You are asked to design a full-wave rectifier DC supply that can supply 5 V with a DC current of 47 mA, similar to the circuit shown below. The DC supply will use a step-down transformer connected to an Australian wall outlet (50 Hz). If the maximum ripple amplitude of the supply is specified at 0.83 V, what is the minimum value of the capacitor C1 (in mF) that is required ? Specify your answer to two decimal places.
Question
You are asked to design a full-wave rectifier DC supply that can supply 5 V with a DC current of 47 mA, similar to the circuit shown below. The DC supply will use a step-down transformer connected to an Australian wall outlet (50 Hz). If the maximum ripple amplitude of the supply is specified at 0.83 V, what is the minimum value of the capacitor C1 (in mF) that is required ? Specify your answer to two decimal places.
Solution
The ripple voltage (Vr) in a full-wave rectifier is given by the formula:
Vr = I / (f * C)
where:
- I is the load current,
- f is the frequency, and
- C is the capacitance.
We are asked to find the value of C when Vr is 0.83V, I is 47mA, and f is 50Hz.
Rearranging the formula to solve for C gives:
C = I / (f * Vr)
Substituting the given values into the formula gives:
C = 47mA / (50Hz * 0.83V)
Converting mA to A and V to mV to match the units gives:
C = 0.047A / (50Hz * 830mV)
Calculating this gives:
C = 0.047 / (50 * 0.83) = 0.00113 F
Converting F to mF (since 1F = 1000mF) gives:
C = 0.00113 * 1000 = 1.13 mF
So, the minimum value of the capacitor C1 that is required is 1.13 mF (to two decimal places).
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