# Need help with boost converter design

1. Dec 2, 2011

### Nat3

I'm trying to understand how to design a boost converter and have been reading about them for the past week or so. I've come a long way and now understand basically how they work, but I have a couple of questions on some things that are confusing me.

The first thing that is confusing me is the equation:

Vout/Vin = 1/(1-D)

I understand how the equation is derived (i.e. I can do the math), but it doesn't really make sense to me because the voltage across an inductor is proportional to the rate of change of current through it; so doesn't it seem like the voltage gain should be related to the frequency and not the duty cycle ? This is one thing that I'm really confusing about.

The other thing I'm unsure of is how to determine the appropriate frequency and inductor values for the circuit. None of the articles I've read about boost converters really explain how to choose these values. Any advice?

I'm basically at the point where I'm pulling my hair out here, so any help would be VERY MUCH appreciated

Last edited: Dec 2, 2011
2. Dec 2, 2011

### DragonPetter

I had the same intuitive question when I was studying this circuit. It makes sense that your voltage is only coming during the switching times, so why is it not dependent on the switching frequency rather than the current on/off ratio.

The fact is that there are other equations in this circuit that are dependent on the switching frequency, like determining the value of the inductor and its physical size because of saturation currents. Also, the stability and "transient" abilities of the supply is dependent on frequency.

I guess the way I have told myself in my mind is that from the equation V = Ldi/dt, we know that for a given period there will only be one switch, so we know that in a given cycle, the switch looks the same at any frequency. The only difference is how long in that period the switch is high and low, which is the duty cycle. With duty cycle, its best to think of it in averages, so a 50% duty cycle means an average current goes through half the time. So at 75% duty cycle a higher average current goes through during the same period. Referring back to V = Ldi/dt, we see that we have more current at a higher duty cycle, and so the di is larger, while the dt stays the same, and so average voltage goes up.

I hope that helps.

3. Dec 3, 2011

### Nat3

I think I understand what your saying. A higher duty cycle means more "on" time, so the current level will rise to a higher level than it would with a lower duty cycle because it has more time to. When the switch is opened, the higher current level will drop in the same amount of time (which is basically instantaneously) as a lower duty cycle/lower current would, and a higher voltage will be induced across the inductor?

So, a higher frequency would mean less "on" time and less "off" time. Does this mean that a higher frequency will result in less induced voltage because the current doesn't have as much time to rise? So a low frequency should be chosen for the boost converter?

4. Dec 3, 2011

### Nat3

I'm still pluggin' away at this, trying to understand.. I'm confused about a number of things, but one of the most confusing has to do with how the equation $$\frac{V_o}{V_i}=\frac{1}{1-D}$$ is derived. I've read Wikipedia's article on boost converters and understand most of that equation's derivation, but I get confused towards the end.

I know the idea behind a boost converter is that the voltage across an inductor is proportional to the rate of change of current. So, to amplify a voltage, the current level flowing through an inductor needs to be quickly changed.

To show that I really have been working hard to understand this, and so you can see where I'm at, here's my work. Hopefully I did the math correctly, but my calculus is a little rusty so let me know if I didn't.. To summarize, I get one equation for when the switch is closed and another for when the switch is open:

$$\Delta I_Lon=\frac{V_i}{L}DT$$
$$\Delta I_Loff=\frac{(V_i-V_o)(1-D)T}{L}$$

Assuming those equations are correct, here's where I'm confused. Wikipedia says:
Then they equate the two current equations. But the thing is, the voltage is being amplified, right? So, since the voltage has been increased, due to conservation of energy mustn't the current be decreased, meaning they will not be equal?

Hopefully someone can clear this up for me, as I there are a few more things I'm confused about and I have to get this done be next Monday

5. Dec 3, 2011

### MisterX

The switching frequency for an ideal boost converter does not affect the average output voltage.
This can be demonstrated by solving the equations for Vout in terms of Vin. The factor of T multiplies each nonzero term, and thus we can remove the T from both terms, since we know T≠0. T is the switching period.

It does affect the oscillations of currents inductor current and output voltage, just not the average voltages. When ID is zero, the output voltage would follow an RC discharge curve. Obviously, during the other part of the cycle, the capacitor must be charged back up to that level.

While the switch is conducting, the inductor current ramps up at a rate proportional to Vi, so obviously the peak to peak value would depend on the total amount of consecutive switch on time, and thus on frequency.

To operate in continuous mode, the inductor current must not be allowed to reach 0 while the current is decreasing. In discontinuous operation the wave form appears clipped off at 0, and from what I've read the output voltage does depend on the switching frequency.

My impression is that switching at tens to hundreds of kHz is typical for many recently designed power converters.

Last edited: Dec 4, 2011
6. Dec 3, 2011

### Mike_In_Plano

The equation you have is based on the assumption that the converter is operating in continuous mode. That means that even though the current in the inductor increases and decreases, there is never a moment in the cycle in which the current drops to zero.

If you take this assumption, that means that current will always be flowing through the switching device (i.e. mosfet) or through the rectifier. Thus the voltage across the inductor will be:

Vin - Vsw, While the switch is on (Ton)
- OR -
Vin - (Vout + Vdiode) when the rectifier is carrying the current (Toff)

If you assume that the switch is perfect (Vsw = 0) and the rectifier is perfect (Vdiode = 0),
then these equations simplify a bit:

Vin, during Ton
Vin - Vout, during Toff.

Now, using the typical assumption that the average voltage across the inductor is zero,

Vin x Ton/Tcycle + (Vin - Vout) x Toff/Tcycle = 0

D is defined as Ton/Tcycle, while (1-D) is Toff/Tcycle. Thus:

Vin x D + (Vin - Vout) x (1-D) = 0

=> Vin x D = (Vout - Vin) x (1-D)
=> Vin x D = Vout x (1-D) - Vin x (1-D)
=> Vin x D = Vout - Vout x D - Vin + Vin x D
=> 0 = Vout - Vout x D - Vin
=> 0 = Vout (1-D) - Vin
=> Vin = Vout (1-D)
=> Vin/Vout = (1-D)
=> Vout/Vin = 1/(1-D)

- Shew...

7. Dec 3, 2011

### gnurf

I think this is the key concept here. There has to be a volt-second equality on the inductor, i.e. Vin x ton = (Vout - Vin) x toff. That means that if you plot the inductor voltage as a function of time, the area that the graph makes above and below the x-axis must be equal. If OP can explain why this must be true, the rest follows easily.