I have been attempting to simulate a DC-DC Boost converter in 'CircuitLab', rather than going down the route of using an IC I have decided to go down the 'traditional route' and use a MOSFET, simply because I need to be able to digitally vary the duty cycle of the switching frequency which will be driven from an Arduino Uno. I have used the equations given from Texas Instruments - However, I'm unsure if this is would still work with my circuit seeing as these calculations are eventually used with an IC.

The requirements for my converter are as follows;

  • Vin = 3v
  • Vout = 14v
  • Iout = 500mA
  • Switching Frequency = 100KHz

Based on the equations given I have been able to form the following calculations;

  • Duty Cycle = 0.83
  • Inductor Ripple Current = 0.47A
  • Inductor = 50uH
  • Maximum Switch Current = 3.17A
  • Output Capacitor = 41uF

When selecting a MOSFET, the main characteristic I looked for was to make it a 'Logic Level ' MOSFET - As the PWM will be driven from an Arduino. The chosen MOSFET is the 'STB55NF06L' with an RDS of VGS = 5v, 27.5A.

In order to decide which diode to use I checked to see if the Non-repetitive Peak Surge Current was > than IOut. Also, I chose a diode with a short reverse recovery time. The selected diode is the '1N5819'.

Since deciding on these components I have tried to run a simulation, but am not getting the results I have desired. The output isn't boosting at all, Vout is dropping to 2.5v. Any ideas on how I can fix this?

Here is my circuit diagram;

Circuit Diagram


I have built the boost converter using the information I have gathered from the answers given and I've run into a slight problem. I have used a 22uH inductor and a 33uH inductor that I had immediately available (FT00765 & FT00766). As these inductors have a fairly low current rating, they were unable to take the current of my circuit - as a result, getting incredibly hot. I understand that I need to purchase an inductor of the same value but a higher current rating. However, what I don't understand is the difference between current rating and current saturation rating. Based on my simulation, the peak current through the inductor is around 16A at the initial boost, and then levelling off at between 2A to 4A in CCM. Therefore, I have been looking at inductors such as; 2300HT-220V-RC with a current rating of 19A, would this be a suitable inductor or do I need to look at an alternative? I have also looked at AIRD-02-220K which also seams like a feasible inductor.

  • \$\begingroup\$ Have you got the duty cycle the right way round? On for 83% of the time? \$\endgroup\$
    – Neil_UK
    Nov 9, 2015 at 15:23
  • \$\begingroup\$ I can barely use that websim to caculate some voltages on a resistive network [and I couldn't figure out how to get any currents from it]. Use something more serious for this, LTspice etc. \$\endgroup\$
    – Fizz
    Nov 9, 2015 at 15:29

2 Answers 2


TLRD: your calculations are fine. I got myself confused by checking them against a different calculator that targeted DCM not CCM. I'm not gonna change the stuff below because Arsenal replied to it, and it might also be useful to you to compare the two modes.

Your inductor looks an order of magnitude too big vs what I get at https://learn.adafruit.com/diy-boost-calc/the-calculator

enter image description here

And it does look like your inductor is far too big:

enter image description here enter image description here

@Arsenal N.B.: I figured out the difference between these two calculators. The adafruit one gives you a DCM (discountinous current mode) solution, whereas TI's gives a CCM solution.

enter image description here enter image description here

Which one is preferable depends on various factors. Basically at high output currents you want CCM to lower the losses and stresses on the components, but DCM has better transient response. A lot more about that can be read here, including a side-by-side power efficiency calculation example.

  • \$\begingroup\$ Have you had a look at the inductor ripple current with that 5µH inductor? It should be in the order of 4A to 5A which is way too high (it should be in the order of 30% of the peak input current, if I'm not mistaken). A value of 29µH would be more appropriate. \$\endgroup\$
    – Arsenal
    Nov 9, 2015 at 17:04
  • \$\begingroup\$ @Arsenal: twas DCM vs CCM. \$\endgroup\$
    – Fizz
    Nov 9, 2015 at 17:32
  • \$\begingroup\$ Oh yeah, there is a ">" symbol in that table, so it's the lower boundary solution... Nice update though. \$\endgroup\$
    – Arsenal
    Nov 9, 2015 at 18:13
  • 1
    \$\begingroup\$ @Thaitan: With the built-in models LTspice is like 100-times faster than with subcircuit-based ones like ST's. I've cut the sim time to 5ms with ST's model because it was very slow. You do get a bit less overshoot with ST's model though, especially in DCM. imgur.com/a/eqVmP Also you should add (as I did) an ESR value for the output cap, otherwise the ripple will be unrealistically low; I used 0.05ohms in all the simulations. \$\endgroup\$
    – Fizz
    Nov 9, 2015 at 19:18
  • 1
    \$\begingroup\$ @Thaitan I've updated my answer to include my calculations for the inductor value. \$\endgroup\$
    – Arsenal
    Nov 9, 2015 at 21:09

When I open your schematic, I can see that you are only simulating 100µs or 10 PWM cycles. Your voltage will not get there in that amount of time.

So instead try simulating 15 ms (depends on the design):

Simulation results for 15 ms

Let's try a ballpark calculation of what happens (this is really a back of the envelope type calculation):

You have a peak current of 3.17 A, which will go to the output capacitor for 17% of the 100 kHz (of course it will drop, but I'll neglect that just now).

So there will be a charge of 17%/100 kHz*3.17 A = 5.4 µC transferred to the output. If we assume a constant current load of 500 mA (which 28 Ohms are not), over one cycle it will take 500 mA/100 kHz = 5 µC from the output. So we get a total of 0.4 µC per cycle charged. The voltage increases by 0.4 µC/47 µF = 8.5 mV per cycle.

To get from the starting point of 2.5 V to 13 V you'd need at least 1250 cycles or 12,5 ms.

Well it turned out faster, but that's a rough sketch on what you can expect.

One thing which seems a bit odd is the inductor value of 50 µH - it seems like the TI application note uses a slightly different approach to the calculation than the one I got from LT - thanks to RespawnedFluff for checking out the values. Below is my calculation, which sort of is obsolete in that case.

Normally you go for a ripple current of 20% to 40% of the peak inductor current. (that value is a compromise as explained here)

So we go from here: $$I_{Peak} = I_{avg} + \frac{\Delta I_L}{2}$$ and $$\Delta I_L = 30\% I_{Peak}$$

which leads us to:

$$\Delta I_L = \frac{0.3 I_{avg}}{0.85}$$

And with:

$$I_{avg} = \frac{V_o}{\eta V_i} I_{out}$$

We arrive at:

$$\Delta I_L = \frac{0.3 V_o I_{out}}{0.85 \eta V_i}$$

For the ideal case (\$\eta = 1\$) we get a result of 0.824 A as ripple current. Roughly double of what you calculated.

With that and the equation given in the application note of TI:

$$ L = \frac{V_i(V_o-V_i)}{\Delta I_L f_s V_o}$$

I get a value of 28.6 µH. (which is not a common value, next value would be 33 µH or 27 µH)

  • \$\begingroup\$ He calculated \$\Delta I_l\$ correctly according to the TI document he used (eq. 6), but he opted for the lower value. \$ 0.2 I_{out} \frac{V_{out}}{V_{in}} = 0.2 \times 0.5 \times \frac{14}{3} = 0.4666A \$. TI also says that first 0.2 factor can be in the range 0.2 to 0.4. \$\endgroup\$
    – Fizz
    Nov 9, 2015 at 21:23
  • \$\begingroup\$ @RespawnedFluff that's true, haven't checked the numbers... My calculations are based on a LT seminar I had last week on DCDC converters. Seems like that approach leads to different values (when I use 20% I end up with 0.55 A and 42 µH). \$\endgroup\$
    – Arsenal
    Nov 9, 2015 at 21:37
  • \$\begingroup\$ With the TI formulas, I get 50.5uH. But after plugging that into the sim, the real ripple was about 0.5A. With the LT value of 42 µH I get about 0.56A in simulation. So LT's formulas are more accurate. \$\endgroup\$
    – Fizz
    Nov 9, 2015 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.