Author Topic: Voltage question?  (Read 22206 times)

Offline Leslie

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Re: Voltage question?
« Reply #30 on: September 08, 2010, 11:25:12 PM »
They are tiny.


Is 61mm(L) x 57mm(W) x 15.5mm(H)


Fully encapsulated and by what im reading load protected in case you trim it at a higher volt than the lit can handle the load at this higher volt..

A lttle more expensive than the LM2576 model but that chip is reliable at 28v they work very good.


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Offline Leslie

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Re: Voltage question?
« Reply #31 on: September 09, 2010, 04:57:47 AM »
^%= desired percentage of change output.

Radj+=  (VO(100+^%) - (100+2^%)
         (   -----------     ----------   ) = Kohm
                 1.225              ^%

Ive done this above eaquasion but there are some blanks.

My converter is 3.3v.

For mine I want 6.6 V out.

So is this 200% * VO or 100% more than VO the equation is not explained well enough.

Monkey could you give us a clue exactly what  the definition of ^% for this equation..  The rest is a breeze other than this one undefined.

By standard this should be 200% for double but the method explained.

And what of the down Radj is this indicated as EG 80% for 4/5ths.

It must be.


So.

100+200=300*3.3=990

1.225X200=245

990/245=4.04

-

100+2*200=500/200=2.5

RADJ=  4.04-2.5 = 1.54K

I think a 5 k Vpot is in order. This seems reasonable if my calculations are correct.
« Last Edit: September 09, 2010, 05:00:56 AM by 317537 »

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Offline Sundsvall

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Re: Voltage question?
« Reply #32 on: September 09, 2010, 06:35:25 AM »
If the standard output voltage is 2V and the desired is 5V, shouldn't it be like this.
5/2=2,5 (250%)
Peter
« Last Edit: September 09, 2010, 06:37:45 AM by Sundsvall »
Midsummer sun = up 02:54   down 22:51   angle 51,0° :)
Midwinter sun =    up 09:19   down14:18   angle 4,2° :(
Mean annual temperature = 3,1°C

Offline Sundsvall

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Re: Voltage question?
« Reply #33 on: September 09, 2010, 06:46:34 AM »
I'm sorry for my blindness, I didn't have to use an example :-[

If the standard output voltage is 3,3V and the desired is 6,6V.
6,6/3,3=2 (200%)
Peter

Midsummer sun = up 02:54   down 22:51   angle 51,0° :)
Midwinter sun =    up 09:19   down14:18   angle 4,2° :(
Mean annual temperature = 3,1°C

Offline Leslie

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Re: Voltage question?
« Reply #34 on: September 11, 2010, 10:04:05 AM »
Ahh thank for that Peter.

You nailed it I left out the 1.225*^% in my text and used this as my equasion from the post.

1.225x200


So we have it in text form

Radj+=  (VO(100+^%) - (100+2^%)
         (   -----------     ----------   ) = Kohm
              1.225^%              ^%

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Offline Leslie

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Re: Voltage question?
« Reply #35 on: September 11, 2010, 10:27:05 AM »
Nearly nailed it.  Just a little bit out me thinks.

VO must equal the voltage the converter is default at, not the desired output = 3.3v not 6.6v.

100+200=300

300x3.3=990

1.225X200%= 245

990/245= 4.04

-
100+(2x200)=500

500/200=2.5

4.04-2.5=1.54k

Oh I gots it right despite my equasion being wrong written.





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Offline Leslie

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Re: Voltage question?
« Reply #36 on: September 11, 2010, 10:51:00 AM »
Yes peter very smart, you have the equation method correct however it say VO= regulated voltage prior to adjustment not after.

This presidents a new interesting problem for the 2v module at 5v as this returns  = negative 0.12

This is IMO 120 ohms below 1 k.

1000-120= 880- ohms.

This should be correct.  But I thiumk we are ready to own this DC to DC converter.


« Last Edit: September 11, 2010, 10:54:08 AM by 317537 »

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Offline Leslie

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Re: Voltage question?
« Reply #37 on: September 11, 2010, 11:11:28 AM »
This is very interesting as the resistor values used are very similar to the 2576.  But his must be the same circuit inside the IC but in PCB with a larger output fet.  But the references is set at 1.225v and feedback divider must have a way to block by VO when the resistor is connected from VO to trim to by pass R2.

R2 in this diagram is this trim resistor we speak of with the LM2576adj and the DC converter is comparable maybe.

Or maybe parallel over the R1 to force R2 to a higher divider point.







 
« Last Edit: September 11, 2010, 11:15:17 AM by 317537 »

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Offline Sundsvall

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Re: Voltage question?
« Reply #38 on: September 13, 2010, 08:00:45 AM »
Yes peter very smart, you have the equation method correct however it say VO= regulated voltage prior to adjustment not after.

This presidents a new interesting problem for the 2v module at 5v as this returns  = negative 0.12

This is IMO 120 ohms below 1 k.

1000-120= 880- ohms.

This should be correct.  But I thiumk we are ready to own this DC to DC converter.




Didn’t get that prior to adjustment, but now I’d put one more word to my knowledge. :) I’m not sure that the negative answer is as you suggest. My guess is that if you get a return with 0,88, that would be 880 ohm. I don’t have a clue what the negative answer could be but maybe we have to consider that 0 ohm is the limit for increasing the voltage.

Isn't the input voltage limited to 18 - 36 V?
The output voltage prior to adjustment. Isn't that the output voltage before you put anything there like for this one 2 V?

Peter
« Last Edit: September 13, 2010, 08:23:12 AM by Sundsvall »
Midsummer sun = up 02:54   down 22:51   angle 51,0° :)
Midwinter sun =    up 09:19   down14:18   angle 4,2° :(
Mean annual temperature = 3,1°C

Offline Leslie

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Re: Voltage question?
« Reply #39 on: September 21, 2010, 04:03:49 AM »
Well I recieved these regulators and they only adjust from 3.3 to 4.7v then they cutout and you have to reconnect and disconnect them.

So again it out of your usage with the 24v model.

I can use two of these outputs in series no problem with a parallel input as they are very isolated.  I have tested this and ive set them top 2.4v for the red lights and the other to 4.2v and run it ontop to make 6.6v for my new 5watt leds.

They use a bit of power with no load.  Not like the lm2676.  Unless youre draining more than 3 amps out of this regulator your DC converter should be better for the task.

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