# Switch-mode power supply, or maybe linear?

Discussion in 'Power Electronics' started by saripitu, Jul 12, 2011.

1. ### saripitu

26
0
Jul 12, 2011
Hi,

I am trying to design a buck (step down) dc dc converter from 12 V to deliver 9 V and 0.6 A.
I have calculated values of 2.7uH for the inductor and 100nF for the Caparcitor with a transistor of 4MHz freq. The diode has a 0.7V drop. But im not sure if these will deliver the power I need. Maybe some1 with access to a circuit simulator could help me??? or some1 that has built one of these before?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
It would be interesting to know how you've calculated those values, perhaps it's for a particular chip, or a particular circuit, but they're not general principles.

Have you considered a linear regulator? If you don't have experience with this sort of stuff, a linear regulator is probably easier. Added to that, it will not be significantly less efficient than a switchmode regulator for the voltages you're looking at.

There are some three terminal 9 volt regulators that can do what you require very simply.

3. ### saripitu

26
0
Jul 12, 2011
Thanks for the recomendation, I will try the voltage regulator!
I calculated the values using the buck equations, and putting the values I had in. These equations give you the minimal values of the components to keep the system in a continuous mode right? But I don't know how much it affects the output using larger values...
Ive looked everywhere online and for a good price, the max output current is 0.5 A and the circuit I need to power up has is its specs 0.6 A. Its a router, so I dont know if it will still be transmitting perfectly if I apply o.5A instead of 0.6.

Thanks again!

4. ### MagicMatt

70
0
Jun 15, 2011
The chances are your router will not function correctly. If it requires 0.6A and your supply can only provide 0.5A, then you're going to see a voltage drop that may affect things either instantly or long term.

You need a 9V DC 0.6A power supply - go buy and 9V power supply that has a regulated output of 0.6A or more, cost would be about \$10 max. It'll be cheaper than trying to make something... if you raid eBay you might even get one for \$1.

5. ### saripitu

26
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Jul 12, 2011
I have seen a few that gives me that power from mains but the idea is to be able to connencted to my boat power supply, 12 V or 24V and I cannot find anything from these... :S it would be great if I can make it for less than 5 pounds...

6. ### MagicMatt

70
0
Jun 15, 2011
Ahh, so your input voltage isn't 12V anyway! Yes it's rated 12V, but I know most boats can actually be anywhere from 11.5V to 14.8V, so a linear regulator is the best route.

Just run the whole thing through one of these. 23p each, but I'm afraid you'll have to buy 10.
http://uk.rs-online.com/web/p/linear-regulator/7140666/

You may want some sort of heat sink on there. You can buy one, or as I did, lop a chunk off an old PC CPU heat sink, as that was free (you can even get one out a broken PC at your local tip if you ask nicely).

7. ### saripitu

26
0
Jul 12, 2011
Ok! I will try the voltage regulator! Thanks for the advice!

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
Well, the issue is, if you change the frequency, the required inductance changes (and 4MHz is a really high frequency). The high frequenct will cause all sorts of problems for the device switching the power, the recovery characteristics of the diode, and the filter capacitor.

Please show us where you "plugged in" these values and I'll try to show you a more sensible set of values.

However I think you'll be fine with the linear regulator. Contrary to advice given, a wide range of input voltages is actually a reason to use a switchmode regulator over a linear regulator. However the power is low, and the additional complexity of a switchmode converter is hard to justify.

There are small commercial switchmode power supplies that are made, here are some examples. You'll find the same sort of thing on eBay.

9. ### saripitu

26
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Jul 12, 2011
http://www.daycounter.com/Calculators/Switching-Converter-Calculator.phtml

This is the link I used to confirm the values for the components that I had calculated (I normally mess up the equations).

I have been told that the voltage regulators are old designed devices and they are not exactly the most stable component, thus the use of capacitors around it to control it. So if i can avoid it it would be great.

Cant open the link youve given me! But the ones ive seen in ebay and similar do not deliver the right current, either too much or too little...

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
The link still works for me <shrug>. On that page, you should perhaps target a frequency of around 100khz.

Too much current is not going to be a problem. That's what they *can* deliver, not what they *will*.

Getting a supply capable of about 50% more current than you need is a good idea as it means it won't be stressed when operating.

11. ### saripitu

26
0
Jul 12, 2011
Thanks for the info!! then, if I pick a converter that delivers up to 1A to power up the router (spec 9V 0.6A) is it going to damage it? if so, how do I make sure it doesnt go higher than 0.6A?

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
Think about how your water is supplied. If it's from a dam, what prevents the whole dam from emptying when you turn on the tap?

The answer is that your tap doesn't need the entire dam full of water, so only some normal amount comes out.

Voltage is supplied, current is demanded.

Your power supply provides the voltage and can supply UP TO a certain current.

The device attached for a given voltage will demand a certain current. As long as the amount demanded is less than the capacity of the power supply to supply all is well.

13. ### saripitu

26
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Jul 12, 2011
OK! I understand now! thanks!

14. ### climatex

37
0
Jul 14, 2011
From 12V to 9V with 0.6A?

Grab three silicon rectifiers, e.g. 1N4007, and connect them in series with your 12v supply. It's the simplest way to degrade the output voltage.

15. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
Perhaps you should read what the source of the "12V" is, and consider the change in voltage drop with load. Putting both together will illustrate that this is a very poor solution in most cases.

16. ### saripitu

26
0
Jul 12, 2011
Also, i would like to add reverse polarity protection to the system. I have been told that the easiest way would be a diode and a fuse in series both in parallel to the system. But Im not sure if this would work???

..............................................................................
............................diode.......fuse.............................
................¦------------¦<¦---------[¦-----¦]---------¦..................
................¦..........................................¦..................
................¦.........______________........¦...................
12V---------------¦.......................¦--------------------9V....
..........................¦.......................¦............................
..........................¦.....DC/DC........¦...........................
......-----------------¦_____________¦-------------------.........
...............................................................................

(Sorry for the abstract image but you get the idea right?) :S

Or would it be a better idea to put the fuse and the diode in parallel with the load???

Thanksss

Last edited: Jul 15, 2011
17. ### climatex

37
0
Jul 14, 2011
Just put a standard rectifier diode in series in your circuit at the very beginning, with the voltage and current rating above your circuit needs. It won't conduct at reversed polarity. And no fuses blown either

18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
No that's wrong.

You place the fuse in series with the 12V coming in, and then place a diode across the 12V, after the fuse, and reverse biased. If the power is connected backward, the diode conducts and the fuse blows.

This often leads to the death of the diode too, unless you use a very large diode. You could also put a small resistor (say 0.47R) in series with the fuse. That would drop less than 0.25 volts at the normal operating current, but would limit the current to around 24A before the fuse blows. If your diode can handle single pulses of this size, it will survive.

Note that your fuse may also have enough resistance to perform this function.

There is another thread around here at the moment showing how such a reverse polarity diode copes without protection. Let's say that it performs it's task admirably -- once.

19. ### saripitu

26
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Jul 12, 2011
Sorry steve!! I must be the most annoying member!! i dont quite understand... if the fuse is in series with the 12V input, wont it blow up as soon as some power comes in??

I belive is something like this???

12Vin--------------[|-----|]-------------->>>......
.............................................|.................
............................................---................
.............................................^.....diode....
............................................---................
.............................................|................
---------------------------------------------------------------

Also, what rating would the fuse have??

Last edited: Jul 15, 2011
20. ### climatex

37
0
Jul 14, 2011
Why keep things simple when we're able to complicate them further.