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I’m trying to build a digitally controlled conventional boost converter and implementing a basic Maximum Power Point Tracking (MPPT) algorithm to extract the maximum possible power from the panel under various lighting and shading conditions. Part of the circuit as you know, involves a Boost converter. Due to me not having a lab available due to the pandemic, I therefore need some theoretical guidance in trying to figure this out. The completed circuit looks like this. Which includes a Voltage and current measuring circuit due to the ARM board. Complete circuit

But for the mean time, I’m trying to build my boost converter shown here;

Boost converter circuit

I’m using a 220uF capacitor, 470ohm ceramic load resistor, MOSFET(datasheet), a Diode 1A 40V(Datasheet) and the 10W solar panel (Datasheet)

Therefore, my question is, how would I be able to work out the inductor value and some kind of rationalle for the PWM frequency?

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  • \$\begingroup\$ Define your voltage ripple. Then work out the current ripple. You can calculate the inductor value from the switching frequency. A smaller inductor will result in a faster changing current. This yields a larger ripple. \$\endgroup\$
    – user110971
    Commented Apr 15, 2020 at 15:58
  • \$\begingroup\$ So this is how I worked it out. imgur.com/a/gSHsP5M Any, ideas if what I did makes any sense or im heading in the wrong direction? Also how do I choose what PWM frequency would be good? \$\endgroup\$
    – Q.T.π
    Commented Apr 15, 2020 at 16:54
  • \$\begingroup\$ Fix the link to the diode please. \$\endgroup\$
    – Andy aka
    Commented Apr 15, 2020 at 17:31
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    \$\begingroup\$ Also, is your load a fixed 470 ohm resistor as per your top diagram? Ditto 220 uF for the capacitor? I also note that the MOSFET you have chosen is not a great choice for a 20 volt panel - its on-resistance is about 0.5 ohm - the device is more economically suited to a panel voltage over 50 volts. I also note the 1 ohm sense resistor and again this seems much too high for the limited voltage supply from the panel. Your maximum efficiency is seriously compromised by this. \$\endgroup\$
    – Andy aka
    Commented Apr 15, 2020 at 17:51
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    \$\begingroup\$ some simulation tools may help experiment with the design without having to build parts, such as schmidt-walter-schaltnetzteile.de/smps_e/aww_smps_e.html, or QUCS. \$\endgroup\$
    – scorpdaddy
    Commented Apr 15, 2020 at 19:30

1 Answer 1

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Part of the circuit as you know, involves a Boost converter. Due to me not having a lab available due to the pandemic, I therefore need some theoretical guidance in trying to figure this out.

For a DCM boost converter (the type most regularly used): -

Fact 1 – The inductor in a boost converter has to despatch sufficient energy during each switching cycle to keep the load at the desired voltage.

Fact 2 – The cyclical energy despatched, multiplied by the switching frequency, equals the power required by the load to raise its voltage from the input voltage level to the required output voltage. This is the “power lift”.

An Example:

  • Input voltage is 100 volts.
  • Desired Output voltage is 150 volts at 0.6 amps.
  • \$F_{SWITCH}\$ = 100 kHz

Power and Energy Calculations:

  • The power uplift is (150 -100) x 0.6 = 30 watts
  • The energy uplift is 30 watts ÷ \$F_{SWITCH}\$ = 300 µJ

Assume that the input voltage could be as low as 90 volts: -

  • Raising 90 volts to 150 volts at 0.6 amps is a power uplift of 36 watts.

Inductance calculation:

enter image description here

Boost inductor calculator:

enter image description here

A duty cycle of 25% was chosen in the above to ensure that the circuit didn't slip into CCM (different formulas). With the charge period being 25% of the time and the discharge period being about 37% of the time this leaves a hold-time of about 38% of the cycle (a bit of spare room).

Simulation output:

enter image description here

Simulation circuit:

enter image description here

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  • \$\begingroup\$ This is the minimum inductor value, right? AFAIK, higher values will create less ripple? \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Jul 3, 2020 at 9:11
  • \$\begingroup\$ No, the inductor can be smaller in value but duty cycle (D) would also be smaller. The ripple voltage (?) in the picture is quite high due to me choosing a small value for output capacitor so the simulation would settle in a reasonable length of time. \$\endgroup\$
    – Andy aka
    Commented Jul 3, 2020 at 9:15

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