Need help understanding boost converter

Discussion in 'General Electronics Chat' started by Robotics Guy, Dec 3, 2011.

  1. Robotics Guy

    Thread Starter New Member

    Mar 11, 2011
    I'm trying to understand how a boost converter works and could really use some help :)

    There are a number of things that I'm confused about, but one of the biggest sources of confusion is 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 :rolleyes: 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_0)(1-D)T}{L}

    So, assuming those equations are correct, here's where I'm confused. Wikipedia says:

    Then they equate the two 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?

    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 :eek:

    Thanks for your help.
  2. JMac3108

    Active Member

    Aug 16, 2010
    Robotics Guy,

    Yes, you are correct about the voltage being increased and the current being decreased - conservation of energy, right? :) But you are thinking about input and output currents and voltages, and the equations being equated are for the inductor current. Think of it this way...

    A fundamental property of an inductor is that you can't change the current through it instantaneously. So the current right before you close the switch, and the current right after you close the switch are equal.

    Another way to look at it ...

    Current is flowing through the inductor through the closed switch to ground. When the switch is opened, the inductor begins to source current in order to maintain the current at the same level. It is able to do this because of the energy stored in its magnetic foeld. The point, again, is that at the moment the switch opens, the current is the same as just before it opened.

    Let me know if this helps. If not, I'll try to think of another way to explain it. Switching power supplies can be difficult to get your head around at first, but once you understnad them they are really a lot of fun to work with.
  3. Robotics Guy

    Thread Starter New Member

    Mar 11, 2011
    I see what you're saying, and it makes sense that the current flowing through the inductor before the switch is opened is equal to the current flowing through it immediately after it's opened, but only for a brief instant. Wikipedia says:

    One of the things that's confusing me about the boost converter, as I mentioned, is that it seems like due to conservation of energy, if the voltage is increased then the sum of the current flowing out of the inductor must be less than what flowed into it, right?
  4. jimkeith

    Active Member

    Oct 26, 2011
    At higher output voltages, the inductor discharges more quickly--this results in a lower average current out--the charge time for the inductor remains constant, so you can see that as the output voltage increases the ratio of discharge time /charge time decreases.
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
    This is the key to understanding it: the instantaneous current in the inductor may not change at the instant of switching between the input and output circuits, but the average current in a higher voltage output circuit will always be less.

    This not only a question of rates of decay of current. Actually, if only a small percentage boost is obtained, in some configurations where the coil voltage is added to the input voltage, the decay may be slower.

    The coil however only delivers current for part of the time, according to the duty cycle and whether it is operating in continuous conduction (current not given time to decay to zero), ir discontinuous conduction (current decays to zero). This means that the load current averaged over a whole cycle will be lower than the current flowing out of the coil during the part of the cycle when it feeds the load.
  7. crutschow


    Mar 14, 2008
    By definition the current into and out of an inductor has to be the same. There's no other place for the current to go. But the current does change, increasing when the switch is closed and voltage is being applied to the inductor, and decreasing when the switch is open and supply current is being supplied to the load.

    Energy is being stored in the inductor magnetic field when the current is increasing and energy is released to the output load when the current is decreasing. The energy stored and the energy released are equal (assuming an ideal circuit).