inductor current rating

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http://www.ti.com/lit/an/slva001e/slva001e.pdf?keyMatch=tl494&tisearch=Search-EN


In the above link there is an example curcuit which is 32 volt dc to 5 volt 10 Amper output.

ıt ıs calculatıng an inductor 140 uH. It ıs ok but the thing here i have never understood is is there such an inductor in the market?

ım checking the ınductor suppliers and they offer like very low Isat values for 140 uH. So how to find that inductors?

what can we say about the 140 uH inductor ratings should be ?
 
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There may or may not be such an inductor on the market, but there ARE plenty of cores available, and the required data and formulas are widely published, see the Ferroxcube catalogue among others.

Do a few sums and add wire as required.

What you will usually find is that for energy storage coils like the buck converter you mention you will need a gapped core (The energy is mostly stored in the flux in the gap, and having the gap lowers the Al value but raises the saturation current).
Be careful about having the winding too close to the gap, fringing fields in the area cause lots of excess heating in windings placed too closely over the gap, also Litz wire is sometimes a good thing.

Winding your own magnetics is really not that hard (Use a E core design rather then a toroid if you need more then about 20 turns or so, toroids are a ballache to wind if you don't have the special tools), ETD39/49/59 are fairly home construction friendly.

Regards, Dan.
 
ıt ıs calculatıng an inductor 140 uH.

Well the problem here is that the example circuit shown is completely obsolete...

- It uses old slow crummy BJTs
- Therefore, it switches at a ridiculously low 20kHz
- Therefore, it needs a huge inductor

Besides, it doesn't have syncronous rectification, so the diode losses are going to be important, and it is probably going to have lots of losses in the huge inductor, and in the slow crummy switching BJT.

It will also probably need a heatsink... lol.

Here's a little more modern version :

LTC3851 - Synchronous Step-Down Switching Regulator Controller - Linear Technology

Use good modern MOSFETs, surface mount obviously, with PCB cooling, each MOSFET will dissipate less than 1W. You can run it at 200-400kHz and use a very small inductor, which will be off the shelf, cost $1 unit price, like Bourns SRR1280 shielded surface mount inductors for example.
 
the Ecore style all have a "gap" in the middle leg and this "gap" is covered by many turns of the winding.

How important this is depends on gap length, but you can space the inner layer of winding away from the gap by adding layers of kapton tape or similar inder the middle part of the winding (This of course increases leakage in a transformer).

I had problems with it running a 2.5mm gap in a resonant transformer with a single turn primary wound over the secondary running 1KVa for a few seconds at a time in a pulse power application. The secondary winding right over the gap was cooking, it would be less of a problem with a smaller gap.

Agree that a 20Khz non syncronous buck switcher is like '1980 called and wants its power supply back', we can do far better today (200KHz is routine and 1MHz and up not unheard of).

Regards, Dan.
 
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TDK Inductors for Power Circuits Wound Ferrite VLBseries e.g. VLB7050HT-R15M (150nH, 35A)
http://product.tdk.com/en/catalog/datasheets/inductor_commercial_power_vlb_en.pdf

It is 150nH, not µH, that's a 1000x difference.

Why would anyone want to build the original circuit ? Just the heatsink for the slow crummy BJT will cost more than the total BOM for a modern solution, and what you get for free is the EMI radiated by this large mass of metal stuck on a switching semiconductor, not to mention the huge coil and large PCB track loop areas which are always going to leak a lot more than a small shielded inductor and tight SMD layout.
 
It is 150nH, not µH, that's a 1000x difference.

Why would anyone want to build the original circuit ? Just the heatsink for the slow crummy BJT will cost more than the total BOM for a modern solution, and what you get for free is the EMI radiated by this large mass of metal stuck on a switching semiconductor, not to mention the huge coil and large PCB track loop areas which are always going to leak a lot more than a small shielded inductor and tight SMD layout.

Im not planning to build this curcuit. The reason i have asked this question is just what you are saying. the inductor in this example curcuit is not practical and not a good choice.

Im not good at calculating and choosing the inductors. so it seemed to me that this inductor is too big in size. so it is not practical.

i just wanted to make sure that im not missing anything. for big value inductors with high current ratings i need very big inductors in size and /or with air gap also. how about the metarial type?

and also can anybody make me a calculation for choosing an inductor for lets say 140 uH and 10 amper. the calculation and then choosing the right core according to this calculation.
 
OK !

All you need is v=L di/dt

For a buck switching converter with these parameters :

Vin 30V
Vout 5V
Iout 10A

Duty cycle d = 5V/30V = 0.166

Let's pick a switching frequency F=20kHz.
Period is T = 1/F = 50µs.
Top MOS conducts for T*d = 8.3µs
Bottom MOS (or diode) conducts for T*(1-d) = 41.6 µs

We can design for a ripple current of 3A on the inductor. This means the current will be a sawtooth, minimum 8.5A, maximum 11.5A.

When top MOS is ON :
During 8.3µs we have 25V across the inductor, we want the current to increase by 3A.
v=L di/dt, therefore L = V dt / di, so L= 70µH

When top MOS is OFF :
During 41.6µs we have 5V across the inductor, we want the current to increase by 3A.
same formula, L= 70µH again.

Note this inductor is smaller than the one in the schematic. Their ripple current could be lower.

This is still too small :
Invalid Request

Since a 70µH 10A inductor is impractical, using modern parts, we can switch at 300kHz (15x faster) which makes the inductor 15x smaller, ie, 4.7µH.

4.7µH 10A is easy to find, cheap 14x14mm SMD inductor. Here's one.

Invalid Request

> for big value inductors with high current ratings i
> need very big inductors in size and /or with air gap also.
> how about the metarial type?

Check out this discussion :

inductance - How to wind a toroid for 170 uH Inductor - Electrical Engineering Stack Exchange

For a DC-DC, large value inductor means :

- more turns, more copper wire length : either higher resistive losses, or more expensive (more copper), or both
- lots of core material (since it has a maximum flux density) : more expensive, higher hysteresis losses in core
- larger bulk and weight : annoying, complicates routing, leaks more EMI, indirectly more expensive, detaches/breaks your board in case of shock, etc
 
OK !

All you need is v=L di/dt

For a buck switching converter with these parameters :

Vin 30V
Vout 5V
Iout 10A

Duty cycle d = 5V/30V = 0.166

Let's pick a switching frequency F=20kHz.
Period is T = 1/F = 50µs.
Top MOS conducts for T*d = 8.3µs
Bottom MOS (or diode) conducts for T*(1-d) = 41.6 µs

We can design for a ripple current of 3A on the inductor. This means the current will be a sawtooth, minimum 8.5A, maximum 11.5A.

When top MOS is ON :
During 8.3µs we have 25V across the inductor, we want the current to increase by 3A.
v=L di/dt, therefore L = V dt / di, so L= 70µH

When top MOS is OFF :
During 41.6µs we have 5V across the inductor, we want the current to increase by 3A.
same formula, L= 70µH again.

Note this inductor is smaller than the one in the schematic. Their ripple current could be lower.

This is still too small :
Invalid Request

Since a 70µH 10A inductor is impractical, using modern parts, we can switch at 300kHz (15x faster) which makes the inductor 15x smaller, ie, 4.7µH.

4.7µH 10A is easy to find, cheap 14x14mm SMD inductor. Here's one.

Invalid Request

> for big value inductors with high current ratings i
> need very big inductors in size and /or with air gap also.
> how about the metarial type?

Check out this discussion :

inductance - How to wind a toroid for 170 uH Inductor - Electrical Engineering Stack Exchange

For a DC-DC, large value inductor means :

- more turns, more copper wire length : either higher resistive losses, or more expensive (more copper), or both
- lots of core material (since it has a maximum flux density) : more expensive, higher hysteresis losses in core
- larger bulk and weight : annoying, complicates routing, leaks more EMI, indirectly more expensive, detaches/breaks your board in case of shock, etc

thanx so much for the explanation. i know how to design and calculate a buck converter.

i had asked how to calculate the inductor.
how to choose a toroid ,E , EI ... whatever its shape. then how many turns you have to wire in order to get desired inductor value.

lets say we need 140 uH 10 A inductor to wire. so how to determine the parameters like shape, size, metarial ...
 
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