# Boost Converter Basics

Every circuit we’ve talked about so far has some input voltage $V_{in}$, which is used to power the rest of the circuit. The important criteria to note is that whatever our power source is, we only go down from here. No matter what fancy combo of R, L, C, transistor, we put in, they all take some voltage to turn on. But what if we want to generate some higher voltage - if you give me 5V, how can I give you 10V back out? The answer to that is the **boost converter**.

Now unfortunately, this circuit is going to be one of those where we make some “intuitive” (read: known after the fact) guesses and then validate their correct-ness.

Let’s derive how this gets us to some stepped up voltage. Our ideal switch (S1) has two states - open and closed. Since it’s ideal, closed will be a perfect short circuit, and open is a perfect open circuit. In the open state, we have one voltage loop going across the outside:

$$ V_{in} = V_L + V_D + V_{R||C} $$

Where the last term is the combined parallel impedance of the resistor and capacitor. Let’s start making some assumptions! For a DC circuit, the impedance of a capacitor is near infinity, so we can say $ V_{R||C} \approx V_R = V_{out}$ and ignore the capacitor. We can also buy very nice diodes that require only a super small voltage, so we can say $V_D \approx 0 $. Rewriting this version:

$$ V_L = V_{in} - V_{out}$$

Now let’s do the other half. With the switch closed, we short out the right half of the circuit, so there’s not much left:

$$ V_L = V_{in} $$

Now here’s the big trick: this circuit “should” be a DC-DC converter, so we should only be seeing DC at all times. Maybe different DC, but just DC nonetheless. Therefore, the inductor should have no voltage drop - at DC, an inductor has zero resistance. With some neat math concerning on/off times, we can get the following conclusion:

$$ V_{L,on} + V_{L,off} = 0 $$

$$ T_{on} V_{L,on} + T_{off} V_{L,off} = 0 $$

$$ DT_{total} V_{L,on} + (1-D)T_{total} V_{L,off} = 0 $$

$$ D V_{L,on} + (1-D) V_{L,off} = 0 $$

$$ D (V_{in}) + (1-D) (V_{in} - V_{out}) = 0 $$

$$ D V_{in} + V_{in} - V_{out} - DV_{in} + DV_{out} = 0 $$

$$ V_{in} = (1-D) V_{out} $$

$$ \dfrac{V_{out}}{V_{in}} = \dfrac{1}{1-D} $$

Where $D$ represents the percent of time we spent on in one time period. We have a special name for $D$ - the duty cycle of a wave. It’s represented as a decimal value - being high/low for equal times is $D=0.5$, always on is $D=1$, and always off $D=0$. Proving things in math is alright, but let’s do it in simulation.

```
* Ideal Boost Converter
vsw n3 0 pulse(0 5 0 1u 1u 20u 40u)
vin n1 0 DC 10
l1 n1 n2 0.1mH
sw1 n2 0 n3 0 switch1 ON
d1 n2 n4 mod1
c1 n4 0 10uF
r1 n4 0 100
.tran 0.1u 5m
.options list node post
.end
```

We have a decent switching voltage for our very good ideal source - only about 5% of the signal is spent in transition. The default diode isn’t great, the voltage drop is about 2V, so that’ll really mess up our assumptions in the off case. Let’s see what we get.

Not too bad! If everything were ideal, we’d go from 10V to 20V smoothly, but in the real world nothing is quite so simple. We clearly have some huge startup transient to get the gears rolling, but it fades off pretty quickly. We don’t really lock in at 20V, instead, we’re much higher at 28V. This makes sense though, as the large diode voltage drop will make us push a bit more. Additionally, since we have real rise and fall times, our duty cycle increases slightly too. Since it’s an inverse relationship between duty cycle and output, all it takes is a small input change to get a large output change!

Overall, we’ll live with it. There’s loads of optimizations to be done on the voltage ripple, the capacitor/inductor values, the actual switching implementation, but it’s enough for a first effort.