Series inverter output frequency with commutation dead time In a series inverter, let 0.5T be the natural half-oscillation time of the load commutation (each conduction interval), and let T_off be the dead time between turning off one thyristor and turning on the next. What is the output frequency f in terms of T and T_off?

Introduction / Context:Series inverters rely on LC commutation to produce quasi-sinusoidal current bursts through the load. Between successive half-cycles, a dead time T_off is inserted to allow device recovery and safe commutation. This dead time stretches the electrical period and lowers the output frequency compared to the ideal LC period.

Given Data / Assumptions:

• Each conduction interval spans a natural half-oscillation = 0.5T.
• After each half-cycle, there is a dead time T_off before the opposite device conducts.
• The output period consists of two half-cycles plus two dead times.

Concept / Approach:

One full cycle requires positive half-cycle conduction, a dead time, negative half-cycle conduction, and another dead time. Summing durations yields the total period; the reciprocal gives frequency.

Step-by-Step Solution:

Verification / Alternative check:

In the ideal limit T_off → 0, f → 1/T, which matches the LC natural frequency period. Increasing T_off reduces f, as expected.

Why Other Options Are Wrong:

• 1/(T + T_off) and 1/(0.5T + T_off) omit one of the dead times.
• 1/(2T + 2T_off) double counts the oscillation period.

Common Pitfalls:

Confusing the natural LC period with the practical output period that includes mandatory recovery times.

Final Answer:

f = 1 / (T + 2T_off)