Electrical – Forward biasing of a BC547 transistor doesn’t turn on LED

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.

As Olin said, circuit shown in #1 and #3 are completely open loop. So the bias stability of the circuit is less and it can even lead to thermal runaway. The bias stability can be improved by including a negative feedback mechanism in these circuits. Circuits #2 and #4 does that.

Negative feedback in circuit #2: Assume that the collector current increases. This leads to increase in voltage across \$R_E\$. So the voltage at emitter increases. Since \$V_{BE}\$ remains almost constant, the voltage at base also increases. This leads to decrease in base current as an effect the collector current also reduces. So there exists a negative feedback to stabilize the operating point.

$$I_C\uparrow I_ER_E\uparrow V_E\uparrow V_B\uparrow I_B\downarrow I_C\downarrow $$

Negative feedback in circuit #4: Assume that the collector current increases. This leads to increase in voltage across \$R_C\$. So the voltage at collector decreases. This leads to decrease in base current as an effect the collector current also reduces. So here also there is a negative feedback to stabilize the operating point.

$$I_C\uparrow (I_C+I_B)R_C\uparrow V_C\downarrow I_B\downarrow I_C\downarrow $$

PS: The negative feedback affects the ac signal also which will reduce the gain of the amplifier. To avoid that a bypass capacitor is usually connected in parallel to \$R_E\$ in circuit #2.