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Step-down chopper with DC motor load

Short answer

A step-down chopper controls a DC motor's speed by varying the duty cycle k to change average armature voltage V0 = kVs. In ON state, source supplies armature; in OFF state, current freewheels through diode. Steady-state speed ω = (kVs − IaRa)/(Kφ).

Explain the operation of a step down chopper with dc motor as load.

A step-down (Type-A) chopper controls the speed of a separately excited dc motor by varying the average armature voltage V0=kVsV_0 = kV_s. The motor armature is an RLE load: resistance RaR_a, inductance LaL_a and back EMF EbE_b.

Circuit

  +----CH----+-------------+
  |          |             |
  |          |            Ra
 Vs        FD ^            La
  |          |            (Eb)  motor
  |          |             |
  +----------+-------------+

Modes of operation

Mode 1: CH ON (0<t<Ton0 < t < T_{on}). Source connects to the armature, v0=Vsv_0 = V_s:

Vs=Raia+Ladiadt+EbV_s = R_a i_a + L_a\frac{di_a}{dt} + E_b

The current rises exponentially from IminI_{min} to ImaxI_{max}; energy is stored in LaL_a.

Mode 2: CH OFF (Ton<t<TT_{on} < t < T). The armature current freewheels through FD, v0=0v_0 = 0:

0=Raia+Ladiadt+Eb0 = R_a i_a + L_a\frac{di_a}{dt} + E_b

The current decays from ImaxI_{max} to IminI_{min}. If LaL_a is small or kk is low, it can reach zero before the next ON period (discontinuous conduction); then v0=Ebv_0 = E_b for the remaining time.

 v0  Vs |____      ____
        |    |    |    |
      0 |____|____|____|____ t
 ia Imax|   /\        /\
        |  /  \      /  \
    Imin| /    \____/    \__ t
         Ton Toff

Steady-state relations (continuous conduction)

V0=kVs=Eb+IaRaEb=Kϕ ωω=kVs−IaRaKϕ\begin{aligned} V_0 &= kV_s = E_b + I_a R_a \\ E_b &= K\phi\,\omega \\ \omega &= \frac{kV_s - I_a R_a}{K\phi} \end{aligned}

So speed is controlled by the duty ratio kk: increasing kk increases V0V_0 and the speed. Torque T=KϕIaT = K\phi I_a depends on the load current.

Current limits:

Imax=VsRa⋅1−e−kT/Ta1−e−T/Ta−EbRa,Imin=VsRa⋅ekT/Ta−1eT/Ta−1−EbRaI_{max} = \frac{V_s}{R_a}\cdot\frac{1-e^{-kT/T_a}}{1-e^{-T/T_a}} - \frac{E_b}{R_a}, \quad I_{min} = \frac{V_s}{R_a}\cdot\frac{e^{kT/T_a}-1}{e^{T/T_a}-1} - \frac{E_b}{R_a}

with Ta=La/RaT_a = L_a/R_a. Conduction is continuous if Imin>0I_{min} > 0.

Features

  • Operates in the first quadrant only (motoring, forward); braking needs a Class B or C chopper.
  • High chopping frequency or an added series inductor reduces current ripple and torque pulsation.
  • Smooth, efficient (no rheostat loss) control; widely used in battery vehicles, forklifts and traction.

✓ Answer checked by Claude Opus 5.5 (high effort)

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