- 2074 Asoj · 2+5 marks
Chopper definition; single-phase half bridge inverter
Short answer
A chopper is a static DC-DC converter using duty cycle control to vary average output voltage. A half-bridge inverter uses two switches and a centre-tapped DC supply to generate a square-wave output of ±Vs/2; feedback diodes handle reactive load current.
Define chopper circuit. Explain the working principle of single phase half bridge inverter, with the necessary circuits and waveforms.
Chopper
A chopper is a static switch that converts a fixed DC voltage into a variable DC voltage (DC-to-DC converter). It connects and disconnects the load from the source periodically; the average output is controlled by the duty cycle k = Ton/T, e.g. Vo = k·Vs for a step-down chopper.
Single-phase half-bridge inverter
A half-bridge inverter uses two switches (S1, S2 with antiparallel diodes D1, D2) and a centre-tapped DC supply (two equal sources Vs/2 or two equal capacitors). The load is connected between the switch midpoint A and the supply centre point O.
+ +-----------+
Vs/2 | S1 D1
− | |
O---[LOAD]--A
+ | |
Vs/2 | S2 D2
− +-----------+
Working:
- 0 < t < T/2: S1 ON, S2 OFF. The upper source drives current through S1 → A → load → O, so vAO = +Vs/2.
- T/2 < t < T: S2 ON, S1 OFF. The lower source drives current O → load → A → S2, so vAO = −Vs/2.
- S1 and S2 must never conduct together (short circuit of the supply); a small dead time is used.
- With RL load, current lags the voltage. After S1 turns off, the load current continues through D2 (feeding energy back to the lower source) until it reaches zero, then S2 conducts. Similarly D1 conducts after S2 turns off. Hence D1 and D2 are called feedback diodes.
Waveforms:
S1 gate |‾‾‾‾‾|_____|‾‾‾‾‾|____
S2 gate |_____|‾‾‾‾‾|_____|‾‾‾‾
vo +Vs/2+-----+ +-----+
| | | |
0 ---+-----+-----+-----+---> t
−Vs/2 +-----+ +--
0 T/2 T
io (RL) /‾‾‾\ /‾‾‾\
(exponential rise and fall, lags vo)
D1 S1 D2 S2 (conducting device)
Output relations:
- Square wave of amplitude Vs/2; RMS value Vo = Vs/2.
- vo = Σ (2Vs/nπ) sin nωt, n = 1, 3, 5, …
- Fundamental RMS = 2Vs/(√2·π) = 0.45·Vs.
- Frequency f = 1/T set by the gating signals.
It needs only two switches but gives half the output of a full bridge and requires a centre-tapped supply.
✓ Answer checked by Claude Opus 5.5 (high effort)
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