1 / 43

Week 3

Week 3. Capacitance & Inductance Transitions. Graph: y = e x Graph : y = e -x Graph : y = 1 - e -x. Try your hand at graphing these functions. y = e x. y = e x. y = e -x. y = 1 - e -x. e x y 2 1 0.5 2 2 0.75 2 3 0.875 2 4 0.9375 2 5 0.96875 2 6 0.984375 2 7 0.9921875

edythe
Download Presentation

Week 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Week 3 Capacitance & Inductance Transitions

  2. Graph: y = ex Graph: y = e-x Graph: y = 1 - e-x Try your hand at graphing these functions

  3. y = ex y = ex

  4. y = e-x

  5. y = 1 - e-x e x y 2 1 0.5 2 2 0.75 2 3 0.875 2 4 0.9375 2 5 0.96875 2 6 0.984375 2 7 0.9921875 2 8 0.99609375 2 9 0.998046875 2 10 0.999023438

  6. First-order System Has the same equation • y() is the final value. • y(0+) is the initial value. • Tau,, is the time constant.

  7. For RC C in Farads For LC L in Henrys Time Constant

  8. Capacitance is the ability of a body to store an electrical charge.  CapacitanceC is given by Gives the voltage/current relationship

  9. An RC circuit is one where you have a capacitor and resistor in the same circuit. Suppose we have the following circuit: RC Circuit – Initial Conditions Initially, the capacitor is UNCHARGED (q = 0) and the current through the resistor is zero. A switch (in red) then closes the circuit by moving upwards. The question is: What happens to the current and voltage across the resistor and capacitor as the capacitor begins to charge as a function of time? Which path do you think it takes? VC Time(s)

  10. Voltage Across the Resistor - Initially If we assume the battery has NO internal resistance, the voltage across the resistor will be the EMF. e VResistor t (sec) After a very long time, Vcap= e, as a result the potential difference between these two points will be ZERO. Therefore, there will be NO voltage drop across the resistor after the capacitor charges. Note: This is while the capacitor is CHARGING.

  11. Current Across the Resistor - Initially Imax=e/R t (sec) Since the voltage drop across the resistor decreases as the capacitor charges, the current across the resistor will reach ZERO after a very long time. Note: This is while the capacitor is CHARGING.

  12. Voltage Across the Capacitor - Initially e Vcap t (sec) As the capacitor charges it eventually reaches the same voltage as the battery or the EMF in this case after a very long time. This increase DOES NOT happen linearly. Note: This is while the capacitor is CHARGING.

  13. Current Across the Capacitor - Initially Imax=e/R t (sec) Since the capacitor is in SERIES with the resistor the current will decrease as the potential difference between it and the battery approaches zero. It is the potential difference which drives the value for the current. Note: This is while the capacitor is CHARGING.

  14. The graphs we have just seen show us that this process depends on the time. Let’s look then at the UNITS of both the resistance and capacitance. Time Domain Behavior Unit for Resistance = W = Volts/Amps Unit for Capacitance = Farad = Coulombs/Volts

  15. The “Time” Constant It is clear, that for a GIVEN value of "C”, for any value of “R” it effects the time rate at which the capacitor charges or discharges. Thus the PRODUCT of R and C produce what is called the CIRCUIT Capacitive TIME CONSTANT. We use the Greek letter, Tau, for this time constant. Another way to express Farads as Coulombs/Volt Another way to express R The question is: What exactly is the time constant?

  16. The “Time” Constant The time constant is the time that it takes for the capacitor to reach 63% of the EMF value during charging.

  17. Let’s test our function e 0.98e 0.95e 0.86e 0.63e Steady State Transient State 0.63e 0.86e 0.95e 1RC 2RC 31RC 4RC 0.98e εis the full voltage of the source Applying each time constant produces the charging curve we see. For practical purposes the capacitor is considered fully charged after 4-5 time constants( steady state). Before that time, it is in a transient state.

  18. Charging Functions Charge and voltage build up to a maximum… …while current fades to zero Likewise, the voltage function can be divided by another constant, in this case, “R”, to derive the current charging function. Now we have 3 functions that allows us to calculate the Charge, Voltage, or Current at any given time “t” while the capacitor is charging.

  19. Suppose now the switch moves downwards towards the other terminal. This prevents the original EMF source to be a part of the circuit. Capacitor Discharge – Resistor’s Voltage e At t =0, the resistor gets maximum voltage but as the capacitor cannot keep its charge, the voltage drop decreases. VResistor t (sec)

  20. Capacitor Discharge – Resistor’s Current Similar to its charging graph, the current through the resistor must decrease as the voltage drop decreases due to the loss of charge on the capacitor. I=e/R IResistor t (sec)

  21. Capacitor Discharge – Capacitor's Voltage The discharging graph for the capacitor is the same as that of the resistor. There WILL be a time delay due to the TIME CONSTANT of the circuit. In this case, the time constant is reached when the voltage of the capacitor is 37% of the EMF.

  22. Capacitor Discharge – Capacitor’s Current Similar to its charging graph, the current through the capacitor must decrease as the voltage drop decreases due to the loss of charge on the capacitor. I=e/R Icap t (sec)

  23. The bottom line to take away…. Time to charge 63% = time constant “tau” =τ = RC Charge and voltage build up to a maximum… When charging a capacitor …while current fades to zero. When discharging a capacitor All three fade away during discharge.

  24. Capacitor Circuit Operation

  25. Capacitor Circuit Operation • LINEAR Caps Follow the Capacitance Law; in DC Recall the Circuit Representation • Where • Q  The CHARGE STORED in the Cap, Coulombs • C  Capacitance, Farad • Vc DC-Voltage Across the Capacitor • Discern the Base Units for Capacitance • The Basic Circuit-Capacitance Equation

  26. “Feel” for Capacitance • Recall Capacitor Law • Pick a Cap, Say 12 µF • Solving for Vc • Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage • Caps can Be DANGEROUS! • Now Assume That The Cap is Charged to hold 15 mC • Find V c

  27. Forms of the Capacitor Law • The time-Invariant Cap Law • If vC at − = 0, then the traditional statement of the Integral Law • Leads to DIFFERENTIAL Cap Law • The Differential Suggests SEPARATING Variables • If at t0, vC= vC(t0) (a KNOWN value), then the Integral Law becomes • Leads to The INTEGRAL form of the Capacitance Law

  28. Capacitor Integral Law • Thus a Major Mathematical Implication of the Integral law • Express the VOLTAGE Across the Cap Using the INTEGRAL Law • The Voltage Across a Capacitor MUST be Continuous • An Alternative View • The Differential Application • If i(t) has NO Gaps in its i(t) curve then • Even if i(y) has VERTICAL Jumps: • If vC is NOT Continuous then dvC/dt → , and So iC → . This is NOT PHYSICALLY possible

  29. Capacitor Differential Law • Thus a Major Mathematical Implication of the Differential Law • Express the CURRENT “Thru” the Cap Using the Differential Law • A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit • If vC= Constant Then • Cap Current • Charges do NOT flow THRU a Cap • Charge ENTER or EXITS The Cap in Response to Voltage CHANGES • This is the DC Steady-State Behavior for a Capacitor

  30. Charges do NOT flow THRU a Cap • Charge ENTER or EXITS The Capacitor in Response to the Voltage Across it • That is, the Voltage-Change DISPLACES the Charge Stored in The Cap • This displaced Charge is, to the REST OF THE CKT, Indistinguishablefrom conduction (Resistor) Current • Thus The Capacitor Current is Called the “Displacement” Current Capacitor Current

  31. Capacitor Summary From Calculus, Recall an Integral Property The Circuit Symbol Note The Passive Sign Convention Now Recall the Long Form of the Integral Relation • Compare Ohm’s Law and Capacitance Law • Capacitor Ohm The DEFINITE Integral is just a number; call it vC(t0) so

  32. Consider Finally the Differential Application Capacitor Summary cont • Some Implications • For small Displacement Current dvC/dt is small; i.e, vC changes only a little • Obtaining Large iC requires HUGE Voltage Gradients if C is small • Conclusion: A Capacitor RESISTS CHANGES in VOLTAGE ACROSS It

  33. CONCLUSION: Capacitance

  34. The Inductor • Second of the Energy-Storage Devices • Basic Physical Model: Circuit Symbol

  35. Physical Inductor • Inductors are Typically Fabricated by Winding Around a Magnetic (e.g., Iron) Core a LOW Resistance Wire • Applying to the Terminals a TIME VARYING Current Results in a “Back EMF” voltage at the connection terminals Some Real Inductors

  36. Inductance Defined • Where the Constant of Proportionality, L, is called the INDUCTANCE • L is Measured in Units of “Henrys”, H • 1H = 1 V•s/Amp • Inductors STORE electromagnetic energy • They May Supply Stored Energy Back To The Circuit, But They CANNOT CREATE Energy From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law For a Linear Inductor The Flux Is Proportional To The Current Thru it

  37. Inductance Sign Convention • Inductors Cannot Create Energy; They are PASSIVE Devices • All Passive Devices Must Obey the Passive Sign Convention

  38. Inductor Circuit Operation Separating the Variables and Integrating Yields the INTEGRAL form Recall the Circuit Representation In a development Similar to that used with caps, Integrate − to t0 for an Alternative integral Law Previously Defined the Differential Form of the Induction Law

  39. Drill Problem 4-13, pp. 260-261. The 0.05F in P4-13 is initially charged to 8v. At t = 0, a 20v source is connected. Determine the expressions for: I(t) and vc(t) for t > 0 I(t)=1.2e-2t vc(t) = 20-12e-2t

  40. Summary

  41. Circuit Behavior of Reactive Components

  42. CONCLUSION: Inductance

More Related