# Inductor Value for TPS6107

#### John Miller

Joined Feb 14, 2018
3
Hello,

I am trying to find the inductor value for a circuit involving a TPS6107.
http://www.ti.com/product/TPS61070/datasheet/application_and_implementation#SLVS5103348 I used this link,section 11.2.2.2 for my calculations.

My values are
Output Voltage = Vout = 5v
Initial Voltage = Vbat = 3.7v
Desired Current = Io = 380mA
Switching Frequency = f = 1200 kHz

I begin by trying to find the "maximum average inductor current", equation 3, which they say is equal to Io * (Vout/(Vbat*.8)). Plugging my values in, I get 641.89mA.

Now for the next part, I compute L with equation 4. I assume that the maximum average inductor current is the "ripple current",delta I sub(L), because in the link, equation 3 is supposed to represent the maximum average inductor current but I believe the symbol does not load correctly as only the subscript L is shown. So I use 640 mA, w respect to sig figs, as my ripple current. I plug my my values in for equation for 4 and get .00000125 = L.

My questions are:

Was I right to assume the maximum average inductor current is the ripple current?

For equation 3, Vbat is multiplied by .8 but the efficiency of my booster is .9. Should .9 be there instead or is there no relation?

#### ebp

Joined Feb 8, 2018
2,332
I don't know what that weird sub-L is supposed to be. The average inductor current in a boost converter is equal to the average current from the input power supply. The 0.8 appears to be an efficiency estimate fudge factor that is conservative. It will slightly overestimate the required current rating for the inductor for most real circuits using that part. It is very unlikely that the overestimate will be of any practical concern. The current rating of most real inductors is not a "hard" number, but rather something based on some assortment of parameters. It is often based on a reduction of inductance due to the average current's effect on the core material. If the inductor is gapped ferrite, more care is required. If it any sort of "powdered" material there is usually lots of room to play. You don't get a lot of choice n off-the-shelf inductors in any case.

The actual inductance is almost always selected based on the magnitude of peak-to-peak ripple current that will be allowed. Unless some specific factor dictates, this allowable ripple is a rule-of-thumb thing. P-P ripple of 20% - 30% of the average inductor current is a common target.

What it is going to come down to is that you estimate the average current, pick an allowable ripple current, go to the catalogs to see what you can buy off the shelf, pick something that looks close, then do the calculations with the values of the part you selected to see if everything is OK and acceptable. For example, if the average current is 600 mA and the p-p ripple is chosen as 120 mA and that yields an inductance of 4 µH, you may find inductors with a choice of current ratings of perhaps 500 mA and 700 mA in values like 3.3, 4.7, 5.6 µH - in small inductors you won't find 4 µH off the shelf. You might in something rated for 10 or 20 A. The 500 mA rated part might be perfectly OK, especially if your original efficiency estimate was overly conservative, but you have to critically evaluate it, which means a good data sheet and a clear understanding of all the hooks. If you are only building a few and neither cost not board area is critical, go with the 700 mA rated part. I'd probably pick the 4.7 µH part, but perhaps the 3.3. If you have two inductors both rated at 700 mA and nominally 4.7 µH, one with a powdered iron core is going to actually have notably different properties from one with a different powdered iron formulation or "sendust" or moly-permalloy (MPP) core material. Powdered iron is cheap. MPP is expensive. If you are going to build a million of something, spending a thousand dollars of engineering time to save one cent per inductor is worthwhile.

#### John Miller

Joined Feb 14, 2018
3
I don't know what that weird sub-L is supposed to be. The average inductor current in a boost converter is equal to the average current from the input power supply. The 0.8 appears to be an efficiency estimate fudge factor that is conservative. It will slightly overestimate the required current rating for the inductor for most real circuits using that part. It is very unlikely that the overestimate will be of any practical concern. The current rating of most real inductors is not a "hard" number, but rather something based on some assortment of parameters. It is often based on a reduction of inductance due to the average current's effect on the core material. If the inductor is gapped ferrite, more care is required. If it any sort of "powdered" material there is usually lots of room to play. You don't get a lot of choice n off-the-shelf inductors in any case.

The actual inductance is almost always selected based on the magnitude of peak-to-peak ripple current that will be allowed. Unless some specific factor dictates, this allowable ripple is a rule-of-thumb thing. P-P ripple of 20% - 30% of the average inductor current is a common target.

What it is going to come down to is that you estimate the average current, pick an allowable ripple current, go to the catalogs to see what you can buy off the shelf, pick something that looks close, then do the calculations with the values of the part you selected to see if everything is OK and acceptable. For example, if the average current is 600 mA and the p-p ripple is chosen as 120 mA and that yields an inductance of 4 µH, you may find inductors with a choice of current ratings of perhaps 500 mA and 700 mA in values like 3.3, 4.7, 5.6 µH - in small inductors you won't find 4 µH off the shelf. You might in something rated for 10 or 20 A. The 500 mA rated part might be perfectly OK, especially if your original efficiency estimate was overly conservative, but you have to critically evaluate it, which means a good data sheet and a clear understanding of all the hooks. If you are only building a few and neither cost not board area is critical, go with the 700 mA rated part. I'd probably pick the 4.7 µH part, but perhaps the 3.3. If you have two inductors both rated at 700 mA and nominally 4.7 µH, one with a powdered iron core is going to actually have notably different properties from one with a different powdered iron formulation or "sendust" or moly-permalloy (MPP) core material. Powdered iron is cheap. MPP is expensive. If you are going to build a million of something, spending a thousand dollars of engineering time to save one cent per inductor is worthwhile.

Thank you so much man I really appreciate the help, cleared up a lot for me.

Cheers

#### Colin55

Joined Aug 27, 2015
519
There is a lot more to selecting an inductor than uH. You need to buy a whole lot of inductors and try them and you will be amazed.
The DC resistance is important and the material of the core and the size of the core and the number of turns.