I need help on a 12v-220v Boost converter design. the converter is to have a feedback system that keeps the output voltage steady even when the input voltage drops. I have done a lot of reading and designs using a NE555 as the control system but it seems all my calculations are wrong. I am able to get a 250v output but if I add a load to it the output voltage drops drastically. I'm really confused. I need to use it for an inverter project in an exhibition next month. So please I need help.

I will attach the screen shot of the simulation on proteus 8.

Screen shot of the initial design i have. I added a zener diode at the output to limit it to 220v because it was giving me about 250v with no load.

  • \$\begingroup\$ "...the output voltage drops drastically." Then your inductor is too small. \$\endgroup\$ Aug 19, 2015 at 21:31
  • 2
    \$\begingroup\$ I would also try placing a simple resistive load at the output, then measuring the voltage across that instead of plugging directly into a DMM. \$\endgroup\$
    – RYS
    Aug 19, 2015 at 21:51
  • \$\begingroup\$ What is your frequency you are switching at? Are you sure the output of your 555 is a push/pull output or is it an open collector output? \$\endgroup\$
    – KyranF
    Aug 19, 2015 at 22:36
  • \$\begingroup\$ I agree that you need a resistive load for a more accurate simulation result \$\endgroup\$
    – KyranF
    Aug 19, 2015 at 22:36
  • \$\begingroup\$ @KyranF: NE555 is specified as push-pull, so I would be very surprised if it was OC. \$\endgroup\$ Aug 20, 2015 at 0:09

2 Answers 2


I am able to get a 250v output but if I add a load to it the output voltage drops drastically.

A standard boost regulator circuit transfers energy each switching cycle. It isn't equivalent to a transformer and so loading effects change the output voltage considerably. For instance, during the period when the transistor is on, the inductor is effectively grounded and current rises linearly to some value. You have "charged" the inductor with energy: -

Energy = \$\dfrac{L.I^2}{2}\$

When the transistor opens, this energy has nowhere to go but through D1 to the output capacitor and load. If you store 0.1 mJ and then release it to the load/cap and you do this 100,000 times per second, you are actually forcing a power of 10 watts to the output. Clearly if there is no load current, the output voltage keeps rising (due to charging the cap) and after a short period of time you get smoke.

To turn this system into a voltage regulator requires that the duty cycle be controlled by an error amplifier. In other words, to keep the output regulated on virtually an open circuit requires that the duty cycle is almost zero. As load current increases, the duty cycle MUST increase to maintain good voltage regulation.


No you can't build a 12v to 220v booster circuit like that because the voltage ratio for input to output is more than 18. at least not with the simple basic booster circuit that you are having there. Common power ratio for practical booster circuit is 2-3, which means, if you have 12v input, the maximum practical output voltage would be 12x3=36v; For your project, you really need a transformer circuit. Turn your 12vdc to 12vac using your 555 chip, then step-up the voltage using transformer.


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