Design a battery charger for safe operation in a damp garage environment to use

Question

Design a battery charger for safe operation in a damp garage environment  to use for charging your 12 VDC car battery.

Design specifications include:
(a) Input is a 110rms VAC. (VP=110 x sqrt(2)) at 60 Hz from a three wire service that meets the National Electrical Code.
(b) Output is a nominal 12 volts VDC at the cathode of the diode.
(c) Specify a resistor, R, to limit the maximum battery charging current to 10 amperes into the 12 volt car battery.
(d) There is no ripple voltage design specification. Explain why this is unnecessary in this application.
(e) The battery charger case is metal.
(f) Assume a diode with VF = 0.7 V
(g) The fuse in the primary circuit is to protect the power supply from a short-circuit at the battery terminals either from total battery failure or accidentally short circuiting the charging cable to ground. (For example, dropping a wrench across the battery terminals).
(h) A voltage regulator is not required

Your Design must include:

Well-labeled circuit diagram including the identification of the incoming “hot”, “neutral”, and “ground wires (U.S. standards) including the National Electric Code color coding of these wires and also show the correct color-coded wiring for a standard grounded duplex receptacle and plug.

i. Key design equations and supporting calculations
ii. Component specifications including:
       · Transformer-turns ratio
       · Diode-current and power ratings
       · Your assessment and short discussion as to whether the laboratory 1N4001 diodes could be used
       · Value for R1 serving as a current limiting resistor
       · Current rating of a fuse in the primary circuit to protect the power supply against a short circuit condition at the battery terminal.

Explanation / Answer

Let turn ratio between primary and secondary is 1:1.

Input voltage to secondary is Vi=1002 sin(wt)

Diode will conduct only when Vi is greater than (12 +Vf)v, and current will start following in the secondary circuit.

i= current in the circuit= (vi-0.7-12)/R

imax= maximum current = (1002-0.7-12)/R

Given imax=10 A

Then. R=(1002-12.7)/10=12.87 ohm

When vin< 12.7 v , then diode is open circuit , so current through resistor is zero.

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