Great discussion I learned a lot. Before going to sleep on the bed, I would like to tell some points

@Irving you have put the screw terminal in the output of the supply which i don't want. I want both the power jack and the screw terminal to be used on the input side. I want to use header pins for output side.

 

@Irving
can i try to connect the components on the board as you shown in your circuit diagram? Is it safe?
I will connect heatsink as you advised

Of course its safe... as long as you get things the right way round!

Do you know how to calculate the resistor for the LED?

Don't connect anything to the output until you've checked the output voltage. Do you have a multimeter?

What heatsink do you have? Can you post a picture?

 

Great discussion I learned a lot. Before going to sleep on the bed, I would like to tell some points

@Irving you have put the screw terminal in the output of the supply which i don't want. I want both the power jack and the screw terminal to be used on the input side. I want to use header pins for output side.

Thats fine, get a copy of KiCAD and redraw it

 

Hello,

You can get Kicad over here:
https://www.kicad.org/
Max collected some tutorials:
Useful Links for learning KiCad

Bertus

 

Do you know how to calculate the resistor for the LED?

Basic 5mm LED with a red lens. It has a typical forward voltage of 2.0V and a rated forward current of 20mA.

12-volt power source and a red LED with the desired current of 20 mA;

The formula becomes Resistor = (12-2)/0.02 = 500 ohms

 

Basic 5mm LED with a red lens. It has a typical forward voltage of 2.0V and a rated forward current of 20mA.

12-volt power source and a red LED with the desired current of 20 mA;

The formula becomes Resistor = (12-2)/0.02 = 500 ohms

Good, but don't forget if you're powering from 12v AC it's going to be nearer 15vDC (unless you reduce it to 9v AC input), so you might want to go for 10mA @ 12v DC, it'll be plenty bright enough.

 

@Irving I have a doubt, I have voltage regulator L7805CV that rated 1.5A.

Is this possible to use IN4007 diode's to make full wave Rectifier bridge for L7805CV Or should I use different diode's to make full wave Rectifier bridge.

 

Hello,

The 1N4007 is rated for an avarage current of 1 Amp.
The 1N5408 is rated for an avarage current of 3 Amp.

Bertus

 

@Irving I have a doubt, I have voltage regulator L7805CV that rated 1.5A.

Is this possible to use IN4007 diode's to make full wave Rectifier bridge for L7805CV Or should I use different diode's to make full wave Rectifier bridge.

As @bertus said, the 1N4007 are rated 1A, but to be specific that's 1A average current @ 75degC ambient. so as long as you keep 1A as your peak output and keep the leads long (as they are the heat-sink) they'll suffice for now. Also, as you identified, the L7805CV is "up to 1.5A", however that's under very specific conditions including proper heat-sinking and limited input voltage, typically 2 - 2.5v above the output. It's not a recommended operating condition for the device - look at the data-sheet, you won't find 1.5A referred to in the recommended operating conditions other than it's mentioned in the context of a reduced output regulation quality factor.

If you never draw the full 1A continuously it makes little difference - you stated you didn't want to buy new parts and this was an experimental prototype. So my advice is stick with what you have, recognise its not going to do 1A continuously & limit yourself to 500mA - 750mA average current draw, which is more than enough for the majority of PIC or Arduino projects. You can overthink these things to the point you never do anything...

 

Thats fine, get a copy of KiCAD and redraw it

@Irving

I use proteus to make circuit diagram. yes I said I don't want to buy new component's but considering the situations. I've just changed only diodes in my circuit diagram. I would try to get 1N5408 from local shop

 

I use proteus to make circuit diagram. yes I said I don't want to buy new component's but considering the situations. I've just changed only diodes in my circuit diagram. I would try to get 1N5408 from local shop

Fair enough, it's your call. Though 5401s would do, there's no need to pay a premium for 1000v rated when 100v will do.

Incidentally, when drawing electrolytic or polarised capacitors it's useful to use the alternative polarised symbol as in my schematic to show polarity, or put a + sign on the appropriate lead.

 

Hello,

What is going to feed the powersupply?
Keep in mind that, when using AC input, the voltage on the capacitor will be about 1.414 times the AC voltage.

Bertus

 

Hello,
What is going to feed the powersupply?

I have two power sources

1. 12v 1 A SMPS
2. 5v 1A Mobile charger

@Irving , @bertus

Have you designed your own power supply for your microcontroller or do you use a ready-made power supply from the market?

What are the ratings of your power supply?

 

Basic 5mm LED with a red lens. It has a typical forward voltage of 2.0V and a rated forward current of 20mA.

12-volt power source and a red LED with the desired current of 20 mA;

The formula becomes Resistor = (12-2)/0.02 = 500 ohms

Are you sure the 20mA is the operating current of the LED? I think it’s the absolute maximum current. LEDs usually operate at less than the current specified in the datasheet. I’d calculate the resistor based on 10mA. It still will be bright. Then, I’d use a 1K resistor.
(12-2)/0.01 = 1KΩ

 

hi Push,
I am puzzled why you are requesting a design for a 5V 1A power supply, when you you already have a 5V 1A, also 12V 1A PSU.??

E

 

I have two power sources

1. 12v 1 A SMPS
2. 5v 1A Mobile charger

@Irving , @bertus

Have you designed your own power supply for your microcontroller or do you use a ready-made power supply from the market?

What are the ratings of your power supply?

Be very careful with mobile phone chargers. They are not always a good option.

In general, with PSUs being plentiful and cheap, its not usually worth it for standard outputs such as 5 or 12v. Even for special requirements its not hard to find something, or add an additional buck converter or linear regulator on the output. For example, I needed +/-15v @ 3A plus 10v @ 100mA for a project. The main PSU was two off-the-shelf bricks from eBay - I couldn't even buy the parts that cheap, and a LM317 adjustable regulator set for 10v with three resistors.

 

If you want a cheap bench supply for experimenting, find an old dead PC and liberate the power supply. That will give you +5, +12 and possibly -5 and -12 depending on age. A switch between the PS_ON (normally green) connection and ground (black) will let you turn the outputs on and off cleanly.

 

hi Push,
I am puzzled why you are requesting a design for a 5V 1A power supply, when you you already have a 5V 1A, also 12V 1A PSU.??

E

@ericgibbs It is very important for an embedded engineer to have good knowledge of hardware and software. As student, I think this is the right time to improve skills. I wish I also want to make my own development board. There are many development boards available in the market. Where the power circuit is also inbuilt along with the microcontroller circuit. If I buy a board from the market, I might not learn much about its hardware. But if I try to make it by myself then I can learn a lot.

Some people's opinion may differ from me, they suggest to buy board and practice on it by writing coding. I think it might just improve their coding skills but I don't think they have that much good knowledge about hardware.

if I find any useful components in any old equipment, then I take it out. And I use it where I need. My benefit from this is that if I make a mistake, I only get to learn something from my mistake. No more money is wasted to purchase new components. I buy components only when it is more necessary.

My purpose for creating this thread was to build a power supply for my board. I shown my best attempt to build my a power supply
And I have got guidance and suggestions in a very good way. I am grateful to all of you. Specially I would like to thank @Irving , he spent a lot of time to teach me basics in the thread

still i have some doubts i am tring to figure out them if i don't understand then i will definitely come back

Good Night

 

My purpose for creating this thread was to build a power supply for my board. I shown my best attempt to build my a power supply

hi P1,
Your answer has covered the question I asked.
Wanting to design your own PSU is a worthwhile exercise, most engineers go thru the 'build a power supply' stage.

My advice would be to build the PSU as a modular project ie: build the input first stage, test and debug and then the next stage.

E

 

I want to make universal power supply that should take input between 12V- 5V DC. If I make a 12V supply, I can also take 5 volts with same circuit.

The block diagram in post #1 will *not* work if the 12 V input is reduced to 5 V. The 7805 regulator has a minimum voltage requirement for its input-to-output differential voltage, sometimes called the dropout voltage; you can look up the value on the datasheet. Also, there is a voltage drop across the diode bridge.

Each diode has a forward voltage (Vf) of approx 0.7 V, and two diodes are conducting at any given time, so that is a 1.4 V minimum voltage drop between the input to the circuit and the input to the 7805 regulator.

Finally, there is the filter capacitor. With an AC source voltage, such as the secondary of a small power transformer with a 12 Vac output, there will be an AC component to the votlage across the bulk filter capacitor, called the ripple voltage. As the supply's output current increases, so will the magnitude of the AC ripple voltage across the bulk filter capacitor. With a 1000 uF capacitor, the peak-to-peak ripple voltage at 1 A output current will be approx. 8 V. That is a lot. Here are some numbers ...

Input = 12 Vac = 17 Vpeak
Minus two diode drops = 17 - 1.4 = 15.6 Vp
Minus the ripple voltage at 1 A = 15.6 - 8 = 7.6 V (negative peak)
Minus the 7805 typical dropout voltage of 2 V = 7.6 - 2 = 5.6 V

So the design will work with a 12 Vac input, but just barely. With a 12 Vdc input, there is no ripple and you don't need the diode bridge.

Note, the ripple voltage is based on a linear approximation that is way easier to do in my head. Your results will vary.

ak