Designing a 11V,550mA DC power Supply:

[TipleA]

How Can I design a power supply circuit with:

Input voltage: 100-240Vac
Output voltage: 11Vdc , 550mA
Using Full-wave Rectifier Filter and (Zener Diode circuit) Regulator in the design.
Dear Friends, What should be the Ratings of the Components ?

 

[BradtheRad]

If you are only permitted to use a full-wave rectifier, and a zener diode...

You are dropping a lot of volts, so there needs to be a robust volt-dropping resistor.

Each diode in the full-wave rectifier needs to withstand the peaks of 240 VAC. This is 340V.

When the supply is minimum (100 VAC), peaks will be 140V.

Although the zener diode may be rated 11 V for low current levels, it may also carry high current which may elevate its operating voltage higher than that.

The load calculates to be 20 ohms (11/.55).

A safety resistor is customary above the zener diode. Whatever its value, it must drop as much as 329 V while carrying more than 550 mA. Its rating will end up needing to be on the order of a few hundred watts (even though it will not be carrying that entire amount every moment).

You want the zener diode to carry some amount of current at all expected supply voltages. Maybe 25 mA minimum (5% of the load current) when the supply is 100 VAC (peaks at 140VDC). It will need to be rated 305 mW minimum.

When the supply is the maximum expected (240 VAC), a whole new round of calculations is needed.

 

[TipleA]

If you are only permitted to use a full-wave rectifier, and a zener diode...

You are dropping a lot of volts, so there needs to be a robust volt-dropping resistor.

Each diode in the full-wave rectifier needs to withstand the peaks of 240 VAC. This is 340V.

When the supply is minimum (100 VAC), peaks will be 140V.

Although the zener diode may be rated 11 V for low current levels, it may also carry high current which may elevate its operating voltage higher than that.

The load calculates to be 20 ohms (11/.55).

A safety resistor is customary above the zener diode. Whatever its value, it must drop as much as 329 V while carrying more than 550 mA. Its rating will end up needing to be on the order of a few hundred watts (even though it will not be carrying that entire amount every moment).

You want the zener diode to carry some amount of current at all expected supply voltages. Maybe 25 mA minimum (5% of the load current) when the supply is 100 VAC (peaks at 140VDC). It will need to be rated 305 mW minimum.

When the supply is the maximum expected (240 VAC), a whole new round of calculations is needed.

Thanks For Answer, But I was in haste on that day I assumed Input supply only 240V and Modeled the Required Supply in Proteus, I am going to attach a Doc File that show some of my calculations , The only purpose is That as You are Very senior and man of Knowledge , You may Comment on my model and Add your valueable Advice!!

 

[BradtheRad]

I examined your report on the design of your power supply. You used the correct systematic approach.

I believe it will work as you intend.

Here are my observations.

1.

Regarding the transformer, are you also required to calculate its watt rating? Ampere rating?

2.

When you calculate the value of the smoothing capacitor, it would be appropriate to write out the figures you inserted in the equation, to obtain the answer 10 mF.

I agree you can use a smaller value than 10,000 uF. You prefer to use 220 uF. It would be appropriate to state what amount of ripple this gives you, in the simulation at least.

3.

You calculated the safety resistor to have a value of 2 or 3 ohms. This appears suitable for operation in most circumstances.

It will also be appropriate to calculate the amount of Watts this resistor will need to withstand.
Also to state what you recommend its watt rating should be.

4.

The same applies to the zener diode. What amount of watts must it endure? What power rating should be chosen for the zener?

5.

Notice that the lower ohm value of the safety resistor, the more current goes through it, and also through the zener diode. As a result both components will waste more power as heat. What if you were to reduce the amount of current going through them? They would not waste as much power.

How far can you carry this, so there will be a minimum of power wasted?

6.

In real life, the house voltage can vary by as much as 5%. Your power supply will probably work as you designed for normal house voltage. However if it were to drop 5%, how would this affect operation? Would the supply voltage still be regulated?

If it were to rise 5%, would any component be exposed to overmuch current?

 

[TipleA]

Yup You are very right ,I did make that mistake of Assuming Quite a high Ripple value, that is clearly against the soul of a Regulated Power Supply,
But only Theoratically ,
Paractically thanks to My Proteus work, My Simulation Ran quite good and I got The Best Results and Yes the lowest Power losses.

 

[BradtheRad]

To be able to construct a power supply to specs, will always be a useful skill.

One of the most useful, and most rewarding, projects I ever made was a dual adjustable DC supply. And much less expensive than a storebought one.