Electronic Circuits and Projects Forum

You have not specified the output voltage that u require.
U need a step down transformers,two capacitors(one across the rectifier and the other across the IC that u will use for the desired output voltage).U will also need a bridge rectifier at the output of the transformer.The capacitors will do the required filtering.
Just try it out.
Byeeeee,
Peeyush

Hi, thanks for the reply. so i will need a capacitor at the output of the rectifier to filter off the ripples. any ideal wat value should i use? should i need a resistor to form a RC LPF. the input to the full bridge rectifier will be a 48V 50Hz signal. Any protection circuit should i use to protect my regulator IC ?

cheers!
skyy

Originally Posted by skyymanz

Hi, thanks for the reply. so i will need a capacitor at the output of the rectifier to filter off the ripples. any ideal wat value should i use? should i need a resistor to form a RC LPF. the input to the full bridge rectifier will be a 48V 50Hz signal. Any protection circuit should i use to protect my regulator IC ?

cheers!
skyy

No a resistor will just cause wasted power and voltage drops.

Your bulk capacitance can be chosen sufficiently large to reduce your ripple down to sufficient levels. If it is higher frequency noise you are concerned about, snubbers on the rectifier diodes can help. But you havnt said what all you care about for your design.

Also, most newer regulator IC's have built in protection circuits for overvoltage. Look at its schematic from the datasheet.
Otherwise, you can use a zener diode to protect from occasional high line voltage conditions that might over voltage the input.

If you are seriously concerned about protection, it is best to just make sure you get that built into the regulator. The vendors know how to protect their own devices best.

Go for National LM 738 .Its a godd high current rating Regulator ic(5A).

Hi,

i am concerned for a good regulated dc signal. i have source for the regulator and its quite ok. but wat value of capacitor should i use to minimised the ripple from the output of the amplifier? a value of 1000uF good enough??

cheers!
SKyy

With 48V ac input the bridge rectifier output with capacitor give - without load - almost 70V dc. Regulator chips with HV suffix accept only 60V input.
Still need the required regulated output voltage and current for proper answer.

Originally Posted by skyymanz

Hi,

i am concerned for a good regulated dc signal. i have source for the regulator and its quite ok. but wat value of capacitor should i use to minimised the ripple from the output of the amplifier? a value of 1000uF good enough??

cheers!
SKyy

For a FULL WAVE rectification scheme
A first order approximation to what you should use is:

Vripple (p-p) = (Vpeak) / [ 2 * F * Cmin * Req ]

Where,

Vripple is peak to peak ripple voltage.
Vpeak is peak voltage of the sinewave (not RMS!)
F is frequency of operation i.e. 50Hz
Cmin is minimum value of capacitor for the given ripple voltage.
Req is equivalent resistance (load) seen by capacitor. Iload etc.. Estimate maximums.

The equation doesn't account for diode drops but it won't affect your answer too much. In this equation set your desired ripple voltage and solve for the Cmin. Then add 10% margin to that value then pick the nearest standard value. As you can see, higher load current => higher ripple voltage. Lower Capacitance => higher ripple. Lower frequency of operation => higher ripple.