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The Projects Forum Working on an electronics project and would like some suggestions, help or critiques? If you would like to comment or assist others with their projects, this is the place to do it.

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  #1  
Old 11-07-2012, 08:16 PM
slapshot136 slapshot136 is offline
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Default how to size an inductor?

I want to build this



only I don't have either the 1N5824 or the 33uH, how important are they? I do have some regular diodes that I can replace the 1N5824 with (.7vDrop, 3A) and some inductors, but I don't have any way of measuring the size of the inductors that I have (I pulled them from old electronics and they aren't marked) - I tried building it without the diode and an unknown inductor, but found that I needed to add a small ceramic capacitor on the output to keep it at 5v, otherwise it dropped when I added a load. I don't quite understand why this was, so I tried a couple other caps thinking maybe the 220f was bad, but the result was the same - it seems to prefer tiny ceramic caps?

short version:

does the inductor matter/can it be replaced/resized or compensated for?

does the diode matter/can it be replaced with a regular diode?

why is adding a small ceramic cap needed to keep the output voltage stable?

P.S. - I was thinking the diode was to provide an output path for feedback from a large motor turning into a generator via its momentum, but my application is to power a computer rather than motors
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  #2  
Old 11-07-2012, 08:22 PM
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Hello,

I can not see any image or schematic.

Please upload the image or schematic to the AAC server.
Here is how to do this:
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Bertus
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  #3  
Old 11-07-2012, 08:34 PM
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A 1N5824 is a Schottky diode, so using an ordinary silicon rectifier is pretty much not going to work as intended. Without a schematic I can't tell you about what the inductor needs to be, but the fact that you're going ahead building things at random pretty much tells me that facts and reality are secondary concerns. What do you care what we all think?
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Old 11-07-2012, 11:18 PM
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sorry about that, I meant to attach this:



P.S. - can I edit my original post to include this somehow?

Quote:
Originally Posted by Papabravo View Post
but the fact that you're going ahead building things at random pretty much tells me that facts and reality are secondary concerns. What do you care what we all think?
im trying to build it with what I have available, but want to know why the stuff is there and if there are alternatives
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Old 11-07-2012, 11:25 PM
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Go here http://www.ti.com/product/lm2596, read the datasheet, and see for yourself if the choice of the components is important or not. You will find that some substitutes really won't work properly.
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Old 11-07-2012, 11:37 PM
slapshot136 slapshot136 is offline
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Quote:
Originally Posted by kubeek View Post
You will find that some substitutes really won't work properly.
is there any way of knowing other than trying? from the datasheet there I see that they used a 68uH inductor instead - is it better go with a bigger one or a smaller one? or both but add the smaller one after the feedback wire?


Quote:
Originally Posted by Papabravo View Post
A 1N5824 is a Schottky diode, so using an ordinary silicon rectifier is pretty much not going to work as intended.
what was it's intended use?
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Old 11-07-2012, 11:45 PM
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If you had read the datasheet whole, you would notice that right on the next page starts the design procedure, which takes you through selecting all the components according to your needs.
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Old 11-08-2012, 01:46 PM
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An SMPS works by opening and closing a switch on one side of an inductor. When the switch is closed the higher(input) voltage flows through the inductor and charges the output capacitor. When the feedback line detects the output voltage rising it opens the switch AND the inductor does what....??

The answer to the question will explain the purpose of the diode.

We'll give you the answer if you make an attempt to find it on your own.
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Old 11-08-2012, 04:59 PM
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Quote:
Originally Posted by Papabravo View Post
When the feedback line detects the output voltage rising it opens the switch AND the inductor does what....??
the inductor would attempt to maintain the current flow by converting stored magnetic energy into electrical current - but wouldn't the capacitor also do this, to a much higher magnitude? and considering I want a 5v power supply rather than an x amp power supply, isn't the capacitor the only component that I need?

anyways, the inductor would raise the voltage in an attempt to keep the current constant, and thus charge the capacitor by a small amount (let me put in some numbers and check that the capacitor can handle the energy)

let I = 3, V = 5
energy stored in inductor = .5*33*10^-6*3^2 = 0.1485 mJ
energy stored in capacitor = .5*220*10^-6*5^2 = 2.75 mJ
inductor energy discharged into capacitor + original capacitor energy = 2.75+0.1485 ~ 2.9 mJ
2.9 mJ = .5 * 220*10^-6*(v)^2 => v= 5.13v

going with a capacitor even slightly >5v would allow it to absorb the excess energy from the inductor (although I have a 16v one, leaving plenty of extra room there)

I think that the IC would attempt to use as little of the feedback current as possible, to keep it efficient, and only track the voltage as best it can, and notice quickly if the voltage drops ever so slightly under 5v, and thus charge the capacitor back up to 5v, although at this point the inductor would charge itself up first since it is directly in the path of the current, prior to the capacitor (causing more voltage slag/ripple depending on the inductors size)

as for the diode, I am still puzzled by it - if it were specified to have a breakdown voltage of 5~6v or so, it would make sense, but the way it is now still does not make sense to me.. nor do I know why it needs to be a schottky diode as opposed to a regular one, but from what I searched they just seem to be a better diode in general (more "ideal") and faster, which would make it better if it was an over-voltage safeguard (which is what I think it is meant to be), or is it because the capacitor wouldn't be able to charge up as fast as the inductor discharges, despite it being several times larger, and the voltage is expected to go into the ~40-50v range? if that is the case, wouldn't it make more sense to have a fast-charging capacitor across the inductor, or is my scale way off and the schottky diode is much faster than what a capacitor can charge up to, and it's acceptable for the voltage to go from 0 to 40v prior to the capacitor, and let the "slow" capacitor keep a relatively constant 5v after it by itself?

I guess I have another question now - why can't the chip detect the output voltage from its output pin? why does it need to travel through an inductor first? perhaps that is the reason for the inductor, although exactly how the voltage regulator works is a bit beyond me..

Quote:
Originally Posted by kubeek View Post
If you had read the datasheet whole, you would notice that right on the next page starts the design procedure, which takes you through selecting all the components according to your needs.
it gives me "ideal" components - it doesn't help me much in terms of "what if" - that's like playing with lego but only by following the instruction manual 100% and never deviating - you don't actually learn much or make anything interesting or even learn why it was designed a certain way as opposed to different alternatives (heck you won't even know alternatives exist)

Last edited by slapshot136; 11-08-2012 at 05:26 PM.
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  #10  
Old 11-08-2012, 05:31 PM
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Components are seldom used spuriously in a circuit. If it's in there it is likely needed for proper circuit operation. Random substitution of components is unlikely to result in a working circuit.

Your understanding of the regulator operation is not correct. In a switching regulator, the diode, inductor and output capacitor are central to its operation. You can't have a switching regulator without those. The inductor stores the energy by increasing its current when the switch is ON. When the switch turns off the inductive current wants to keep moving. This forward biases the diode which keeps the current flowing and transfers the inductive energy to the capacitor. The controller generates a high frequency PWM signal to control the switch. The PWM duty cycle is controlled by the feedback from the output voltage.

Read this for more info. Also you might try reading the data sheet for the LM2596. If you want to do "what if" then you need to understand that. It's not Legos.
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