transistor voltage drop
You need a high-side switch to get this to work, as in this answer 


Q3 shifts the 3.3V level to your 4.7V (rather than the 9V shown) and R2 supplies base current for Q2. R3 prevents leakage in Q3 from being amplified by Q2. 

Pick R2 to give about 1/10 to 1/20 of the load current. R3 can be a few times larger than R2 and you can make R1 of similar value to R3.

Q2 should saturate when on and give you tens to hundreds of mV drop from the 4.7V rail depending on the transistor and the load current. If you substitute a logic-level p-channel MOSFET for Q2, you could make the drop even lower under favorable conditions. For example, a 5m\$\Omega\$ Rds(on) power MOSFET would have a drop of only 50\$\mu\$V at 10mA.