3.3V to >8V

Papabravo

Joined Feb 24, 2006
21,228
You can use a switching regulator in the boost configuration. There are many chips from Linear Technology, National Semiconductor and others that will help you in this quest.
 

SgtWookie

Joined Jul 17, 2007
22,230
It would help us to help you, if you will tell us what your 3.3v power supply is, how much current is available, and how much current that you need at your output voltage.

Keep in mind that to get 8v at a given current from 3.3v, it will require at least 2.43 as much current from the 3.3v supply. Also, the boost conversion will not be 100% efficient; 90% efficient is about the best you can hope for.

So, to get 1A current output at 8V, you would need (8v*1A/90%)/3.3v=2.7A from the 3.3v supply - that is with a really efficient DC-DC converter.
 

kingdano

Joined Apr 14, 2010
377
So, to get 1A current output at 8V, you would need (8v*1A/90%)/3.3v=2.7A from the 3.3v supply - that is with a really efficient DC-DC converter.
in other words, bank on having at least 3A available on the 3.3V supply - have to leave room for error.

if i were designing it - i would aim for 4A - i am a bit of a worry-wart though.
 

SgtWookie

Joined Jul 17, 2007
22,230
Over-capacity is almost always a good thing.

The 90% figure is if a really, really good boost circuit were used. It would be safer to assume more like 75% to 80% efficiency, and then add on a 20% safety margin. That would give you a 3.88A input current requirement; pretty close to your estimate.
 

kingdano

Joined Apr 14, 2010
377
Over-capacity is almost always a good thing.

The 90% figure is if a really, really good boost circuit were used. It would be safer to assume more like 75% to 80% efficiency, and then add on a 20% safety margin. That would give you a 3.88A input current requirement; pretty close to your estimate.
:D

hack math FTW!

i said jeez...whats the closest integer to 2x the calculation?

:D


also for anyone lurking - over-capacity is not good for biasing components (current limit resistors, input biasing of opamps etc) Sgt and I are talking about things like power budgeting, power dissipation ratings etc...CAPACITOR VOLTAGE RATINGS <<<<<<------ thats a good one to know...rule of thumb i was taught was to take your voltage rail and double it for a capacitor if at all humanly possible.


/rant
 

SgtWookie

Joined Jul 17, 2007
22,230
Well, after you've been doing it for awhile, the math gets to be pretty automatic, and you can "ballpark it" without having to set pencil to paper (or finger to calculator...)

But for those who aren't so familiar with such things, it's nice to have some kind of formula to plug numbers into. That way you can literally "do it by the numbers". ;)
 

jamjes

Joined May 10, 2010
35
I always thought capacitors were pretty inert things until I saw how they explode if wired up wrongly :/ So yes, the double voltage thing for capacitors is the least you can do :D Oh, and connect them up right :)
 

Thread Starter

okin

Joined May 9, 2010
12
It would help us to help you, if you will tell us what your 3.3v power supply is, how much current is available, and how much current that you need at your output voltage.

Keep in mind that to get 8v at a given current from 3.3v, it will require at least 2.43 as much current from the 3.3v supply. Also, the boost conversion will not be 100% efficient; 90% efficient is about the best you can hope for.

So, to get 1A current output at 8V, you would need (8v*1A/90%)/3.3v=2.7A from the 3.3v supply - that is with a really efficient DC-DC converter.
I need 40mA on my output and available is 800mA.
 

SgtWookie

Joined Jul 17, 2007
22,230
Okin,
Where are you located? In the USA?

Please put your general location in your user profile.
You can get there by clicking on the "User CP" link near the top of the page, and then "Edit Your Details". There is a spot for "Location" towards the bottom. Click the "Save Changes" button to store.

If you are in the States near a Radio Shack, we can whip something up quickly that you can use.
 

Thread Starter

okin

Joined May 9, 2010
12
Okin,
Where are you located? In the USA?

Please put your general location in your user profile.
You can get there by clicking on the "User CP" link near the top of the page, and then "Edit Your Details". There is a spot for "Location" towards the bottom. Click the "Save Changes" button to store.

If you are in the States near a Radio Shack, we can whip something up quickly that you can use.
Much, much thank you and sorry because I am late with replay message.

Unfortunately I live in Croatia in Europe.
 

Jani-Jan

Joined May 20, 2010
4
i am also new in this great forum .
hi all
if you need step up dc to dc converter then use mc34063
for schematic you can use its data sheet its so simple.
 

SgtWookie

Joined Jul 17, 2007
22,230
I ran through the calculations for a similar device as the MC34063, but the efficiency was terrible due to operating from such a low voltage.

See the attached. You will probably have to substitute other transistors for the devices I used. I really have no idea what parts you can find/get where you are.

You may have to wind the transformer yourself. I suggest a ferrite toroid. You can find them in junked equipment. For example, computer power supplies have toroids in them.
 

Attachments

retched

Joined Dec 5, 2009
5,207
Unfortunately I live in Croatia in Europe.
I guess you dont like the place? Or is it just difficult to get parts?

That could be oppourtunity knocking. Time to open up shop. Become a Croation distributor of electronic components.


But to your need for a converter, his may do the trick for you. It will take 2.6v to 5.5 v input and convert up to 32v.

I would also check other LED booster drivers. This one is overkill, perhaps.. but you should be able to find them that boost to lower voltages.

3.3v is a standars voltage for uC these days, so there are other dc-dc converters for taking these voltages and boosting them for white LED use.

http://pdfserv.maxim-ic.com/en/ds/MAX8595Z-MAX8596Z.pdf
 

SgtWookie

Joined Jul 17, 2007
22,230
OK, wait a minute.

The circuit I posted is a voltage regulator. If you want to power LEDs with it, then it needs to be changed to a current regulator.

That will be easy to do; actually the circuit will be more simple than it is now.

Let me know.
 

Thread Starter

okin

Joined May 9, 2010
12
I ran through the calculations for a similar device as the MC34063, but the efficiency was terrible due to operating from such a low voltage.

See the attached. You will probably have to substitute other transistors for the devices I used. I really have no idea what parts you can find/get where you are.

You may have to wind the transformer yourself. I suggest a ferrite toroid. You can find them in junked equipment. For example, computer power supplies have toroids in them.
much thanks. I will try your advice. It is not a big problem for components here in Croatian. I can order everything but I must wait few days.
 
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