i am traying to design 12vdc regulated supply but i am not surge what rating of component transformer,diode and capactior
if input volatage 230vac, o/p12vdc and load 20ohms

 

I'd have to say....start reading.

SMPS

 

I'd start with a transformer rated for about 14VAC - 16VAC on the secondary winding with a rating of 1A RMS or higher. Bridge rectifier rated at least 60V and 2A (higher is better). Filter cap should be 25V rated at about 5,000 - 10,000 uF.

 

12V into 20 ohms is only 0.6A. Then use a 7812 1A regulator IC.

 

thanks for responce

but how can i chose rating of component
230v-transformer+diode+capacitor-12v
transformer- power rating
max. voltage/current of transformer ?
min. voltage/current of transformer ?
diode-power rating
-max./min voltage rating of diode ?
-max/min current rating of diode ?
capacitor-power rating
max/min voltage of capacitor ?
max/min capacitance of capacitor ?
power supply-efficiency ?
yet no body has define clearly and exact plz guide and explain with suitable example

 

Post #3 answers many of your questions.
Common sense answers the remainder of your questions.

 

yet no body has define clearly and exact plz guide and explain with suitable example

Find the datasheet for the 7812 integrated circuit. It's a 3-pin component that will regulate the voltage of your supply to 12v. Very simple and handy. That datasheet will show several application examples that will help you, in addition to what's been suggested already.

Working backwards, the 12v regulator will require a supply of ~14v DC (unregulated and rippled) minimum, in order to maintain 12V at its output. About 1.4V are lost in the full wave bridge rectifier, raising your target to over 15V AC before the rectifier. But a rectifier and capacitor filter arrangement behaves like a peak detector, and will give you a higher DC voltage than the AC voltage RMS specification of the transformer. A transformer rated to supply 14v AC under load will provide, once rectified and filtered, maybe 16V DC or more when not loaded. But the voltage will drop quickly under load, so if you need sustained current, don't go lower.

 

A 10VAC transformer has a loaded peak output root-2 times higher at 14.14V. A full-wave bridge rectifier drops about 1.8V and ripple not smoothed by a pretty big filter capacitor drops about 0.3V. So the loaded output voltage will be about 12.04VDC. Without a load and with a small load current then the output voltage will be higher.

 

I'd start with a transformer rated for about 14VAC - 16VAC on the secondary winding with a rating of 1A RMS or higher. Bridge rectifier rated at least 60V and 2A (higher is better). Filter cap should be 25V rated at about 5,000 - 10,000 uF.

sir why you choose 14vac -16ac , bridge rectifier- 60v, capacitor-25v
give specific reason

 

sir my purpose is not only build this circuit but also i want to learn concept
can you explain me with example thanks

It is not a simple derivation. There are rules of thumb used:

FWB rectifier circuit, transformer secondary current rating should be at least 1.8X DC output current.

Transformer secondary voltage rating is more complicated: peak voltage is not actually 1.4X voltage rating (loaded) because FWB conducts current in very narrow pulses that flattens off the voltage peaks making it non sinusoidal, reducing the DC voltage some.

 

sir why you choose 14vac -16ac , bridge rectifier- 60v, capacitor-25v
give specific reason

Experience. Hard to summarize.

 

This looks rather like a student trying to get a model answer to an assignment along the lines of :

"Design a power supply to meet the following requirements, explaining how the component values are determined..."

Or am I just being too suspicious?

 

There are two main ways to make a regulated voltage source, Switch Mode Power Supplies (SMPS) or Linear Voltage regulated.

If you have an AC source and want a AC output SMPSs are better as the transformer physical sizing is reduced by (normally) increasing the freq. This allows it to fit within a plug socket, think of an Apple iPhone charger.

If you want to reduce a DC voltage, using a potential diver circuit loses a lot of power in the resistors, hence you use a Darlington Pair of (normally) NPN bipolar transistors.

The circuitry is nowadays combined into a single chip LT14xxx etc. If you look at this data sheet you will see there are many example circuits in the application notes. You can look up the output resistance of the circuit, but you can assume the full output voltage will be applied to the circuit and thus the current flow is based on the downstream circuit.

http://docs-europe.electrocomponents.com/webdocs/078f/0900766b8078f50d.pdf

If you want to add an auto current limiter to the circuit, simply add a resistor in series that will generate 0.7v at what ever current you want to limit the circuit at. Use that 0.7v to reduce the Iadj to 0A using a NPN bipolar.

Hope this helps.

Ste