Hi there,
i am student of electronics engineering and want to build a power supply, as far as voltages are concerned i know how to calculate output voltages but i'm still wondering how do we calculate output current of a power supply when we're going to built it. lets say if i want to build 15 volt and 5 amp power supply i can deal with 15 volts but how do i calculate those 5 amperes? any suggestions?

 

Welcome to AAC!

A thread belongs to the OP (original poster). Trying to take over someone elses thread is called hijacking, which is not allowed at All About Circuits. I have therefore given you a thread of your very own.

 

Ohm's law I presume?

If you have 15 volts supplied, then the circuit will try to take as much current as it needs to have 15 volts.

You can design a current limiter, but you cannot control both voltage and current.

When you see a power supply rated a 220V @5A, (for example), you are just reading the maximum safe value for the components. You might get 220V @5.2A, but you're pushing the power supply, and it could get dangerous.

 

First thing you need, is a stepdown transformer capable of supplying approx 3 amps more than your projected needs......all your circuitry downstream needs to "eat" too
The excess current only assures that your end - use product will have enough current available for whatever you are feeding with it to draw...........

A fuse or small circuit breaker, and a switch on the power cord hot line. Nobody is immune to snafu shortcircuits, smoke and sparks........

Next, a full-wave bridge rectifier...with at least twice your needed voltage as its PIV spec. 25 amp...........Radioshack sells a nice one. If you expect large current draw, park it on something like a computer CPU heatsink, with a little pancake fan for insurance.

Regarding V-reg - I have had good luck running 6 LM317's in parallel for added amperage, though a beefy heatsink is paramount if you go that route. I have gone with the series-pass transistor method, tho' those TO-3 t'sistors can get expensive, and I'm a cheapskate
Consumer level electronic strobe units for example...draw an initial pulse in excess of 8 amps off their 4 AA pack.....Studio master / slave units are something else again.

Ripple filtration circuit schemes are plentiful, depending on how clean you need the DC output to be.

It's a good idea to lean toward overkill on your component specs. unlike some "foreign" built equipment that uses bare minimum, and fails all to easily, usually taking your project with it.

 

Yeah, the basic error here is that a power supply provides a voltage at UP TO as much as 5 amps. The load decides how much current it needs.

 

Actually i wanted to ask how do i know that my supply is of 5 amps that was a question and thanks all for you support.

 

Design the power supply with a pencil before you buy parts.
Buy a transformer that is large enough to provide 5 amps.
Buy rectifiers that are large enough to provide 5 amps.
Buy output transistors that are large enough to provide 5 amps.
Install capacitors that are large enough to supply a 5 amp regulator.
etc.

 

ok, if its the mains, it'll be the preverbial 13 amp ring main supply. it's worked out for you at the otherend, and the bits in meter box on the wall, the rest is via the fuse, 3 amp will blow at 5 amp for sure etc..

regaurding battery supplies, the 'potential volt/amp' is in the amp capacity, i.e. 1500mah 40amp or 100amp leisure battery etc etc.

there's a part two, the 'release' a battery is capabal of, a car battery is the easiest to exsplain, 800 amp crank, or realease, or start up capacity, is how much it can release momentarily, like a peak voltage/amapage, if you continue to start the car endlesly/continualy drawing at the cark rateing something will pop.. usualy the battery in abscence of a weeker component/wire down the line.

your components have a wattage limit, continuous or peak, the peak will have a time value to it, exseed it and something goes pop, or boom inthe case of a battery exsploiding.
(dont forget ac or sine wave etc stuff is impedence not resistance...the change of volt/current warms up more than a static flat line current, again in plain english AC is a different wattage ability to DC).

hense a fuse is a 'safety' compoinent, usualy rated to with in the tolerences of the device/curcuit. if the fuse wont blow then the next compinent in line is usualy for the chop, along with this n that through out a curcuit or all the bits or ic's etc as well if you really boobed up.

a resister of a 1/4 wat on dc will start smoking if you slowly turn up the supply, crank it up to max and it'll simply burn out and posably a couple othercompoinents along with it.

voltage drop current and volts... ohmes law applied to watts.

wire, as you'll know, a teeny thin bit of three strands will make a luverly heater coil for ooo 3 or 4 seconds conected to a car battery and simply aloud to pass the electrons, put the right size resister in there the wire doesnt burn out, but.. what watt wil the resister need to be to disapate the heat of the voltage drop accross it and the current flowing. stick in a 5 or 10 amp wire the flow is more acomading and allows it 'easier' for want of simplicity without going thermal meltdown.

amps is a figurative 'thing' ish, they say only the current the bulb takes etc, hmm i've had some strange tomes watching stuff pop, the ohmes law was correct, the figures checked, but it seemed a the more current 'available' it can sometmes mean more compoinents or a higher wattage 'current limit' is required.. just a one off here and there though.

regaurding how do you know what 'amp' is acuauly 'there', if your battery only has 50mah in it, your not going to blow any fuses under narmal useage.. thats 50mah per hour capacity, 100 mah for 30 second, 200 mah for 15 seconds.. 400mah for 7.5 seconds etc etc, tis posable to create more than is there figurativly speeking but wear goggles lol..BANG! lol. so the potential stinger at 13 amps in the mains can be very hi on a momentary pulse but something would blow in a narmal fused curcuit, hopefully not involving a person or the curcuit board.

if your question was how can you 'predict it' in a scientistical fasion, ahhh well... some of the greatest nminds played about with that one to come up with things like ohms law, Kirchy babey and his mates had there say too lol. i think even good old Einstein had a few points to make.


get your watts calcs rong, there you'll have your fuse...fast blow diodes for example, thin wire melts and starts a fire..or you'll touch it to see if that transister is actualy smoking..owch! tsssssssss! branded lol.(it peels off eventualy lol).

 

Actually i wanted to ask how do i know that my supply is of 5 amps that was a question and thanks all for you support.

You know your power supply can provide 5A if you build it with parts that can do so.

So an LM317 can only handle 1.5A. So if you want to be able to output 5A, you would need an external pass transistor to ensure you can supply 5A.

It sounds like to me that you don't understand Ohms law. It's really quite simple.

Voltage is also known as "potential difference". It might make more sense if you use "potential difference". So think of it as a diving board or something. The higher you go up the faster you go when you jump off it. Same thing with voltage. The larger the potential difference you set (between voltage and ground), the more charge that flows (current) through a resistance.

I= V/R

The current in a circuit is dictated by the voltage of the power supply and the resistance of the load.

 

You get a 3Ω 100W resistor and connect it across the 15V power supply.
If the voltage still reads 15V then you have a 15V@5A supply.

 

i'm really sorry but i'm yet to understand a single thing

 

Actually i wanted to ask how do i know that my supply is of 5 amps that was a question and thanks all for you support.

You design it so that it is capable of 5 amps.

If you buy a transformer that can deliver 5 amps after the rectification process, use rectifiers that can deliver 5 amps without smoking, use capacitors that can filter the DC to provide 5 amps, and use a regulator chip that can supply 5 amps, you will get up to as much as 5 amps when you need 5 amps.

Look at the top of the "Chat" page and find, "Ohm's Law for noobies". Read it and report back here whether it was helpful, too simple to be bothered with, or whatever you think of it. That wil give us an idea of what educational level you are at.

 

i understand ohm law resistance times current is voltage. but actually i m kind week in practicals i dont get what to do. as you said buy capacitors that can filter dc to 5 amp now how do i come to know that which value of capacitor is gonna filter 5 amps?

 

√2 C Er F = I

You start by choosing a transformer that will provide more than 15 volts after rectification. Suppose you choose a transformer rated at 15 V RMS. The peak voltage will be 1.414 X the RMS value. The rectifiers will use up a bit of voltage and the regulator will need some headroom in the voltage department.

15 X 1.414 = 21.21 volts peak
Subtract about 1.5 volts for the loss of two rectifiers (if you use a full wave bridge configuration).
Subtract about 2 volts that the regulator chip will use as headroom
Subtract the 15 volts you want as the output.
You have 2.71 volts leftover.
5 amps/ √2 (2.71) 120 Hz = 10,859 uf

Edit: These numbers might be a bit off because they depend on the rectifiers you purchase and assume that the power line voltage available is proper and steady at your house. This is the basic procedure but, professional designers also allow for power line voltages from 105 to 125 RMS. That being a bit more complicated, I left it out. You might want to use a high frequency switching power supply. That being a lot more complicated, I left it out.

 

This chart shows that a full wave (4 diode) bridge circuit will provide .62 times the current rating of the transformer. To get 5 useful amps, you must buy a transformer rated for 5/.62 = 8.07 amps

 

This chart shows that a full wave (4 diode) bridge circuit will provide .62 times the current rating of the transformer. To get 5 useful amps, you must buy a transformer rated for 5/.62 = 8.07 amps

thank you very much. i really appriciate all the members that they'are always willing to help.
i have one more question what is the difference b/w KVA and KW some devices are rated as KW and some KVA like transformers and generators why?

 

It's hard to know exactly which part you don't know, and how to ask for it.

anyway..leaving off the killo part, a volt amp is a watt if the voltage and current are in phase. As soon as anything gets inductive or capacitive, the phase shift between applied voltage and the resulting current starts to increase. This makes the actual power applied to the load become less than what you're giving it in volt-amps.

In the practical world that I live in, amp meters read amps and volt meters read volts. As long as my volt amps do not exceede the transformer, I'm not going to smoke the transformer. The load might use 2 amps at 24 volts (RMS) and that's 48 VA, but it only gets the use of 40 watts because of a phase lag in its inductance. That doesn't bother me. My transformer is safe.

Labeling transformers in VA helps idiot-proof them...not talking about the ingenuity of some idiots.