Intermediate electrical engineering: 50 circuits and electronics concepts

All 50 cards

1. 1
   Thevenin’s theorem
   Any linear two-terminal network can be replaced by a voltage source $V_{\text{Th}}$ in series with a resistance $R_{\text{Th}}$

   $V_{\text{Th}}$ is the open-circuit voltage across the terminals.
2. 2
   Norton’s theorem
   Any linear two-terminal network can be replaced by a current source $I_N$ in parallel with a resistance $R_N$

   $I_N$ is the short-circuit current. $R_N = R_{\text{Th}}$ and $I_N = V_{\text{Th}}/R_{\text{Th}}$.
3. 3
   Finding the Thevenin resistance
   Turn off every independent source (short voltage sources, open current sources) and find the resistance seen from the terminals

   With dependent sources, use $R_{\text{Th}} = V_{\text{oc}}/I_{\text{sc}}$ or apply a test source.
4. 4
   Maximum power transfer theorem
   A load receives the most power when $R_L = R_{\text{Th}}$, and then $P_{\max} = \dfrac{V_{\text{Th}}^2}{4R_{\text{Th}}}$

   Efficiency at that point is only 50%: the source resistance dissipates as much power as the load.
5. 5
   Superposition principle
   In a linear circuit, any voltage or current is the sum of the contributions from each independent source acting alone, with the others turned off

   It doesn’t apply to power, which depends on the square of current or voltage.
6. 6
   Nodal analysis
   Pick a reference node, write KCL at every other node with the node voltages as unknowns, and solve the equations

   A voltage source between two non-reference nodes makes them a supernode.
7. 7
   Mesh analysis
   Assign a current to each mesh (a loop with no other loop inside it), write KVL around each one, and solve for the mesh currents

   It works for planar circuits. A current source shared by two meshes makes them a supermesh.
8. 8
   Time constant of an RC circuit
   $\tau = RC$

   With $R$ in ohms and $C$ in farads, $\tau$ is in seconds: $10\ \mathrm{k\Omega}$ and $100\ \mathrm{\mu F}$ give $1\ \mathrm{s}$.
9. 9
   Time constant of an RL circuit
   $\tau = \dfrac{L}{R}$
10. 10
    Step response of a first-order circuit
    $x(t) = x(\infty) + \big[x(0^+) - x(\infty)\big]\,e^{-t/\tau}$

    $\tau$ is $RC$ or $L/R$, using the Thevenin resistance seen by the capacitor or inductor.
11. 11
    How far a first-order response gets after one and five time constants
    About 63% of the way to its final value after $\tau$, and over 99% after $5\tau$

    $1 - e^{-1} \approx 0.632$ and $1 - e^{-5} \approx 0.993$
12. 12
    Quantities that can’t change instantly
    The voltage across a capacitor and the current through an inductor

    A jump would need infinite current ($i = C\,dv/dt$) or infinite voltage ($v = L\,di/dt$).
13. 13
    Damping cases of a series RLC circuit
    Overdamped if $\alpha > \omega_0$, critically damped if $\alpha = \omega_0$, underdamped (oscillating) if $\alpha < \omega_0$

    For a series RLC, $\alpha = \dfrac{R}{2L}$ and $\omega_0 = \dfrac{1}{\sqrt{LC}}$.
14. 14
    Phasor
    A complex number giving the amplitude and phase of a sinusoid at a known frequency: $V_m\cos(\omega t + \phi) \to V_m\angle\phi$

    Phasors turn the differential equations of AC circuits into algebra with complex numbers.
15. 15
    Impedance of a resistor, an inductor and a capacitor
    $Z_R = R$, $Z_L = j\omega L$, $Z_C = \dfrac{1}{j\omega C}$

    $\omega = 2\pi f$ is the angular frequency and $j = \sqrt{-1}$. The magnitudes $\omega L$ and $1/(\omega C)$ are the reactances.
16. 16
    Phase of the current in an inductor and a capacitor
    In an inductor the current lags the voltage by 90°; in a capacitor it leads by 90°

    Mnemonic: ELI the ICE man (voltage E before current I in an inductor L; I before E in a capacitor C).
17. 17
    Resonant frequency of an LC or RLC circuit
    $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$

    At $f_0$ the inductive and capacitive reactances are equal and cancel.
18. 18
    Series vs parallel RLC at resonance
    Series: the impedance is at its minimum (just R) and the current peaks. Parallel: the impedance is at its maximum and the current drawn is smallest
19. 19
    Quality factor of a series RLC circuit
    $Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

    The bandwidth is $f_0/Q$, so a higher $Q$ means a sharper, narrower peak.
20. 20
    Real (average) power in an AC circuit
    $P = V_{\text{rms}} I_{\text{rms}} \cos\theta$, in watts

    $\theta$ is the phase angle between voltage and current. Only resistance consumes real power.
21. 21
    Reactive power
    $Q = V_{\text{rms}} I_{\text{rms}} \sin\theta$, in volt-amperes reactive (var)

    It flows back and forth between the source and the inductors and capacitors, doing no net work.
22. 22
    Apparent power and the power triangle
    $S = V_{\text{rms}} I_{\text{rms}}$, in volt-amperes (VA), with $S^2 = P^2 + Q^2$

    As complex power, $\mathbf{S} = \mathbf{V}\mathbf{I}^* = P + jQ$ using RMS phasors.
23. 23
    Power factor
    $\mathrm{PF} = \dfrac{P}{S} = \cos\theta$

    It is lagging for inductive loads such as motors (current lags voltage) and leading for capacitive loads.
24. 24
    Power factor correction
    Adding capacitors in parallel with an inductive load so they supply its reactive power locally

    The real power is unchanged, but the line current and its losses fall.
25. 25
    Line and phase voltage in a balanced wye (star) connection
    $V_L = \sqrt{3}\,V_{\phi}$, and the line current equals the phase current

    The three phases are $120^\circ$ apart; a $230\ \mathrm{V}$ phase voltage gives about $400\ \mathrm{V}$ between lines.
26. 26
    Line and phase current in a balanced delta connection
    $I_L = \sqrt{3}\,I_{\phi}$, and the line voltage equals the phase voltage
27. 27
    Total power in a balanced three-phase load
    $P = \sqrt{3}\,V_L I_L \cos\theta$

    $V_L$ and $I_L$ are line values and $\theta$ is the phase angle of each phase’s load. It equals $3V_{\phi}I_{\phi}\cos\theta$.
28. 28
    Ideal transformer: voltage and current ratios
    $\dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}$ and $\dfrac{I_1}{I_2} = \dfrac{N_2}{N_1}$

    Power in equals power out, so stepping the voltage up steps the current down.
29. 29
    Impedance seen through an ideal transformer
    $Z_{\text{in}} = \left(\dfrac{N_1}{N_2}\right)^2 Z_L$

    This is how a transformer matches a load to a source, as an audio output transformer does for a loudspeaker.
30. 30
    Why power is transmitted at high voltage
    For the same power, a higher voltage needs less current, and line losses scale with $I^2R$

    Transformers make stepping the voltage up and down efficient, one reason the grid uses AC.
31. 31
    Ideal op-amp rules with negative feedback
    No current flows into either input, and the output adjusts so the two input voltages are equal (a virtual short)
32. 32
    Gain of an inverting amplifier
    $\dfrac{V_{\text{out}}}{V_{\text{in}}} = -\dfrac{R_f}{R_{\text{in}}}$

    The inverting input sits at virtual ground, so the input impedance is $R_{\text{in}}$.
33. 33
    Gain of a non-inverting amplifier
    $\dfrac{V_{\text{out}}}{V_{\text{in}}} = 1 + \dfrac{R_f}{R_g}$

    $R_f$ runs from the output to the inverting input and $R_g$ from the inverting input to ground.
34. 34
    Voltage follower (unity-gain buffer)
    An op-amp with its output tied to the inverting input and the signal on the non-inverting input; its gain is 1

    Its very high input impedance and low output impedance let a weak source drive a heavy load.
35. 35
    BJT currents in the active region
    $I_C = \beta I_B$ and $I_E = I_B + I_C$

    $\beta$ (also written $h_{FE}$) is the current gain, often around 100.
36. 36
    BJT regions of operation
    Cut-off: both junctions reverse-biased. Active: base–emitter forward, base–collector reverse. Saturation: both forward-biased

    Amplifiers use the active region; switches toggle between cut-off and saturation.
37. 37
    n-channel MOSFET regions of operation
    Cut-off: $V_{GS} < V_{th}$. Triode: $V_{GS} > V_{th}$ and $V_{DS} < V_{GS} - V_{th}$. Saturation: $V_{GS} > V_{th}$ and $V_{DS} \ge V_{GS} - V_{th}$

    Beware the name clash: MOSFET amplifiers work in saturation, unlike BJT amplifiers.
38. 38
    MOSFET drain current in saturation
    $I_D = \tfrac{1}{2}\mu_n C_{ox}\dfrac{W}{L}(V_{GS} - V_{th})^2$

    The square law, ignoring channel-length modulation. $W/L$ is the channel’s width-to-length ratio.
39. 39
    Universal logic gates
    NAND and NOR: either one alone can build any Boolean function
40. 40
    De Morgan’s laws
    $\overline{A \cdot B} = \overline{A} + \overline{B}$ and $\overline{A + B} = \overline{A} \cdot \overline{B}$
41. 41
    XOR gate
    Outputs 1 when its inputs differ: $A \oplus B = A\overline{B} + \overline{A}B$

    It adds two bits without the carry, so it gives the sum output of a half adder.
42. 42
    Two’s complement
    To negate a binary number, invert every bit and add 1

    An $n$-bit two’s complement number ranges from $-2^{n-1}$ to $2^{n-1} - 1$, so 8 bits cover −128 to 127.
43. 43
    D flip-flop
    On each active clock edge the output Q takes the value of the input D, and holds it until the next edge
44. 44
    Combinational vs sequential logic
    Combinational outputs depend only on the present inputs; sequential outputs also depend on stored state

    Adders and multiplexers are combinational; registers and counters are sequential.
45. 45
    Cutoff frequency of a first-order RC filter
    $f_c = \dfrac{1}{2\pi RC}$

    At $f_c$ the output is $1/\sqrt{2}$ of the passband level, which is $-3\ \mathrm{dB}$.
46. 46
    Decibels for voltage and for power
    $20\log_{10}\dfrac{V_2}{V_1}$ for voltage, $10\log_{10}\dfrac{P_2}{P_1}$ for power

    Doubling the voltage is about $+6\ \mathrm{dB}$; doubling the power is about $+3\ \mathrm{dB}$.
47. 47
    Roll-off of a first-order filter
    $20\ \mathrm{dB}$ per decade (about $6\ \mathrm{dB}$ per octave)

    Each extra order adds another $20\ \mathrm{dB}$ per decade.
48. 48
    Fourier series
    A periodic signal written as the sum of a DC term and sinusoids at the fundamental frequency and its integer multiples (harmonics)
49. 49
    Fourier series of a square wave
    Only odd harmonics, with amplitudes falling as $1/n$: $x(t) = \dfrac{4A}{\pi}\displaystyle\sum_{n\ \text{odd}} \dfrac{\sin(n\omega_0 t)}{n}$

    For a square wave switching between $+A$ and $-A$, rising at $t = 0$.
50. 50
    Fourier series of even and odd signals
    An even signal has only cosine terms (plus DC); an odd signal has only sine terms