ee292: fundamentals of eceb1morris/ee292/docs/slides15.pdfΒ Β· rl example β’current before switch...
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![Page 1: EE292: Fundamentals of ECEb1morris/ee292/docs/slides15.pdfΒ Β· RL Example β’Current before switch π0β =0 β’KVL around loop π β π β π =0 π +πΏ π
π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/1.jpg)
http://www.ee.unlv.edu/~b1morris/ee292/
EE292: Fundamentals of ECE
Fall 2012
TTh 10:00-11:15 SEB 1242
Lecture 15
121016
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/2.jpg)
Outline
β’ Review General RC Circuit
β’ RL Circuits
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/3.jpg)
General 1st-Order RC Solution
β’ Notice both the current and voltage in an RC circuit has an exponential form
β’ The general solution for current/voltage is: β« π₯ β represents current or voltage β« π‘0 β represents time when source switches β« π₯π - final (asymptotic) value of current/voltage
β« π β time constant (π πΆ)
β’ Find values and plug into general solution
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/4.jpg)
Example
β’ Solve for π£π(π‘)
β« π£π = ππ steady-state analysis
β« π£π 0+ = 0 no voltage when switch open
β« π = π πΆ equivalent resistance/capacitance
β’ π£π π‘ = ππ + 0 β ππ πβπ‘/(π πΆ) = ππ β ππ πβπ‘/(π πΆ)
β’ Solve for ππ(π‘)
β« ππ = 0 fully charged cap no current
β« ππ 0+ =ππ βπ£π(0+)
π =
ππ β0
π =
ππ
π
β’ ππ π‘ = 0 +ππ
π β 0 πβπ‘/(π πΆ) =
ππ
π πβπ‘/(π πΆ)
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ππ
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/5.jpg)
RC Current β’ Voltage
β« π£π π‘ = ππ β ππ πβπ‘/(π πΆ)
β’ Current
β« ππ =ππ βπ£π(π‘)
π = πΆ
ππ£π(π‘)
ππ‘
β« ππ = πΆππ
π πΆπβπ‘/(π πΆ)
β« ππ =ππ
π πβπ‘/(π πΆ)
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ππ
http://www.electronics-tutorials.ws/rc/rc_1.html
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/6.jpg)
First-Order RL Circuits β’ Contains DC sources, resistors, and a single inductance
β’ 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
β« Use differential forms for I/V relationships for inductors and capacitors
3. Assume solution of the form πΎ1 + πΎ2ππ π‘ 4. Substitute the solution into the differential equation to
determine the values of πΎ1and π 5. Use initial conditions to determine the value of πΎ2 6. Write the final solution
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/7.jpg)
RL Example β’ Current before switch
β« π 0β = 0
β’ KVL around loop
β« ππ β π π π‘ β πΏππ π‘
ππ‘= 0
β« π π‘ +πΏ
π
ππ π‘
ππ‘=
ππ
π
Notice this is the same equation form as the charging capacitor example
β’ Solution of the form
β« π π‘ = πΎ1 + πΎ2ππ π‘
β’ Solving for πΎ1, π
β« πΎ1 + πΎ2ππ π‘ +πΏ
π πΎ2π ππ π‘ =
ππ
π
πΎ1 =ππ
π
1 +πΏ
π π = 0 β π = β
π
πΏ
β’ Solving for πΎ2
β« π 0+ = 0 = ππ
π + πΎ2πβπ‘π /πΏ
β« 0 =ππ
π + πΎ2π0
β« πΎ2 = βππ
π
β’ Final Solution
β« π π‘ =ππ
π β
ππ
π πβπ‘π /πΏ
β« π π‘ = 2 β 2πβ500π‘
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/8.jpg)
RL Example β’ π π‘ = 2 β 2πβ500π‘
β’ Notice this is in the general form we used for RC circuits
β« π =πΏ
π
β’ Find voltage π£(π‘)
β« π£π = 0, steady-state short
β« π£ 0+ = 100
No current immediately
through π , π£ = πΏππ(π‘)
ππ‘
β’ π£ π‘ = 100πβπ‘/π
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/9.jpg)
Exercise 4.5
β’ Initial conditions
β’ For π‘ < 0 β« All source current goes
through switched wire
β« ππ π‘ = ππΏ π‘ = 0 π΄
β« π£ π‘ = ππ π‘ R = 0 V
β’ For π‘ = 0+ (right after switch)
β« ππΏ π‘ = 0 Current canβt change
immediately through an inductor
β« ππ π‘ = 2 A, by KCL
β« π£ π‘ = ππ π‘ R = 20 V
β’ Steady-state β« Short inductor
β’ π£ π‘ = 0 β« Short circuit across inductor
β’ ππ = 0 β« All current through short
β’ ππΏ = 2 π΄ β« By KCL
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/10.jpg)
Exercise 4.5
β’ Can use network analysis to come up with a differential equation, but you would need to solve it
β’ Instead, use the general 1st-order solution
β’ Time constant π
β« π =πΏ
π =
2
10= 0.2
β’ Voltage π£(π‘) β« π£ π‘ = 0 + 20 β 0 πβπ‘/0.2 = 20πβπ‘/0.2 V
β’ Current ππΏ(π‘), ππ (π‘) β« ππΏ π‘ = 2 + 0 β 2 πβπ‘/0.2 = 2 β 2πβπ‘/0.2 A β« ππ π‘ = 0 + 2 β 0 πβπ‘/0.2 = 2πβπ‘/0.2 A
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/11.jpg)
RC/RL Circuits with General Sources
β’ Previously,
β« π πΆππ£π(π‘)
ππ‘+ π£π(π‘) = ππ
β’ What if ππ is not constant
β« π πΆππ£π(π‘)
ππ‘+ π£π(π‘) = π£π (π‘)
β« Now have a general source that is a function of time
β’ The solution is a differential equation of the form
β« πππ₯(π‘)
ππ‘+ π₯(π‘) = π(π‘)
β« Where π(π‘) is known as the forcing function (the circuit source)
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π£π (π‘)
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/12.jpg)
General Differential Equations
β’ General differential equation
β’ πππ₯(π‘)
ππ‘+ π₯(π‘) = π(π‘)
β’ The solution to the diff equation is
β’ π₯ π‘ = π₯π π‘ + π₯β(π‘)
β’ π₯π π‘ is the particular solution
β’ π₯β(π‘) is the homogeneous solution
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/13.jpg)
Particular Solution
β’ πππ₯π π‘
ππ‘+ π₯π π‘ = π(π‘)
β’ The solution π₯π π‘ is called the forced response
because it is the response of the circuit to a particular forcing input π π‘
β’ The solution π₯π π‘ will be of the same functional
form as the forcing function
β« E.g.
β« π π‘ = ππ π‘ β π₯π π‘ = π΄ππ π‘
β« π π‘ = cos ππ‘ β π₯π π‘ = π΄cos ππ‘ + π΅sin(ππ‘)
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/14.jpg)
Homogeneous Solution
β’ πππ₯β π‘
ππ‘+ π₯β π‘ = 0
β’ π₯β π‘ is the solution to the differential equation when there is no forcing function
β’ Does not depend on the sources
β’ Dependent on initial conditions (capacitor voltage, current through inductor)
β’ π₯β π‘ is also known as the natural response
β’ Solution is of the form
β’ π₯β π‘ = πΎπβπ‘/π
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/15.jpg)
General Differential Solution
β’ Notice the final solution is the sum of the particular and homogeneous solutions
β« π₯ π‘ = π₯π π‘ + π₯β(π‘)
β’ It has an exponential term due to π₯β(π‘) and a term π₯π π‘ that matches the input source
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/16.jpg)
Second-Order Circuits
β’ RLC circuits contain two energy storage elements
β« This results in a differential equation of second order (has a second derivative term)
β’ This is like a mass spring system from physics
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π =ππ](https://reader036.vdocuments.us/reader036/viewer/2022071000/5fbc2a75389a23292e61ab7b/html5/thumbnails/17.jpg)
RLC Series Circuit
β’ KVL around loop
β« π£π π‘ β πΏππ π‘
ππ‘β π π‘ π β π£π π‘ = 0
β’ Solve for π£π π‘
β« π£π π‘ = π£π π‘ β πΏππ π‘
ππ‘β π π‘ π
β’ Take derivative
β«ππ£π π‘
ππ‘=
ππ£π π‘
ππ‘β πΏ
π2π π‘
ππ‘2 β π ππ π‘
ππ‘
β’ Solve for current through capacitor
β’ π π‘ = πΆππ£π π‘
ππ‘
β’ π π‘ = πΆππ£π π‘
ππ‘β πΏ
π2π π‘
ππ‘2 β π ππ π‘
ππ‘
β’π2π π‘
ππ‘2 βπ
πΏ
ππ π‘
ππ‘+
1
πΏπΆπ π‘ =
1
πΏ
ππ£π π‘
ππ‘
β’ The general 2nd-order constant coefficient equation
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