When setting up a transistor as a switch is there any difference between putting the load on the collector or emitter?
As far as I can see the only difference is in calculating the Vbe i.e. calculating what voltage is required to turn the transistor to saturation because of the voltage drop across the load
Answer
The most important difference is that for the common collector (that's the one with the load on the emitter side) you'll need a higher drive voltage. While for the common emitter 0.7 V is already enough, for the common collector the voltage must be 0.7 V + the voltage across the load.
Suppose your load is a 12 V relay, and you also supply 12 V to the collector. If you want to control that by a 5 V microcontroller then that 5 V is the maximum you can supply to the base. The emitter will be 0.7 V lower, that's 4.3 V, which is too low to activate the relay. The voltage can't go higher, because then there wouldn't be no base current anymore. So if the load voltage is higher than the control voltage you can't use common collector.
Also different is how you calculate base current. Suppose you apply 5 V on the base, the load on the emitter's side is 100 Ω and the transistor's \$h_{FE}\$ is 150. Maybe you would expect the current to be 4.3 V/100 Ω = 43 mA. That won't be the case. A base current of \$I_B\$ will cause 150 \$\times\$ \$I_B\$ through the 100 Ω resistor, not \$I_B\$. Therefore the created voltage \$V_E\$ = 150 \$\times\$ \$I_B\$ \$\times\$ 100 Ω. So the resistance seen by the base current is \$ {R_E}^{'} = \frac{V_E}{I_B} = \frac{150 \times I_B \times 100 \Omega}{I_B} = 150 \times 100 \Omega = 15 k\Omega\$.
So that 100 Ω resistor will cause a base current of only \$\frac{5V -0.7V}{15 k\Omega}\$ = 290 µA.
That's why you often won't need a base resistor in common collector configuration. You will need one though if the load consists of LEDs for instance, because contrary to the resistor these will cause a more or less constant voltage drop.
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