The calculation is as follows:



Relay has a 400Ohm coil, so draws 12V / 400Ohm = 30mA.
To fully turn-on the transistor BC337, which means a minimal collector emitter saturation voltage, a base current of 1/10 of collector current, means 3mA must flow. As the base emitter voltage is about 0.7V when turning on the transistor and 0.7V / 1.2kOhm = 580µA is already flowing through the lower 1k2 resistor, then between base of BC337 and collector of BC556 a 1k2 resistor must be inserted.

Why?
As the collector emitter voltage of BC556 can be neglected when being turned-on, 5V - 0.7V is dropping across this (upper) 1k2 resistor, resulting in a current of 4.3V / 1.2kOhm = 3.58mA which just equals the sum of 3mA base current and 580µA current flowing through the lower 1k2 resistor.

What is the benefit of the lower 1k2 resistor?
Above we have discussed, what the situation should be when fully turning on the relay. We have designed the circuit using a much higher base current than is actually needed to turn-on the relay, because we have driven the transistor into saturation to achieve a very low collector emitter saturation voltage. For this we need a base current which is much higher than from what would be needed when considering the current gain of transistor.

Current gain of transistor is often specified for a collector emitter voltage of about 5V. Current gain of BC337 is specified for a collector emitter voltage of 1V (because it's a typical switching transistor) and is about 250 for a collector current of 30mA. Means, we can turn-on the BC337 resulting in a collector emitter voltage of 1V, when we drive a current of about 30mA / 250 = 12µA into its base.

But this means, that the transistor is only partially turned-on. Also, the current gain can drastically differ from part to part. Such a low base current in fact is totally unsuited to drive a relay, but it tells, what possibly could happen, if we are not carefully turning-off all the base current, when we want the relay to be turned-off!

And that's the point where the lower base resistor comes into play: As the transistor needs always about 0.7V to turn-on, the 1k2 resistor from base to signal ground will draw 580µA, as calculated above. So, applying 0.7V to the base will cause a current flow of 580µA + 12µA, although only 12µA will flow into the base of transistor. But becasue this current must also flow through the upper 1k2 resistor, again about 0.7V will drop at this resistor. So, to turn-on the BC337 a minimum potential at collector of BC556 of about 1.4V and a minimum current of about 580µA is needed. And this voltage/current will even only result in a partial turning on of BC337 as discussed above.

So, any sort of noise must at least be higher than 1.4V and must be able to deliver at least 580µA, before the BC337 is in danger to be turned on. Without this lower 1k2 base resistor only about 0.7V and a current of 12µA would be needed to do this. So, this additional base resistor drastically increases the noise immunity of the relay driver, by adding a certain and well defined threshold for turning on the transistor and by this the relay.

Kai