Half Wave Rectifier as a Battery Charger PDF

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Half Wave Rectifier...

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Half-Wave Rectifier as a Battery Charger A half-wave rectifier circuit is the simplest possible circuit that can be used as a battery charger. One such circuit is given in Figure 1, where R and VB are the internal resistance and the voltage at the time of charging of the battery.

Figure 1: A half-wave batter-charger circuit If the single-phase supply voltage is given as vs ( t ) = Vm sin(ωt ) V, the diode begins to conduct only when it is forward biased. This condition happens at ωt = α such that α = sin −1 ( VB / Vm ) The diode will continue to conduct until ωt = β , where β = π − α . In other words, the

diode conducts for α ≤ ωt ≤ π − α and the diode current is

i( t ) =

v s ( t) − VB R

Example: _____________________________________________________________ A single-phase 240-V, 50-Hz source is used to charge a 240-V battery. The battery voltage at the time of charging is 180 V. If the internal resistance of the battery is 0.5 Ω, determine the current supplied by the source. Calculate the charge supplied to the battery during each cycle. Sketch the supply voltage, output voltage, and the charging current. Solution The source voltage is given as

v s ( t ) = 240 2 sin(ωt ) = 339.411sin(ωt ) V where

ω = 2 πf = 100 π = 314.16 rad/s

Guru/PEUCR/HWBC/ April 4, 2006

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A half-wave battery charger

The diode begins to conduct when α = sin −1 (180 / 339.411) = 32.03o (0.559 rad ) The diode will continue its conduction until β = π − α = 147.97 o

( 2.583 rad)

The output voltage and the input (supply) voltage waveforms are shown in Figure 2.

Figure 2: The output and the input voltage

Figure 3: Charging current in the battery

The current supplied by the supply voltage, for α ≤ ωt ≤ β , is i( t ) =

339.411sin(ω t ) − 180 = 678.822 sin( ωt ) − 360 A 0 .5

Note that the maximum current in the diode is 318.822 A.

Guru/PEUCR/HWBC/ April 4, 2006

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A half-wave battery charger

The charge delivered to the battery during each cycle is β

Q = ∫ (678.822 sin(θ) − 360 ) dθ α

= 678.822[cos(α ) − cos(β )] − 360(β − α ) = 422.33 C During the initial charging cycles when the battery voltage is about 180 V, the charge transferred to the battery during each cycle is 422.33 C. As the battery is being charged, its voltage increases. As voltage increases so does α , but β decreases. Therefore, with each charging cycle the conduction period (β − α) of the diode decreases. The battery will continue to charge until the battery voltage becomes equal to the maximum voltage of the input cycle because this circuit does not protect the battery from overcharging. There are better circuits that not only limit the maximum current in the diode but also provide a safeguard for the battery from overcharging. The average and rms currents in the battery are computed as β

I DC =

1 (678.822 sin( θ) − 360 )d θ = 67.242 A 2π ∫α β

I RMS =

1 (678.822 sin(θ) − 360 )2 dθ = 130.473 A 2 π ∫α

The dc power supplied to the battery is

P B = 180 × 67.242 = 12.104 kW The power dissipated by the internal resistance of the battery is P R = (130.473) × 0 .5 = 8.512 kW 2

The apparent power supplied by the source is S INPUT = VS( RMS) I S( RMS) = 240 × 130.473 = 31 .314 kVA Thus, the power factor, ratio of the total power output to the apparent power input, is

pf =

PR + PB 12.104 + 8 .512 = = 0 .658 SINPUT 31.314

The peak-inverse-voltage across the diode is PIV = 180 + 339.411 = 519.411 V

Guru/PEUCR/HWBC/ April 4, 2006

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A half-wave battery charger...