Adding power to a circuit and voltage regulation
The current drawn depends on the LOAD  (whatever the LM7805 is driving).  It's gonna output 5V regardless the input voltage.
What changes is the POWER DROPPED in the regulator.   Instead of dropping 15V-&gt;5V, it has to drop 20V-&gt;5V   Which means the POWER consumed IN THE REGULATOR will increase about 50% (Load power is unchanged)
This all comes from Ohms Law...  That old  &quot;V=IR&quot; thing.    Power is  P=I*V, so if you're increasing &quot;V&quot; and &quot;I&quot; remains constant  (and it will here because the load is not changing and still &quot;sees&quot; a 5V source) then &quot;P&quot; must also increase.
More power = More heat, so you do need to be careful you are not overloading the regulator and that you (if appropriate) provide adequate heatsinking.
You didn't say anything about the load so can't comment further.