Current source for boost converter
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.