跳到主要内容

电路理论(英文班)知识汇总

提示
  • 带单位,写 t>0t>0 (如有)
  • When solving Thevenin equialent/Norton equialent, determine the positive / negative voltage. If there is a voltage source across the resistor we want to obtain the Thevein equivalent(or there is a current source through the branch), we can only find VthV_{th} or INI_{N} ,while the RthR_{th} or RNR_{N} is 00 or \infty
  • power is different from energy
  • saturate 饱和
  • DC 直流
  • ideal transformer: When the voltage is not given in the problem, it can be set by yourself.
  • fraction 分数 / 比例
  • w=2πfw=2\pi f
  • Note the positive and negative of voltage and current!!!

DC circuits

  • A branch ( Number bb )is a single two-terminal element in an electric circuit. A node( Number nn ) is the point of connection between two or more branches. A loop ( independent loops ll ) is a closed path in a circuit. ( n1n-1 is the number of tree branch.)
b=l+n1b=l+n-1
  • Delta to Wye
R1=RbRcRa+Rb+RcR_{1}=\frac{R_{b}R_{c} }{R_{a}+R_{b}+R_{c} }

(自己在对面,两面夹击,除以所有)

  • Wye to Delta
Ra=R1R2+R2R3+R3R1R1R_{a}=\frac{ {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1} }}{R_{1} }

(所有相乘,除以自己,得到对面)

  • Nodal Analyses
[G11G12G1NG21G22G2NGN1GN2GNN][v1v2vN]=[i1i2iN]\left[ \begin{matrix} {G_{1 1} } & {G_{1 2} } & {\cdots} & {G_{1 N} } \\ {G_{2 1} } & {G_{2 2} } & {\cdots} & {G_{2 N} } \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {G_{N 1} } & {G_{N 2} } & {\cdots} & {G_{N N} } \\ \end{matrix} \right] \left[ \begin{matrix} {v_{1} } \\ {v_{2} } \\ {\vdots} \\ {v_{N} } \\ \end{matrix} \right]=\left[ \begin{matrix} {i_{1} } \\ {i_{2} } \\ {\vdots} \\ {i_{N} } \\ \end{matrix} \right]

GkkG_{kk} Sum of the conductances connected to node k Gkj=GjkG_{kj}=G_{jk} Negative of the sum of the conductances directly connectiong nodes k and j vkv_{k} Voltage of node k iki_{k} Sum of all independent current sources directly connected to node k, current entering the node treated as positive

  • Mesh Analyses
[R11R12R1NR21R22R2NRN1RN2RNN][i1i2iN]=[v1v2vN]\left[ \begin{matrix} {R_{1 1} } & {R_{1 2} } & {\cdots} & {R_{1 N} } \\ {R_{2 1} } & {R_{2 2} } & {\cdots} & {R_{2 N} } \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ {R_{N 1} } & {R_{N 2} } & {\cdots} & {R_{N N} } \\ \end{matrix} \right] \left[ \begin{matrix} {i_{1} } \\ {i_{2} } \\ {\vdots} \\ {i_{N} } \\ \end{matrix} \right]=\left[ \begin{matrix} {v_{1} } \\ {v_{2} } \\ {\vdots} \\ {v_{N} } \\ \end{matrix} \right]

RkkR_{kk} Sum of the resistances in mesh k Rkj=RjkR_{kj}=R_{jk} Negative of the sum of the resistances in common with mesh kk and jj iki_{k} Unknown mesh current for mesh kk in the colockwise direction vkv_{k} Sum taken clockwise of all independent voltage sources in mesh kk , voltage rise as positive(Current flows from the positive terminal)

  • Super Node & Super Mesh
  • Network theorems
  • Superposition principle
  • Source transformation
  • Thevenins(open-circuit voltage)/Norton(short-circuit current) theorems
RN=RThR_{N}=R_{Th} IN=VThRThI_{N}=\frac{V_{Th} }{R_{Th} }
  • maximum power transfer:
RL=RThR_{L}=R_{Th}
  • Wheatstone bridge
  • op amp equivalent circuit
  • basic op amp circuits
  • With an ideal op-amp, the output voltage changes immediately when the voltage at the input changes, and for this reason the current can potentially reach infinite values.
  • current through a capacitor---like conductors---voltage can't change instantly
i=Cdvdti=C \frac{dv}{dt}
  • voltage across an inductor---like resistors---current can't change instantly
v=Ldidtv=L \frac{di}{dt}
  • energy stored in a capacitor 12Cv2\frac{1}{2}Cv^2; in an inductor 12Li2\frac{1}{2}Li^2 ,energy stored is 12L(i(t)2i(0)2)\frac{1}{2}L(i(t)^2-i(0)^2)
  • Please note that Thevenin equivalent circuits do not generally exist for circuits involving capacitors and resistors.
  • D-W/W-D of capacitor & inductor: find the inverse of capacitor
  • Thevenin equivalent of circuits involving capacitors and resistors, we can get CThC_{Th}
  • For capacitors: v(0)+=v(0)v(0)^+=v(0)^-
  • For inductors: i(0)+=i(0)i(0)^+=i(0)^-
  • Complete response:
Complete response=natural reponse+forced response=transient response+steadystate responseComplete\ response=natural\ reponse+forced\ response=transient\ response+steady-state \ response
  • τ\tau : for RC, τ=RC\tau=RC , for RLRL , τ=LR\tau=\frac{L}{R}
  • Singularity functions:
    • unit step function u(t)u(t) : u(t)={0,t<01,t>0u ( t )=\left\{\begin{matrix} {0,} & {\qquad t < 0} \\ {1,} & {\qquad t > 0} \\ \end{matrix} \right.
    • unit impulse function: δ(t)={0,t<0Undefined,t=00,t>0\delta( t )=\begin{cases} {0,} & {\qquad t < 0} \\ {\mathrm{U n d e f i n e d},} & {\qquad t=0} \\ {0,} & {\qquad t > 0} \\ \end{cases}Meaningful only when integrating
    • unit ramp function: r(t)={0,t0t,t0r ( t )=\left\{\begin{matrix} {0,} & {\qquad t \leq0} \\ {t,} & {\qquad t \geq0} \\ \end{matrix} \right.
  • wave representation: not to have negative coefficients in singularity functions and Note the coefficients on the axes
  • τ=RC/LR\tau=RC / \frac{L}{R}
  • step response(short-cut approach)
x(t)=x()+[x(t0+)x()]e(tt0)/τx ( t )=x ( \infty)+[ x ( t_{0}^{+} )-x ( \infty) ] e^{-( t-t_{0} ) / \tau}
  • Short-cut approach can only be applied to voltage across capacitor.
  • Sometimes can't use this method(in op amp), so we need to use straightforward approach.
  • finding initial values: looking at voltage across the capacitors and current through the inductors, and get other values
  • source free RLC series:
    • quadratic equation: s2+RLs+1LC=0s^2+\frac{R}{L}s+\frac{1}{LC}=0
    • natural frequencies:
      • s1=α+α2w02s_{1}=-\alpha+\sqrt{ \alpha^2-w_{0}^2 }
      • s2=αα2w02s_{2}=-\alpha-\sqrt{ \alpha^2-w_{0}^2 }
    • neper frequency: α=R2L\alpha=\frac{R}{2L} .
    • resonant frequency / undamped natural frequency: w0=1LCw_{0}=\frac{1}{\sqrt{ LC } }
    • s2+2as+w02=0s^2+2as+w_{0}^2=0
    • i(t)=A1es1t+A2es2ti(t)=A_{1}e^{ s_{1}t }+A_{2}e^{ s_{2}t } , and A1A_{1}, A2A_{2} determined from the initial values( i(0)i(0) and di(0)dt\frac{di(0)}{dt})
  • three types of solutions
    • overdamped case α>w0\alpha>w_{0} ,two roots are negative and real
      • decays and approaches zero as it increases
      • i(t)=A1es1t+A2es2ti(t)=A_{1}e^{ s_{1}t }+A_{2}e^{ s_{2}t }
    • critically damped case α=w0\alpha=w_{0} , s1=s2=αs_{1}=s_{2}=-\alpha
      • i(t)=(A2+A1t)eαti(t)=(A_{2}+A_{1}t)e^{ -\alpha t }
    • underdamped case α<w0\alpha<w_{0}
      • s1=α+jwds_{1}=-\alpha+jw_{d} , s2=αjwds_{2}=-\alpha-jw_{d}
      • wd=w02α2w_{d}=\sqrt{ w_{0}^2-\alpha^2 }
      • i(t)=eαt(B1coswdt+B2sinwdt)i(t)=e^{ -\alpha t }(B_{1}\cos w_{d}t+B_{2}\sin w_{d}t)
  • source free RLC parallel
    • α=12RC\alpha=\frac{1}{2RC}
    • w0=1LCw_{0}=\frac{1}{\sqrt{ LC } }
    • solving the constant by getting v(0)v(0) and dv(0)dt\frac{dv(0)}{dt}
  • step response: vss/iss(t)=v/i()=Vs/Isv_{ss} / i_{ss}(t)=v / i(\infty)=V_{s} / I_{s}
    • x(t)=xt(t)+xss(t)x(t)=x_{t}(t)+x_{ss}(t)
  • Simple method: Simplify as much as possible to RLC series or RLC parallel or directly use Thevenins theroem.

AC circuits

Fundamentals

  • VmV_{m} amplitude
  • cos(wt+ϕ)ϕ\cos(wt+\phi)\to \angle \phi
  • sin(wt+ϕ)ϕ90\sin(wt+\phi)\to \angle{\phi-90^\circ}
  • dvdtjwV\frac{dv}{dt} \Leftrightarrow jwV
  • vdtVjw\int v \, dt \Leftrightarrow \frac{V}{jw}
  • inductor: V=jwLIV=jwLI , and II lags VV
  • capacitor: V=IjwCV=\frac{I}{jwC} , and VV lags II
  • Y=G+jB=1R+jXY=G+jB=\frac{1}{R+jX}
  • other: YΔY-\Delta , Kirchhoff's law is same as DC circuits
  • superposition theorem: different frequencies

Power analysis

Be careful to distinguish between amplitude and RMS values!

  • instantaneous power: p(t)=v(t)i(t)p(t)=v(t)i(t)
  • average power: P=12VmImcos(θvθi)P=\frac{1}{2}V_{m}I_{m}\cos(\theta_{v}-\theta_{i})
    • θv=θi\theta_{v}=\theta_{i}
    • P=12VmIm=12Im2R=12I2RP=\frac{1}{2}V_{m}I_{m}=\frac{1}{2}I_{m}^2R=\frac{1}{2}\mid I\mid^2R
  • maximum average power transfer (so a factor of 12\frac{1}{2} precedes the equation)
    • ZL=ZThZ_{L}=Z^*_{Th}
  • when XL=0X_{L}=0, RL=ZThR_{L}=\mid Z_{Th}\mid
  • rms: Xrms=1T0Tx2dtX_{rms}=\sqrt{ \frac{1}{T}\int _{0}^T \,x^2 dt }
    • Irms=Im2I_{rms}=\frac{I_{m} }{\sqrt{ 2 } }
    • Vrms=Vm2V_{rms}=\frac{V_{m} }{\sqrt{ 2 } }
    • if V/IV / I has constant like c+dsinc+d\sin \dots , the Vrms/Irms=c2+d22V_{rms} / I_{r ms}=\sqrt{ c^2+\frac{d^2}{2} }
  • apparent power(VA):
    • S=12VIS=\frac{1}{2}VI^*
    • S=VrmsIrmsS=V_{rms}I^*_{rms}
    • S=Irms2Z=Vrms2ZS=I^2_{rms}Z=\frac{V^2_{rms} }{Z^*} (注意什么时候带共轭什么时候不带共轭)
    • When answering the question, S should be written in the form P+jQP+jQ .
    • QQ is the reactive power
  • power factor pf=PS=cos(θvθi)pf=\frac{P}{S}=\cos(\theta_{v}-\theta_{i})
    • leading: current leads voltage
    • lagging: voltage leads current
    • power factor correction
      • phasor diagram:
      • power factor correction
      • C=Qcw2Vrms=P(tanθ1tanθ2)wVrms2C=\frac{Q_{c} }{w^2V_{r ms} }=\frac{P(\tan\theta_{1}-\tan\theta_{2})}{wV^2_{rms} }
      • QL=Vrms2XLQ_{L}=\frac{V^2_{r ms} }{X_{L} }
  • reactive power: Q=Im(S)=Irms2X=VrmsIrmssin(θvθi)Q=Im(S)=I^2_{r ms} X=V_{r ms}I_{r ms}\sin(\theta_{v}-\theta_{i})
    • Q=0Q=0 resistive loads
    • Q<0Q<0 capacitive loads(leading pf)
    • Q>0Q>0 inductive loads(lagging pf)
    • power triangle: using QQ , not pf, that means, leading/lagging depends on QQ in the figure
  • Z=R2+X2|Z|=\sqrt{ R^2+X^2 }

Three-Phase Circuits

  • in three-phase circuits, VpV_{p} is rms value of the phase voltages
  • phase sequence: time order in which voltage pass through their respective maximum values
    • positive: Van,Vbn,VcnV_{an},V_{bn},V_{cn} counterclockwise, a leads b leads c
      • abc sequence
    • negative: Van,Vbn,VcnV_{an},V_{bn},V_{cn} clockwise
      • acb sequence
  • phase xx: line-to-neutual xx
  • line xx: line-to-line xx
  • YY connection of the source:
    • VL=3VPV_{L}=\sqrt{ 3 }V_{P}
    • Vab=3Van30V_{ab}=\sqrt{ 3 }V_{an}\angle 30^\circ
  • Δ\Delta connection of the load:
    • IL=3IPI_{L}=\sqrt{ 3 }I_{P}
    • Ia=3IAB30I_{a}=\sqrt{ 3 }I_{AB}\angle -30^\circ
  • power (remind to ×3\times3 )
    • S=3Sp=3VpIp=3Ip2Zp=3Vp2ZpS=3S_{p}=3V_{p}I_{p}^*=3I_{p}^2Z_{p}=\frac{3V_{p}^2}{Z_{p}^*}
    • Zp=Zpθ\mathbf{Z_{p} }=Z_{p}\angle\theta
    • S=3VpIpθ=3VLILθ\mathbf{S}=3V_{p}I_{p}\angle\theta=\sqrt{ 3 }V_{L}I_{L}\angle \theta
  • power loss of transmission lines in three-phase system and single-phase system
    • single-phase: IL=PLVLI_{L}=\frac{P_{L} }{V_{L} }
      • Ploss=2IL2R=2RPL2VLP_{loss}=2I_{L}^2R=2R \frac{P_{L}^2}{V_{L} } (2 wires)
    • three-phase: IL=PL3VLI_{L}=\frac{P_{L} }{\sqrt{ 3 }V_{L} }
      • Ploss=3IL2R=RPL2VL2P_{loss}=3I_{L}^2R=R \frac{P_{L}^2}{V_L^2}
  • no need to get conjuction when doing ΔY\Delta-Y .

Magnetically Coupled Circuits

Be careful with the dot convention!!!

  • voltage ratings of transformers are usually specified in rms values
  • mutual voltage(eg)
    • v1=M12di2dtv_{1}=M_{12} \frac{di_{2} }{dt}
  • M=kL1L2M=k\sqrt{ L_{1}L_{2} }
  • dot convention
    • enter the dotted terminal \to positive at the dotted terminal
    • leave the dotted terminal \to negatifve at the dotted terminal
  • induction in series
    • aiding connection: L=L1+L2+2ML=L_{1}+L_{2}+2M
    • opposing connection: L=L1+L22ML=L_{1}+L_{2}-2M
  • Transformer
    • From two KVL: Zin=R1+jwL1+w2M2R2+jwL2+ZLZ_{in}=R_{1}+jwL_{1}+ \frac{ {w^2M^2} }{R_{2}+jwL_{2}+Z_{L} }
    • ZR=w2M2R2+jwL2+ZLZ_{R}=\frac{w^2M^2}{R_{2}+jwL_{2}+Z_{L} }
  • I-V relationships
[V1V2]=[jωL1jωMjωMjωL2][I1I2]\left[ \begin{matrix} { {\bf V}_{1} } \\ { {\bf V}_{2} } \\ \end{matrix} \right]=\left[ \begin{matrix} {j \omega L_{1} } & {j \omega M} \\ {j \omega M} & {j \omega L_{2} } \\ \end{matrix} \right] \left[ \begin{matrix} { {\bf I}_{1} } \\ { {\bf I}_{2} } \\ \end{matrix} \right]
  • change to T circuit: (using mesh method)
[V1V2]=[jω(La+Lc)jωLcjωLcjω(Lb+Lc)][I1I2]\left[ \begin{array} {c} { {\mathrm{V}_{1} }} \\ { {\mathrm{V}_{2} }} \\ \end{array} \right]=\left[ \begin{array} {c c} { {j \omega( L_{a}+L_{c} )} } & { {j \omega L_{c} }} \\ { {j \omega L_{c} }} & { {j \omega( L_{b}+L_{c} )} } \\ \end{array}\right] \left[ \begin{array} {c} { {\mathrm{I}_{1} }} \\ { {\mathrm{I}_{2} }} \\ \end{array} \right] La=L1ML_{a}=L_{1}-M Lb=L2ML_{b}=L_{2}-M Lc=ML_{c}=M
  • change to Π\Pi circuit (using nodal method)
[I1I2]=[1jωLA+1jωLC1jωLC1jωLC1jωLB+1jωLC][V1V2]\left[ \begin{array} {c} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{array} \right]=\left[ \begin{array} {c c} { {\frac{1} {j \omega L_{A} }+\frac{1} {j \omega L_{C} }} } & { {-\frac{1} {j \omega L_{C} }} } \\ { {-\frac{1} {j \omega L_{C} }} } & { {\frac{1} {j \omega L_{B} }+\frac{1} {j \omega L_{C} }} } \\ \end{array} \right] \left[ \begin{array} {c} { {\mathbf{V}_{1} }} \\ { {\mathbf{V}_{2} }} \\ \end{array} \right] LA=L1L2M2L2MLB=L1L2M2L1ML_{A}={\frac{L_{1} L_{2}-M^{2} } {L_{2}-M} } \, \qquad L_{B}={\frac{L_{1} L_{2}-M^{2} } {L_{1}-M} } LC=L1L2M2ML_{C}=\frac{L_{1} L_{2}-M^{2} } {M}
  • ideal transformers
    • V2V1=N2M1=n\frac{V_{2} }{V_{1} }=\frac{N_{2} }{M_{1} }=n
    • v1v_{1} , v2v_{2} both positive at dotted terminal-- +n+n otherwise-- n-n
    • I1I_{1} , I2I_{2} both enter or leave-- n-n otherwise-- +n+n
    • S1=S2S_{1}=S_{2}
    • reflected impedance Zin=ZLn2Z_{in}=\frac{Z_{L} }{n^2}

Two-Port Networks

  • current: always in the two-port, transmission: I-I
  • impedance [z][z] , admittance [y][y] ,hybrid [h][h] , inverse hybrid [g][g] , transmission [T][T] , inverse transmission [t][t]
[V1V2]=[z][I1I2],[I1I2]=[y][V1V2],[V1I2]=[h][I1V2]\left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ { {\mathbf{V}_{2} }} \end{matrix} \right]=[ \mathbf{z} ] \left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{matrix} \right], \qquad\left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ { {\mathbf{I}_{2} }} \end{matrix} \right]=[ \mathbf{y} ] \left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ {\mathbf{V}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} { {\mathbf{V}_{1} }} \\ {\mathbf{I}_{2} } \\ \end{matrix} \right]=[ \mathbf{h} ] \left[ \begin{matrix} { {\mathbf{I}_{1} }} \\ {\mathbf{V}_{2} } \\ \end{matrix} \right] [I1V2]=[g][V1I2],[V1I1]=[T][V2I2],[V2I2]=[t][V1I1]\left[ \begin{matrix} {\mathbf{I}_{1} } \\ {\mathbf{V}_{2} } \\ \end{matrix} \right]=\left[ \mathbf{g} \right] \left[ \begin{matrix} {\mathbf{V}_{1} } \\ {\mathbf{I}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} {\mathbf{V}_{1} } \\ {\mathbf{I}_{1} } \\ \end{matrix} \right]=\left[ \mathbf{T} \right] \left[ \begin{matrix} {\mathbf{V}_{2} } \\ {-\mathbf{I}_{2} } \\ \end{matrix} \right], \qquad\left[ \begin{matrix} {\mathbf{V}_{2} } \\ {\mathbf{I}_{2} } \\ \end{matrix} \right]=\left[ \mathbf{t} \right] \left[ \begin{matrix} {\mathbf{V}_{1} } \\ {-\mathbf{I}_{1} } \\ \end{matrix} \right]
  • reciprocal
    • z12=z21z_{12}=z_{21}
    • y12=y21y_{12}=y_{21}
    • h12=h21h_{12}=-h_{21}
    • g12=g21g_{12}=-g_{21}
    • ΔT=1\Delta_{T}=1
    • Δt=1\Delta_{t}=1
    • has dependent sources are not reciprocal
  • relationship
[y]=[z]1,[g]=[h]1,[t][T]1[ {\bf y} ]=[ {\bf z} ]^{-1}, \qquad[ {\bf g} ]=[ {\bf h} ]^{-1}, \qquad[ {\bf t} ] \neq[ {\bf T} ]^{-1}
  • problems
    • determine parameters
    • Thevenin equivalent-solve quadratic equation

Interesting problems

  • 7.50 t=0t=0R3R_{3} is shorted
  • when i0=0i_{0}=0, the source can be look like a wire.
  • 13.47 using mesh method.