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 0.68 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 0.68 V, what is the minimum value of the capacitor C1 (in mF) that is required ? Specify your answer to two decimal places.
Solution
To calculate the minimum value of the capacitor C1, we can use the formula for the ripple voltage of a full-wave rectifier, which is given by:
Vr = I / (f*C)
where: Vr is the ripple voltage, I is the load current, f is the frequency of the AC supply, and C is the capacitance.
We can rearrange this formula to solve for C:
C = I / (f*Vr)
Substituting the given values:
C = 47 mA / (50 Hz * 0.68 V)
Note that we need to convert mA to A and V to mV to match the units:
C = 0.047 A / (50 Hz * 680 mV)
C = 0.047 / (50 * 0.68) F
C = 0.00138235294 F
Converting Farads to milliFarads (1 F = 1000 mF):
C = 1.38235294 mF
So, the minimum value of the capacitor C1 required is approximately 1.38 mF (rounded to two decimal places).
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