Booster Converter Question

So i've been learning about boost converters, and it makes sense. From what I understand in the schematic below when the switch closes the inductor receives a large amount of current (since we have no load). So the current shoots sky-high (don't want to leave the switch closed too long!)

So once we open the switch, the voltage rises because the inductor resists the change in current, so the voltage goes higher to resist this (I still get confused on "why".....I wish there was a good "water" analogy for it)

So switch opens back up, the voltage is higher while the magnetic field collapses. Lets pretend the supply was originally 10v, and doing the switch open/close/open brings it up to....20v for instance.

Is that permanent? I mean lets pretend the "load" was a capacitor for instance, would it store 20v? I mean surely the inductor can't be keeping it at 20v forever right?

Also side question, if you had a super strong capacitor for the "load" could you store that high voltage in it? (like lets pretend we had a 200v capacitor) could you store 200v in it and then release it somehow like a HUGE charge all at once? (obviously it would be a bad idea ofc)

Thanks!

I sort of do not like the figure. There is always a capacitor on the output side. As you mentioned, since the current through the diode is discontinuous, there needs to be an energy storage to keep the load current flowing when the switch is closed and the inductor is "charging up".

From what I understand in the schematic below when the switch closes the inductor receives a large amount of current (since we have no load). So the current shoots sky-high (don't want to leave the switch closed too long!)

The inductor current rises increases with time. Right, it can increase very quickly, amperes in microseconds. That's why the switches operate at very high frequencies (now up to MHz for small converters ~100W and 10-100 kHz for >100 W). So yes, the current ripple should be regulated by turning off the switch and letting the current pass through the diode to the output capacitor.

So once we open the switch, the voltage rises because the inductor resists the change in current, so the voltage goes higher to resist this (I still get confused on "why".....I wish there was a good "water" analogy for it)

Once the switch opens, the diode turns on as the current must flow somewhere. A diode that conducts current has zero voltage across it so the voltage across the inductor is defined by the difference between the voltage across the output cap and the input voltage- using the Kirchhoff's voltage law.

Is that permanent? I mean lets pretend the "load" was a capacitor for instance, would it store 20v? I mean surely the inductor can't be keeping it at 20v forever right?

The voltage across the inductor is always defined by voltage sources. Since it is not shown in your figure, look at this figure:

In the on state the inductor voltage is defined by the input voltage only.

In the off state the inductor voltage is defined by the difference between the input and the output voltages.

The inductor will continuously keep charging and discharging to keep the output voltage at a specified value. In steady-state (constant load), the average amount of energy coming from the inductor to the output capacitor will be equal to the average energy drawn by the load.

Also side question, if you had a super strong capacitor for the "load" could you store that high voltage in it? (like lets pretend we had a 200v capacitor) could you store 200v in it and then release it somehow like a HUGE charge all at once? (obviously it would be a bad idea ofc)

Of course, ideally this boost converter can charge a capacitor to any voltage. In reality, the boost ratio is capped at ~10x due to power switch/diode ratings but that's not important. As long as the diode can block 200V and the capacitor can sustain 200V operation, there is nothing but the load and the control system to prevent even higher voltages.

If you had a very large capacitor, it would take a long time to charge and to discharge. But the power converter operation would be the same. Since capacitor voltage depends on how much current is used to charge it and discharge it, the inductor current and the load are the ones that actually set how quickly the capacitor charge is changed. Obviously, all loads have defined current. Similarly, the inductor, the diode, and the power switch have a rated and maximum current as well. The same goes for the output capacitor.

• Your figures show a buck-boost, rather than a standard boost. Jan 11, 2016 at 7:49

This is usually the sort of diagram used to explain a boost converter: -

And, in many ways it misses the point of the control of the switch.

If you want (say) 20V on the output, that means you desire a certain power to be dissipated in the resistor load. It is relevant to think of it in these terms because, in each switching cycle, you are storing energy in the inductor and then releasing it into the load. The number of times you do this per second IS the power delivered to the load. There is NO inherent voltage regulation.

So the signal flow is: -

Energy (inductor) ---> power (frequency) ---> resistor ---> voltage

It is by no means a voltage regulator - it is a power regulator and this is a really big deal about boosters that makes the "invisible" control system so profoundly important to appreciate. The main way of controlling the energy (given a constant input voltage) is by using the duty cycle of the switch and therefore the invisible control system HAS to measure the output voltage and adjust the duty cycle accordingly.

A lot of systems on light loads will drop the frequency massively because they are unable to adequately run at less than (say) 0.1% duty cycle. Some will operate in burst mode when the load is disconnected - there is an awful lot more to the control system than meets the eye.

If the load resistor goes open circuit, the duty cycle becomes (it has to become) zero. If the input voltage doubles the duty cycle has to quarter to maintain the same energy flow to the output load.

So once we open the switch, the voltage rises because the inductor resists the change in current, so the voltage goes higher to resist this (I still get confused on "why".....I wish there was a good "water" analogy for it)

There is a reasonable water analogy as follows :

You have a pipeline from a mountain lake down to your home. The mountain lake is 2 miles higher altitude than your house. The mountain lake is going to be our battery. The pipeline, which could have coils in it will be our inductor.

At your place, at the end of the pipe, there will be a TEE fitting, two separate openings. This is the node where the transistor (switch) and diode meet.

One of the pipe openings has a spring loaded door being held shut by the diode. The other opening has a door that you can easily open or close and lock, this will be our transistor.

The pipe is filled with standing water and both doors are closed.

Open the transistor door, and the water spills out onto the ground, mostly wasted. But water begins to flow in the pipe faster and faster. Maybe 20 miles per hour velocity of water current. Aha, current !

That moving water has kinetic energy : $(1/2) * mass *velocity^2$

Now, slam the transistor door closed and lock that door.

All of that stored kinetic energy in the moving water wants to make a tsunami (high pressure, or voltage), and will try to break open your transistor door.

It is fortunate that you supplied a "diode door", spring loaded. The spring loaded door opens, providing a path for all of that moving water. The moving water (energy) is what you will use, and also store in your capacitor, to smooth out the big gulp of water energy.

Energy stored in an inductor is $(1/2) * L * I^2$ similar to the kinetic energy above.