# how to size an inductor?

Discussion in 'The Projects Forum' started by slapshot136, Nov 7, 2012.

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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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Hello,

I can not see any image or schematic.

Here is how to do this:
Attachments and Images

Bertus

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3. ### PapabravoAAC Fanatic!

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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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sorry about that, I meant to attach this:

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

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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5. ### kubeekAAC Fanatic!

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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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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?

what was it's intended use?

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7. ### kubeekAAC Fanatic!

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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.

8. ### PapabravoAAC Fanatic!

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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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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..

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)

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10. ### crutschowAAC Fanatic!

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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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ah, I see - the diode is needed so the inductor has a complete circuit to work with, but I still fail to see the benefit of it being a special schottky diode, wouldn't a regular diode perform the same function?

can I simplify this to mean that if the feedback voltage is < 5v, the switch is turned on at the next possible cycle, and that if it's >5v, then the switch is opened at the next possible cycle?

is the inductor there for the purpose of preventing the full input voltage from ever getting to the output? that would make sense if it's because the capacitor is unable to regulate that well enough, and that would explain why the feedback would need to be after the inductor if the direct output of the SMPS is always either 0 or max, regardless of the capacitor

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12. ### kubeekAAC Fanatic!

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Schottkys typically have faster reverse recovery than pn diodes and have lower forward drop, 0.2v compared to 0.6, so you overall get better efficiency of the supply.

13. ### PapabravoAAC Fanatic!

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You're almost there. The voltage from the inductor when the switch turns off is in the negative direction which will tend to discharge the capacitor. The main purpose of the diode is to clamp that excursion to (GND - 0.2V). Now the feedback line can monitor the output and turn the switch back on when the voltage dips just below the desired output.

The 0.2V forward voltage drop of the diode AND it's current handling capability AND it's reverse breakdown voltage are all important features of this essential component.

I don't know if you can get a SPICE model for this part but simulating the circuit would be a great way to try alternative parts without making copious quantities of magic smoke.

Good Luck

14. ### crutschowAAC Fanatic!

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Not totally clear. The inductor current flow is from the switch towards the capacitor thus the voltage at the switch side goes negative, turning on the diode so the current continues to flow, but the voltage at the capacitor side of the inductor stays positive. The inductor continues to charge (not discharge) the capacitor when the switch is off.

Last edited: Nov 8, 2012
15. ### crutschowAAC Fanatic!

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Not exactly. In a PWM controller the error voltage does not change the duty-cycle on a cycle-by-cycle basis. The error amplifier response frequency is much slower than the PWM frequency (to meet feedback loop stability requirements) so the PWM duty-cycle changes at a relatively slow rate compared to its frequency.
The purpose of the inductor is to store energy and smooth the output of the PWM switching output. If you had only the capacitor there would be huge current spikes and high power dissipation in the switch transistor when it turns on. That would reduce the efficiency to be no better than a linear regulator. This would defeat the main purpose of a switching regulator to change voltage levels with high efficiency.

16. ### PapabravoAAC Fanatic!

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The capacitor cannot continue to charge indefinitely as the voltage on the switch side of the inductor drops toward GND. As it goes 0.2V below GND and turns on the diode the voltage on the capacitor will have a low impedance path to (GND - 0.2V). Of course it cannot do this instantaneously but it will do this.

17. ### crutschowAAC Fanatic!

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Of course the capacitor can't charge indefinitely, but it will continue to charge until the energy in the inductor is transferred to the capacitor and thus the inductor current drops to zero. At that point the voltage on the diode will rise to equal the output voltage but, being reversed biased, no further current flows. This happens in converters operating in the discontinuous mode. (See this).

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what would happen if the diode was replaced with a capacitor? would it charge up along with the inductor, and then be able to provide a place for it to dissipate the energy on the side that is directly connected to the output, thus avoiding the ripple that would be caused when the voltage goes under the diode's threshold at the cost of a bit of efficiency from charging the capacitor?

19. ### crutschowAAC Fanatic!

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Dissipate what energy? All the energy stored in the inductor, less the small resistive and diode loss, is already being transferred to the capacitor. The output ripple is caused by the increase and decrease in the inductor current as the switch is turned on and off.

Replacing the diode with a capacitor would give a large drop in efficiency, not just a little, since you would be charging that capacitor with a large surge of current from the switch which wastes half the energy*. If it were a practical circuit, someone would have already used it.

I suggest before you think of all kinds of "what ifs" you first build the circuit and understand the circuit that works. Then play with it to your hearts content. For that a simulator such as LTspice is a great tool. Saves a lot of soldering and the generation of all the magic smoke your circuits are likely to exhibit.

*Charging a capacitor from any resistive source, such as a switch, gives energy stored on the capacitor of 1/2 CV$^{2}$ whereas the energy taken from the source voltage is CV$^{2}$.

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