EE292: Fundamentals of ECEb1morris/ee292/docs/slides15.pdf · RL Example âĒCurrent before switch...
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EE292: Fundamentals of ECE
Fall 2012
TTh 10:00-11:15 SEB 1242
Lecture 15
121016
Outline
âĒ Review General RC Circuit
âĒ RL Circuits
2
General 1st-Order RC Solution
âĒ Notice both the current and voltage in an RC circuit has an exponential form
âĒ The general solution for current/voltage is: âŦ ðĨ â represents current or voltage âŦ ðĄ0 â represents time when source switches âŦ ðĨð - final (asymptotic) value of current/voltage
âŦ ð â time constant (ð ðķ)
âĒ Find values and plug into general solution
3
Example
âĒ Solve for ðĢð(ðĄ)
âŦ ðĢð = ðð steady-state analysis
âŦ ðĢð 0+ = 0 no voltage when switch open
âŦ ð = ð ðķ equivalent resistance/capacitance
âĒ ðĢð ðĄ = ðð + 0 â ðð ðâðĄ/(ð ðķ) = ðð â ðð ðâðĄ/(ð ðķ)
âĒ Solve for ðð(ðĄ)
âŦ ðð = 0 fully charged cap no current
âŦ ðð 0+ =ðð âðĢð(0+)
ð =
ðð â0
ð =
ðð
ð
âĒ ðð ðĄ = 0 +ðð
ð â 0 ðâðĄ/(ð ðķ) =
ðð
ð ðâðĄ/(ð ðķ)
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ðð
RC Current âĒ Voltage
âŦ ðĢð ðĄ = ðð â ðð ðâðĄ/(ð ðķ)
âĒ Current
âŦ ðð =ðð âðĢð(ðĄ)
ð = ðķ
ððĢð(ðĄ)
ððĄ
âŦ ðð = ðķðð
ð ðķðâðĄ/(ð ðķ)
âŦ ðð =ðð
ð ðâðĄ/(ð ðķ)
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ðð
http://www.electronics-tutorials.ws/rc/rc_1.html
First-Order RL Circuits âĒ Contains DC sources, resistors, and a single inductance
âĒ Same technique to analyze as for RC circuits 1. Apply KCL and KVL to write circuit equations 2. If the equations contain integrals, differentiate each
term in the equation to produce a pure differential equation
âŦ Use differential forms for I/V relationships for inductors and capacitors
3. Assume solution of the form ðū1 + ðū2ðð ðĄ 4. Substitute the solution into the differential equation to
determine the values of ðū1and ð 5. Use initial conditions to determine the value of ðū2 6. Write the final solution
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RL Example âĒ Current before switch
âŦ ð 0â = 0
âĒ KVL around loop
âŦ ðð â ð ð ðĄ â ðŋðð ðĄ
ððĄ= 0
âŦ ð ðĄ +ðŋ
ð
ðð ðĄ
ððĄ=
ðð
ð
Notice this is the same equation form as the charging capacitor example
âĒ Solution of the form
âŦ ð ðĄ = ðū1 + ðū2ðð ðĄ
âĒ Solving for ðū1, ð
âŦ ðū1 + ðū2ðð ðĄ +ðŋ
ð ðū2ð ðð ðĄ =
ðð
ð
ðū1 =ðð
ð
1 +ðŋ
ð ð = 0 â ð = â
ð
ðŋ
âĒ Solving for ðū2
âŦ ð 0+ = 0 = ðð
ð + ðū2ðâðĄð /ðŋ
âŦ 0 =ðð
ð + ðū2ð0
âŦ ðū2 = âðð
ð
âĒ Final Solution
âŦ ð ðĄ =ðð
ð â
ðð
ð ðâðĄð /ðŋ
âŦ ð ðĄ = 2 â 2ðâ500ðĄ
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RL Example âĒ ð ðĄ = 2 â 2ðâ500ðĄ
âĒ Notice this is in the general form we used for RC circuits
âŦ ð =ðŋ
ð
âĒ Find voltage ðĢ(ðĄ)
âŦ ðĢð = 0, steady-state short
âŦ ðĢ 0+ = 100
No current immediately
through ð , ðĢ = ðŋðð(ðĄ)
ððĄ
âĒ ðĢ ðĄ = 100ðâðĄ/ð
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Exercise 4.5
âĒ Initial conditions
âĒ For ðĄ < 0 âŦ All source current goes
through switched wire
âŦ ðð ðĄ = ððŋ ðĄ = 0 ðī
âŦ ðĢ ðĄ = ðð ðĄ R = 0 V
âĒ For ðĄ = 0+ (right after switch)
âŦ ððŋ ðĄ = 0 Current canât change
immediately through an inductor
âŦ ðð ðĄ = 2 A, by KCL
âŦ ðĢ ðĄ = ðð ðĄ R = 20 V
âĒ Steady-state âŦ Short inductor
âĒ ðĢ ðĄ = 0 âŦ Short circuit across inductor
âĒ ðð = 0 âŦ All current through short
âĒ ððŋ = 2 ðī âŦ By KCL
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Exercise 4.5
âĒ Can use network analysis to come up with a differential equation, but you would need to solve it
âĒ Instead, use the general 1st-order solution
âĒ Time constant ð
âŦ ð =ðŋ
ð =
2
10= 0.2
âĒ Voltage ðĢ(ðĄ) âŦ ðĢ ðĄ = 0 + 20 â 0 ðâðĄ/0.2 = 20ðâðĄ/0.2 V
âĒ Current ððŋ(ðĄ), ðð (ðĄ) âŦ ððŋ ðĄ = 2 + 0 â 2 ðâðĄ/0.2 = 2 â 2ðâðĄ/0.2 A âŦ ðð ðĄ = 0 + 2 â 0 ðâðĄ/0.2 = 2ðâðĄ/0.2 A
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RC/RL Circuits with General Sources
âĒ Previously,
âŦ ð ðķððĢð(ðĄ)
ððĄ+ ðĢð(ðĄ) = ðð
âĒ What if ðð is not constant
âŦ ð ðķððĢð(ðĄ)
ððĄ+ ðĢð(ðĄ) = ðĢð (ðĄ)
âŦ Now have a general source that is a function of time
âĒ The solution is a differential equation of the form
âŦ ðððĨ(ðĄ)
ððĄ+ ðĨ(ðĄ) = ð(ðĄ)
âŦ Where ð(ðĄ) is known as the forcing function (the circuit source)
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ðĢð (ðĄ)
General Differential Equations
âĒ General differential equation
âĒ ðððĨ(ðĄ)
ððĄ+ ðĨ(ðĄ) = ð(ðĄ)
âĒ The solution to the diff equation is
âĒ ðĨ ðĄ = ðĨð ðĄ + ðĨâ(ðĄ)
âĒ ðĨð ðĄ is the particular solution
âĒ ðĨâ(ðĄ) is the homogeneous solution
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Particular Solution
âĒ ðððĨð ðĄ
ððĄ+ ðĨð ðĄ = ð(ðĄ)
âĒ The solution ðĨð ðĄ is called the forced response
because it is the response of the circuit to a particular forcing input ð ðĄ
âĒ The solution ðĨð ðĄ will be of the same functional
form as the forcing function
âŦ E.g.
âŦ ð ðĄ = ðð ðĄ â ðĨð ðĄ = ðīðð ðĄ
âŦ ð ðĄ = cos ððĄ â ðĨð ðĄ = ðīcos ððĄ + ðĩsin(ððĄ)
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Homogeneous Solution
âĒ ðððĨâ ðĄ
ððĄ+ ðĨâ ðĄ = 0
âĒ ðĨâ ðĄ is the solution to the differential equation when there is no forcing function
âĒ Does not depend on the sources
âĒ Dependent on initial conditions (capacitor voltage, current through inductor)
âĒ ðĨâ ðĄ is also known as the natural response
âĒ Solution is of the form
âĒ ðĨâ ðĄ = ðūðâðĄ/ð
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General Differential Solution
âĒ Notice the final solution is the sum of the particular and homogeneous solutions
âŦ ðĨ ðĄ = ðĨð ðĄ + ðĨâ(ðĄ)
âĒ It has an exponential term due to ðĨâ(ðĄ) and a term ðĨð ðĄ that matches the input source
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Second-Order Circuits
âĒ RLC circuits contain two energy storage elements
âŦ This results in a differential equation of second order (has a second derivative term)
âĒ This is like a mass spring system from physics
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RLC Series Circuit
âĒ KVL around loop
âŦ ðĢð ðĄ â ðŋðð ðĄ
ððĄâ ð ðĄ ð â ðĢð ðĄ = 0
âĒ Solve for ðĢð ðĄ
âŦ ðĢð ðĄ = ðĢð ðĄ â ðŋðð ðĄ
ððĄâ ð ðĄ ð
âĒ Take derivative
âŦððĢð ðĄ
ððĄ=
ððĢð ðĄ
ððĄâ ðŋ
ð2ð ðĄ
ððĄ2 â ð ðð ðĄ
ððĄ
âĒ Solve for current through capacitor
âĒ ð ðĄ = ðķððĢð ðĄ
ððĄ
âĒ ð ðĄ = ðķððĢð ðĄ
ððĄâ ðŋ
ð2ð ðĄ
ððĄ2 â ð ðð ðĄ
ððĄ
âĒð2ð ðĄ
ððĄ2 âð
ðŋ
ðð ðĄ
ððĄ+
1
ðŋðķð ðĄ =
1
ðŋ
ððĢð ðĄ
ððĄ
âĒ The general 2nd-order constant coefficient equation
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