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Boost regulator relation; SMPS block diagram

Short answer

In boost regulator, during on-time inductor voltage Vs causes current rise, off-time voltage Va−Vs causes fall; equating Vs·kT = (Va−Vs)(1−k)T gives Va = Vs/(1−k). An SMPS uses rectified AC, high-frequency switching, a HF transformer, and output rectifier with PWM feedback to maintain constant output voltage efficiently.

Derive the relation of average output voltage with input voltage in boost regulator with necessary circuit diagram and wave forms. Briefly explain about switch mode power supply with block diagram.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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