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I'm building a boost converter that utilizes a 555 timer switching at 60 kHz at a 96% duty cycle which drives an IRF3205 MOSFET to power a hand wound inductor (calculated to have ~330 uH) with a 6A10 diode and a 400 V, 80 uF capacitor at 5 volts. My goal is to achieve 360 volts.

The problem I'm having is that the circuit consistently maxes out at around 80 volts and attempting to increase the voltage e.g increasing the inductance of the inductor by adding an iron rod, adding another inductor in series, increasing the duty cycle of the 555 timer and even increasing the voltage has no significant effect on the voltage. I can only lower the voltage using these methods.

I think it may be because of the diode limiting the current as replacing the 6A10 diode with a higher resistance one seems to decrease the voltage to ~60 and further attempts to mess with that also fail.

I do not have any more diodes that I can use and so I am wondering if it is current available to the inductors which is limiting the output before I buy any more diodes.

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  • \$\begingroup\$ What load resistance is across the capacitor? \$\endgroup\$
    – Andy aka
    May 25, 2020 at 9:01

2 Answers 2

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The problem I'm having is that the circuit consistently maxes out at around 80 volts

Consider that the IRF3205 might be a limiting factor: -

enter image description here

Then consider that the 6A10 diode will be next in-line for limiting the voltage: -

enter image description here

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  • \$\begingroup\$ Correct if I am wrong, but is the switching mosfet connected at all to the output voltage in a boost converter (I don't think so)? I reckon the maximum drain souce voltage the MOSFET experiences, is given by the supply voltage plus the voltage across the inductor during the off phase. \$\endgroup\$
    – vtolentino
    May 25, 2020 at 12:12
  • \$\begingroup\$ @vtolentino the peak voltage at the MOSFET drain is the peak voltage delivered by the inductor to the output capacitor (ignoring the diode drop which makes it 0.7 volts bigger). So if the output capacitor is 50 volts then the peak voltage at the drain is 50.7 volts. \$\endgroup\$
    – Andy aka
    May 25, 2020 at 12:19
  • \$\begingroup\$ Looking from the load side, it makes sense. \$\endgroup\$
    – vtolentino
    May 25, 2020 at 12:53
  • \$\begingroup\$ I just switched out the MOSFET for a proper fast switching one and it is working great! I can get up to 360 volts when increasing the input voltage to 10v! \$\endgroup\$
    – katu
    May 25, 2020 at 19:34
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The maximum achievable output voltage with a boost converter depends on the supply voltage, efficiency and duty cycle according to the following:

$$V_{out, (max)} \approx \dfrac{V_{in,(min)}\cdot n}{1-D}=\dfrac{V_{in,(min)}^2\cdot n}{V_{in,(min)}-\Delta I_L \cdot f_{SW} \cdot L}$$

Where:

\$n\$ is the efficiency

\$D\$ is the duty cycle

\$\Delta I_L\$ is the inductor's ripple current

\$f_{SW}\$ is the switching frequency

\$L\$ is the inductance

In your case, considering a conservative efficiency of \$86\%\$ and an input voltage of \$5V\$:

$$V_{out, (max)} \approx \dfrac{5V \cdot 85\%}{1-96\%}=106V$$

A more detailed description can be found in this application note from Texas Instrument.

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  • \$\begingroup\$ Thanks, I'm still left wondering where frequency plays a role in this. \$\endgroup\$
    – katu
    May 25, 2020 at 6:25
  • \$\begingroup\$ @ballpoint The duty cycle can be written in terms of the switching frequency and the characteristics of the inductor. I have updated my answer. \$\endgroup\$
    – vtolentino
    May 25, 2020 at 6:47

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