[SOLVED] Power Electronic need help solving this :( - I_in_rms value.

[jokerx]

dear ppl: i have problem with this problem...ive searched all internet and my books..cant find answer to this or formula...

An ideal buck-boost converter with a switching frequency fsw = 200 kHz is
operating with a duty cycle d = 33.333 % and delivers an output power of
P = 50.00 W into a purely resistive load. Its input voltage is Vin = 20.00 V and
the inductor is L = 10.00 μH. All components can be assumed to be ideal and
the converter is operating in continuous conduction mode. What is the RMS
value of the input current Iin?

 

[KerimF]

It is 4.33 A
Now your turn to find out why

 

[jokerx]

Thanks it was helpful

 

[KerimF]

Since it is an ideal case then P_in = P_out
P_in = V_in * I_average = P_out

You know V_in and P_out

OK...
I_average = P_in / V_in = 50 / 20 = 2.5 A (but this is not the RMS value)

The input current is a pulsed type with a duty cycle Dc=1/3.

I_peak * Dc = I_average
I_peak = I_average / Dc
I_peak = 2.5 * / (1/3) = 2.5 * 3 = 7.5 A


Be tuned...

 

[jokerx]

ok thanks aloot

 

[KerimF]

I_rms = SQRT [ (I_peak * I_peak) * Dc ]

 

[jokerx]

Man your the greatest

 

[KerimF]

That is why the lower the duty cycle is, the higher the Irms becomes for a constant average current. This is important to know since a wire is heated more if its current is pulsed than being DC even if both have the same average value.

Note:
The formula: I_rms = SQRT [ (I_peak * I_peak) * Dc ] is good if I_peak is constant during the on-state of the duty cycle.
If it varies then we will need to use... integral which you will learn later I guess

 

[jokerx]

aha..i learned more here with you than with the teacher ...can you give me a hint where i can find info about these kind of things...and How would i solve it IF ive been asked to find the IC capacitor RMS current?

 

[KerimF]

I wrote this from memory. I was graduated 35 years ago
In general if the current is a function of time:
I = f(t)
We can find its RMS in a period of time (t1 to t2) approximately by dividing this period to N equal intervals for example so that the current I(t) is almost constant during each interval.
Let us assume:
I1 in interval 1
I2 in interval 2
I3 in interval 3
etc...

Irms = SQRT [ (I1^2 + I2^2 + I3^2 + ..... In^2) / N ]

Later you will learn that if we let N be very big (infinity), the sum may be called the integral of the function between t1 and t2 which is a course by itself

 

[jokerx]

i see . thanks alot.. what if i had to look for Ic capacitor RMS ..would i calculate the same way?

 

[KerimF]

Sorry I didn't get what you mean by Ic ... how is it generated? Is it periodic?
Ic here is like one asks for Ir resistor RMS !
Perhaps a simple circuit can clarify what you have in mind.

 

[jokerx]

The rms value of the Ic through capacitor

 

[KerimF]

For an ideal buck-boost:
V_out = V_in * Dc / ( 1 - Dc)
V_out = 10 * 2/3 / ( 1 - 2/3) = 20 V

Since P_out = P_in and P_out = V_out * I_load:
I_load = P_in / V_out
I_load = 100 / 20 = 5 A

We can devide one period by 3; equal intervals
In 2 intervals, the capacitor is discharged by the load with a current 5 A
In 1 interval, the capacitor should be charged with a current so that:
I_chg * t_chg = I_dis * t_dis
or
I_chg = I_dis * t_dis / t_chg
But the ratio t_dis / t_chg = 2 / 1 = 2
I_chg = 5 * 2 = 10 A

So for 2 intervals the current is Ic1 = 5 A
And for 1 interval the current is Ic2 = 10 A
Their signs (positive and negative) are not important here since we will take their square.

Ic_rms = SQRT { [ (Ic1^2)*2 + (Ic2^2) * 1 ] / 3 }
Ic_rms = SQRT { [ (5^2)*2 + (10^2)*1 ] / 3 }
Ic_rms = SQRT { [ 50 + 100 ] / 3 } = SQRT(50)
Ic_rms = 7.07 A

I hope it is clear.

 

[jokerx]

yep thanks alot.. i found the same value as u