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I've come across DC boost converter and told myself - why not to try it as well? So I started building a simple circuit which includes BD243C NPN Transistor, ATTiny13A programmed to give some pulse to the Base of the transistor, 16 uH inductor, 82 uF 450 wv electrolytic capacitor and IN4007 diode to prevent current flowing from capacitor back to the circuit.

I calculated the needed duty cycle to have 30V on the output from 5V input which was around 0.83. Then I wanted the pulse width (10 ms period), I got the result of 8,333 ms. The remaining 1,667 ms should indicate the time there's no pulse happening.

So where is the problem? When I powered my circuit using 5V battery, voltage rises to 7V, if I wanted to have for example 10V, I would have to wait for almost an hour to have 10V on the output and if I wanted more, I would have to wait more than one hour. I don't know exactly where the problem is.

Before when I tried this circuit for the first time with 1ms period and I wanted 10V on the output from 5V, it worked. According to calculations, I had 10V on the output, which was okay, but it took like 10 minutes to have 10V, but it was at least someting for a start. And with 10ms, it doesn't even charge up to 10V, not even to 30V.

Here's the schematic of my circuit. enter image description here

Would be nice if anyone gave me an advice what should be fixed.

Thanks in advance

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  • \$\begingroup\$ V1 grounded to the MCUs 0 volts would help significantly plus adding a series base resistor. \$\endgroup\$ – Andy aka Jul 2 '19 at 17:24
  • \$\begingroup\$ Ah, thanks for your reply, sir! I have already added 10k Ohm base resistor. If I may ask, what is MCU? \$\endgroup\$ – MerryGR Jul 2 '19 at 17:28
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    \$\begingroup\$ Then don’t show half a circuit and expect people to guess you have added a base resistor. MCUs is your micro. \$\endgroup\$ – Andy aka Jul 2 '19 at 17:31
  • \$\begingroup\$ @MerryGR The MCU is U1. \$\endgroup\$ – John D Jul 2 '19 at 17:31
  • \$\begingroup\$ @ThePhoton sorry, I forgot to put it there.. \$\endgroup\$ – MerryGR Jul 2 '19 at 17:40
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Let's calculate what happens when we turn your transistor on for 8.3 ms.

We know that inductors have the constituent equation

$$\frac{dI}{dt} = \frac{V}{L}$$

So when we apply 5 V to one side of L1 and ground the other side, we get

$$\frac{dI}{dt} = \frac{5\ {\rm V}}{16\ {\rm\mu H}}= 312,500\ \frac{\rm A}{\rm s}$$

And after 8.3 ms, the current should ramp to

$$(312,500\ \frac{\rm A}{\rm s})(8.3\ {\rm ms})=2,593\ {\rm A}$$

This is much more than your 5 V supply is likely to be able to supply, and much more than your inductor is able to carry without vaporizing.

You need to design your circuit for much, much smaller current ripple through the inductor.

You could look for a 16 mH inductor able to support 3 A current (which will be big and expensive). You will likely need to increase your load capacitance by a few orders of magnitude as well to achieve acceptable voltage ripple.

Or you could use a much higher switching frequency. Typical switching frequencies for this kind of circuit are 50 kHz to 5 MHz, so at least 500 times higher than your 0.1 kHz.

You might need to use a built-for-purpose switching controller instead of a general-purpose microcontroller to achieve the switching frequency you need.

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  • \$\begingroup\$ Thank you, I needed a proper explanation and you provided me with it. :) Many thanks! Maybe I'll stick to 10V for now, but at least I understood the principle behind it. \$\endgroup\$ – MerryGR Jul 2 '19 at 19:03

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