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Old 02-28-2012, 03:47 AM #1
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Default New Bucking Driver IC - Will it work?

http://www.fairchildsemi.com/ds/FA/FAN4603.pdf

I stumbled upon this today.

Me and RHD were having a discussion if this IC would be viable or not, as the voltage "appears" to be fixed according to the datasheet....

However, based off of other IC's I've seen in the same category,(specifically this one; http://www.fairchildsemi.com/ds/FA/FAN5350.pdf) each different voltage output has to do with what the voltage reference is on the inside of the chip. If you want 1.87 volts output, then the voltage reference inside is 1.87 volts.

For example, if I took the ever reliable LM317, and took its adjust pin and without ANY external components, just attached it to the Vout pin, I would get 1.25 volts out. Simple enough. If I changed that internal voltage reference to 1.8 volts, I'd get 1.8 volts out.

Basically, that's the same exact thing they're doing with the IC listed above, though the neat thing about it is it has ALL the components necessary inside of the module already. It's a truly monolithic switching regulator. The only two things external that are needed are a Current sense resistor, and a capacitor.

However, RHD seems to believe the IC has an internal voltage divider, making the FB pin useless... I really don't believe this is the case, as why would they even bother to put the FB pin in place then? It doesn't make sense. So I think they just change the internal voltage reference inside of the chip to alter the output voltage! If we got the 1 volt reference version, then the dropout of the regulator would be only 1 volt. And fairchild even states in the datasheet "Other voltage outputs available upon request" which may mean we can get some even lower versions if we ask nicely!

Which is why I believe this IC would be a perfect choice for the Open Source Buck version of the switching drivers we've started to create. Yes, the output current is limited to 600 mA for ONE driver. Since these things are so damn tiny, it would be trivial to add another driver to the same board and get even more current out.

6 volts input limit you say? Well, this driver fills in the gap for red laser diodes. The boost driver is great for driving blues, and violets; however, it can't drive reds though!

What are your thoughts?


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Old 02-28-2012, 04:03 AM #2
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Default Re: New Bucking Driver IC - Will it work?

I don't necessarily know that it has an internal voltage divider.

I simply believe that these are fixed output voltages modules, and that you won't be able to use them as constant current sources for LDs.
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Old 02-28-2012, 04:04 AM #3
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Default Re: New Bucking Driver IC - Will it work?

So can it be used in constant current mode? I can't find it on the datasheet.

And 600mA is kind of a bummer. If you can parallel 3 on the same board it would be pretty sweet. It's 2.5mm (wide) x3 = 7.5mm, plus a little extra space for traces. If you could fit all the caps and stuff on the other side of the board, sweet.
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Old 02-28-2012, 05:05 AM #4
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Default Re: New Bucking Driver IC - Will it work?

All that's needed to change it to CC mode is an external resistor... Just because it's not in the datasheet doesn't mean it's impossible to do.

You can technically change ANY switching/linear regulator to CC mode as long as there's a FEEDBACK or ADJUST pin on it!

Anyway, I ordered some samples. They'll be her in a few days, and I'll report back here with my findings.

Here's hoping.
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Old 02-28-2012, 05:47 AM #5
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Default Re: New Bucking Driver IC - Will it work?

it seems the voltage is fixed... I don't think you can do much about that but what do I know ... good luck
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Old 02-28-2012, 10:54 AM #6
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Default Re: New Bucking Driver IC - Will it work?

I don't think my explanation was clear enough?....

The feedback pin is basically the same thing as the ADJ pin on an LM317. However, instead of 1.25 volts reference inside of the LM317, the IC I listed has one that changes with the output voltage.

So, if I want 1 volt out, then the voltage reference is one volt.

That's how they get "fixed" voltage outputs. They tell you to just tie the feedback pin right to Vout, like my example I gave in the original post.

But, if you ignore those instructions, and put a current sense resistor, then you get CC mode. I'm positive this is how this IC works. Without any doubts. You would then calculate the output current by using I = 1.00 / R where 1 is the voltage reference inside.
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Old 02-28-2012, 12:58 PM #7
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Default Re: New Bucking Driver IC - Will it work?

Le Quack - you're really positive about this, but you're just guessing. You're not providing any actual citation or reference for why you believe this to be true.

What value resistor would you use? I tried to explain to you that there isn't a special kind of resistor that acts as a "current sense resistor" regardless of where it's used. A current sense resistor is only so named because the circuit it is used in provides for that functionality. There's nothing to indicate that your IC exposes that functionality.
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Old 02-28-2012, 03:49 PM #8
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Default Re: New Bucking Driver IC - Will it work?

Well, I don't think it will be that way you say.... Why would they make a fixed and adjustable version of voltage regulators then... hmmm.

the fixed output is gotten by a resistor voltage divider inside the IC (after the Reference Voltage)..

So you wouldn't be able to get more than the fixed voltage the IC is meant to put out. You could use a current sense resistor I suppose, but not like you say.....

but like this:



I think this should work wired like this and you will use the fixed output voltage in your formula for the current sense resistor.

So you will have a constant current but efficiency will drop, also why would you want to use that, and 1.8V wouldn't be of much use in this hobby....

I say drop the idea of using that IC..

if I am wrong please somebody more fluent correct me


Quote:
Originally Posted by Le Quack View Post
I don't think my explanation was clear enough?....

The feedback pin is basically the same thing as the ADJ pin on an LM317. However, instead of 1.25 volts reference inside of the LM317, the IC I listed has one that changes with the output voltage.

So, if I want 1 volt out, then the voltage reference is one volt.

That's how they get "fixed" voltage outputs. They tell you to just tie the feedback pin right to Vout, like my example I gave in the original post.

But, if you ignore those instructions, and put a current sense resistor, then you get CC mode. I'm positive this is how this IC works. Without any doubts. You would then calculate the output current by using I = 1.00 / R where 1 is the voltage reference inside.
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Old 02-28-2012, 04:44 PM #9
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Default Re: New Bucking Driver IC - Will it work?

Well, we can't be sure one way or the other - the only way to go about finding out about this (and I think Quack may be right) is to test it. Maybe it won't work, and then we're only out a little bit of money ^_^
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Old 02-28-2012, 05:14 PM #10
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Default Re: New Bucking Driver IC - Will it work?

I think Le Quack is wrong in believing that the output is ranging from 1.0 to 1.8V ..

as much as I can see in the datasheet there is a different IC for each of the voltage values. And it's fixed. It's not floating for any of those ICs.

see this quotes:




just pointing out

also what RHD said is right:

Quote:
However, RHD seems to believe the IC has an internal voltage divider, making the FB pin useless... I really don't believe this is the case, as why would they even bother to put the FB pin in place then? It doesn't make sense. So I think they just change the internal voltage reference inside of the chip to alter the output voltage! If we got the 1 volt reference version, then the dropout of the regulator would be only 1 volt. And fairchild even states in the datasheet "Other voltage outputs available upon request" which may mean we can get some even lower versions if we ask nicely!
FB pin is existing in every switching voltage regulator whether it is a fixed or and adjustable version. the difference between the two is this :



in the adjustable version you put your own voltage divider based on the internal reference voltage with external components to set the output voltage,

in the fixed version it's already inside the IC.. so nothing much to do about it... You can't access the internal reference voltage like Le Quack wants to.
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Old 02-28-2012, 06:09 PM #11
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Default Re: New Bucking Driver IC - Will it work?

Well, even if it was the case of the internal voltage divider, (which I still am skeptical about) then the output would still be adjustable. It would still be able to output constant current, because there's still accessible feedback.

Let me try to explain;

If the internal reference voltage is 1.25 volts, then the op amp wants to see 1.25 volts on the FB side to maintain a certain output, correct?

If the current sense resistor has 2.5 volts on it, and the voltage divider "divides" the voltage by 2, then the op amp would see 1.25 volts on the FB side, and thus maintain regulation.

The design I'm looking at anyway would include an op amp to help lower losses in the resistor anyway...so it wouldn't be an issue with or without the internal voltage divider regardless!

Just as long as the FB pin is there, it can be changed to CC mode.

And why am I skeptical about the voltage divider aspect? Because, if you checked the other link I gave in the original post to the other switching regulator, it's "fixed" too, but they use different voltage references within the chip to get different outputs. It seems to me that it's the exact same with this IC.


And yes, RHD I understand switching topologies quite well, and understand what it means to make something constant current.
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Old 02-28-2012, 06:38 PM #12
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Default Re: New Bucking Driver IC - Will it work?

Well,

I guess you have to try it then.... I might be wrong
I will wait for your tests eagerly

but why do you need a 1.8V tops?

What are you planning on driving with that IC?
Isn't some other a better choice?

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Old 02-28-2012, 06:40 PM #13
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Default Re: New Bucking Driver IC - Will it work?

I never said I was using the 1.8 volt version.

The 1 volt version would have the least default dropout out of the other ones.

And the reasoning behind this IC is simply because of SIZE. The PCB could be microsized, and everything would still be on there.

I'll try some tests out once I get them in the mail...I already ordered some samples. P:
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Old 03-02-2012, 03:43 AM #14
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Default Re: New Bucking Driver IC - Will it work?

Ah, I did testing, and RHD was right.

The output is always 1 volt, no matter how you change the feedback method. It'd be a great IC if it was adjustable, but alas, they are not.

Disappointment.
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Old 03-02-2012, 04:33 AM #15
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Default Re: New Bucking Driver IC - Will it work?

Well, good try, man.
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Old 03-02-2012, 06:41 AM #16
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Default Re: New Bucking Driver IC - Will it work?

Quote:
Originally Posted by Le Quack View Post
Ah, I did testing, and RHD was right.

The output is always 1 volt, no matter how you change the feedback method. It'd be a great IC if it was adjustable, but alas, they are not.

Disappointment.
yep that's what I expected too +1 for your efforts though
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