What are the real world limitations of using a transistor as a switch?
In that configuration, the transistor is an emitter follower/ common collector. As such, the maximum voltage the load will ever see is your 3.4V - 0.7V = 2.7V. Since that's about half what the load expects, it will only take half the current.

Your circuit needs to be common emitter, more like this.







simulate this circuit &ndash; Schematic created using CircuitLab

HOWEVER: With the load currents you are talking about the base current will be a lot with a simple transistor, likely more than whatever you have driving this can supply. You would be wise to replace Q1 with a suitable N-Channel MOSFET instead.

Obviously, whichever part you chose needs to be able to carry the load current indefinitely without burning out.

Note D1 is a fly-back diode. It is required so the transistor does not kill itself when it turns off.

Leaving the base open

Not a good idea for any transistor circuit. Leakage current from collector becomes base current. That's why good designs include some form of biasing resistor to hold the base at the right "parked" voltage.