I am trying to step 3V up to 9V via a boost converter, the idea being I have the current but not the voltage to drive a DC motor. I used this this website to calculate the parts values I would need, and assembled the boost converter on my breadboard. I noticed two things:

  1. Increasing the duty cycle with which I was PWMing the transistor barely increased the voltage. I was led to believe it would do so more greatly.

  2. The capacitor would charge until I provided a load across it, at which point it would reach an equilibrium. This makes sense to me (the capacitor charges when the switch/transistor is off, and discharges if there is a load when the switch is on), but placing the H-Bridge that drives my motor in parallel with the load resistor results in an immediate discharge of the capacitor and the voltage drops from 9V to 4.5V, not enough to power the motor. Placing the H-Bridge in series with the load resistor does not result in it being powered but as if the load was disconnected.

I unfortunately do not remember the model number of the diode off the top of my head and won't be able to get to the lab to check it until later in the day. It is some type of fast switching Schottky diode.


simulate this circuit – Schematic created using CircuitLab

Am I doing something wrong in particular, or is this just something a boost converter cannot do? Should I be using a different circuit of some type?

  • 2
    \$\begingroup\$ Show us the schematic diagram of the circuit you ended up with. Show the values for each component that you arrived at. \$\endgroup\$ – Dave Tweed Apr 16 '14 at 15:10
  • \$\begingroup\$ Is there a recommended tool for drawing up schematics to post around here? \$\endgroup\$ – tnecniv Apr 16 '14 at 15:15
  • 1
    \$\begingroup\$ There's a built-in schematic tool. It shows a diode, pencil, and a resistor. It's right next to the add image button. \$\endgroup\$ – horta Apr 16 '14 at 15:15
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    \$\begingroup\$ Yes, edit your post, position the cursor where you want the schematic to appear, and hit control-M. This will bring up the CircuitLab editor in your browser. \$\endgroup\$ – Dave Tweed Apr 16 '14 at 15:16
  • \$\begingroup\$ Updated with schematic \$\endgroup\$ – tnecniv Apr 16 '14 at 17:50

If you have the current but not the voltage to drive your dc motor you might be disappointed. You might have the ability (in your battery) to supply 1 amp at 3 volts but, to power a 1 amp motor with 9 volts means your battery will need to supply maybe 3.5 amps into the boost regulator.

Power in equals power out plus losses and to obtain 9 volts at 1 amp requires a power in of 9 watts plus about 15% extra for the boost converter losses. That takes the input current up to about the 3.5 amp level. Can your battery sustain this level I wonder?


This is not kind of question that can be answered precisely, but I will try to give some suggestions.

If circuit parts are calculated correctly, and you still have such voltage drop - it may be caused by:

  • too small coil (in size) - you may need thicker wire
  • power source can;t keep up with motor power requirements
  • wrong capacitor type, it should be low-ESR (normal capacitor can't charge quick enough due to its internal resistance)
  • damaged or worn out capacitor

If you add more information about your power source, motor, real schematics, information about coil size - maybe someone will answer more precisely.

  • \$\begingroup\$ This isn't an answer really; it's a series of comments and questions. \$\endgroup\$ – Andy aka Apr 16 '14 at 19:14
  • \$\begingroup\$ @Andyaka I changed it. Now better? \$\endgroup\$ – Kamil Apr 16 '14 at 19:37

There are two parameters that matter in the boost converter: Duty cycle, and switching frequency. (And a third: whether it's in continuous or discontinuous conduction mode)

You are saying that the capacitor "drains" when you supply the load -- this means that you're not switching the converter fast enough to re-"fill" the capacitor.

The reason for this may be that there's too high resistance in your inductor, or your transistor can't sink enough current (typically, power N-channel MOSFETs are used,) or your battery source has too high internal resistance.

A 2N2222 is not a power transistor. Also, you don't indicate how much current your oscillator will source. Given that you're using a BJT, the amount of current through the transistor is directly proportional to the amount of current out of the oscillator (up to the saturation point of the BJT.) Without knowing more, I would suspect using a higher-performance transistor with a higer switching rate (say, 20 kHz) and a lower-resistance inductor would improve the performance you see.

Also, the schematic shows a sine generator. You want to use a square wave for a boost converter.

Hook up a scope to various points in your circuit, to measure how it's doing, and this will tell you where it goes wrong.

  • \$\begingroup\$ I did try it with an n-channel Power MOSFET first (an IRF630), but it resulted in about a 3V output across the capacitor. Perhaps it's the inductor then. I will try some other inductors when I go to the lab later, I guess. \$\endgroup\$ – tnecniv Apr 16 '14 at 18:23

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