Choppers transfer power from one DC circuit to another by converting a fixed voltage to a variable voltage. They have various applications including speed control of DC motors, positional control in robotics, regulated power supplies, inputs to inverters, and solar power conversion. A step-down chopper, or buck converter, uses a controlled switch and freewheeling diode to reduce the output voltage below the input voltage. The duty cycle of the controlled switch determines the output voltage. Ripple in the inductor current is minimized by using a higher switching frequency and larger inductor.

1. CHOPPERS
 Transfer power from DC circuit to another DC circuit.
 Converts fixed voltage to variable voltage.

2. Application of Chopper
1.Speed control of DC Motor.
2.Positional control in robotics.
3.Regulated power supply (SMPS)
4.Input to Inverter.
5.Solar Power Converter.

3. Step Down Chopper
CH: - fully controlled switch
Operation: - Periodic
Time period: - T
Duty cycle: - Ton/T
Ton = DT
CH: - ON Vo=Vs
CH: - OFF Io=0/Vo=0
t
t
t

4. Buck Converter
S1 :- Fully Controlled Switch
-MOSFET(high frequency)/IGBT/SCR+Commutation Circuit
S2:- Uncontrolled Switch
:- provides free wheeling action (due to L)
DC Motor load
-Step down chopper with a freewheeling
diode.

5. Buck Converter
SW:-ON (0<t<DT)
Vo = Vs
Vs = 𝑖𝑅 + 𝐿
𝑑𝑖
𝑑𝑡
+ 𝐸
L :- charging
SW :- OFF (DT<t<T)
0 = 𝑖𝑅 + 𝐿
𝑑𝑖
𝑑𝑡
+ 𝐸
L:- discharging

6. 𝐼𝑚𝑎𝑥 =
𝑉𝑠
𝑅
1−𝑒
−𝑇𝑜𝑛
𝜏
1−𝑒
−𝑇
𝜏
−
𝐸
𝑅
𝛕 = time constant = L/R
𝐼𝑚𝑖𝑛 =
𝑉𝑠
𝑅
𝑒
𝑇𝑜𝑛
𝜏 −1
𝑒
𝑇
𝜏−1
−
𝐸
𝑅
T = time period
(Valid when 𝛕 ≤ T)
∆I = ripple in inductor current = Imax – Imin
Ripple must be as low as possible
Maximum ripple in IL occurs if D = 0.5 (Buck Converter)
∆I = Vs/4fL (assume 𝛕>>T)To reduce the ripple, f and L must be
high∆𝐼 =
𝑉𝑠
𝑅
1−𝑒
−𝑇𝑜𝑛
𝜏
1−𝑒
−𝑇
𝜏
−
𝑒
𝑇𝑜𝑛
𝜏 −1
𝑒
𝑇
𝜏−1
=
𝑉𝑠
𝑅
(1−𝑒
−𝑇𝑜𝑛
𝜏 )
(1−𝑒
−𝑇
𝜏 )
−
𝑒
−𝑇
𝜏 (𝑒
𝑇𝑜𝑛
𝜏 −1)
(1−𝑒
−𝑇
𝜏 )
=
𝑉𝑠
𝑅
1−𝑒
−𝑇𝑜𝑛
𝜏
1−𝑒
−𝑇
𝜏
(1 − 𝑒
−𝑇
𝜏 ∗ 𝑒
𝑇𝑜𝑛
𝜏 )
=
𝑉𝑠
𝑅
1−𝑒
−𝑇𝑜𝑛
𝜏
1−𝑒
−𝑇
𝜏
(1 − 𝑒
−𝑇𝑜𝑓𝑓
𝜏 )

7. Discontinuous Conduction
• Current can become 0 for finite duration of time.
• Boundary :- When current IL=0 only for an instant
of time.
• For boundary :- Imin =0
•
𝑉𝑠
𝑅
𝑒
𝑇𝑜𝑛
𝜏 −1
𝑒
𝑇
𝜏 −1
= E/R
• 𝑒
𝑇𝑜𝑛
𝜏 − 1 =
𝐸
𝑉𝑠
𝑒
𝑇
𝜏 − 1
• 𝑇𝑜𝑛 = 𝜏𝑙𝑛 1 +
𝐸
𝑉𝑠
𝑒
𝑇
𝜏 − 1