# How to guarantee steady state conditions on boost converter

I am teaching myself about boost converters, and I understand the operating principles- to a point. Specifically, I understand what happens to the conductor during the switch-closed portion of the cycle. My understanding:

As the switch-closed circuit has basically no load, the voltage source will continue to drive the current through the inductor higher and higher, at a rate that depends on the inductance. This is what the inductance equation dictates, V=L(di/dt). Thus the voltage across the inductor will remain at the voltage provided by the source during the switch-closed portion of the cycle. The inductor accumulates energy.

Then, the switch opens, and the load "downstream" of the inductor suddenly becomes much higher, because now the circuit includes whatever resistance is offered by the high-voltage component the converter is hooked up to.

So now the inductor sees a drop in current because of this new load, and responds with a voltage in the direction of the current to compensate. This voltage is proportional to the rate at which the current drops when the switch opens.

Now to my question: what is that rate? How do I find it?

In order for the system to be in steady state, the current must, in the remaining portion of the cycle, drop to what it was at the beginning of the last cycle, before the switch closes again. If the current doesn't drop fast enough, then the inductor won't dissipate all the energy it gained during the switch-closed portion, and so it will gain more and more energy every cycle. That doesn't seem ideal.

But I feel like I'm missing some key piece of information about the switch-open portion that would let me solve for what the output voltage (or at least di/dt) will be. I know how to find what it should be- it's whatever it takes to go from Imax to Imin in the allowed time. But I don't know how to prove that it will be, if indeed it will be. And if it's not guaranteed to be, what variables (load resistance, duty cycle, etc.) would I have to manipulate to get it there?

• Perhaps this is why many converters specify a minimum load. Oct 4, 2021 at 16:11

Picture from here which also provides a good explanation.

When the switch is closed you have Vin across the inductor leading to an upslope of the inductor current of $$\Vin=L*di/dt.\$$

When the switch is open, you have $$\Vout-Vin\$$ (approximately, subtract the diode drop if non-synchronous to be more accurate) across the inductor. Just apply $$\Vout-Vin=L*di/dt\$$ to get the slope of the inductor current, and solve for the time it takes to get back to the starting point for energy balance.

Note that many converters switch modes at light loads into PFM, so that the switching frequency drops down to improve efficiency and possibly avoid the need for a minimum load.

Think of it this way...the current isn't going to flow from a source to a higher voltage load without some convincing. Enter the inductor, which stores current like a capacitor stores charge.

When the switch is closed, the inductor goes from input to ground. Little power is lost, but the current in the inductor builds up.

Now, the trick the inductor does is that once the current is flowing, it won't stop abruptly. In a way, it's kind of like it has momentum. So when you switch the inductor back to the load, it's going to continue pushing current for some time against that voltage drop.

The rate at which the current builds up during the "switch on" period is, as you point out, proportional to Vin. During the "switch off" period, the rate at which the current builds up is proportional to Vin-Vout (possibly with a diode drop included if it's not synchronous). For a boost design, of course, Vin-Vout is negative, and the current will thus diminish during this phase instead of building up.

The duty cycle on your switch will be determined by these two rates of change of current.

Then, the switch opens, and the load "downstream" of the inductor suddenly becomes much higher, because now the circuit includes whatever resistance is offered by the high-voltage component the converter is hooked up to.

Not quite. The current that the inductor accumulated during the charge phase then starts to ramp down towards zero. The load resistance is largely irrelevant because it's the output capacitor that is the dominant component here. Think of it without an external load and you are charging a capacitor. If you think about it this way you'll get a clearer picture because, as with any boost converter, to avoid large output ripple voltages you MUST have a capacitor that dominates the external load.

So now the inductor sees a drop in current because of this new load, and responds with a voltage in the direction of the current to compensate. This voltage is proportional to the rate at which the current drops when the switch opens.

The inductor current falls at a pretty much fixed rate (without examining minutia). That inductor current falls towards zero amps. The voltage across the inductor is literally the difference between input supply voltage and the output voltage (neglecting the diode drop of 0.6 volts) and, we can therefore say that the rate of current falling is embodied in the same formula you used earlier: -

$$V_{OUT} - V_{IN} = -L\cdot \dfrac{di}{dt}$$

Now to my question: what is that rate? How do I find it?

Hopefully that should be clear.

Here's a web based calculator you can use to examine the currents: -

Calculator and explanations can be found on my site here.

In order for the system to be in steady state, the current must, in the remaining portion of the cycle, drop to what it was at the beginning of the last cycle, before the switch closes again. If the current doesn't drop fast enough, then the inductor won't dissipate all the energy it gained during the switch-closed portion, and so it will gain more and more energy every cycle. That doesn't seem ideal.

This is why you will never see a boost converter without a supervisory circuit that aims to keep the output voltage constant at the demanded level. If the output is constant then exactly the right level of energy feeding into the converter is pushed to the output to power the load (once the initial transient has settled down and, of course, ingoing switching inefficiencies)