Is biasing required when transistor work as switch?

Biasing might only be required for small signal switching below 1V. Small input signals might not always have enough voltage to cross the 0.7V B-E potential barrier. Then a slight offset voltage (biasing) must be used to bring the signal closer to the critical decision voltage of 0.7V across the B-E junction.

Note that I would refer to the B-E junction to be 0.7V rather than just the Base since it's not actually the Base that requires the 0.7V but it's the potential across the junction. Although the Base voltage is the same as the Base-Emitter voltage when the Emitter is connected to the ground, it's not always the case.

Biasing is not necessary for switching signals larger than 1V since the amplitude is enough even with Base resistor to raise the B-E junction potential from 0V to above 0.7V if correct value for base resistor is used.

It is however very important that a Base resistor Rb is provided to limit the current flowing from the input through the B-E junction to the ground. Transistors are best protected when the control is done with the correct current and not voltage, given that the correct voltages are applied. This resistor should be determined by the forward bias current gain (B) of the transistor, supply voltage, input switching signal voltage (Vin) and the load resistance RL.

Example: If one use a NPN transistor and connect the Emitter to the ground to switch a basic load via the collector. The load resistance (RL) = 100R, Supply voltage (Vs) = 10V, transistor forward bias current gain (B) = 100, and input switching voltage (Vin) = 5V, then the following method may be used
First consider:

Collector current:
Ic=(Vs - Vce)/RL

Base current:
Ib=(Ic*10)/(B) (The factor 10 is used for deviations in hfe value, it is a safety factor)

Vin-Vbe=Rb*Ib

Thus:
Rb= [B.RL.(Vin-Vbe)]/((Vs-Vce)*10)
Rb= [100.100.(5-0.7)]/((10-0.1)*10)
Rb= 434R
N.A.V.=470R


More about using transistor as a switch for DC supplies

Switching lights or relays on by using the Collector of a NPN is okay, but the method is not advised at all for more complex circuits. Beware of using the method to turn on other electronic systems or circuit boards. In this method the circuit will float at Vs when off, which can also damage other equipment dependant on them. It will also never be properly grounded when on. That may lead to erratic response in digital equipment or increased circuit noise in analog.
For turning on circuit boards or complex devices, the alternative in a NPN to turn the circuit on via its positive supply with the Emitter. This is not a good idea either since the 0.7V drop might cause supply voltage to be to low and increase line resistance. The supply voltages will drop and not be constant for changes in supply current.

Rather use PNP transistor, supply positive voltage to the Emitter and supply the load with the Collector. Pull the Base voltage low to create a 0.7 volt drop across the Base-Emitter junction. The calculations work almost the same as above except the Vin is now with reference to the Vs and not ground since we are pulling down. But this will ensure less resistance from the transistor when on and less volt drop, approximately 0.1V. Since the load is now controlled through it's positive supply. The load remains properly grounded for all conditions. But in this method one would have to pull the Base voltage low through a Base resistor aswell. To turn the device on a second NPN transistor is often used at it's Base to perform that function if conventional Vin=On and ground = Off input states are used. This method will result in a two transistor switch.

ANSWER: yes it is not required since the transistor must be forced to saturate to be used as a switch. However the current cannot be to the extent of adding a voltage forward drop to the transistor satureted comndition. as a rule a forced beta of ten is used