I have a boost converter that with load 22K, it produces 120V with input 12V. the inductor is 220uH and the capacitor is 150uF. i need it for ultrasonic pulser with frequency of 1-10 MHz, the load of the pulser is 2.2 Kohm. i use arduino as switching controller of the mosfet in boost converter. the arduino will get the output voltage and adjust PWM (31 KHz) with PI system to calculate duty cycle. the problem is, when the output voltage is not attached to the pulser load, it gives exact 120V, but attached, it drops down to 70V, to maintain the output 120V, the input need to be increased to 24V, but i want it to be 12V, what i have done are

  1. make the KP and KI of PI higher, not work
  2. increase the inductor value, not work

how can i maintain the output? as far as i know, that the gain need to be higher, but how?

the second question is, how to calculate the value of capacitor and inductor? for the capacitor rating, is it necessary to match the capacitor rating with output voltage or the input voltage?

could i use a cascade boost converter? for an example using XL6009 to boost my 12V to 28V and then feed it to the boost converter?

this is my schematic

enter image description here

  • \$\begingroup\$ Huh, this is the 100 thousand question on this site. Cool. \$\endgroup\$ Commented May 7, 2018 at 7:33
  • \$\begingroup\$ Is your shift key broken? \$\endgroup\$
    – winny
    Commented May 7, 2018 at 9:28

1 Answer 1


With 12 volts, 220 uH and 31 kHz you can analyse the energy transferred per cycle. At 50% duty, the time period is 16.13 us and this allows the current in the inductor to ramp up to 880 mA. This current and 220 uH implies a stored energy of 85 uJ. Barring losses, this energy is transferred 31,000 times per second to the load and output capacitor.

This is a power delivery of 2.64 watts at a duty of 50%.

If the load is 2.2 kohm and you want 120 volts across it then it requires 6.55 watts. So clearly you need more like 80% duty to achieve your aims. But this relies on perfect components with minimal losses and that won't happen in reality.

It's likely you are pushed significantly over the 80% duty requirement and this starts to be bothersome for boost converters. I see you have 4 inductors in series/parallel so, maybe just use two in parallel to reduce the inductance to 110 uH and the current will ramp to 1760 mA at 50% duty. This will then store an energy of 170 uJ and deliver a power of 5.28 watts i.e. a significant step closer.

But you have to choose inductors that don't get close to saturation at this higher current.


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