Hi, I'm prepering for a switched-mode converters exam and this old exam question is bothering me. I'm talking about e) and f) subquestions. I can't get my head around it, despite I think it's really easy. Do you calculate it with I_in=I_L=(I_out/D')? What are the right answers for these?
Easy Inductor Current Figure problem
- Thread starter LeoT
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The average voltage across an inductor must be zero, i.e. the average current derivative is zero. So if I_L goes from 0A to 3A during one switching period, is the I_out 1,5A? Or is it higher (1,8A) due to the duty ratio being 0,6?
Hi,
Yes the average voltage is zero and that is often helpful, but another way of looking at it is to simply equate two values of voltage and two values of time thusly:
v1*t1=v2*t2
That makes it simple to see that if v2 is different than v1, then t2 must be different than t1 in order to maintain the 'balance'.
An example is when energizing a relay coil. We energize a coil with a 12 volt DC voltage for 2 seconds. How long does it take to completely discharge the inductor with a 2 volt negative voltage (or just a 2v clamp)?
Well, since the volt seconds charging is 24 and the discharge voltage is 2 volts, that means we need to balance this:
24=2*t2
Solving for t2 we get:
t2=12
and now we have balance:
24=24
See how simple that is?
But yes a consequence of this is that the average voltage is zero and that helps in problems too.
As to your question about the current, the current is usually assumed to be triangular, so do you know what the average of a triangular wave shape is?
Note the voltage across the inductor is very nearly rectangular so we call it rectangular for a simplification.
Also note that the volt seconds is sometimes thought of as energy so it's an energy balance. Whatever gets stored during charge gets removed during discharge.
If you play around with (in units) V*s=J/A you see more into this.
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This is Homework Help, not Homework Do, so we give help not answers.
What are you confused about?
Sorry for asking so directly. This isn't my homework, it's an old exam question which I want to know the right answer to know if I got it right or not as there are no answers provided for these.
Question e) If we assume that the converter is a Boost Converter, what is the average output current?
The Boost Converter has two states:
State #1 - The switch is ON
State #2 - The switch is OFF
So, what can you tell us about the Output Current when the Switch in ON vs when the Switch is OFF ?
Neither one. It is true that the inductor current goes from 0A to 3A during one switching period. And notice that in the boost converter the inductor is in series with the input source hence the inductor average current must be equal to input average current I_in_avg = IL_avg=?
On the other hand in the boost converter the load is in series with the diode, hence the average output current must be equal to the average diode current Id_avg = I_out_avg=?
And for your example the diode current will look like this:
Is that clear?
So the output current is flowing only when the switch is OFF.
I'm not 100% sure but I think I undestood it now. When the switch is off is when output current is happening and I_D slope is the same as I_out slope. The I_out average is 1,2 A.
Wrong, do the math again.
No, usually we have a big capacitor at the output. Hence during the ON time ( Switch is ON but diode is OFF), this output capacitor provides a current to the load. And in the steady-state condition, we know that the average current through a capacitor is zero (I_charge = I_discharge) and this is why Id_avg = I_out_avg is true in steady-state condition.
Hi,
Not sure what circuit you are referring to there, but you know what the average of a triangle is so you can use that to determine the average current because we always assume a triangle for the full cycle or part of the cycle. I'll give examples.
First, the average of a triangle shape wave that spans the entire switching cycle is Vpk/2 where Vpk is the peak. So if you have a triangle with peak of 5 units that goes from 0 seconds to 2 seconds and the switching cycle is 2 seconds then the average is 5/2 which is 2.5 units.
Now in some converters the current does not flow for the entire cycle, so you have to split the calculation into two parts: the part with the triangle and the part with zero. For example, say we have a switching cycle of 2 seconds and the triangle is there for 0 to 1 seconds with peak of 5 units, and from 1 to 2 seconds the wave is zero.
First we do the triangle, and since the triangle part is 5 units high the average of that part alone is 5/2 as before which is 2.5 units again. Since the time the triangle is 'on' is 1/2 the time of the entire cycle, we have to divide that by 2 so we get 2.5/2 which is 1.25 units. So the average is 1.25 units.
For another example, say the switch cycle is 3 seconds and we have the following wave:
0 units for 0 to 1 seconds,
triangle with peak 6 units spanning from 1 to 2 seconds,
0 for 2 to 3 seconds.
Starting with the triangle, we get 6/2 which is 3 units.
Now we have two areas where it is zero and they are for 2/3 of the time, so the triangle is 'on' for only 1/3 of the time, so we multiply by 1/3 and we get:
3*1/3=1 unit
So the average here is just 1 unit.
As a final example, say we have a cycle that has two triangle shape waves and one zero portion. The switch cycle is 3 seconds, and we have the following:
triangle from 0 to 1 seconds with peak 1,
triangle from 1 to 2 seconds with peak 3,
zero from 2 to 3 seconds.
The average of the first triangle is 1/2 units.
The average of the second triangle is 3/2 units.
The duty cycle of the first is 1/3, so we get (1/2)*(1/3)=1/6 units.
The duty cycle of the second is 1/3, so we get (3/2)*(1/3)=1/2 units.
The sum is 1/6+1/2=1/6+3/6=4/6=2/3 or 0.666667 units.
So the average of this dual triangle wave is 0.666667 units.
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That makes perfect sense. So as stated in the question, we are talking about a Boost-converter circuit and using Boost-converters equations, I got 0,6A for the output current and 1,5A for the input. For example I get the same 0,6A from (3A/2)*0,4.
Yes, you got the correct answer.
