# Calculating values for a DC step-up circuit

As a learning exercise, I am attempting to make a step-up DC converter (5V to 12V). The basic design I'm following is this:

Source: http://www.daycounter.com/LabBook/BoostConverter/Boost-Converter-Equations.phtml

From the source above, I've derived an equation for a value of L1 which will avoid an unstable output voltage (I'm ignoring Vtransistor since it seems to be negligible for my components):

$$L \ge {V_{in} (V_{out} + V_{diode} - V_{in}) \over {f (V_{out} + V_{diode}) \min(I_{load})}}$$

I'm using an IRF520 MOSFET and an IN4007 diode, with an Arduino providing the PWM source (running at 62.5kHz), so I believe this fills in as:

$$L \ge {5\text{V} \times (12\text{V} + 1.1\text{V} - 5\text{V}) \over {62500\text{Hz} \times (12\text{V} + 1.1\text{V}) \times 0.0001\text{A}}} = {40.5\text{V}^2 \over {81.875 \text{V} \text{A} / \text{s}}} = 0.495\text{H}$$

(to guarantee the 100µA minimum load, I'm putting a 100kΩ resistor in series with the diode, although I'm not sure this is the right thing to do since it will be dividing the output voltage. I guess that once I have something for this to drive I'll have to recalculate the value for the specific circuit)

So that gives me a number, but 0.495H seems like a huge value, and I'd rather not accidentally produce 100V if I mess up the PWM duty cycle.

I have two questions: does this look at all reasonable? and is there some way I can calculate the maximum obtainable voltage from a given inductor in this setup?

• It would be better to use a schottky diode instead of the 4007: it has a much lower voltage drop, so better efficiency Commented Jan 15, 2017 at 18:53

Boost converters store energy in an inductor in one half cycle then, in the other half cycle, they dump some or all of that energy into the capacitor. If you have a constant input supply voltage and a switching MOSFET that has zero ohms on-resistance, the energy taken by the inductor is dependant only on the time the transistor is "on" and the inductance value.

If this energy is stored and dumped many times per second, the transfer of energy becomes a transfer of power and, that transfer of power requires "something" (a load) on the output that dissipates that power or takes it away.

If the "load" doesn't "use" enough power, the output voltage rises until it does. This usually means that if you don't have a load, the voltage on the output rises to a point where the output capacitor fails or the transistor fails or the diode fails.

Don't use a 1N400x diode because it has a really crappy reverse recovery time.

Your load is only 0.1 mA and that means you need either a very large inductance or an extremely small duty cycle so, try to be realistic and take (say) 10 mA through a fixed resistor - this will reduce the inductor's value by 100 times.

Having a fixed value of resistor in parallel with your actual load means the load conditions are largely "fixed" and therefore, it will be easier to control and more stable.

A boost converter is a power converter and not a voltage regulator - it needs help to regulate by having a fairly stable load and a control system that is always monitoring the output and adjusting the duty cycle accordingly.

This is why you see many boost designs (including fly-back topology) that state the minimum load conditions.

• So this would mean putting a 1kΩ resistor from the +12V rail to the ground instead of in series with the diode, and using a 4.95mH inductor (or 4.7mH since that's what I have). I already have a voltage divide in place to monitor the output and adjust the duty cycle, and in fact that's where my question on maximum voltage comes from: if I get those calculations wrong it may raise the voltage when it should be lowering it, and quickly get out of hand! But it sounds like you're saying whichever inductor I use this would always lead to a blown component, rather than "saturating" it as I'd assumed.
– Dave
Commented Jan 15, 2017 at 17:25
• Also how do you know that the 1N400x has a poor reverse recovery time? I checked the datasheet but couldn't see any value given. Did you calculate it from other values somehow or is it prior knowledge about the component?
– Dave
Commented Jan 15, 2017 at 17:28
• There are data sheets that will tell you the value but generally speaking if something isn't proudly boasted on a data sheet there is a good reason. Commented Jan 15, 2017 at 17:32
• .. and even if it's proudly given on a datasheet, check against other devices (you can often sort device tables on distributor's websites). For example, people tend to use 1970's operational amplifiers, because they have "excellent drift, low power, high common mode rejection, high bandwidth", according to their 1970's datasheets... Commented Jan 15, 2017 at 17:35