Suppose this is the biasing process of an NPN transistor.
simulate this circuit – Schematic created using CircuitLab
The battery to the left forward biases the emitter diode because its +Ve terminal is connected to the the P-type base and -Ve terminal to the n-type emitter.
But how does the right battery reverse bias the collector diode?
I get the +Ve side of this battery is connected to n-type collector, but its -Ve side is also connected to the N-type emitter.
The voltage from V1 is divided between R1 and the B-E junction of the transistor. Since that junction is a forward-biased diode, the voltage drop across it is only about 0.65V; the rest of the voltage appears across R1, setting the magnitude of the base current.
The voltage from V2 is divided among R2, the C-B junction of the transistor, and the B-E junction. Since we've already established that the last one of those is 0.65V, we can say that V2 - 0.65V is divided across R2 qand the C-B junction. As long as this quantity is positive (i.e., V2 is greater than 0.65V), the C-B junction is reversed-biased.
Let's say the base is at approximately 0.5 v potential above ground and causing a collector current of (say) 1mA. If V2 is at 10 volts with respect to ground (emitter) and R2 is 1kohm (for instance) then there will be 9V at the collector.
With 9V at the collector and 0.5 volts at the base, the CB junction is reverse biased by 8.5 volts.
Both batteries share a terminal. So that the collector-base diode is reverse biased, \$V_C\$ must be greater than \$V_B\$, as
$$ V_{CB} = V_C - V_B $$
Viewed another way, since the emitter terminal is grounded directly, must \$V_C > V_{BE}\$
For the right battery to reverse bias the collector diode( or collector base junction as it is called), the potential at the collector must exceed that at the base. Otherwise, the transistor is no longer in active mode.
Note that it is the collector base junction which is reverse biased in active mode.
