Introduction to Power Electronics

A Silicon Controlled Rectifier (SCR) is a four-layer PNPN, three-terminal (anode A, cathode K, gate G) unidirectional switch. It stays OFF in the forward direction until a gate pulse (or breakover voltage) turns it ON, and then it stays ON (latches) until the anode current falls below the holding current.

Structure and two-transistor model

The PNPN structure is split into a PNP transistor Q1 (P1-N1-P2) and an NPN transistor Q2 (N1-P2-N2) sharing the middle layers.

   A                       A
   |                       |
  [P1]                    Q1 (PNP)  emitter at A
  [N1] ---+               |  \
  [P2]    | split  ==>    |   +--- B1 = C2
  [N2]    |               |  /
   |      |          C1 = B2 --+--- G
   K                          Q2 (NPN)
                              |
                              K   (Q2 emitter)

Collector of Q1 drives the base of Q2, and collector of Q2 drives the base of Q1, forming a positive feedback (regenerative) loop.

Mathematical expression

For each transistor, IC = α·IE + ICBO.

IC1 = α1·IA + ICBO1
IC2 = α2·IK + ICBO2
IK  = IA + IG
IA  = IC1 + IC2   (KCL at the N1-P2 region)

IA = α1·IA + ICBO1 + α2(IA + IG) + ICBO2

IA = (α2·IG + ICBO1 + ICBO2) / (1 − (α1 + α2))

Working

• Forward blocking: with small IG, α1 and α2 are small, α1 + α2 << 1, and IA is a tiny leakage current. J2 is reverse biased and blocks the voltage.
• Turn-on: a gate current increases IK and so α2. As current grows, α1 and α2 rise. When α1 + α2 → 1, the denominator approaches zero and IA rises sharply, limited only by the external load. Both transistors saturate and the device latches ON.
• After turn-on, the gate loses control; the SCR stays ON as long as IA > holding current IH.
• Reverse blocking: with cathode positive, J1 and J3 are reverse biased; only a small leakage flows until reverse breakdown.

V-I characteristic

          IA
           ^       IG2>IG1>IG0=0
           |      |  on-state
           |      |
       IL -|..../ |
       IH -|.../  |
           |   \  \    \
           |    \  \    \  forward
-VBR       |     IG2 IG1 IG0  blocking
 ----+-----+--------------------> VAK
     |     |   VBO2  VBO1  VBO0
     |reverse
     |blocking

• Reverse blocking region: small leakage until reverse breakdown VBR.
• Forward blocking region: small leakage until forward breakover voltage VBO.
• Forward conduction region: after breakover, voltage drops to about 1–2 V and current is set by the load.
• Effect of gate current: larger IG lowers the breakover voltage (IG2 > IG1 > IG0 gives VBO2 < VBO1 < VBO0). With large enough IG, the SCR behaves almost like a diode.
• Latching current IL: minimum anode current needed just after triggering to keep it ON. Holding current IH (< IL): minimum current to stay ON.

A chopper is a DC-to-DC converter that switches a fixed DC supply ON and OFF to give a variable average DC output.

Given: Vs = 220 V, R = 10 Ω, Vch = 2 V, f = 1 kHz, duty cycle k = 0.5.

During ON time the load voltage is Vs − Vch = 218 V; during OFF time it is 0.

vo
218 |----+    +----+    +---
    |    |    |    |    |
  0 +----+----+----+----+----> t
    |<kT>|
    |<--T=1ms->|

a) Average output voltage

Va = (1/T)∫₀^(kT) (Vs − Vch) dt = k(Vs − Vch)
   = 0.5 × (220 − 2) = 0.5 × 218
Va = 109 V

b) RMS output voltage

Vo,rms = [ (1/T)∫₀^(kT) (Vs − Vch)² dt ]^(1/2)
       = √k · (Vs − Vch)
       = √0.5 × 218
Vo,rms = 154.15 V

c) Chopper efficiency

Output power:

Po = (1/T)∫₀^(kT) vo²/R dt = k(Vs − Vch)²/R
   = 0.5 × 218² / 10 = 2376.2 W

Input power (source current = (Vs − Vch)/R during ON):

Pi = (1/T)∫₀^(kT) Vs·(Vs − Vch)/R dt
   = k·Vs(Vs − Vch)/R
   = 0.5 × 220 × 218 / 10 = 2398 W

Efficiency:

η = Po/Pi = (Vs − Vch)/Vs = 218/220 = 0.9909

Answer: Va = 109 V, Vo,rms = 154.15 V, η = 99.09 %

Inverters

An inverter is a power electronic circuit that converts DC into AC of desired voltage and frequency. Thyristors, MOSFETs or IGBTs are switched in sequence so that the load sees an alternating voltage. Examples: single-phase half-bridge and full-bridge inverters, used in UPS, solar systems and AC motor drives.

Chopper calculation

Given: Vs = 220 V, R = 10 Ω, Vch = 2 V, f = 1 kHz (T = 1 ms), k = 0.5, so kT = 0.5 ms.

Output voltage is (Vs − Vch) during 0 < t < kT and 0 during kT < t < T.

a) Average output voltage

Va = (1/T)∫₀^(kT) (Vs − Vch) dt
   = (kT/T)(Vs − Vch) = k(Vs − Vch)
   = 0.5 × 218
Va = 109 V

b) RMS output voltage

Vo,rms = √[ (1/T)∫₀^(kT) (Vs − Vch)² dt ]
       = √k · (Vs − Vch)
       = 0.7071 × 218
Vo,rms = 154.15 V

c) Chopper efficiency

Po = (1/T)∫₀^(kT) (Vs − Vch)²/R dt = k(Vs − Vch)²/R
   = 0.5 × 218²/10 = 2376.2 W

Pi = (1/T)∫₀^(kT) Vs·i dt,  i = (Vs − Vch)/R
   = k·Vs(Vs − Vch)/R
   = 0.5 × 220 × 218/10 = 2398 W

η = Po/Pi = (Vs − Vch)/Vs = 218/220

Answer: Va = 109 V, Vo,rms = 154.15 V, η = 99.09 %

The only loss is the 2 V switch drop, so efficiency is very high, which is the main advantage of switching (chopper) control over resistive control.

Choppers

A chopper is a static DC-to-DC converter that converts a fixed DC voltage into a variable DC voltage by switching a semiconductor device ON and OFF at high frequency. The average output is Vo = k·Vs (step-down), where k = Ton/T is the duty cycle. Choppers are used in DC motor drives, battery vehicles and SMPS.

Single-phase full-bridge inverter

A full-bridge inverter uses four switches (S1–S4, SCRs/IGBTs with antiparallel diodes D1–D4) and one DC source Vs to produce an AC square wave of ±Vs across the load.

     +Vs
  +---+-----------+
  |   |           |
  |  S1 D1       S3 D3
  |   |    LOAD   |
  |   A---[ R ]---B
  |   |           |
  |  S4 D4       S2 D2
  |   |           |
  +---+-----------+
     0 (−)

Working:

1. 0 < t < T/2: S1 and S2 are ON. Current flows +Vs → S1 → A → load → B → S2 → negative. vAB = +Vs.
2. T/2 < t < T: S3 and S4 are ON. Current flows +Vs → S3 → B → load → A → S4 → negative. vAB = −Vs.
3. Switches in the same leg (S1–S4 or S3–S2) are never ON together, or the supply would be short-circuited.
4. With an RL load, current lags; when switches change over, the stored energy returns to the source through the diodes (D3, D4 or D1, D2) before the current reverses. These are feedback diodes.

Waveforms (resistive load):

gate S1,S2 |‾‾‾‾‾|_____|‾‾‾‾‾|_____
gate S3,S4 |_____|‾‾‾‾‾|_____|‾‾‾‾‾
           0   T/2    T
 vo   +Vs  +-----+     +-----+
           |     |     |     |
      0 ---+-----+-----+-----+---> t
           |     |     |     |
     −Vs         +-----+     +---

Output relations:

• Output is a square wave of amplitude Vs (twice that of a half-bridge with the same supply).
• RMS output voltage Vo = Vs.
• Fourier series: vo = Σ (4Vs/nπ) sin nωt for n = 1, 3, 5, …; fundamental RMS = 4Vs/(√2·π) = 0.9·Vs.
• Frequency is set by the switching rate; output voltage can be varied by PWM.

Given: Vm = 200 V, forward breakover voltage VBO = 100 V at IG = 1 mA, RL = 100 Ω.

The SCR fires when the instantaneous supply reaches VBO: Vm sin α = VBO.

i) Firing angle

sin α = VBO/Vm = 100/200 = 0.5
α = 30°

ii) Conduction angle

With zero holding current, the SCR conducts from α to 180°:

θc = 180° − α = 180° − 30° = 150°

iii) Average current (RL = 100 Ω)

Iav = (1/2π)∫α^π (Vm/RL) sin θ dθ
    = (Vm / 2πRL)(1 + cos α)
    = (200 / (2π × 100))(1 + cos 30°)
    = 0.3183 × 1.8660
Iav = 0.594 A

iv) Average current with IH = 200 mA

Now the SCR turns OFF when the load current falls to IH before the end of the half cycle:

i = (Vm/RL) sin θ = 2 sin θ  A
2 sin θ = 0.2  →  sin θ = 0.1
θ_off = 180° − 5.74° = 174.26°

(At turn-on, i = 2 sin 30° = 1 A > IH, so the SCR latches normally.)

Iav = (1/2π)∫α^θoff 2 sin θ dθ
    = (2/2π)[cos 30° − cos 174.26°]
    = (1/π)(0.8660 + 0.9950)
Iav = 0.592 A

New conduction angle = 174.26° − 30° = 144.26°.

Answer: α = 30°, conduction angle = 150°, Iav = 0.594 A (IH = 0) and Iav ≈ 0.592 A (IH = 200 mA, conduction 144.26°)

Chopper

A chopper is a static switch that converts a fixed DC voltage into a variable DC voltage (DC-to-DC converter). It connects and disconnects the load from the source periodically; the average output is controlled by the duty cycle k = Ton/T, e.g. Vo = k·Vs for a step-down chopper.

Single-phase half-bridge inverter

A half-bridge inverter uses two switches (S1, S2 with antiparallel diodes D1, D2) and a centre-tapped DC supply (two equal sources Vs/2 or two equal capacitors). The load is connected between the switch midpoint A and the supply centre point O.

   +  +-----------+
 Vs/2 |          S1  D1
   −  |           |
      O---[LOAD]--A
   +  |           |
 Vs/2 |          S2  D2
   −  +-----------+

Working:

1. 0 < t < T/2: S1 ON, S2 OFF. The upper source drives current through S1 → A → load → O, so vAO = +Vs/2.
2. T/2 < t < T: S2 ON, S1 OFF. The lower source drives current O → load → A → S2, so vAO = −Vs/2.
3. S1 and S2 must never conduct together (short circuit of the supply); a small dead time is used.
4. With RL load, current lags the voltage. After S1 turns off, the load current continues through D2 (feeding energy back to the lower source) until it reaches zero, then S2 conducts. Similarly D1 conducts after S2 turns off. Hence D1 and D2 are called feedback diodes.

Waveforms:

 S1 gate |‾‾‾‾‾|_____|‾‾‾‾‾|____
 S2 gate |_____|‾‾‾‾‾|_____|‾‾‾‾
 vo +Vs/2+-----+     +-----+
         |     |     |     |
   0  ---+-----+-----+-----+---> t
   −Vs/2       +-----+     +--
         0    T/2    T
 io (RL)   /‾‾‾\     /‾‾‾\
 (exponential rise and fall, lags vo)
     D1  S1   D2  S2   (conducting device)

Output relations:

• Square wave of amplitude Vs/2; RMS value Vo = Vs/2.
• vo = Σ (2Vs/nπ) sin nωt, n = 1, 3, 5, …
• Fundamental RMS = 2Vs/(√2·π) = 0.45·Vs.
• Frequency f = 1/T set by the gating signals.

It needs only two switches but gives half the output of a full bridge and requires a centre-tapped supply.

Two-transistor analogy of SCR

The PNPN layers of an SCR are split into a PNP transistor Q1 (P1-N1-P2) and an NPN transistor Q2 (N1-P2-N2). The collector of each transistor drives the base of the other, so they form a regenerative loop.

      A
      |
     Q1 (PNP)
      |   \
      |    +-- IB1 = IC2
     IC1   |
      |    |
  G --+-- Q2 (NPN)
           |
           K

IC1 = α1·IA + ICBO1
IC2 = α2·IK + ICBO2,   IK = IA + IG
IA  = IC1 + IC2
IA  = (α2·IG + ICBO1 + ICBO2) / (1 − (α1 + α2))

When gate current raises α1 + α2 towards 1, IA rises sharply (limited only by the load) and the SCR latches ON.

TRIAC firing angles

Given: Vs = 230 V (rms), 50 Hz, VBO(DIAC) = 25 V, C = 0.6 μF, R = 2 kΩ to 20 kΩ.

The capacitor voltage lags the supply. The TRIAC fires when vC reaches the DIAC breakover voltage:

Xc = 1/(2πfC) = 1/(2π × 50 × 0.6×10⁻⁶) = 5305.2 Ω
Vc = Vs·Xc/√(R² + Xc²)   (rms)
φ  = tan⁻¹(R/Xc)         (lag of vC behind vs)
vC = √2·Vc·sin(ωt − φ)
Fires when √2·Vc·sin(α − φ) = VBO
α  = φ + sin⁻¹( VBO / (√2·Vc) )

Minimum firing angle (R = 2 kΩ):

Z  = √(2000² + 5305.2²) = 5669.6 Ω
Vc = 230 × 5305.2/5669.6 = 215.21 V,  peak 304.36 V
φ  = tan⁻¹(2000/5305.2) = 20.66°
sin⁻¹(25/304.36) = 4.71°
α_min = 20.66° + 4.71° = 25.37°

Maximum firing angle (R = 20 kΩ):

Z  = √(20000² + 5305.2²) = 20691.7 Ω
Vc = 230 × 5305.2/20691.7 = 58.97 V,  peak 83.40 V
φ  = tan⁻¹(20000/5305.2) = 75.14°
sin⁻¹(25/83.40) = 17.44°
α_max = 75.14° + 17.44° = 92.59°

Answer: minimum firing angle ≈ 25.4°, maximum firing angle ≈ 92.6° (DIAC/TRIAC gate loading on the RC network neglected).

Importance of freewheeling diode

A freewheeling diode (FD) is connected across an inductive load (cathode to the positive end) in a rectifier or chopper.

• When the switch turns OFF, the load inductance tries to keep current flowing. The FD gives this current a path, so the inductor does not produce a large L·di/dt voltage spike that could damage the switch.
• Stored energy in the inductor is delivered to the load instead of back to the source, improving efficiency and power factor.
• Load current becomes smoother and can be continuous.
• In controlled rectifiers it prevents the output voltage from going negative, raising the average output voltage.

Step-up chopper

A step-up chopper gives an average output voltage greater than the input, Vo > Vs.

       L     iL            D
 +Vs--uuuu--+----------->|---+-----+
            |                |     |
            CH (switch)      C   LOAD  Vo
            |                |     |
 −  --------+----------------+-----+

Principle:

1. CH ON (0 < t < Ton): The inductor is connected directly across Vs. Current rises linearly and L stores energy. The diode is reverse biased; the capacitor supplies the load.

L·di/dt = Vs   →   ΔI = Vs·Ton / L

2. CH OFF (Ton < t < T): The inductor current cannot change suddenly, so the inductor voltage reverses and adds to Vs. Current flows through D to the load and capacitor. Load voltage = Vs + L·di/dt > Vs.

L·di/dt = Vs − Vo  (negative, current falls)
ΔI = (Vo − Vs)·Toff / L

Expression for output voltage (steady state, rise = fall):

Vs·Ton/L = (Vo − Vs)·Toff/L
Vs(Ton + Toff) = Vo·Toff
Vo = Vs·T/Toff = Vs/(1 − k),   k = Ton/T

Waveforms:

CH    |‾‾‾‾|__|‾‾‾‾|__|‾‾‾‾|__
iL      /|\  /|\  /|\  (rises when ON,
       / | \/ | \/ |    falls when OFF)
vL  +Vs ‾‾‾‾      ‾‾‾‾
        −(Vo−Vs) __    __

Example: k = 0.5 gives Vo = 2Vs; k = 0.75 gives Vo = 4Vs. As k → 1, Vo theoretically → ∞, but in practice losses limit it. Used in regenerative braking of DC motors and boost converters.

Given: vs = 50 sin(2π·50t), so Vm = 50 V; α = 30°; R = 100 Ω; half-wave controlled rectifier with resistive load.

Average load voltage

Vdc = (1/2π)∫α^π Vm sin ωt d(ωt)
    = (Vm/2π)(1 + cos α)
    = (50/2π)(1 + 0.8660)
Vdc = 14.85 V
Idc = Vdc/R = 0.1485 A

RMS load voltage

Vrms = [ (1/2π)∫α^π Vm² sin² ωt d(ωt) ]^(1/2)
     = (Vm/2)·√[ (π − α)/π + sin2α/(2π) ]
     = 25 × √[ 0.8333 + 0.1378 ]
     = 25 × 0.9855
Vrms = 24.64 V
Irms = 0.2464 A

Efficiency (rectification efficiency)

Pdc = Vdc²/R = 14.85²/100 = 2.205 W
Pac = Vrms²/R = 24.64²/100 = 6.070 W
η = Pdc/Pac = (Vdc/Vrms)²
  = (14.85/24.64)² = 0.3633
η = 36.33 %

Ripple factor

Form factor FF = Vrms/Vdc = 24.64/14.85 = 1.659
RF = √(FF² − 1) = √(1.659² − 1)
RF = 1.324

Answer: Vdc = 14.85 V, Vrms = 24.64 V, η = 36.33 %, ripple factor = 1.324

A TRIAC (Triode AC switch) is a three-terminal, bidirectional thyristor that can conduct in both directions. Its terminals are MT1, MT2 and gate G. It is equivalent to two SCRs connected in inverse parallel with a common gate.

        MT2
         |
    +----+----+
    |         |
   SCR1 ->   <- SCR2    (inverse parallel)
    |         |
    +----+----+
         |       G
        MT1 -----+

Working principle:

• When MT2 is positive with respect to MT1, a gate pulse (positive or negative) turns ON the left PNPN path and current flows MT2 → MT1.
• When MT1 is positive with respect to MT2, a gate pulse turns ON the other path and current flows MT1 → MT2.
• Once ON it latches until the current falls below the holding current, which happens naturally at each zero crossing of AC.
• It can be triggered in four modes (MT2 +/−, gate +/−); sensitivity is highest in quadrants I+ and III−.

V-I characteristic is identical in the first and third quadrants: blocking until breakover (reduced by gate current), then a low on-state voltage.

Applications: light dimmers, fan speed regulators, heater control, AC motor speed control, AC static switches. It is usually triggered by a DIAC.

Limitations: lower voltage/current and dv/dt ratings than SCRs, and suitable mainly for 50/60 Hz.

Variable output voltage in inverter

Yes. The output voltage of an inverter can be varied by:

1. Varying the DC input voltage using a controlled rectifier or chopper before the inverter.
2. Pulse width modulation (PWM) inside the inverter (most common): single-pulse, multiple-pulse or sinusoidal PWM. Changing the pulse width (modulation index) changes the RMS output while the frequency stays fixed.
3. Controlling the AC output with an AC voltage controller or a tapped transformer.
4. Series/phase-shift connection of two inverters, where varying the phase shift between them changes the resultant voltage.

Average output voltage of SCR full-wave rectifier

Consider a centre-tapped full-wave controlled rectifier with resistive load (a bridge gives the same result).

      T1
  +---|>|---+
  )         |
  ) vs      +---[ R ]---+
  )==CT-----------------+
  ) vs      |
  )         |
  +---|>|---+
      T2

• Positive half cycle: T1 is forward biased and fired at ωt = α; it conducts until π.
• Negative half cycle: T2 is fired at π + α and conducts until 2π.
• Load voltage repeats every π.

vo   |  /‾\    /‾\
     | /   \  /   \
     |/  |  \/  |  \
   --+---+--+----+---+--> ωt
     0   α  π  π+α  2π

Derivation:

Vdc = (1/π)∫α^π Vm sin ωt d(ωt)
    = (Vm/π)[−cos ωt]α^π
    = (Vm/π)[−cos π + cos α]
Vdc = (Vm/π)(1 + cos α)

• At α = 0, Vdc = 2Vm/π (same as uncontrolled full-wave rectifier).
• At α = π, Vdc = 0.
• Average load current Idc = Vm(1 + cos α)/(πR).

RMS value for reference:

Vrms = (Vm/√2)·√[ (π − α)/π + sin2α/(2π) ]

Given: Vs = 230 V (rms), 50 Hz, DIAC breakover VBO = 20 V, C = 0.5 μF, R = 5 kΩ to 15 kΩ.

Assumption: before the TRIAC fires it is OFF, so the RC network across the TRIAC sees almost the full supply (lamp resistance is small compared with R and Xc). DIAC current loading on C is neglected.

Method

The capacitor voltage lags the supply by φ = tan⁻¹(ωRC). The TRIAC fires when vC reaches the DIAC breakover voltage.

Xc = 1/(2πfC) = 1/(2π × 50 × 0.5×10⁻⁶) = 6366.2 Ω
Vc = Vs·Xc/√(R² + Xc²)        (rms)
φ  = tan⁻¹(R/Xc)
α  = φ + sin⁻¹( VBO/(√2·Vc) )

Minimum firing angle (R = 5 kΩ)

Z  = √(5000² + 6366.2²) = 8095.0 Ω
Vc = 230 × 6366.2/8095.0 = 180.88 V
Vc(peak) = √2 × 180.88 = 255.80 V
φ  = tan⁻¹(5000/6366.2) = 38.15°
sin⁻¹(20/255.80) = 4.48°
α_min = 38.15° + 4.48° = 42.63°

Maximum firing angle (R = 15 kΩ)

Z  = √(15000² + 6366.2²) = 16295.0 Ω
Vc = 230 × 6366.2/16295.0 = 89.86 V
Vc(peak) = √2 × 89.86 = 127.08 V
φ  = tan⁻¹(15000/6366.2) = 67.00°
sin⁻¹(20/127.08) = 9.06°
α_max = 67.00° + 9.06° = 76.06°

Answer: minimum firing angle ≈ 42.6°, maximum firing angle ≈ 76.1°

So the lamp conducts from about 137° (brightest) down to about 104° (dimmest) in each half cycle.

Two-transistor analogy of SCR

An SCR (PNPN) can be viewed as a PNP transistor Q1 (P1-N1-P2) and an NPN transistor Q2 (N1-P2-N2) connected so that the collector of each feeds the base of the other.

      A
      |
     Q1 (PNP)
      |   \
      |    +-- IB1 = IC2
     IC1   |
      |    |
  G --+-- Q2 (NPN)
           |
           K

IC1 = α1·IA + ICBO1
IC2 = α2·IK + ICBO2
IK  = IA + IG,   IA = IC1 + IC2
IA  = α1·IA + α2(IA + IG) + ICBO1 + ICBO2
IA  = (α2·IG + ICBO1 + ICBO2) / (1 − (α1 + α2))

With IG = 0, α1 + α2 is small and only leakage flows (forward blocking). A gate current increases α2, current increases, α1 rises, and when α1 + α2 → 1 the anode current becomes very large (limited by the load). The regenerative action latches the SCR ON, and the gate loses control.

Numerical

Given: R = 50 Ω, Vs = 120 V rms, so Vm = 120√2 = 169.71 V; SCR drop VT = 1.5 V. Taking the peak load voltage as Vm − VT = 168.21 V:

(a) Firing angle for Iav = 1 A

Iav = (Vm − VT)(1 + cos α)/(2πR)
1   = 168.21 (1 + cos α)/(2π × 50)
1 + cos α = 314.16/168.21 = 1.8677
cos α = 0.8677
α = 29.8°

(If the 1.5 V drop is subtracted over the conduction interval exactly, α ≈ 29.0°; the difference is small.)

(b) Average power to the load

Power in a resistor depends on the RMS value:

Vrms = ((Vm − VT)/2)·√[ (π − α)/π + sin2α/(2π) ]
     = 84.10 × √[ 0.8344 + 0.1373 ]
     = 82.90 V
P = Vrms²/R = 82.90²/50
P ≈ 137.5 W

Answer: α ≈ 29.8°, average load power ≈ 137.5 W (the DC component alone is Vdc·Idc = 50 V × 1 A = 50 W).

A step-down chopper gives an average output voltage lower than the input: Vo = k·Vs, where k = Ton/T. With an RL (or RLE) load, a freewheeling diode carries the load current when the switch is OFF.

       CH
 +Vs --/ --+--------+
           |        R
           FD       L
          (↑)       E (back emf, may be 0)
           |        |
 −   ------+--------+

Working

Mode 1 (CH ON, 0 < t < Ton): Load connected to Vs, FD reverse biased. Load current rises exponentially from I1 to I2.

Vs = R·i + L di/dt + E
i(t) = (Vs − E)/R · (1 − e^(−t/τ)) + I1·e^(−t/τ),  τ = L/R
At t = Ton: i = I2

Mode 2 (CH OFF, 0 < t' < Toff): Inductor current continues through FD; load voltage = 0. Current decays from I2 to I1.

0 = R·i + L di/dt + E
i(t') = I2·e^(−t'/τ) − (E/R)(1 − e^(−t'/τ))
At t' = Toff: i = I1

Waveforms

vo  Vs |‾‾‾‾|____|‾‾‾‾|____
       0   Ton   T
io  I2 |   /\     /\
       |  /  \   /  \
    I1 | /    \_/    \_   (continuous current)
iCH    | /|      /|        (rising part)
iFD    |    \_      \_     (falling part)

Expressions (steady state)

Solving the two modes with i(T) = i(0) = I1:

I2 = (Vs/R)(1 − e^(−kT/τ))/(1 − e^(−T/τ)) − E/R
I1 = (Vs/R)(e^(kT/τ) − 1)/(e^(T/τ) − 1) − E/R

Average output voltage and current:

Vo = (1/T)∫₀^(Ton) Vs dt = k·Vs
Io = (Vo − E)/R = (k·Vs − E)/R

Peak-to-peak ripple current:

ΔI = I2 − I1
   = (Vs/R)·(1 − e^(−kT/τ))·(1 − e^(−(1−k)T/τ))
     / (1 − e^(−T/τ))

ΔI is maximum at k = 0.5:

ΔImax = (Vs/R)·tanh(R/(4fL)) ≈ Vs/(4fL)  when 4fL >> R

So a larger inductance or higher chopping frequency reduces the ripple. Used in DC motor speed control (with E as motor back emf) and battery-operated vehicles.

Drawbacks of DIAC

• It has no gate terminal, so its switching point cannot be controlled; it turns on only when the applied voltage reaches its fixed breakover voltage (about 30 V).
• It is a low-power device; it cannot handle large load currents, so it is used only as a trigger device for TRIACs/SCRs.
• It has a negative resistance region and a fairly large voltage drop after breakover, and its breakover voltage varies from device to device.

Operating principle of TRIAC

A TRIAC is a bidirectional, three-terminal thyristor (MT1, MT2, gate G), equivalent to two SCRs in inverse parallel with a common gate. It is a five-layer device (N-P-N-P-N between MT1 and MT2), with an N region also near the gate.

        MT2
         |
    +----+----+
    |         |
   SCR1 ->   <- SCR2
    |         |
    +----+----+
         |      G
        MT1 ----+

• MT2 positive w.r.t. MT1: the P1-N1-P2-N2 path is forward biased. A gate pulse turns it ON and current flows MT2 → MT1.
• MT2 negative w.r.t. MT1: the P2-N1-P1-N4 path is forward biased. A gate pulse turns it ON and current flows MT1 → MT2.
• Once ON it latches until the current falls below the holding current (at every AC zero crossing).

V-I characteristic:

              I
              ^     on (quadrant I)
              |    /
              |   |
  -VBO        |   |
 ----|--------+---|-----> V(MT2-MT1)
     |        |  +VBO
     |        |
    /         |
on (quadrant III)

Increasing gate current reduces the breakover voltage in both directions, just like an SCR.

Why four-quadrant operation

Both the MT2 voltage and the gate current can be of either polarity, giving four triggering modes:

Since the TRIAC can be triggered with any of the four combinations of anode (MT2) and gate polarity, i.e. in all four quadrants of the V–IG plane, it is called a four-quadrant device. In practice, modes I+ and III− are preferred because they need the least gate current.

Given: 220 V (rms), 50 Hz supply; RL = 15 Ω; C = 1 μF; R = 2 kΩ to 20 kΩ; DIAC breakover VBO = 12 V.

Assumptions: RC branch is across the supply, DIAC current loading on C is neglected, and the TRIAC drop is neglected.

Firing angle and conduction angle

The capacitor voltage lags the supply by φ = tan⁻¹(ωRC). The TRIAC fires when vC = VBO.

Xc = 1/(2πfC) = 1/(2π × 50 × 1×10⁻⁶) = 3183.1 Ω
Vc = Vs·Xc/√(R² + Xc²)
α  = tan⁻¹(R/Xc) + sin⁻¹( VBO/(√2·Vc) )
Conduction angle = 180° − α

R = 2 kΩ (maximum conduction):

Z  = √(2000² + 3183.1²) = 3759.3 Ω
Vc = 220 × 3183.1/3759.3 = 186.28 V, peak 263.44 V
φ  = tan⁻¹(2000/3183.1) = 32.14°
sin⁻¹(12/263.44) = 2.61°
α_min = 34.75°  →  conduction = 145.25°

R = 20 kΩ (minimum conduction):

Z  = √(20000² + 3183.1²) = 20251.7 Ω
Vc = 220 × 3183.1/20251.7 = 34.58 V, peak 48.90 V
φ  = tan⁻¹(20000/3183.1) = 80.96°
sin⁻¹(12/48.90) = 14.20°
α_max = 95.16°  →  conduction = 84.84°

Load voltage waveforms

Max conduction (α = 34.75°)
vL |    /‾\
   |   |   \
 --+---+----+----------+----> ωt
   0  35°  180°  |    /
                 |   |   (same in −ve half,
                 +‾‾‾     from 214.75° to 360°)

Min conduction (α = 95.16°)
vL |      |\
   |      | \
 --+------+--+-------+---> ωt
   0    95°  180°    |  /
                     | |  (from 275.16° to 360°)
                     +‾

In each half cycle the load voltage is zero until α and then follows the supply sine wave up to the zero crossing.

Average power at maximum conduction (α = 34.75°)

VL,rms = Vs·√[ (π − α)/π + sin2α/(2π) ]
       = 220 × √[ 0.8069 + 0.1491 ]
       = 220 × 0.9778 = 215.11 V
P = VL,rms²/RL = 215.11²/15
P = 3084.7 W

Answer: conduction angle ranges from 84.84° (R = 20 kΩ) to 145.25° (R = 2 kΩ); power at maximum conduction ≈ 3.08 kW (full-wave power would be 220²/15 = 3226.7 W).

Single-phase full-bridge inverter

A full-bridge inverter converts DC into AC using four switches (S1–S4, with feedback diodes D1–D4) and a single DC supply Vs. The load is connected between the midpoints A and B of the two legs.

       +Vs
   +----+------------+
   |    |            |
   |   S1 D1        S3 D3
   |    |   LOAD     |
   |    A---[RL]-----B
   |    |            |
   |   S4 D4        S2 D2
   |    |            |
   +----+------------+
       −Vs (0)

Operation:

1. 0 < t < T/2: S1 and S2 ON → vAB = +Vs.
2. T/2 < t < T: S3 and S4 ON → vAB = −Vs.
3. S1–S4 (or S3–S2) in the same leg must never be ON together.
4. With RL load, current lags. When S1, S2 turn OFF, the load current continues through D3, D4, returning energy to the supply, until it reverses and S3, S4 take over (and similarly for the next half).

Waveforms:

 S1,S2 |‾‾‾‾‾|_____|‾‾‾‾‾|____
 S3,S4 |_____|‾‾‾‾‾|_____|‾‾‾‾
 vo +Vs+-----+     +-----+
       |     |     |     |
     0-+-----+-----+-----+---> t
       |     |     |     |
   −Vs       +-----+     +---
       0    T/2    T
 io      /‾‾‾\       /‾‾‾\
 (RL) __/     \_____/     \_

• Output RMS voltage = Vs; fundamental RMS = 4Vs/(√2π) = 0.9Vs.
• vo = Σ (4Vs/nπ) sin nωt, n = 1, 3, 5, …
• Output is twice that of a half-bridge with the same supply, and no centre tap is needed.

Methods of controlling output power in DC-to-DC conversion

The average output Vo = (Ton/T)·Vs = k·Vs is varied by changing the duty cycle:

1. Constant frequency (PWM) control: T fixed, Ton varied. Most common; easy filtering.
2. Variable frequency (FM) control: Ton (or Toff) fixed, T varied. Harmonics vary with frequency, so filtering is harder.
3. Current-limit (hysteresis) control: switch is turned OFF when load current reaches a maximum and ON when it falls to a minimum, keeping current within a band.

Can two complementary BJTs make an SCR?

Yes, in principle. An SCR behaves like a PNP and an NPN transistor connected in a regenerative loop (two-transistor analogy): the collector of the PNP drives the base of the NPN and the collector of the NPN drives the base of the PNP.

   A
   |
  PNP (Q1) emitter
   |  \
   |   +--- base Q1 = collector Q2
  C1   |
   +---+--- G (base Q2)
       |
      NPN (Q2)
       |
       K

A small gate current turns Q2 ON, Q2 turns Q1 ON, and Q1 keeps Q2 ON, so the pair latches like an SCR (IA = (α2·IG + ICBO1 + ICBO2)/(1 − α1 − α2)). Such a discrete latch is used in low-power circuits, but it cannot replace a real SCR in power work because discrete BJTs have lower voltage blocking, poorer surge and dv/dt capability, and need careful biasing.

Turning OFF of SCR (commutation)

Once ON, an SCR cannot be turned off by the gate. To turn it OFF, the anode current must be reduced below the holding current IH, and then a reverse voltage must be maintained for at least the turn-off time tq (about 10–100 μs) so that the junctions recover their blocking ability.

Methods:

1. Natural (line) commutation: In AC circuits the current falls to zero at the end of each half cycle and the supply reverses, so the SCR turns off automatically. Used in AC controllers and controlled rectifiers.
2. Forced commutation: In DC circuits current never becomes zero naturally, so an extra circuit (L, C and an auxiliary SCR) forces it to zero:
   • Class A: self commutation by resonating load
   • Class B: self commutation by LC circuit
   • Class C: complementary commutation (C or L-C switched by another load-carrying SCR)
   • Class D: impulse commutation by an auxiliary SCR
   • Class E: external pulse source
3. Reducing anode current / opening the circuit: increasing load resistance or interrupting the anode circuit so that IA < IH.
4. Reverse biasing: applying a reverse voltage across anode and cathode.

 iA  |‾‾‾‾‾‾\
     |       \___  IH  ___
     |           \_____/      reverse recovery
 vAK |____________
     |            |___ reverse bias ≥ tq ___|‾‾ forward
                   <------- tq ------->

The circuit turn-off time must be greater than the device turn-off time tq; otherwise the SCR turns ON again when forward voltage reappears.

Classification of chopper on the basis of power flow (quadrant of operation)

So by power flow, choppers are unidirectional (Class A, B) or bidirectional (Class C, D, E). By output level they are also classed as step-down (Vo < Vs) and step-up (Vo > Vs).

Principle of step-up converter with resistive load

       L     iL           D
 +Vs--uuuu--+----------->|---+-----+
            |                |     |
            S (switch)       C     R   Vo
            |                |     |
 −  --------+----------------+-----+

1. Switch ON (0 < t < Ton): The inductor is placed directly across Vs; current rises linearly and energy is stored in L. The diode is reverse biased, and the capacitor supplies the resistive load.

vL = Vs  →  ΔI = Vs·Ton/L

2. Switch OFF (Ton < t < T): Inductor current cannot change suddenly; its voltage reverses and adds to Vs. Current flows through D into C and R. Load voltage is now higher than Vs.

vL = Vs − Vo (negative)  →  ΔI = (Vo − Vs)·Toff/L

3. Steady state: current rise during ON equals fall during OFF:

Vs·Ton = (Vo − Vs)·Toff
Vo = Vs·T/Toff = Vs/(1 − k),  k = Ton/T

Load current Io = Vo/R and, ignoring losses, input current Is = Io/(1 − k).

Waveforms:

S    |‾‾‾‾‾‾|___|‾‾‾‾‾‾|___
iL   |  /‾‾\   /‾‾\
     | /    \_/    \_   rises ON, falls OFF
vo   |‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾  ≈ Vs/(1−k), small ripple

Example: Vs = 12 V, k = 0.6 → Vo = 12/0.4 = 30 V. Since 0 < k < 1, Vo is always greater than Vs.

Given: IG = 1.25 mA, forward breakover voltage VBO = 110 V, applied voltage 220 V, RL = 150 Ω, IH = 0.

Assumption: the 220 V applied voltage is taken as the peak value (Vm = 220 V), as is usual in this type of textbook problem.

a) Firing angle

The SCR fires when the supply reaches VBO:

Vm sin α = VBO
sin α = 110/220 = 0.5
α = 30°

b) Conduction angle

With zero holding current, it conducts until 180°:

θc = 180° − 30° = 150°

c) Average output voltage

Vav = (1/2π)∫α^π Vm sin θ dθ = (Vm/2π)(1 + cos α)
    = (220/2π)(1 + 0.8660)
    = 35.01 × 1.8660
Vav = 65.34 V

d) Average current

Iav = Vav/RL = 65.34/150
Iav = 0.4356 A ≈ 435.6 mA

Answer: α = 30°, conduction angle = 150°, Vav = 65.34 V, Iav = 0.436 A

If 220 V is instead taken as RMS (Vm = 311.13 V): α = 20.70°, conduction angle = 159.30°, Vav = 95.84 V, Iav = 0.639 A.

Classification of choppers

1. By output voltage level:
   • Step-down chopper (buck): Vo = k·Vs < Vs
   • Step-up chopper (boost): Vo = Vs/(1 − k) > Vs
   • Step-up/down chopper (buck-boost)
2. By quadrant of operation (direction of Vo and Io):
   • Class A: first quadrant (+V, +I)
   • Class B: second quadrant (+V, −I), regenerative
   • Class C: two-quadrant, I and II (current reversible)
   • Class D: two-quadrant, I and IV (voltage reversible)
   • Class E: four-quadrant
3. By commutation: voltage-commutated, current-commutated, load-commutated, impulse-commutated.

Step-down chopper with RL load

       CH
 +Vs --/ --+--------+
           |        R
           FD       L
           |        E (0 if pure RL)
 −   ------+--------+

Mode 1 – CH ON (0 < t < Ton): Load is connected to Vs; FD reverse biased. Current rises exponentially from I1 to I2:

Vs = R·i + L·di/dt + E
i = I1·e^(−t/τ) + ((Vs − E)/R)(1 − e^(−t/τ)),  τ = L/R

Mode 2 – CH OFF (0 < t < Toff): Inductor energy keeps current flowing through FD; vo = 0. Current decays from I2 to I1:

0 = R·i + L·di/dt + E
i = I2·e^(−t/τ) − (E/R)(1 − e^(−t/τ))

Waveforms (continuous current):

vo Vs |‾‾‾‾|____|‾‾‾‾|____
      0   Ton   T
io I2 |   /\     /\
   I1 |__/  \___/  \___
      (rises: CH)(falls: FD)

Output relations:

Vo = (1/T)∫₀^(Ton) Vs dt = k·Vs
Io = (k·Vs − E)/R
ΔI = I2 − I1, maximum at k = 0.5:
ΔImax = (Vs/R)·tanh(R/(4fL)) ≈ Vs/(4fL)

Higher chopping frequency or larger L gives smoother current. Used for DC motor speed control.