Rectifier Diode Circuit Capacitor

Diodes + full wave rectifier + capacitor ?

it is a full wave rectifier with capacitor filter. (ideal operation of the circuit)

the a.c. input voltage = 230V at 50 Hz.
the load d.c. current is 1.5A.
the power supply is to supply a d.c. voltage of 25V.
the percentage output ripple voltage is to be less than 1%.

determine:
(a) the transformer turns ratio
b) value of the capacitor

Using a bridge rectifier there are two diode voltage drops to consider, and the voltage used is close to the peak voltage using a capacitor input filter.

The transformer voltage under load.
As 25V DC is required with 1% ripple, allow 0.5% extra = 25.125V, so the average is 25V. The ripple is assumed as reasonably symmetrical here, though it isn't really..
Allow additional voltage for the 2 series diodes in the bridge rectifier, = 1.4V
Allow additional voltage for voltage drops in the diodes and wires due to current of 1.5A = 0V for ideal operation.
The total drops are 1.4V so add to the supply voltage, = 26.525V peak.

Assuming a sine wave supply, the RMS value is 26.525V x 0.707 = 18.75V. I will leave the turns ratio for 230V input to you, 230V/18.75V.

The ripple voltage...
The voltage for 1% ripple is 0.25V peak to peak.
The formula is:
Vpp = I_Amps / (2 x freq x C_Farads)
Transposing..
C = I / 2fV
= 1.5A / 2 x 0.25Vpp x 50Hz
= 0.06F
= 60,000 uF
In practice the ripple will tend to be greater because of the equivalent series resistance (ESR) of the capacitor.

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help needed on full-wave rectifier circuit?

http://img169.imageshack.us/img169/6875/lastscan1qk9.jpg

Thew figure shows a full-wave rectifier circuit, with a capacitor filter. The secondary voltage is 100 Vrms, 50 Hz, with C = 100 μF and RL = 10 kΩ. The forward voltage drop across the
diode, VD = 0.7V.
(i) What is the ripple frequency of VL? Calculate the peak-to-peak ripple voltage, Vr(pp) .
(ii) Calculate the dc values for load voltage VL and load current IL.
(iii) Work out the percentage ripple factor at the load. State any assumptions made.

Your help is greatly appreciated. Tks a lot!!!

Secondary voltage is 100vCT, I assume.

Ripple freq is 100Hz, 2x input freq.
Load voltage and ripple voltage is dependent on load current.
I'll assume a load current of 50mA.

Ripple amplitude is calculate by measuring how much the cap discharges in the 10ms period when the diodes are both off. This is only accurate for small values of ripple, otherwise you should use the exponential equation.

CV=IT
V = IT/C = 0.05*10m/100μ = 5 volts P-P
Quite high. If it were my design I'd increase the C.
You realize this is rough calculation.

DC voltage is about 50 * 1.4 -0.7-2.5 (half the ripple)
about 67 volts.
Percent ripple is 5/67 = 7%