I made a boost converter with 1 1mH inductor, a mosfet, a 100uF capacitor rated for 400v, a 20 ohm 1/2 watt resistor, and an ICM7555 cmos timer.

Using previous knowledge and wikipedia pages on the subject, I assumed that my volt6age should be brought from a 5v input to 10 volts. I assumed that the inductor effectively became a battery in series with the source. After seeing a ridiculous (and continuously rising) voltage, I became suspicious.

What is going on? I not getting 10 volts, but I am rather getting a continuously rising voltage. My mosfet is rated for 55 volts so I haven't been testing above that, but it rises steadily until I disconnect it at 50 volts. If anyone could take the time to explain what is going on, that would be great.

*NOTE : My 555 timer is connected with unknown capacitors (small) as C and a 1k as Rb and a 10k as Ra.

Circuit diagram:

enter image description here

EDIT : For clarification, I changed "but it rises steadily until it reaches 50 volts" to "it rises steadily until I disconnect it at 50 volts"

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    \$\begingroup\$ You should post your schematic, otherwise we can only guess... \$\endgroup\$
    – anrieff
    Commented Mar 7, 2017 at 2:21
  • \$\begingroup\$ Ok I will draw one up. Give me a bit. \$\endgroup\$ Commented Mar 7, 2017 at 2:22
  • \$\begingroup\$ Each pulse causes a flyback pulse current that steps the voltage so much for each pulse depending on size of Cap and current pulse.. You can change it to a same current rated 450V Transistor switch and drive base current with 10% of coil current based on V/DCR. I would expect coil in the 1 Ohm range so 555 can drive base direct with 10Ohms or so with emitter grounded. It may get warm. \$\endgroup\$ Commented Mar 7, 2017 at 2:35
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    \$\begingroup\$ You have no feedback mechanism in that circuit. How is it supposed to know when it's reached the voltage you want to keep it below and stop it rising further? \$\endgroup\$
    – Finbarr
    Commented Mar 7, 2017 at 7:22
  • 2
    \$\begingroup\$ The easiest way to add feedback is to use an IC dedicated to the task of running a boost converter, which often gives you other useful features like shutoff and even switching modes. Failing that, a comparator that tests the output against your desired voltage and keeps the 555 reset provides a more primitive mechanism. \$\endgroup\$
    – Finbarr
    Commented Mar 9, 2017 at 9:05

3 Answers 3


I can't even guess where you got the notion that the inductor would act like a battery. That's not at all what inductors do. Stop making stuff up. Go back and learn the basics.

What is happening is the current builds up in the inductor when the switch is on. When the switch turns off, that current will continue to flow in the short term.

A useful way to think of inductors electronically (not how the physics works) is that they give inertia to current. You apply a fixed voltage to a inductor, and the current will ramp up linearly. The slope is proportional to the voltage (how hard you push) and inversely proportional to the inductance (how much inertia the current has). When you reverse the voltage, the current will then ramp down linearly.

If you try to stop the current abruptly, it won't let you. The inductor will make as high a voltage as it takes to keep the current going immediately. This high voltage will cause the current to ramp down fast, but this still takes finite time. This is just like trying to stop a moving mass instantly. It will make a high force, which then slows down the mass quickly. The faster you try to stop the mass, the higher the force will be. That may require damaging whatever is trying to stop the mass.

Your inductor is making little squirts of current thru the diode every time you try to shut off its current. The current can't go thru the transistor anymore, so the diode is the only option. It takes whatever voltage the cap is at to allow this current. The higher this voltage, the quicker the current in the inductor will ramp down. You should be able to notice that the voltage on the cap builds up more slowly as it gets higher.

When the cap voltage gets to the maximum voltage the transistor can handle, there is now a second path for current to take. That is thru the transistor, which won't be good for it.

Again, go and learn about inductors and boost converters. There should be much out there.

  • \$\begingroup\$ This was extremely well explained. I will definitely try to learn more about both inductors and boost convertors. I got the idea that the inductor would act like a battery only in the fact that the inductor will create a current source + voltage source. You explained exactly what I didn't understand, and did it in a perfect way. \$\endgroup\$ Commented Mar 9, 2017 at 1:07

Boost regulator IC use feedback to control the voltage. When the voltage gets too high the IC backs off the duty cycle and when the voltage gets too low the duty cycle is increased. Some boost regulator ICs will even switch operating modes depending on load.

The boost converter you put together is open loop, meaning there is no feedback. Without feedback nothing controls the duty cycle to regulate the voltage. The open loop setup is very load dependent. The larger the load the lower the voltage.

What you can do is use a potentiometer to make the duty cycle of the 555 adjustable and then tune the converter to the load. The only problem would be is that the load would have to be constant.


An inductor does not "behave like a battery".

With a capacitor (or to some extent a battery, though batteries are highly non-linear), the faster the change in voltage the higher the current. With an inductor the opposite is true, the faster the change in current the higher the voltage.

At any given time a traitional boost converter can be in one of three states (things are different in "synchronous rectification" converters).

  • Charging: the inductor is connected to the incoming power supply by the transistor and the current in the inductor is increasing. No current is supplied to the output capacitor.
  • Discharging: the inductor and the incoming power supply are connected in series and the current in the inductor is decreasing, current is supplied to the output capacitor through the capacitor.
  • Non-conducting: there is no current flow in the inductor.

The ratio of input voltage to output voltage depends on the ratio of charging time to discharging time.

Under high load the current in the inductor never drops to zero, it is always either charging or discharging. Therefore the output voltage is set by the duty cycle with which the transistor is switched.

However under light or load this is no longer the case. The current in the inductor can't drop below zero. So there is a period where the inductor is neither charging nor discharging. Each cycle now imparts a fixed amount of energy to the inductor and that energy will be transfered to the output side. If there is no load and the components are ideal then the output voltage will keep rising forever (though the rate of rise will slow down).


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