Neglecting the losses of any kind thus assuming a 100% efficiency, one can thing about a boost convert as a voltage translator with constant power. I mean that Pin = Pout.

$$ P_{in} = V_{in}I_{in}$$
$$ P_{out} = V_{out}I_{out}$$

thus

$$ V_{in}I_{in} = V_{out}I_{out} $$

of rewritten differently:

$$ \frac{V_{in}}{V_{out}} = \frac{I_{out}}{I_{in}} $$

thus if you boost 3.3V to 30V and see a power consumption of 9A, then

$$ \frac{3.3V}{30V} = \frac{I_{out}}{9A} $$

$$ I_{out} = 1A $$

If you designed your boost convert to supply *roughly* 1A then it is working fine. Because we assumed a 100% efficiency, the real available output current would be less than 1A.

Usually you do the computation backward, you know your output current requirement and you can compute what would be the input current.

You are completely driving your MOSFET incorrectly. Firstly there is no postive voltage source that can turn on the device and secondly the BJT will not work to pull-down the gate voltage to zero on alternate half cycles of any PWM signal you might apply.

If you tried connecting the NPN BJT's collector to your micro controllers Vcc then you have the basic means to activate the FET but you still need something that discharges the FET's gate capacitance when deactivating.

You should also look to see if you can find a MOSFET that can be adequately driven with a logic level voltage - if not then you should consider using an intermediary voltage (maybe 8V) to power a driver that can properly switch the FET on and off at the PWM rate.

## Best Answer

I know the questioner has accepted an answer but I don't think it is sufficient as it stands so I'm offering what I consider to be a better take on things. The main problem is that I think the wiki article could be better too. It states that a boost converter is type of SMPSU....

I think the only reasonable form of boost converter used in electronics always contains an inductor and capacitor. Here's a more reasonable diagram shown later on the same wiki article: -

Before the switch closes for the first time it can be seen that the capacitor and resistor are connected to the incoming supply (\$V_I\$) via the diode. This pre-charges the cap to \$V_I\$ minus one diode drop (0.6 V for regular diodes and maybe 0.3 V for a schottky diode). Current flow is approximately \$\dfrac{V_I}{R}\$.

When the switch first closes, the inductor current builds from \$\dfrac{V_I}{R}\$ to a higher value. It should be noted at this point that the diode stops the switch discharging energy from the output capacitor because it is reverse biased.

When the switch opens, the inductor current wants to continue flowing in the same direction and, the easiest path to take is through the diode and this pushes "new" energy into the capacitor (charging it slightly). Some of that energy flows into the resistor/load but the majority charges the capacitor (in conventional applications).

After a few cycles the capacitor voltage is now starting to rise higher than the incoming supply and this is what a boost regulator is trying to do.

Assuming the load resistor is fixed in value and the mark-space ratio remains constant (for the sake of this explanation), there reaches a point when the amount of energy used by the load resistor (per switching cycle) matches the amount of energy liberated from the inductor. Equilibrium is reached and the following can be said: -

Energy stored charging the inductor x switching frequency = power dissipated in load resistor.

This is a simple DCM type of boost converter. CCM converters leave residual energy in the inductor and operate at a higher average inductor current and are beyond the scope of this simple explanation.

For instance if the load was taking 10 W and the switch was switching at 100 kHz, the inductor energy stored (and liberated) = \$\dfrac{10}{100,000}\$ = 100 \$\mu J\$. Of course there will be a little loss in the diode but, that energy equation allows you to calculate the size of the inductor needed and the duty cycle of the switch.

If the load resistance increases, in order to maintain the same voltage at the output, the duty cycle of the switch should lower and, if load resistance decreases the duty cycle of the switch should increase to maintain the same voltage at the output.