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This is a buck converter circuit I am analyzing:
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:
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?
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:
Since the minimum current is greater than zero, the input current will be trapezoidal, with the calculated values:
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:
and its time response, which is the sum of all the filtered harmonics:
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.
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.
