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This is a buck converter circuit I am analyzing: enter image description here Here the input voltage is 5 V and output 3.3 V. I am interested to calculate following parameter,

Input Inductor peak current

I know the basic equation for inductor ripple current is: enter image description here

Q1) Can I use this formula for calculating input ripple current?

Q2) What will be the case if the input inductor is replaced with ferrite baed? I mean how can calculate the input ripple current?

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  • \$\begingroup\$ Which method would you understand? time or frequency domain or impedance ratios of harmonic switch spectrum>? current is almost the same mean but THD is different. \$\endgroup\$ – Sunnyskyguy EE75 May 10 '18 at 13:10
  • \$\begingroup\$ What is the f switch rate?? This makes a big difference and which bead did you have in mind? \$\endgroup\$ – Sunnyskyguy EE75 May 10 '18 at 16:11
  • \$\begingroup\$ @Tony, any method is fine wit me. Can take the switching frequency as 2.25 MHz \$\endgroup\$ – vt673 May 10 '18 at 17:44
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If there were no components at the input, the current drawn by the buck converter would have been a triangular, or trapezoidal one (depending on its configuration), and the formula for it would have been the one you gave (ideally).

But the input is now a lowpass LC filter, of a characteristic frequency of \$\frac{1}{2\pi\sqrt{LC}}\$. Whether it's an ideal filter, or not, the transfer function is known (simple LC lowpass, with or without R). Since the input current can also be deduced (you know the topology and the value of it), you can also deduce the harmonic content, and apply it to the transfer function.

For example, here's a simple buck example (the IC in your example works at MHz range, for the sake of time, I chose a smaller switching frequency). L=22\$\mu\$H, C=22\$\mu\$F, RL=3.3\$\Omega\$ => IOUT=1A

$$\Delta_{I_L}=\frac{(5-3.3)*0.666}{22e-6*100e3}=0.515$$ $$I_{MAX}=I_{OUT}+\frac{\Delta_{I_L}}{2}=1.257$$ $$I_{MAX}=I_{OUT}-\frac{\Delta_{I_L}}{2}=0.743$$

Here's how it looks:

Iout

Since the minimum current is greater than zero, the input current will be trapezoidal, with the calculated values:

Isw

And, since the input filter is an ideal one, its transfer function is:

$$H(s)=\frac{\frac{1}{LC}}{s^2+\frac{1}{LC}}$$ $$f_p=\frac{1}{2\pi\sqrt{1e-6*10e-6}}=50.33\text{kHz}$$

The switching frequency is 100kHz, the filter has a -40dB/dec slope, the harmonic content of the input current can be found out, so here is the filtered input current's FFT:

fft

and its time response, which is the sum of all the filtered harmonics:

filt

If there was no input filter, there would have not been any filtering, so the current would have been unchanged. Also, this type of current can be analytically expressed as a Fourier series, and, given the known input filter's transfer funcion, the attenuation of each harmonic can be determined, then mathematically deduced, but it's time-consuming and one of the reasons FFTs and simulators exist.

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  • \$\begingroup\$ The IC uses a half FET bridge with half diode bridge across FETs , so the waveforms will not match yours, but general comments ok. \$\endgroup\$ – Sunnyskyguy EE75 May 10 '18 at 16:50
  • \$\begingroup\$ @TonyStewartolderthandirt True, I only read the quick description to find out what frequency I should use, in rest I just assumed a simple case-scenario. But now that I know this, it's one more reason to think OP should not follow the "mathematical" path to FFT, for sanity reasons... \$\endgroup\$ – a concerned citizen May 10 '18 at 17:12
  • \$\begingroup\$ Any comments for ferrite bead instead of input inductor? \$\endgroup\$ – vt673 May 11 '18 at 4:53
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    \$\begingroup\$ @vt673 They're usually nH, or tens of nH, unless the bead comes with a turn or two (I've seen cases), in which case the value will be more, but still not that much. They also have less series resistance. Which means the cutoff frequency will shift upwards and it will not filter out the current as much as the "regular" LC. The peak resosnance will also, probably, be more pronounced, and if that happens at the switching frequency, instead of attenuation you'll get amplification. The thing to remember, you have a filter, so you're interested in cutoff, rather than a specific type of inductor. \$\endgroup\$ – a concerned citizen May 11 '18 at 5:02
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This is in addition to other answers.

This can be modelled as an LCLC filter with the 2nd L switched with a square pulse and a resistive load.

But in reality, each reactive part will have an ESR value that affects power dissipation and ripple voltage and current.

The primary ripple L current will be approximately the same as the secondary L current except the rise time of primary current will depend on that part's L/R=T ratio for conducted noise back to the source, which may or may not be relevant.

The power out may be viewed from an energy conversion and loss perspective by looking at the DCR of each coil and ESR of each cap with an output of 3.3V@1A or 3.3W at 3.3 Ohms.

The impedance of L2 of 2.2 uH at 2.25MHz is around 33 Ohms or 10 times the minimum load R.

The impedance of L1 of 1uH is about 15 Ohms, so a Ferrite Bead that is rated at 10 Ohms at 1MHz may be considered similar but not exactly the same, depending on rated current, SRF , saturation level etc as it will draw high currents charging up C1.

C1 must also have ultra low ESR due to the high edge currents to minimize power dissipation in a tiny part and must be lower than the big FET switch which is also low RdsOn.

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