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I need to boost 3.3V all the way up to 12.0V.

Current requirements are rather small, I only need to power an SSD1309 OLED driver (from pg7):

Common maximum sink current: 40mA

Almost all of the available high-quality boost converters are providing way more current than this, so I decided to use the smallest converter I can get: MP3438:

The MP3438 is a highly integrated boost converter with a 1.2MHz fixed frequency and a wide input voltage (VIN) range. The MP3438 starts up from a VIN as low as 2.7V, and can support up to a 2A switching current limit with integrated, low RDS(ON) power MOSFETs.

This is their recommended typical application converting 3.3V to 12V: enter image description here

There is a section on pg16 "Selecting the Inductor":

An inductor is required to transfer the energy between the input source and the output capacitors. A larger-value inductor results in less ripple current and a lower peak inductor current, which reduces the stress on the power MOSFET. However, a larger-value inductor is physically larger, has a higher series resistance, and has a lower saturation current. For most designs, the inductance can be estimated with Equation (6):

L = Vin*(Vout-Vin)/(Fsw*Vout*dIl)

Where dIl is the inductor ripple current.

Choose the inductor ripple current to be approximately 20% to 50% of the maximum inductor average current.

Question: what is "the maximum inductor average current"? Is it the 40mA? Or RMS Current (Irms) which is 4.4A for the recommended inductor?

Calculating the above formula with the recommended 3.3uH inductor:

dIL = Vin*(Vout-Vin)/(Fsw*Vout*L)
dIL = 3.3*(12-3.3)/(1200000*12*3.3/1000000)
dIL = 0.60417

This means ripple is 604mA?

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  • \$\begingroup\$ please add the image of led driver and whole lcd module or ref link for similar one. \$\endgroup\$
    – Chr_arj
    Commented May 9, 2023 at 7:36

2 Answers 2

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Question: what is "the maximum inductor average current"?

It is the average inductor current when the current loading on the 12 volt output is greatest. Maybe the boost calculator from my basic website might help a tad: -

enter image description here

I've set the load resistance to 300 Ω to match the 40 mA load current stated in the question.

So, it's running in DCM (discontinuous conduction mode) and, the average inductor current is 145 mA as expected for an ideal circuit. The calculator tells you that the ripple current is 419 mA p-p. The reason why the ripple current value disagrees with the data sheet guidance is because the data sheet assumed the chip would be used in CCM mode.

There's always a more ripply current in DCM because of the inductor current depleting to zero.

The calculator is for an ideal circuit; there will be switching losses, diode volt drops and inductor resistance losses.

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  • \$\begingroup\$ I'm a little bit confused here. 1.) Running in DCM due to low load? Is it the only reason of running in DCM? 2.) "The reason why the ripple current value disagrees with the data sheet guidance is because the data sheet assumed the chip would be used in CCM mode. There's always more ripple current in DCM." - completely lost: this means Ir(DCM)>Ir(CCM), right? But your site gave 419mA DCM, while datasheet formula gave 604mA (CCM as per your assumption?)? \$\endgroup\$
    – Daniel
    Commented May 9, 2023 at 10:39
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    \$\begingroup\$ @Daniel formula 6 is when in CCM and, CCM means a high load current. CCM cannot be sustained at anything like 40 mA load current. Your circuit will run in DCM and you can't use formula 6. \$\endgroup\$
    – Andy aka
    Commented May 9, 2023 at 10:48
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    \$\begingroup\$ So, just to be clear, in the example in my answer, ripple current is always \$I_{PEAK}-I_{MIN}\$ measured in amps peak-to-peak @Daniel \$\endgroup\$
    – Andy aka
    Commented May 9, 2023 at 14:21
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    \$\begingroup\$ @Andyaka FYI the transistor in your diagram is backwards (P-ch). Curious what your thoughts are on compensation; ordinarily I would assume the calculation should use the device's nominal current range, but this device specifies "Adaptive Constant-Off-Time" whatever that means (but not much apparently, as Vcomp is shown directly compared to low side current?). \$\endgroup\$ Commented May 9, 2023 at 14:37
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    \$\begingroup\$ @Andyaka Just one of those boners, you look at it a hundred times and never catch it... it happens. Cheers :) \$\endgroup\$ Commented May 9, 2023 at 17:26
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"the maximum inductor average current" should be the output current (+\$I_{R1}\$) divided by the fraction of time inductor current flows in the direction of the output. Display current may be 128*320 μA, about 40 mA.
I'd assume current limiting to be dissipative and aim for a voltage low in the range of 7-16 V.
For 12(8) V, 40 mA the average current should not be much higher than \$\frac{12(8)}{3.3}\times 40 mA\$: 145(97) mA.

The computation of the ripple current looks correct, note how it is about 30 % of the MP3438's rated current.

With the current as low as this, it may be hard to find an inductor that allows operation in continuous conduction mode (CCM).
This would need a greater inductance than typical application, which may make it bigger in spite of not needing to be rated for 2 A.
Dis-continuous operation (DCM) shouldn't hurt much, but makes eye-balling \$I_{RMS}\$ and \$I_{peak}\$ harder: simulate!

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  • \$\begingroup\$ I wish to know which value's 20-50% shall be the inductor ripple current. Shall it be 20-50% of MP3438's rated output of 2A? Or shall it be 20-50% of the inductor's current 4.4A? \$\endgroup\$
    – Daniel
    Commented May 9, 2023 at 10:33
  • \$\begingroup\$ "would need a greater inductance": how much greater exactly? \$\endgroup\$
    – Daniel
    Commented May 9, 2023 at 10:41
  • \$\begingroup\$ My take is inductor's current. Not rated maximum, but average in application/design. Less than 20 % would be well into the region of diminishing returns. I think 50 % as a limit arbitrary. 100% would be DCM. For low losses at low currents, DCM is fine, but watch the frequency go down and requirements for the output capacitor/filter go up with off-time. \$\endgroup\$
    – greybeard
    Commented May 9, 2023 at 10:53
  • \$\begingroup\$ Minimal inductance for CCM: just set ripple current to twice average (145 mA) and solve. "Proportional" within CCM, but, as Andy aka commented, this formula no longer applies in DCM. \$\endgroup\$
    – greybeard
    Commented May 9, 2023 at 11:17

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