How to calculate appropriate inductor values?

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Lucien

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Hi all,

Does anybody know how to calculate appropriate inductor values for use in various boost circuits? I've been looking at the datasheets for various ICs, and though they give some "recommended values" they don't really give a clear (or any) description of how to calculate them.

Simply following the recommended value's isn't that good an option, since they give several, and usually the provided examples are for quite different applications than driving Luxeons.

I'd be grateful if anyone could give a rough explanation, or point me in the direction of one. Thanks...
 
Using the recommended value is a start, but - if you want the most eff in the switcher, you will need to 'hand-tune' the inductor, this is done by either adding or subtracting windings from the core

An inductor is used to store energy during the cycle and when the in-going current stops, the inductor "snaps-back" with the stored current out of phase

Too low of an inductance will cause the switcher to go into "discontinous mode", that is the current will start going into the core and saturate the core; this is where you would add windings to the core

Too high of an inductance will cause excess resistance on the core and lower the eff of the circuit

Think of an inductor as a bucket to hold water, too small of a bucket and you will spill over water when you are using it, too large of a bucket - and you will waste the storage

To properly gauge the inductor size, you use a hall-effect current probe around one wire of an inductor, during the charge phase - you will see a triangle waveform rising as the current enters the inductor - if you see a "flat" spot in the waveform during the charge phase (MOSFET turned "on"), then the current has saturated the inductor, and you need a larger inductor

Having too many turns on the inductor will cause the other effect, you will have too much resistance and that will lower your eff eff

It also matters on your switcher frequency, a high eff switcher runs at a higher switching speed to allow a smaller inductor and also more responsive

I worked at Intel designing very high power; high eff switchers for 3yrs
 
Mike, thanks for the reply.

When you say a "flat spot" in the waveform, I take it you mean at the top of the wave? It's my understanding that the inductor waveform in a boost circuit should be a triangular wave, so a flat spot at the top would be the current hitting a ceiling. Also, is discontinuous mode "bad"?

(I should mention that I understand qualitatively what happens in the inductor/switch/diode/cap boost circuit just in case anyone might waste time going over that. /ubbthreads/images/graemlins/tongue.gif It's the specifics I'm not sure of)

About using a bigger than necessary inductor, if we could use a inductor with more turns, yet the same total resistance in the wire, would the circuit still be less efficient? I know energy stored in an inductor is proportional to its inductance, so bigger inductor means more energy stored. But since it isn't being dissipated, is it being wasted?

And then there's the formula:

v=L di/dt

which I take to mean that in an inductor, with a fixed voltage applied, the rate of change of current depends on inductance, i.e. what you said about higher frequencies and smaller inductors. I'm not sure though, if this has any relation to efficiency. As far as I'm aware it doesn't, but its what I'm not aware of that's the issue...

Not having a Hall effect probe, I guess I'll be testing various inductors to see what works in my circuit. In it, the peak inductor current is a fixed value set by a series resistor, so I'll probably have to be testing two sets of values instead of just one to optimise my circuit. At the moment its at an uninspiring 50+% efficiency.... /ubbthreads/images/graemlins/icon23.gif
 
Lucien,

Your question is simple. The answer is quite a difficult one. Boost circuits come in many variety or topologies. There's PWM, PFM, Constant on time, Constant off time etc etc etc. Each one has a different set of equations that determines the inductance value.

What you may not be able to determine is if the circuit is even working correctly. By that I mean the control loop for a particular IC might be on a cycle by cycle basis. You really can't tell unless you can monitor the inductor current accurately. Even with the proper components doesn't mean your set and good to go either. Good proper ground layout and high current paths need to be clean, short and done properly for switchers to start working even halfway decent.

What can you do? With simple equipment, it's best to work with switchers with switching frequencies less than 300KHz. These are easier to tame and are less sensitive to ground path and component layout.

Use the free software from the Vendor whenever possible. This will help you optimize your circuit and get you "close" to a working solution. From there, you could use an evaluation board or build one from scratch. That minimizes the number of components you would have to hand tune to get the correct value. The critical components you might have to tweek are the input capacitor value and technology, Inductor, Output capacitor value and technology.

It gets real complicated real fast. For example, some control loops require the output capacitor to have some ESR or equivalent series resistance. These circuits don't behave well with high value ceramic capacitors. Other circuits behave better as the ESR of the output capacitor goes down. So, again, there is not good rule of thumb. The datasheet is your best source of information for the IC you are working with. Consult the manufacturer, they have technical support and you can quiz them with your questions and specifics.

Check out the free software if there is any. National has a web based tool for their ICs. Linear Technology has a free Spice program for their ICs. Maxim doesn't offer much help. Others you have to consult the manufacturer.

One thing that might help you is to use a larger inductor (physical size) for testing purposes. This would help ensure the inductor doesn't saturate. You can then hone in the inductor value, get the circuit working and then work the physical size down and optimize the size and tradeoffs of inductor saturation current and physical size after you know what value you need.

Switching converters are not easy by any means. I can spend several months getting a circuit to work and sometimes I get to the end only to conclude the circuit efficiency isn't good enough and have to start all over with another IC design.

I have some other suggestions if you want to email me, I will expand on prototyping and breadboarding high frequency converters and or help you with your design.

Wayne
 
Hello Lucien,

To get to the heart of the original question, i assume
you want to know more about selecting an inductor for
use in a boost circuit.

I offer some basic guidelines to use when trying to
find a good inductor for a circuit here.

1. Use the smallest core you can get away with.
The losses go up with larger cores.
This means you have to look at the manufacturers specs
and try to find an appropriate size core to fit your watts
output requirement. If you are starting with bargin cores
that you find for low cost in various surplus catalogs, then
try to get one or two of every type they sell so you can try
each and every one.
2. Use the least amount of turns you can get away with.
The losses go up with longer wires due to both ac and dc
effects.
3. Use the largest diameter wire you can fit on the core.
This is usually true, except in cases where the inner core
of the copper isnt used because of skin effects and so wastes
a bit of the copper in the wire. You can also experiment
with using two or even three or four lenghts of smaller diameter
wire to reduce skin effects. This may increase efficiency
in some cases.
4. The requirement for a given inductor value in henries can
be found from the equation given in a previous post:
v=L*di/dt, where 'di' can be approximated by 'delta i' and
'dt' can be approximated by 'delta t' because of the triangular
waveforms usually encountered in converters. This means you
can get away with subsituting constant values in for di and dt
and still get some reasonable results from the use of this
equation. This means:
'v' is the voltage across the inductor.
'di' is the peak current through the inductor.
'dt' is the period of time the inductor has the voltage 'v' across it.
'L' is the value of the inductor in henries.
Usually you cant have a 'di' that is higher then the rating of your
switching transistor, and this value also gives you a way to
approximate your transistors peak voltage drop during the on time.


In closing...

The biggest concern in converter (except for very low power rated units)
design is efficiency. The point of any experiment is to find the
highest efficiency converter you can find using a specific circuit
topology.
For the inductor part of the design this might mean trying:
1. different core sizes
2. different core materials
3. more turns
4. less turns
5. larger diameter wire
6. smaller diameter wire
7. bifilar, trifilar windings

For the other parts of the circuit:
1. For the switching transistor, in the Zetex thread we found
that the Zetex transistors have the lowest voltage drops and
so give the highest efficiency. Comparable designs without
these transistors would have to use MOSFETS.
2. For the output Schottky diode we found that the Zetex 2
amp diode had the lowest voltage drop, and so gave the
highest efficiency.
3. For driving LS LED's, the output ripple has to be
carefully checked in order to avoid prematurely destroying the
device by exceeding its peak current rating.


Wishing you the best of luck with your LED circuits!

Al
 
Hi Wayne,

Thanks for the reply, and the offer of help /ubbthreads/images/graemlins/smile.gif. At the moment I'm working with the ZXSC400, which runs up to 200kHz according to the datasheet, so I don't expect stray EM to be a (significant) problem in the design.

(Mostly I'm doing all this for fun, but also to learn more. It helps that it's party related to the whole flashaholic experience.)

Most likely, when this is done I'll be trying out higher frequency converters. Will be dropping you a line sometime soon. (I'm guessing to your yahoo mail, right?)

Mr Al,

Thanks for the guidelines, it never occurred to me that larger cores would have greater losses (though I believe I understand why).

About the equation:

v = L * di/dt

what would v be during discharge into the output cap? Or would it be a function instead of a parameter?

As for the rest of the parts, I'm pretty much using the same components as in the original ZLT circuit. (I actually spent a month or so reading the thread and picking alternative components. Then I found my choices were short in some way or had already been considered and found lacking /ubbthreads/images/graemlins/tongue.gif)


It may not be easy, but it's loads of fun /ubbthreads/images/graemlins/grin.gif

Lucien
 
Hello again Lucien,

During discharge of the inductor into the cap, the
transistor is off. This means you have a series connection
of four components:
1. the battery
2. the inductor
3. the Schottky diode
4. the cap in parallel with the LS

Also, the inductors voltage reverses (from what it was
when the transistor was on), so you have the inductors
voltage adding to the batteries voltage. This is what
actually provides the 'boost' in 'boost regulator'.
The output voltage is considered constant, at maybe about
3.3 volts (for a 1 watt white LS) because the LS
clamps this voltage to an almost constant value.
The diode drop is usually figured to be about 0.4 volts,
but it looks closer to 0.2 volts for the Zetex 2 amp diode,
so lets use 0.3 volts for the Schottky diode here.
Lets also say the input battery voltage is 2.4 volts.

The above shows that we know approximately every voltage
in the series circuit except the inductors voltage.
This means all we have to do to solve for the inductor
voltage is to add or subtract the other voltages in the
series connection string:
Vb=battery voltage=2.4
Vd=diode voltage=0.3
Vc=cap voltage=LS voltage=3.3
VL=inductor voltage=?

Now, we simply add all the voltages acting as a 'source'
and subtract all the voltages acting as a 'load'. Since
the inductor is now supplying energy to the circuit, it's
acting like a battery which is also a source, so it's
voltage gets added:

Vb+VL-Vc-Vd

Now we apply the law of voltage drops in a series circuit:
"The sum of the voltages in a series circuit equals zero".
This is known as 'Kirchoff's Voltage Law'.

Vb + VL - Vc - Vd = 0

Since we know three of the four quantities, we can easily
find the fourth by substituting the three knowns:

2.4 + VL - 3.3 - 0.3 = 0

Solving for VL (by moving the other three to the other side
of the equation and changing their signs:

VL = 3.3 + 0.3 - 2.4

Now all we do is calculate the value on the right
by adding 3.3 to 0.3 and subtracting 2.4, and we get:

VL = 3.6 - 2.4 = 1.2

or simply:

VL = 1.2 volts.

This means the inductor looks like a small battery of about
1.2 volts during the time the cap is charging up.

Quite simple right?

This discussion assumes the inductors current is a
perfect ramp, but in actual practice for some circuits the
inductor current is very nonlinear and so it is not a ramp.
This simple way of approaching the circuit during turn off
is still useful though in that it provides a starting point
for selecting an inductor and for understanding the operation
of the circuit.


Good luck with your LED circuits,
Al

Note
Edited later: "inductors voltage"
changed to: "inductors current".
The inductors current is a ramp, not the voltage.
 
Question was: "Is discontinuous mode bad ?"

Discontinuous mode occurs when the inductor is saturated and no longer able to absorb any more energy

You will see this on the scope as a "flat-spot" in current while the switch is turned on

Normal:

http://www.inretech.net/pictures/switcher1.JPG

Discontinuous:

http://www.inretech.net/pictures/switcher2.JPG

To properly measure the inductor current, you need to clamp a hall-effect probe around one of the inductor wires, you can not simply add another winding or stick a wire thru the inductor

This mode is "bad" since you are not able to run the switcher at its peak eff because you have saturated the inductor, you can "fix this" by:

) Changing the core material
) Changing the wire size
) Changing the number of turns

On the other hand, if you have too much inductor in the circuit, you will have:

) Slower response time to transients (on ripple mode)
) Resistance losses (voltage lost in core)

I spend 4 yrs designing very high eff Sync convertors for the P2/P3 Motherboards at Intel

)
 
Hmm, it would seem that Intel applies a different meaning to the words 'discontinuous mode' than Linear Tech...

In the Linear Tech datasheets, 'discontinuous mode' is used to describe operation in which the current in the inductor drops to zero for some parts of the converters switching cycle. Saturation of the inductor is dealt with separately from 'discontinuous mode'. Take a look at http://www.linear-tech.com/prod/datasheet.html?datasheet=798
where 'discontinuous mode' operation is suggested as a desirable feature of the component being described, and if you look in the datasheet, this mode is entered at low power levels when the current is well below what the inductor could tolerate.

I don't understand something about your description of what happens when the inductor saturates. Since the inductor can no longer store any energy in its magnetic field, I would presume that the incremental inductance would fall to zero, in which case nothing prevents the current from rising in the coil, until the resistance of the circuit kicks in. I'd expect exactly the opposite of a flat top in current in the primary...what limits the current?

Thanks,
Jon
 
You are correct

Sorry - please excuse my mistake - I got the two mixed up, "discontinuous mode" is the current in the inductor drops to zero for some parts of the converters switching cycle

Even in the best convertors, there are times when the convertor goes into discontinous mode for short periods of time - such as in a Sync convertor without a recovery diode when both FETS are off (to prevent shoot-thru)

When the input voltage is 2x higher than the output voltage in a design, the bottom FET and/or catch-diode will be doing most of the work

Discontinuous mode isn't a large problem unless the excess windings of the inductor start adding resistance to the circuit
 
[ QUOTE ]
MrAl said:
Now, we simply add all the voltages acting as a 'source'
and subtract all the voltages acting as a 'load'. Since
the inductor is now supplying energy to the circuit, it's
acting like a battery which is also a source, so it's
voltage gets added:

Vb+VL-Vc-Vd

Now we apply the law of voltage drops in a series circuit:
"The sum of the voltages in a series circuit equals zero".
This is known as 'Kirchoff's Voltage Law'.

Vb + VL - Vc - Vd = 0


[/ QUOTE ]

Duh-oh /ubbthreads/images/graemlins/twak.gif Learnt that the first year of my uni course. I should have realised that a lot sooner.
 
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