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Question

If T is the time period for a chopper circuit and α is its duty cycle, then the chopping frequency is

The correct answer is

α/T on

Chopper Circuit Fundamentals

A chopper circuit, also known as a DC-to-DC converter, is an electronic device that converts fixed DC input voltage to a variable DC output voltage directly. It works by rapidly switching the DC supply ON and OFF, creating pulses. The frequency at which this switching occurs is called the chopping frequency, and the ratio of the ON time to the total time period is called the duty cycle.

Let's define the key terms related to a chopper circuit operation:

  • Time Period (T): This is the total time for one complete cycle of the chopper's operation. It includes both the time the chopper is ON and the time it is OFF.
  • On-Time (\(T_{on}\)): This is the duration for which the chopper is conducting (turned ON).
  • Off-Time (\(T_{off}\)): This is the duration for which the chopper is non-conducting (turned OFF).
  • Duty Cycle (\(\alpha\)): The duty cycle represents the fraction of the total time period during which the chopper is ON. It is a crucial parameter that determines the average output voltage of the chopper.
  • Chopping Frequency (\(f\)): This is the number of complete cycles (ON and OFF periods) that occur per second. It is the reciprocal of the total time period \(T\).

Understanding the Duty Cycle and Frequency Relationship

The total time period \(T\) for one complete cycle of the chopper circuit is the sum of the on-time and off-time:

\[ T = T_{on} + T_{off} \]

The duty cycle \(\alpha\) is fundamentally defined as the ratio of the on-time to the total time period:

\[ \alpha = \frac{T_{on}}{T} \]

From this definition of duty cycle, we can rearrange the formula to express the total time period \(T\) in terms of the duty cycle \(\alpha\) and the on-time \(T_{on}\). Multiplying both sides by \(T\) and dividing by \(\alpha\), we get:

\[ T = \frac{T_{on}}{\alpha} \]

The chopping frequency \(f\) is defined as the reciprocal of the total time period \(T\). This means it tells us how many cycles occur in one second:

\[ f = \frac{1}{T} \]

Now, to find the chopping frequency in terms of the given parameters \(\alpha\) and \(T_{on}\), we can substitute the expression for \(T\) from the duty cycle definition into the frequency formula:

\[ f = \frac{1}{\left(\frac{T_{on}}{\alpha}\right)} \]

When you divide by a fraction, it's equivalent to multiplying by its reciprocal. Therefore:

\[ f = \frac{\alpha}{T_{on}} \]

Final Chopping Frequency Formula

This derived formula clearly shows the relationship between the chopping frequency, the duty cycle, and the on-time. It means that if you know the fraction of time the chopper is ON (duty cycle) and the actual duration it is ON, you can calculate how many times it switches per second.

Comparing this derived formula with the given options, we find that:

  • Option 1: \(T_{on}/\alpha\) is the total time period \(T\), not frequency.
  • Option 2: \(T_{off}/\alpha\) is not a standard formula for frequency.
  • Option 3: \(\alpha/T_{off}\) is not the correct formula.
  • Option 4: \(\alpha/T_{on}\) matches our derived formula for chopping frequency.

Therefore, the chopping frequency is \(\alpha/T_{on}\).

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Important Questions from Choppers

  1. A step-up chopper is supplied through a source of 200 V and operated at a duty cycle of 50%. Find the average output voltage of the chopper.

  2. Choppers are _______ converters.

  3. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R 

    Assertion A: Chopper converts fixed d.c. input voltage to a controllable d.c. output voltage. The Chopper circuit require forced or load commutation to turn off the thyristors. 

    Reason R: The Chopper circuits are dependent upon the type of commutation and also on the direction of power flow. 

    In the light of the above statements, choose the most appropriate answer from the options given below

  4. A load commutated chopper fed d.c. drive uses 100 V d.c. supply. The maximum chopper frequency is 5 kHz. The value of maximum load current is 100 A. The commutating capacitance will be :

  5. In a lossless buck converter, an average power of 20 W to a load with a regulated 12 V output, while operating at a duty cycle of 0.8 with continuous inductor current. Following data is given :

    (a) The average value of input voltage is given by 15 V
    (b) The average value of input voltage is 16 V
    (c) The average value of input current is 4/3 A
    (d) The average value of input current is 3/4 A

    Which is correct out of the following ?

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