AC to DC circuit design

[adeline]

hi all, im a newbie in circuit design but i had to design a circuit for power supply.

1st of all, i need to design a circuit that allow me to convert 240Vac to 14.2Vdc. i have read on a lot of website which tell me about SMPS, transformer and etc. But i dont really get how to design the circuit at all.

i found a design which more or less fit what i want, but its a 240Vac to 5Vdc circuit, but i dont know what to change to make the output voltage to be 14.2Vdc.
240Vac to 5Vdc

help me pls

 

[ifrythings]

Change the 7805 (regulator) to a 7815 (15V regulator) and make sure you have a transformer that puts out ~20V DC and you'll be good.

BTW this circuit will only do a max of 1A (regulator heat sinked).

If you need more current check out this place.

 

[adeline]

but the 7815 will make my circuit very hot and i dont/cant put a heat sink for it >.<

 

[AndrewT]

Hi,
try an lm317 and set the output to 14.2Vdc if your circuit is that critical om supply voltage.

0.8V extra and your circuit overheats!!!!
What's wrong with your circuit?

240Vac to 14.2Vdc using SMPS involves working with mains side voltages - Don't until you become an expert.

 

[luka]

Hi

You will need heat sink in any kind of regulator, even smps. If you don't need regulation, you can just use trafo + rectifier + cap to get your 14v dc

 

[Ouroboros]

You might find it easier to use a conventional mains transformer + rectifier, then follow it with a simple switching 'buck' regulator, rather than a linear regulator.

Have a look at the LM2595 or similar on the National Semiconductor site.

 

[adeline]

sorry for the late reply >.> my laptop crush on me have to repair it ><

anyway for this design, the input will be 240Vac and the output need to be; output -> Power: 50W; Voltage: 14.2Vdc; Current: 3.5amp.

i cant have heatsink, cant use 78xx series and i need to use a regulator.

im thinking of using a zener diode for the regulator part, but it has been a problem for me to find the resistance needed to stop the current from destroying the diode.

And anyone know the formulas that i need to calculate for this circuit?

 

[jan.didden]

sorry for the late reply >.> my laptop crush on me have to repair it ><

anyway for this design, the input will be 240Vac and the output need to be; output -> Power: 50W; Voltage: 14.2Vdc; Current: 3.5amp.

i cant have heatsink, cant use 78xx series and i need to use a regulator.

im thinking of using a zener diode for the regulator part, but it has been a problem for me to find the resistance needed to stop the current from destroying the diode.

And anyone know the formulas that i need to calculate for this circuit?

For the zener: at full load, the zener should still have, say 10mA, through it. At no load, the zener should be able to absorb the full 3.5Amps. That's a hefty zener: 14.2V, 50W. If you don't want to use a heatsink, you need a fan. Your call.

The series resistor: let's assume your xformer, rectifier and capacitor deliver min 18V. Your resistor needs to absorb 18-14.2=3.8V @ 3.5A ~ 1.1 ohms @ 14W. You see the problems? Shunt regs are notoriously inefficient. You paint yourself in a corner by stating what you don't want before you know the score.

Jan Didden

 

[jaybombalous]

For the zener: at full load, the zener should still have, say 10mA, through it. At no load, the zener should be able to absorb the full 3.5Amps. That's a hefty zener: 14.2V, 50W. If you don't want to use a heatsink, you need a fan. Your call.

The series resistor: let's assume your xformer, rectifier and capacitor deliver min 18V. Your resistor needs to absorb 18-14.2=3.8V @ 3.5A ~ 1.1 ohms @ 14W. You see the problems? Shunt regs are notoriously inefficient. You paint yourself in a corner by stating what you don't want before you know the score.

Jan Didden

Can u even buy a 14 watt resistor? What is the price of one that big? How would u go about getting 1.1 ohms resistance at 14 watts of dissipation in a circuit to begin with?

I understand the math and the fundamentals, now lets talk about real world situations.

 

[bawang]

A simple 1A rated 3 pin regulator and a pass transistor will be able to give you the required 14.2V, 3.5A that you need, assuming that your transformer & rectifiers are sufficient. Heatsinks of sufficient sizes WILL be required for the regulator and pass transistor. Or get a 5A rated regulator..... trim the output voltage by trimming the resistor values of the reference pin....... Look for the LM series regulators at National Semiconductor. What kind of circuit are you driving?

 

[gootee]

Can u even buy a 14 watt resistor? What is the price of one that big? How would u go about getting 1.1 ohms resistance at 14 watts of dissipation in a circuit to begin with?

I understand the math and the fundamentals, now lets talk about real world situations.

You might want to go to mouser.com, digikey.com, and alliedelec.com and order their free catalogs. You need to know what's available, in the real world, and those will be very good references; much better than searching a website, especially when you don't know what 's available. [But even Radio Shack stores have 15-Watt power resistors (or used to).]

Perusing my copy of the latest 1838-page Mouser catalog, I see resistors rated up to at least 1000 Watts.

For 14 Watts, you could use anything rated at some margin above that, depending on how hot you wanted to allow it to possibly get. Of course, if you required it to be non-inductive, the choices would be somewhat more limited.

The common Xicon rectangular-cylinder 15-Watt 5% cement wirewound power resistors are $0.63 (qty 1), for 1 Ohm to 1K Ohms, and $.57 each for qty 10, in the Feb-April 2007 Mouser catalog. The similar 25-Watt ones are less than twice that price, at $1.09 qty 1 and $0.99 qty 10. And the table of standard values for those resistors does include 1.1 Ohms.

The 15W cement power resistors are 49mm x 12.5mm x 11.5mm and the 25W ones are 64mm x 14.5mm x 13.5mm. They also come in 5-Watt and 10-Watt versions, which are smaller and cheaper. All have axial leads. But there are similar 5W, 7W, and 10W versions which have radial leads, for vertical mounting.

But if you want a nice, non-inductive Caddock 15-Watt 1% power film resistor, in a TO-126 case that requires a heatsink, then that would cost $2.78 (qty 1), for 10 Ohms to 1K Ohms, and more for values below 10 Ohms. For a similar 30-Watt Caddock in a TO-220 case, the price would be $3.55 (qty 1).

And there are many other types and sizes of resistors that are commonly available. I also see potentiometers of up to 300 Watts.

You could also make a higher-power resistor out of almost any type of resistor, just by finding some way to get rid of the heat. A relatively-low-power resistor could be capable of dissipating hundreds of watts or more (at least for a while), if it were, for example, placed in the middle of a one-gallon paint can full of oil.

- Tom Gootee

http://www.fullnet.com/~tomg/index.html

-

 

[adeline]

A simple 1A rated 3 pin regulator and a pass transistor will be able to give you the required 14.2V, 3.5A that you need, assuming that your transformer & rectifiers are sufficient. Heatsinks of sufficient sizes WILL be required for the regulator and pass transistor. Or get a 5A rated regulator..... trim the output voltage by trimming the resistor values of the reference pin....... Look for the LM series regulators at National Semiconductor. What kind of circuit are you driving?

im in a project to make a 2.1 speaker, so im in charge of the power supply part..

now to check that my transformer and rectifier are sufficient?

and if i use a 5a rated regulator, how to find out if the resistor value of the reference pin is enough to trim the output voltage?

 

[AndrewT]

Hi Ade,
why do you need a regulator?
Have you considered a transformer/rectifier/smoothing cap PSU?
It works well for all power amplifiers.

 

[bawang]

Hi Adeline, search for LM338 or LM338K (steel cap, TO3 version) of this 5A adjustable regulator. As for the transformer, a 15V, 100VA unit will be more than enough. A cheap bridge rectifier rated at 8A will again be much more than enough, main capacitors at 4700uF, 25V sufficient (all rough estimates). The datasheet on the LM338 will tell you more, including voltage trimming.

 

[Ouroboros]

Why on earth use a linear regulator? It's so easy to replace the linear with a simple switching 'buck' stage which will increase the efficiency a lot, and still keeps the mains transformer / rectifier / capacitor to ensure the safety isolation.

 

[theChris]

IMHO

find the classic regulator circuit. either the 2 transistor simple version or the multi transistor version.

there are two classic circuits that were the main designs of power supplies before switching regulators became popular. both use some type of pass transistor, some form of gain, and some constant voltage reference.

you can even use a lm317 for the gain and voltage reference, with the pass transistor used for power.

imagine attaching the output of a lm317 to the base (via some resistance) of a large, TO3 NPN transistor. the emitter connects to the output cap.

this places the junction within the feedback loop, where the gain of the lm317 will serve to reduce this.

of course the input at low line must exceed the base-emitter voltage plus the dropout voltage of the lm317 at full load with full ripple on the main input cap.

efficiency can then be found by concidering full draw at high line, which will be lower.

 

[paulb]

Why on earth use a linear regulator? It's so easy to replace the linear with a simple switching 'buck' stage which will increase the efficiency a lot, and still keeps the mains transformer / rectifier / capacitor to ensure the safety isolation.

Because linear regulators are much simpler and easier to get working. Also because switching regulators can be very noisy.

 

[gootee]

Because linear regulators are much simpler and easier to get working. Also because switching regulators can be very noisy.

I like linear regulators, very much. I often use them as post-regulators for my switchmode power supplies. 🙂

I suddenly feel an urge to "put in a good word", for the poor, downtrodden SMPS, the apparent "underdog" in the world of audiophile power supplies:

You are right. Some switching regulators can be noisy. But some, if built around ICs like National's LM2611 Cuk-topology switcher, for example, only have about 1mv of ripple. And that's at 1.4 MHz, far above the audio range and easy to filter out, if it even needs to be. That particular IC might not be a good example, in this particular thread, since it can only deliver up to 1.2A. But most of the newer switchmode controller ICs use relatively high frequencies, even compared to those of just five years ago, making their noise and ripple that much easier to filter out than, say, 120 Hz ripple, and often so far above audio frequencies that it wouldn't matter, anyway. The higher frequencies also mean that their inductors can now be quite small, and very cheap.

And if a switcher happens to be the best choice for a particular application's power supply, the power rail noise can essentially be made "as small as desired", by the use of simple LC output filtering followed by a linear post-regulator; even down into the very low tens of microvolts.

As far as linear regulators being simpler and easier to get working: "barely", any more. And NOT in some ways. For one thing, you can go to National.com, or download LTspice (aka "SwitcherCad") from linear.com, and have a switchmode power supply designed FOR you, totally automagically. And THEN you can go to the manufacturer's website, to the switcher IC's product page, and order FREE samples of the IC, PLUS buy a ready-made professionally-tailored "evaluation board" PCB for $25 to $30 (at least in the case of National), to build your power supply on. It doesn't usually get MUCH easier than THAT (or much better, probably, in terms of the quality of what you can end up with).

Also, many newer switcher ICs now claim to require only five external components; about the same as a linear regulator.

And some switchers are now SO EFFICIENT that the _ONLY_ "heatsink" they need is the copper on the PCB! And IIRC, one of the OP's stubborn requirements was/is "NO HEATSINK". (If that really is a "requirement", for some reason, then a linear regulator is already out of the race, anyway.)

And, it certainly appears to be the case that the highest-current unmodified "stock" switcher designs can deliver much higher output currents than the highest-current unmodified "stock" linear regulator designs. So, for output currents above what the LM338 can do, switchers also appear to be "simpler and easier to get working". (And then there's that pesky lack of a negative counterpart for the LM338. What's up with THAT??)

Also, switchers don't get NEARLY as hot, when the difference between the input and output voltages is relatively large. A switcher can give the power BACK, instead of losing it as heat, by giving a larger output current than input current, when the output voltage is lower than the input voltage.

Switchers can also do things that linear regulators just can't do, ever, such as create negative outputs from positive inputs with both referenced to the same ground, and create larger voltages from smaller ones.

OK, putting aside all of that almost-advertising-like "biased" SMPS praise, here's my official "disclaimer": So far, at least, I just don't see any one single type of power supply topology that should always be used, no matter what the application is. So we still always need to try to come up with the "best" power supply design for each individual application; the best "fit", given all of the requirements and constraints. So it's probably a good idea to keep an open mind about ALL of the candidates, so we can successfully implement whatever we might need at the time, having chosen the optimal overall design from the broadest possible range of existing topologies, using the "best" out of _all_ of the available devices, for the application at hand. Sometimes, nothing could be a better choice than a linear regulator. And sometimes the only good choice will be some type of SMPS.

Oh, I almost forgot: Switchers "sound better", too! ;-)

[Lest I start a flame war, I have to say that I was just kidding about switchers "sounding better" (as far as you know). 🙂 ]

Cheers!

- Tom Gootee

http://www.fullnet.com/~tomg/index.html

-

 

[JasonLee]

Aiyah...
Power supply for 2.1 speaker dont need to be regulated power supply lah !

Just get a 230v > 10v-0-10v transformer & 2 diodes & a 2000uf capacitor can liao !
Another idea is to get a 15v laptop power supply from 3rd level SLT. But, make sure of the polarity while it connect to the 2.1 sp.

But, if u realy want to design a 14.2v DC Regulated power supply. Then also go to 3rd level of SLT Kai Chin & get a pc of 2N3055 which is capable to handle at least 7 Amp. With a Zener diode. You can actually make a proper 14.2 V Regulated Power Supply. For project, yes. Commercial ? I doult so !