Switched Power Supplies

A Buck regulator is a switching DC-DC converter whose average output voltage Va is less than the input Vs: Va = k·Vs.

Circuit

        Q (switch)     L    iL
 +Vs ---/ ----+------uuuu----+----+------+
              |              |    |      |
              Dm (↑)         C   LOAD   Va
              |              |    |      |
 −   ---------+--------------+----+------+
  control circuit drives Q (PWM, duty k)

Operation

Mode 1 (Q ON, 0 < t < t1 = kT): Dm is reverse biased. Inductor current rises linearly from I1 to I2; L stores energy and current flows to C and the load.

Vs − Va = L·ΔI/t1   →  t1 = L·ΔI/(Vs − Va)

Mode 2 (Q OFF, t1 < t < T): The inductor current continues through freewheeling diode Dm. Inductor voltage = −Va; current falls from I2 to I1.

Va = L·ΔI/t2   →  t2 = L·ΔI/Va

Waveforms

vQ gate |‾‾‾|____|‾‾‾|____
        0  kT    T
iL   I2 |  /\    /\
     I1 | /  \__/  \__
vC/Va   |~~~~~~~~~~~~~~  small ripple about Va

Derivations

Output voltage: equate the two ΔI expressions:

(Vs − Va)·t1/L = Va·t2/L
(Vs − Va)·kT = Va·(1 − k)T
Va = k·Vs

Ripple current of inductor:

T = t1 + t2 = L·ΔI·Vs / (Va(Vs − Va))
ΔI = Va(Vs − Va)/(f·L·Vs) = Vs·k(1 − k)/(f·L)

Ripple voltage of capacitor: the AC part of iL (triangular) flows into C. Charge added in half a cycle:

ΔQ = (1/2)(T/2)(ΔI/2) = ΔI/(8f)
ΔVc = ΔQ/C = ΔI/(8fC)
ΔVc = Va(Vs − Va)/(8·L·C·f²·Vs)

Critical values for continuous inductor current and voltage:

Lc = (1 − k)R/(2f),   Cc = (1 − k)/(16·L·f²)

Input current Is = k·Ia (ideal). Advantages: high efficiency (about 90 %), only one switch. Disadvantages: input current is pulsating, and no short-circuit protection is inherent.

A Boost regulator is a switching DC-DC converter whose average output voltage is greater than the input: Va = Vs/(1 − k).

Circuit

        L    iL            Dm
 +Vs --uuuu---+----------->|---+-----+
              |                |     |
              Q (switch)       C   LOAD  Va
              |                |     |
 −  ----------+----------------+-----+

Working principle

Mode 1 (Q ON, 0 < t < t1 = kT): The inductor is connected across Vs. Inductor current rises linearly from I1 to I2 and energy is stored. Dm is reverse biased; C supplies the load.

Vs = L·ΔI/t1   →  t1 = L·ΔI/Vs

Mode 2 (Q OFF, t1 < t < T): The inductor voltage reverses and adds to Vs. Current flows through Dm to C and the load; iL falls from I2 to I1.

Vs − Va = −L·ΔI/t2   →  t2 = L·ΔI/(Va − Vs)

Waveforms

Q   |‾‾‾‾‾|___|‾‾‾‾‾|___
    0    kT   T
iL  |    /\     /\        I2 (peak)
    |   /  \   /  \
    |__/    \_/    \_     I1
iD  |      |\_   |\_      (only when Q is OFF)
vC  |~~~~~~~~~~~~~~~~      Va with ripple ΔVc

Expressions

Output voltage:

Vs·t1 = (Va − Vs)·t2
Vs·kT = (Va − Vs)(1 − k)T
Va = Vs/(1 − k)

Peak-to-peak inductor ripple current:

ΔI = Vs·t1/L = Vs·k/(f·L)
   = Vs(Va − Vs)/(f·L·Va)

Capacitor ripple voltage: during t1 the capacitor alone supplies Ia:

ΔVc = Ia·t1/C = Ia·k/(f·C)

Currents: Is = Ia/(1 − k) (ideal, Pin = Pout); I2 = Is + ΔI/2, I1 = Is − ΔI/2.

Critical values:

Lc = k(1 − k)R/(2f),   Cc = k/(2fR)

Advantages: steps up voltage without a transformer; input current is continuous. Disadvantages: high peak current in the switch; output is very sensitive to k as k → 1.

Given: Vs = 10 V, k = 0.45, f = 25 kHz, Ia = 1200 mA = 1.2 A.

Assumption: the values L = 0.1 nH and C = 0.2 nF are not practical (they give a ripple current of 1.8 × 10⁶ A), so they are taken as L = 0.1 mH and C = 0.2 mF (200 μF), the usual values for this problem.

Buck-boost relations (Rashid):

Va  = −Vs·k/(1 − k)
ΔVc = Ia·k/(f·C)
ΔI  = Vs·k/(f·L)

i) Average output voltage

Va = −10 × 0.45/(1 − 0.45) = −4.5/0.55
Va = −8.18 V

(The minus sign shows polarity reversal of the output.)

ii) Peak-to-peak output ripple voltage

ΔVc = 1.2 × 0.45/(25×10³ × 0.2×10⁻³)
    = 0.54/5
ΔVc = 0.108 V = 108 mV

iii) Peak-to-peak ripple current

ΔI = 10 × 0.45/(25×10³ × 0.1×10⁻³)
   = 4.5/2.5
ΔI = 1.8 A

Answer: Va = −8.18 V, ΔVc = 108 mV, ΔI = 1.8 A

A switching regulator is a DC voltage regulator in which a transistor (BJT/MOSFET) is operated as a switch (fully ON or fully OFF) at high frequency (20 kHz – 1 MHz), instead of in the linear region. The output voltage is controlled by the duty cycle k = Ton/T using PWM, and an L-C filter smooths the output.

Vs -->[SWITCH]-->[L-C FILTER]--+--> Vo
         ^                     |
         |                  [divider]
     [PWM CONTROL]<--[ERROR AMP]<--Vref

The error amplifier compares a fraction of Vo with a reference; the PWM controller changes k to keep Vo constant against changes in input or load.

Basic types:

Advantages: high efficiency (80–95 %), since the switch dissipates little power; small size and weight; can step up, step down or invert the voltage.

Disadvantages: output ripple and switching noise (EMI), more complex circuit, slower transient response than linear regulators.

Given: Vs = 12 V, Va = 6 V, R = 450 Ω, ΔVc = 20 mV, f = 20 kHz, ΔI = 0.8 A.

Buck regulator relations:

Va  = k·Vs
ΔI  = Va(Vs − Va)/(f·L·Vs)
ΔVc = ΔI/(8·f·C)

i) Duty cycle

k = Va/Vs = 6/12
k = 0.5 (50 %)

ii) Filter capacitance C

C = ΔI/(8·f·ΔVc)
  = 0.8/(8 × 20×10³ × 20×10⁻³)
  = 0.8/3200
C = 250 μF

iii) Filter inductance L

L = Va(Vs − Va)/(f·ΔI·Vs)
  = 6 × (12 − 6)/(20×10³ × 0.8 × 12)
  = 36/192000
L = 187.5 μH

Check: average load current Ia = 6/450 = 13.3 mA.

Answer: k = 0.5, C = 250 μF, L = 187.5 μH

A Buck-Boost regulator gives an output voltage that can be lower or higher than the input and is of opposite polarity: Va = −k·Vs/(1 − k).

Circuit

        Q (switch)        Dm
 +Vs ---/ ---+-----------|<---+------+
             |                |      |
             L  iL            C    LOAD   Va
             |                |   (− at top)
 −   --------+----------------+------+

Operation

Mode 1 (Q ON, 0 < t < t1 = kT): Current flows from Vs through Q and L. Inductor current rises linearly from I1 to I2; Dm is reverse biased; C supplies the load.

Vs = L·ΔI/t1   →  t1 = L·ΔI/Vs

Mode 2 (Q OFF, t1 < t < T): The inductor keeps the current flowing through C, the load and Dm. Energy stored in L is transferred to the output, which becomes negative. Inductor voltage = Va (negative); current falls from I2 to I1.

Va = −L·ΔI/t2   →  t2 = −L·ΔI/Va

Waveforms

Q   |‾‾‾‾|____|‾‾‾‾|____
    0   kT    T
vL  |+Vs |    |+Vs |
    |    |____|    |____ Va (negative)
iL  |   /\     /\        I2
    |__/  \___/  \___    I1
iD  |     |\_   |\_      (when Q OFF)

Average output voltage

Rise and fall of inductor current are equal in steady state:

Vs·t1/L = −Va·t2/L
Vs·kT = −Va(1 − k)T
Va = −Vs·k/(1 − k)

• k < 0.5: |Va| < Vs (buck)
• k = 0.5: |Va| = Vs
• k > 0.5: |Va| > Vs (boost)

Peak-to-peak ripple current of inductor

ΔI = Vs·t1/L = Vs·k·T/L
ΔI = Vs·k/(f·L)

Also, using t1 + t2 = T:

T = L·ΔI/Vs − L·ΔI/Va = L·ΔI(Va − Vs)/(Vs·Va)
ΔI = Vs·Va/(f·L(Va − Vs))

(Both forms give the same value since Va = −kVs/(1 − k).)

Capacitor ripple (C supplies Ia during t1): ΔVc = Ia·k/(f·C).

Average input current Is = Ia·k/(1 − k) (ideal). The circuit needs only one switch and gives polarity reversal, but the switch and diode carry high peak currents and input current is discontinuous.

Given: Vs = 5 V, Va = 15 V, Ia = 0.5 A, f = 25 kHz, L = 150 μH, C = 220 μF.

Load resistance R = Va/Ia = 15/0.5 = 30 Ω.

Boost regulator relations (Rashid):

Va  = Vs/(1 − k)
ΔI  = Vs(Va − Vs)/(f·L·Va)
Is  = Ia/(1 − k),   I2 = Is + ΔI/2
ΔVc = Ia·k/(f·C)
Lc  = k(1 − k)R/(2f),   Cc = k/(2fR)

(i) Duty cycle

k = 1 − Vs/Va = 1 − 5/15
k = 0.6667 (66.67 %)

(ii) Peak current of inductor I2

ΔI = 5 × (15 − 5)/(25×10³ × 150×10⁻⁶ × 15)
   = 50/56.25 = 0.8889 A
Is = 0.5/(1 − 0.6667) = 1.5 A
I2 = Is + ΔI/2 = 1.5 + 0.4444
I2 = 1.944 A

(Valley current I1 = 1.5 − 0.444 = 1.056 A.)

(iii) Ripple voltage of filter capacitor

ΔVc = 0.5 × 0.6667/(25×10³ × 220×10⁻⁶)
    = 0.3333/5.5
ΔVc = 60.61 mV

(iv) Critical values of L and C

Lc = 0.6667 × 0.3333 × 30/(2 × 25×10³)
   = 6.667/50000
Lc = 133.3 μH

Cc = 0.6667/(2 × 25×10³ × 30)
Cc = 0.444 μF

Since L = 150 μH > Lc and C = 220 μF > Cc, the inductor current and capacitor voltage are continuous.

Answer: k = 0.667, I2 = 1.944 A, ΔVc = 60.61 mV, Lc = 133.3 μH, Cc = 0.444 μF

Inverter

An inverter is a power electronic converter that changes DC into AC of the required voltage and frequency by switching SCRs, MOSFETs or IGBTs in a fixed sequence. Common types are single-phase half-bridge and full-bridge inverters. Uses: UPS, solar power systems, AC motor drives and induction heating.

Buck switching regulator

A Buck regulator steps DC voltage down: Va = k·Vs, where k = Ton/T.

        Q (switch)     L    iL
 +Vs ---/ ----+------uuuu----+----+------+
              |              |    |      |
              Dm (↑)         C   LOAD   Va
              |              |    |      |
 −   ---------+--------------+----+------+

Mode 1 – Q ON (0 < t < kT): Dm is reverse biased; Vs drives current through L to C and the load. iL rises linearly from I1 to I2:

Vs − Va = L·ΔI/t1

Mode 2 – Q OFF (kT < t < T): The inductor current freewheels through Dm. iL falls from I2 to I1:

Va = L·ΔI/t2

Waveforms:

Q  gate |‾‾‾|____|‾‾‾|____
        0  kT    T
vD      |Vs |    |Vs |
        |   |____|   |____ 0
iL  I2  |  /\    /\
    I1  |_/  \__/  \__
Va      |~~~~~~~~~~~~~ (average with small ripple)

Relations:

(Vs − Va)·kT = Va(1 − k)T   →  Va = k·Vs
ΔI  = Va(Vs − Va)/(f·L·Vs)
ΔVc = ΔI/(8·f·C)

The average voltage across the diode (vD) equals Va, since the average inductor voltage is zero. Efficiency is high (about 90 %) because Q is either fully ON or fully OFF.

A Switched Mode Power Supply (SMPS) is a power supply that converts AC or DC input into a regulated DC output by switching a power transistor ON and OFF at high frequency (20 kHz – 1 MHz) and controlling the duty cycle with PWM.

AC->[RECTIFIER]->[HF SWITCH]->[HF  ]->[RECTIFIER]->DC
    [+ FILTER ]  [ (MOSFET)]  [TRAN]  [+ FILTER ]  out
                      ^                    |
                      |                    |
                 [PWM CONTROL]<-[ISOLATION]<+
                                [+ERROR AMP]

Working:

1. AC mains is rectified and filtered to unregulated DC.
2. A MOSFET chops this DC at high frequency.
3. A small high-frequency transformer steps the voltage up/down and gives isolation.
4. The secondary is rectified and filtered to DC.
5. The output is compared with a reference; the PWM controller adjusts the duty cycle to keep the output constant (feedback through an opto-coupler).

Common topologies: flyback, forward, push-pull, half-bridge, full-bridge (and non-isolated buck, boost, buck-boost).

Advantages: high efficiency (70–95 %), small size and weight (high-frequency transformer), wide input range.

Disadvantages: complex circuit, switching noise/EMI, output ripple, needs careful design.

Uses: computers, TV, mobile chargers, LED drivers, telecom equipment.

Boost regulator: average output voltage

        L    iL            Dm
 +Vs --uuuu---+----------->|---+-----+
              |                |     |
              Q (switch)       C   LOAD  Va
              |                |     |
 −  ----------+----------------+-----+

Mode 1 (Q ON, 0 < t < t1 = kT): L is across Vs; iL rises linearly from I1 to I2; Dm off; C feeds the load.

Vs = L·ΔI/t1   →  ΔI = Vs·t1/L

Mode 2 (Q OFF, t1 < t < T): L's voltage adds to Vs; current flows through Dm to C and load; iL falls from I2 to I1.

Va − Vs = L·ΔI/t2   →  ΔI = (Va − Vs)·t2/L

Waveforms:

Q   |‾‾‾‾‾|___|‾‾‾‾‾|___
    0    kT   T
vL  | Vs  |   | Vs  |
    |     |___|     |___ −(Va − Vs)
iL  |    /\     /\        I2
    |___/  \___/  \__     I1

Derivation (equal rise and fall of iL):

Vs·t1 = (Va − Vs)·t2
Vs·kT = (Va − Vs)(1 − k)T
Vs = Va(1 − k)
Va = Vs/(1 − k)

As 0 < k < 1, Va > Vs. Example: k = 0.6 → Va = 2.5 Vs.

Switch mode power supply (SMPS)

An SMPS converts AC or DC input into a regulated DC output by switching a transistor at high frequency and varying the duty cycle (PWM).

AC -->[RECT + ]-->[HF   ]-->[HF  ]-->[RECT + ]--> DC
      [FILTER ]   [SWITCH]  [TRAN]   [FILTER ]   out
                     ^                   |
                [PWM CONTROL]<-[ERROR AMP]<+
                               (via opto-isolator)

1. Mains is rectified and filtered to unregulated DC.
2. A MOSFET switches this DC at 20 kHz – 1 MHz.
3. A small HF transformer changes the level and isolates the output.
4. Output rectifier and LC filter give smooth DC.
5. The error amplifier and PWM controller adjust the duty cycle to hold the output constant.

Advantages: high efficiency (up to 90 %+), small size and weight. Disadvantages: EMI noise and more complex design.

Given: Vs = 14 V, Va = 6 V, R = 500 Ω, ΔVc = 15 mV, f = 20 kHz, ΔI = 0.7 A.

Buck regulator relations (Rashid):

Va  = k·Vs
L   = Va(Vs − Va)/(f·ΔI·Vs)
C   = ΔI/(8·f·ΔVc)
Lc  = (1 − k)R/(2f)
Cc  = (1 − k)/(16·L·f²)

(a) Duty cycle

k = Va/Vs = 6/14
k = 0.4286 (42.86 %)

(b) Filter inductance L

L = 6 × (14 − 6)/(20×10³ × 0.7 × 14)
  = 48/196000
L = 244.9 μH

(c) Filter capacitance C

C = 0.7/(8 × 20×10³ × 15×10⁻³)
  = 0.7/2400
C = 291.7 μF

(d) Critical values of L and C

Lc = (1 − 0.4286) × 500/(2 × 20×10³)
   = 285.71/40000
Lc = 7.143 mH

Cc = (1 − 0.4286)/(16 × 244.9×10⁻⁶ × (20×10³)²)
   = 0.5714/1567347
Cc = 0.3646 μF

Note: L = 244.9 μH is less than Lc = 7.143 mH, so with R = 500 Ω (Ia = 12 mA) the inductor current would actually be discontinuous; C = 291.7 μF is much larger than Cc, so the capacitor voltage is continuous.

Answer: k = 0.4286, L = 244.9 μH, C = 291.7 μF, Lc = 7.143 mH, Cc = 0.365 μF

Given: Vs = 15 V, k = 0.4, f = 20 kHz, L = 120 μH, C = 200 μF, Ia = 1.25 A.

Buck-boost regulator relations (Rashid):

Va  = −Vs·k/(1 − k)
ΔVc = Ia·k/(f·C)
ΔI  = Vs·k/(f·L)

(i) Average output voltage

Va = −15 × 0.4/(1 − 0.4) = −6/0.6
Va = −10 V

The output is 10 V with polarity reversed with respect to the input.

(ii) Peak-to-peak output ripple voltage

ΔVc = 1.25 × 0.4/(20×10³ × 200×10⁻⁶)
    = 0.5/4
ΔVc = 0.125 V = 125 mV

(iii) Peak-to-peak ripple current

ΔI = 15 × 0.4/(20×10³ × 120×10⁻⁶)
   = 6/2.4
ΔI = 2.5 A

Extra check: average input current Is = Ia·k/(1 − k) = 1.25 × 0.4/0.6 = 0.833 A, and peak inductor current I2 = (Is + Ia) + ΔI/2 = 2.083 + 1.25 = 3.33 A.

Answer: Va = −10 V, ΔVc = 125 mV, ΔI = 2.5 A