* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Full PowerPoint
Flip-flop (electronics) wikipedia , lookup
Oscilloscope history wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Radio transmitter design wikipedia , lookup
Analog-to-digital converter wikipedia , lookup
Two-port network wikipedia , lookup
Current source wikipedia , lookup
Power MOSFET wikipedia , lookup
Surge protector wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Valve audio amplifier technical specification wikipedia , lookup
Wilson current mirror wikipedia , lookup
Transistor–transistor logic wikipedia , lookup
Power electronics wikipedia , lookup
Integrating ADC wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Operational amplifier wikipedia , lookup
Voltage regulator wikipedia , lookup
Network analysis (electrical circuits) wikipedia , lookup
Schmitt trigger wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Current mirror wikipedia , lookup
EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross 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 1 EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross RC RESPONSE R Internal Model Vin V out of Logic Gate C Behavior of Vout after change in Vin Vout Vout Vin Vout(t=0) Vin Vout(t=0) 0 0 time 0 0 time EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross RC RESPONSE We said that Vout(t) has the general exponential form Vout(t) = A + Be-t/t Initial value Vout(t=0) = A + B Final value Vout(t→∞) = A But we know Vout(t→∞) = Vin so rewrite Vout(t) = Vin + [Vout(t=0)-Vin]e-t/t We will now prove that time constant t is t = RC Remember, Vout(t) has gone 63% of the way from initial to final value after 1 time constant. EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross RC RESPONSE: Proof of Vout equation R V in V out iR iR = iC Vin Vout (Ohm' s law) R iC = C C dVout (capacitance law) dt But iR = iC ! Vin Vout dVout Thus, =C R dt or dVout 1 = ( Vin Vout ) dt RC EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross RC RESPONSE: Proof We know Is dVout 1 = ( Vin Vout ) dt RC Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC a solution? Substitute: dVout d 1 = Vin [ Vout ( t = 0) Vin ]e t / RC = [ Vout ( t = 0) Vin ]e t / RC dt dt RC = 1 [Vin - Vout (t = 0)]e - t/RC RC 1 1 (Vin - Vout (t)) = [Vin - (Vin [Vout (t = 0) - Vin ]e - t/RC )] RC RC 1 = [Vin - Vout (t = 0)]e - t/RC RC This is the solution since it has the correct initial condition! EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross Example: Draw graph and write equation Assume Vin has been zero for a long time, and steps up to 10 V at t=0. Input node + R Vout C Vin - Assume R = 1KΩ, C = 1pF . Output node ground Ingredients for graph: Vin 1) Initial value of Vout is 0 2) Final value of Vout is Vin=10V 10 Vout 6.3V 3) Time constant RC is 10-9 sec 4) Vout reaches 0.63 X 10 in 10-9 sec Ingredients for equation: Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC => 0 0 1nsec time Vout(t) = 10 - 10e-t/1nsec EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross Charging and discharging in RC Circuits (The official EE40 Easy Method) Method of solving for any node voltage in a single capacitor circuit. 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: 2) The time constant is t = RC. Input node + 3) Find the capacitor voltage before Vin the input voltage changes. This is the initial value for the output, Vout(t=0). R Output node Vout C ground 4) Find the asymptotic value Vout(t→∞), in this case, equals Vin 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. EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross PULSE DISTORTION Vin What if I want to step up the input, 0 0 Vin wait for the output to respond, time Vout 0 0 Vin then bring the input back down for a different response? time Vout 0 0 time EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross PULSE DISTORTION R Input node The pulse width (PW) must be greater than RC to avoid severe pulse distortion. Output node O + C Vin - ground PW = 0.1RC PW = RC 6 5 4 3 2 1 0 PW = 10RC 6 5 4 3 2 1 0 Vout Vout Vout 6 5 4 3 2 1 0 0 1 2 Time 3 4 5 0 1 2 Time 3 4 5 0 5 10 Time 15 20 25 EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross EXAMPLE Suppose a voltage pulse of width 5 ms and height 4 V is applied to the input of the circuit at the right. Sketch the output voltage. Vin R R = 2.5 KΩ C = 1 nF Vout C 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 ms The output increases for 5ms 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. EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross EXAMPLE 4 3.5 3 2.5 2 1.5 1 0.5 00 2 4 6 8 10 Just for fun, the equation for the output is: Vout(t) = 4-4e-t/2ms for 0 ≤ t ≤ 5 ms { 3.44e-(t-5ms)/2.5ms for t > 5 ms EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross APPLICATIONS • Now we can find “propagation delay” tp; 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 tp = - 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. EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross 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! EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross 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 Same -Ix - Vx + + -Vx - Thing! EECS 40 Fall 2002 Lecture 6 W. G. Oldham and S. Ross @#%*ING ASSOCIATED REFERENCE CURRENTS We only need to worry about associated reference currents in 3 situations (here we need to have associated reference current): I 1. 2. 3. Using Ohm’s law in a resistor Using I = C dV/dt in a capacitor Calculating power P=VI + V Otherwise, forget about it! Work with currents and voltage in whatever direction you want! -