Buck boost and inductor calculations

Discussion in 'General Electronics Chat' started by ke5nnt, Apr 5, 2011.

  1. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    I have tried and tried to find the answer to this myself, but I submit to defeat. I'm working with a DC/DC Step-up/Down/Inverting Switching Regulator, the NCP3064 in a step-up configuration. My goal is a voltage input of ~11 to 14V DC with an output of 19.2V at 350mA (can be over 19.2V but not under).

    There is a very handy design tool for the NCP3064 which is an Excel spreadsheet that does calculations for you, found HERE. In this design tool under the "Boost" tab, Section 3 specifies Set ΔIL/IL(avg), then it specifies "it is suggested that ΔIL be chosen to be less than 10% of the average inductor current, IL(avg)."

    I am quite confused on this point. What is ΔIL? and where can it be found on a datasheet for an inductor? I have looked at a number of inductor spreadsheets in an effort to find these values (ΔIL and IL(avg)). I have found Idc (DC Current) and Isat (saturation current), as well as Irms which I'm not sure what that means exactly, I think it's root mean square current.

    Biggest problem I'm having is how to determine the ripple current of a given inductor? Perhaps the ΔIL/IL(avg) is just the tolerance of the inductor?
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    ΔIL is the ripple current through the inductor which is superimposed in ILavg. When they say that ΔIL has to be less than 10% of ILavg, it means that if the average current through the inductor is 1A, then ΔIL<=0.1A.

    Now, Idc specifies the maximum DC current though the inductor.

    Idc=ILavg+ΔIL/2

    Irms is the root mean square current and it determines the power dissipated in the inductor. If ΔIL is small, like in your case, you can assume that Irms=Iavg.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    ΔIL is the change of inductor current. You won't find it on a datasheet, because it's a function of your switching frequency, inductor uH, and how much current your output is.
    As the switching frequency increases, the ΔIL decreases.
    As the current increases, the ΔIL increases.
    Increase the size of the inductor in uH, and the ΔIL decreases.

    There is no spreadsheet for the NCP3064 in particular, but there are for NCP3063, NCP3065, and several others.
     
  4. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Ok, first, thank you for the responses. I have 3 more questions now which should help shine more light on this for me...

    1. How do you determine what the best switching frequency is? Without external components, it should be between 50kHz and 150kHz but I'm not sure where even to start on that.

    2. Assuming the spreadsheet for the NCP3063 is relevant to the 3064, back in the section 3 slot for ΔIL/IL(avg) where the units is a %, this is a value I'm supposed to fill in, but I'm not sure how to determine which value to put. Is it completely up to me? Inductor datasheets are much shorter than I'm used to, I'm not sure what information on them is important for this, if any.

    3. In searching for an inductor, what is the best way to determine what the max DC current value should be? Should I want to keep close to the actual load that the buck boost is supplying (~300mA) in an effort to keep the physical size of the inductor down (somewhat important) or should I shoot for the max load of the buck boost (1.5A)?

    Thanks
     
  5. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    In addition to the questions above (and I hate to bump a thread but this one is getting lost)...

    On the .xls worksheet, I have entered the following, please tell me if I'm doing this properly:

    Design Parameters
    VIN 11V
    Vout 20V
    Iout 0.35A

    Input Conduction Loss Parameters
    Vf 0.5V (diode forward voltage of Schottky Diode
    VSAT (a parameter that is calculated by the .xls) given at 1.4V

    Set ΔIL/IL(avg)
    ΔIL / IL(avg) 10% (based on the given Iripple value of 0.07A and the Inductor Irms value given of 0.7A. 0.07A / 0.70A = 0.1 or 10%
    Iripple is a given value at 0.07A (I assume a ripple current of 70mA?)

    Input Target Frequency
    Target F 150kHz

    The Spreadsheet Calculates key Parameters for the Remaining Component Selection
    Ton/Toff 0.99
    Ton 3316nS
    IL(avg) 0.70A
    Ipk(switch) 0.73A
    RSC 133mOhm (minimum short circuit resistor value)
    L 471uH (rounded to closest available inductor value of 470uH
    Rl (Inductor winding resistance from datasheet) 0.850Ω
    Toff 3μS
    Duty Cycle 50% (max duty cycle for the IC is 83%

    Optimize Cout and Cin for Ripple Voltage
    Cout 220μF
    ESR (Output capacitor ESR) 160mOhm
    Vripple(pk-pk) max 117mV
    Vripple(pk-pk) min 29mV
    Cin 100μF (based on ripple requirements)

    Calculate Resistor Divider for Setting Vout
    Vout 20V
    R2 2.32kOhm
    R1 34.8kOhm

    Optimize Vout with standard R1 values
    R1 (Pick nearest E96 1% R value for R1) 34.0kOhm
    Vout 19.57V

    Iterate CT to achieve desired target frequency
    CT 2.2nF
    Fcalc 150kHz

    Efficiency is 83.42%

    Have I done this correctly please? Also, the questions I asked in the post above are still important to me if anyone can answer those.

    Thank you!
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Generally, the higher your switching frequency, the less the value of inductance you'll need to achieve a given ΔIL/IL(avg). Being able to use a smaller inductor can save on space and cost. However, there may be more losses incurred in the switch itself during the transitions from on to off, or vice versa.

    The smaller the % you fill in there, the larger value of inductance in uH you'll need. If you fill in a large % value, you risk burning up the switch due to excessive peak current, along with inductor saturation.

    If you're going to err, then err on the side of caution. You don't want to be operating the inductor near 100% capacity continuously, as probably won't be very efficient, and might overheat on you.

    It seems to me that we went 'round and 'round with a similar regulator to this one about 6 months ago, and I don't know if you ever got things resolved.
     
  7. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Yes Sir Sarge, you're correct. You are referring to This Thread. Rather than dig up an old thread I decided to start a new one, mostly because it's a different IC, even if it is just a newer version of the same. Also, some parameters of mine have changed and I'm able to get a bit more to the point on a new thread.

    To answer your question, I never did get around to completing the project due mostly to life issues. I've been having a rough year. With that said though, I'm back on it again now which is why I'm digging up old issues. But I think this covers it and I'm pretty sure I've got it figured out now. Now that I have values I can work on selecting components, doing a schematic and working out a board layout. Then I just need to get me a protoboard built and I can give it a shot. Still plenty to do but the hard part is done.

    I owe you and AAC a lot, your feedback is always great. Thanks for all the help!
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    As far as the inductor and the ΔIL / IL(avg), I'd go more around 20.8% which would lower your inductor requirement to 220uH. You'd have more ripple on the output, but how much of a big deal is that with you? The inductor you chose (470uH) would be operating at max DC current, which isn't good.

    Here are a couple of 220uH inductors to look at:
    http://www.mouser.com/ProductDetail...EpiMZZMsg%2by3WlYCkU3FR%2bTyIBZRjxRqAdYlrSXo=
    Not small being toroidal, but comparatively low DC resistance.

    http://www.mouser.com/ProductDetail...EpiMZZMsg%2by3WlYCkU7%2bI2Fr4H6Mu2zKNL1wbJA0=
    That one's rated for 1.3A, so you have a comfortable margin of safety.
    12.5 mm W x 12.5 mm L x 6 mm H - 1/2" square, under 1/4" high is pretty compact.
     
  9. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Aside from a larger (physical size) inductor, is there any adverse effect to staying with a lower ripple current and keeping inductance value at around 470uH? I realize the amp rating on the one I initially selected is lower than ideal, but there are 470uH inductors with larger current ratings. Found a 1.3A one that is about the same physical size ~12mm square. I'm just curious if there's an electrical trade-off between using lower inductance vs. higher ripple current?
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    With a smaller inductor you'll wind up with less overshoot, but more ripple.

    You'd probably be OK with the 470uH 1.3A inductor. The original one you chose would not work very well I'm afraid.
     
  11. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    I wanted to check the calculations on the NCV3063 xls design tool to make sure they came out correctly for the NCP3064.
    I checked the design equations from both datasheets (3063 and 3064) and all of the equations are the same with the exception of one, the calculation for V_{out}. I haven't bothered with this calculation just yet so for the purposes of this post, let's ignore that.
    I used all the same input values from the design tool spreadsheet for my "on paper" math. There are discrepancies so hopefully somebody will be able to tell me what's right.

    From the spreadsheet: \frac{T_{on}}{T_{off}} = 0.99

    Calculation from the datasheet:  \frac{T_{on}}{T_{off}} = \frac{V_{out}+V_{f}-V_{in}}{V_{in}-{V_{SWCE}} where V_{SWCE} is the Darlington Switch Collector to Emitter Voltage Drop from the 3064 datasheet. There may be some small error in the interpolation of the graph given. The Value I came up with was 0.8V. So...

    \frac{V_{out}+V_{f}-V_{in}}{V_{in}-{V_{SWCE}} = \frac{20+0.5-11}{11-0.8} = 0.8873 So, from the spreadsheet you get 0.99 and from the datasheet you get 0.8873.

    ************************************************************

    From the spreadsheet: I_{L(avg)} = 0.70A

    Calculation from the datasheet:I_{L(avg)} = I_{out} (\frac{T_{on}}{T_{off}} +1) = 0.35(0.8873+1) = 0.35(1.8873) = 0.661A

    ************************************************************

    From the Spreadsheet: I_{pk(switch)} = 0.73A

    Calculation from the datasheet: I_{pk(switch)} = I_{L(avg)} + \frac{ΔI_{L}}{2} = 0.661 + \frac{0.07}{2} = 0.696A I guess the "Δ" symbol doesn't work in TEX. the ??I_{L} should be ΔIL.

    ************************************************************

    Finally, from the spreadsheet: R_{SC} = 133mOhm

    Calculation from the datasheet: R_{SC} = \frac{0.20}{I_{PK(Switch)}} = \frac{0.20}{0.696} = 0.287Ohm

    Ok, sorry for all the math, I guess I just don't know which values are correct. I'm pretty sure I did the math correctly, but I'm more inclined to believe the spreadsheet considering it was designed by the IC manufacturer. Still, doesn't make sense to me why the calculations given in the datasheet wouldn't match what's on the spreadsheet.

    Thank you for the help!
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In spreadsheet they use V_{SWCE} = 1.4V and you are using 0.8V.
    But the simplest way is this

    D = 1 - \frac{Vin}{Vout} = 1 - \frac{11}{20} = 0.45


    L_min = \frac{ton*Vin}{2*IL}

    And you can cheek your calculation in this site
    http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html#Aww
    And also you have to remember that NCP3064 is not a classic PWM controller, The NCP3064 is a hysteric (PFM) controller. So this equation may not work in all cases.
     
  13. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    That's an interesting site you linked to there. Thank you for that. I wonder just how accurate it is to the 3064.

    ILcalc.png
     
  14. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    I'm still having trouble deciding what component values to select since I'm getting different values depending on how I calculate them. So I have 2 questions.

    1. I've never used SPICE before, is SPICE something that can simulate this circuit or not?

    2. Will it matter much if my values vary a bit? As in, will it make a big difference if I just choose 1 set of values over the other? See my math-filled post above.

    Thanks.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The site is very helpful and quite accurate.
    But you made the mistake you put L = 0.47H instate of 47μF = 47E-6
    Well of course you can use SPICE for example LTspice

    Witch values are different ??
    Simply use this site
    http://schmidt-walter.eit.h-da.de/smps_e/aww_smps_e.html
    L_min = (3.12us *11V)/(2*0.35A) = 47μH/1A
    The site propose 130μH so you could use 100uH/1A or 220uH/1A
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    He wanted to enter 470uH, which would be 470e-6 or .47e-3.
     
  17. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    And you should also remember that for L=47uH IL_min = 0.1A in CCM mode)
    And for 470uH IL_min = 20mA
     
  18. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    The values for certain components within the circuit. For example, Rsc from the spreadsheet is recommended to be 0.133 Ohm where the calculation from the device datasheet gives a value of 0.287 Ohm. Not sure which value to use.

    The one thing I noticed that caught my eye when comparing the equations provided in the datasheets of the 3063 (of which the spreadsheet is based on) and the 3064 (which I am using) is the equation for V_{out} is the only one that is different between the two devices.

    For the NCV3063: V_{out} = V_{TH}(\frac{R_{2}}{R_{1}}+1)

    For the NCV3064: V_{out} = V_{TH}(\frac{R_{1}}{R_{2}}+1)

    As far as I can tell, V_{TH} isn't even defined on the datasheet, so I'm not sure what value that is. Additionally, I'm sure the reversal of R1 and R2 in the equations would result in completely different values for selecting R1 and R2. You follow?
     
  19. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply use 0.15Ω or 0.22Ω low inductance resistors.

    As for for Vout and Vth
    For for this diagram
    [​IMG]

    V_{out} = V_{TH}(1+\frac{R_{1}}{R_{2}})

    And for Vth look to the datasheet more carefully
    [​IMG]
     
  20. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Oh wonderful! Can't believe I missed that, thank you!

    Dang that works perfectly! V_{TH}(\frac{R_{1}}{R_{2}}+1) = 1.25(\frac{34800}{2320}+1) = 20V
     
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