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:
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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.
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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?
Best Answer
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?