = RC = 2.5  s The output increases for 5  s or 2 time constants. It reaches 1-e -2 or 86% of the final value. 0.86 x 4 V = 3.44 V is the peak value.

11. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 EXAMPLE 0 0.5 1 1.5 2 2.5 3 3.5 4 0246810 Just for fun, the equation for the output is: V out (t) = 4-4e -t/2  s for 0 ≤ t ≤ 5  s 3.44e -(t-5  s)/2.5  s for t > 5  s {

12. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 APPLICATIONS Now we can find “propagation delay” t p ; the time between the input reaching 50% of its final value and the output to reaching 50% of final value. Today’s case, using “perfect input” (50% reached at t=0): 0.5 = e -tp t p = - ln 0.5 = 0.69 It takes 0.69 time constants, or 0.69 RC. We can find the time it takes for the output to reach other desired levels. For example, we can find the time required for the output to go from 0 V to the minimum voltage level for logic 1. Knowing these delays helps us design clocked circuits.

13. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 Adding Voltages in Series In electrical engineering we generally add voltage drops. Example: 1.Go around the path that comprises Vx. Start at + terminal, end at – terminal. 2.Look at the first sign you encounter at each element. If +, add that voltage. If -, subtract. 3.The final sum is Vx. + 5 – 8 – 6 + 8 = -1 And that’s OK!

14. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 JUST PICK A POLARITY In the last slide, we “guessed Vx wrong”; the bottom end was actually higher potential than the top. WE DON’T CARE! If I need to find a voltage or current, I just give it a name and write down a polarity/direction. Whatever I feel like at the time. Then I solve for the unknown. If the voltage/current “doesn’t really go that way”, the answer will be negative. SO WHAT. Just remember how to flip things if you need to: Ix -Ix - Vx + + -Vx - Same Thing!

15. W. G. Oldham and S. RossEECS 40 Fall 2002 Lecture 6 @#%*ING ASSOCIATED REFERENCE CURRENTS We only need to worry about associated reference currents in 3 situations (here we need to have associated reference current): 1. Using Ohm’s law in a resistor 2. Using I = C dV/dt in a capacitor 3. Calculating power P=VI I + V - Otherwise, forget about it! Work with currents and voltage in whatever direction you want!