To design a boost converter for PV panel, Please refer to this publication, section "3.1. Boost Converter Design":
Modeling of Photovoltaic System with modified Incremental Conductance Algorithm for fast changes of irradiance
https://www.hindawi.com/journals/ijp/2018/3286479/
Dear Saad,
Sorry for the first answer which not satify you. however, there are many way to find the the input capasitor. And my 1st answer is also practical solution. But, since your question is spesific for proving :C1>=Io *α²/(0,02*(1-α)*V*Fs), Thus:
The method is charge capacitor C1. You can use simple equation DV= DQ/C1; where D is delta. DQ= Ic.ton/C1. where; Ic is the capacitor current.
You can see my attachment for complete prove of your equation. I already prove the equation that you looking for. Hope this my help you, brother.
Actually, You must concern with the power rating of your converter. Then how many voltage ripple that you permit for your application?
i=C.dV/dt.. is the basic equation you can use.simply used the delta aproximatiOn. It was work! Or you also may concern.. the calculation of output filter of the rectifier. If you used a good rectifier circuit as the input of your boost converter.. I think you don't have to consider the input filter for boost converter.
the input capasitor must handle the minimum voltage to boost.
To obtain the formula you should consicer the converter circuit during on state and off state. The base equation is the derivative form of the current - voltage relation in capacitor. (i = C x dv/dt).
In on state:
C x dv / t_on = i_T
where: C - capacitor capacitance, v - voltage of the capacitor, i_T -mean value of transistor current per switching period, t_on - on state time.
The mean value of transistor current is given by a following formula:
i_T = i_in x D
where: i_in - the value of converter input current, D - value of duty cycle.
In off state:
C x dv / t_off = i_d
where: t_off - off state time, i_d - mean value of diode forward current.
The mean value of diode current is given by a following formula:
i_d = i_in x (1-D).
Now we have two formulas:
C x dv / t_on = i_in x D
C x dv / t_off = i_in x (1-D)
It is possible to replace t_on and t_off with following components D x T_s and (1-D) x T_s respectively. (T_s - switching time period).
after that we have:
C x dv / (D x T_s) = i_in x D
C x dv / ( (1-D) x T_s) = i_in x (1-D)
Now we have to sum up the above equations. A following dependence was obtained:
C x dv / (D x T_s) + C x dv / ( (1-D) x T_s) = i_in x D + i_in x (1-D)
that gives:
C x dv / (D x (1 -D) x T_s) = i_in
taking into accout that T_s = 1/f_s, we obtain:
C = i_in x D x (1-D) / (f_s x dv)
where: f_s switching frequency, dv - maximum value of the input voltage ripples.
It is well known that the maximum value of voltage ripples appears for a duty cycle equal 0.5. Finally a following relation was obtained:
C = i_in / (4 x f_s x dv)
If you want to calculate the value of input current based on known value of output current, you may use the power dependence (100 % efficiency was assumed):
v_in x i_in = v_out x i_out => i_in = i_out x (v_out / v_in).
The exchange of power is made between a current source and a voltage source, thus the boost converter DC DC can be work with Cinput=0, the most important is the calculation of the inductance L and the switch frequency
Taking ito account that the design criterion for capacitors is that the ripple voltage across them should be less than 5%. Hence:
C= Vout* D/(R*f*deltaV_c)
Good luck
Dear Saad,
Sorry for the first answer which not satify you. however, there are many way to find the the input capasitor. And my 1st answer is also practical solution. But, since your question is spesific for proving :C1>=Io *α²/(0,02*(1-α)*V*Fs), Thus:
The method is charge capacitor C1. You can use simple equation DV= DQ/C1; where D is delta. DQ= Ic.ton/C1. where; Ic is the capacitor current.
You can see my attachment for complete prove of your equation. I already prove the equation that you looking for. Hope this my help you, brother.
Brother.. as witten on my note, you should calculate using Fourier transform to get Ic= α I..... eq (2). See again my first note. However, It need your extra work for math, I just give you the final result.But, I'm sure you can do it.
Lucky you, I got the simple way to get directly to : Ic= a.Io/(1-a) .... eq (5).
of course it takes a litle bit calculation, but simpler than the Fourier. See... my notes in attachments. Hope this will help you. :)
OH.. ups. Thanks bro for the correction.May be i need a cup of coffee. It soo late now.
First, I tried to find another approach beside Fourier. The main question was on the Eq. (2) on my first note.
Then I found on internet that formula...and tried to derive from it. Sorry for the mistake.
But, I recheck again. I think... it become the same as the result from Fourier. Bro.
Please check carefully... again.Hope this approach can help you.
But, if you unsatisfy with this method. Just try to calculate using Fourier to get the Eq (2).
Here is the attachment.
What is the maximum magnitude of voltage ripple allowed in input side? Also, if possible share a link on this.
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