Step-down chopper with RL load

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.