Question

The graph between the input voltage ($V_i$) and the output voltage ($V_o$) of a transistor connected in common emitter configuration is shown in the figure. The active, saturation and cutoff regions of the transistor are respectively

• A. I, II and III
• B. II, III and I
• C. II, I and III
• D. I, III and II

Hint:

For a CE transistor switch: Low Input ($V_i$ low) $\rightarrow$ OFF state $\rightarrow$ Cutoff Region (High Output $V_o$). High Input ($V_i$ high) $\rightarrow$ ON state $\rightarrow$ Saturation Region (Low Output $V_o$). The steep transition part in between is the Active Region, used for amplification.

Answer 1

The Correct Option is B

The given graph is the transfer characteristic curve for a common emitter (CE) amplifier, often used as a switch. Let's analyze the regions.
Region I:
In this region, the input voltage $V_i$ is low (typically below the cut-in voltage of the base-emitter junction, $\sim 0.7$V for silicon).
When $V_i$ is low, the base current $I_B$ is approximately zero. This causes the collector current $I_C = \beta I_B$ to also be approximately zero.
With no current flowing through the collector resistor $R_C$, there is no voltage drop across it.
Therefore, the output voltage $V_o = V_{CC} - I_C R_C \approx V_{CC}$, which is a high, constant value.
This region, where the transistor is essentially "off", is called the cutoff region.
Region III:
In this region, the input voltage $V_i$ is high.
This drives a large base current $I_B$.
This, in turn, attempts to drive a very large collector current $I_C = \beta I_B$.
However, the collector current is limited by the external circuit to a maximum of $I_{C,max} \approx V_{CC}/R_C$. When the transistor reaches this limit, it is said to be saturated.
In saturation, the voltage drop across the transistor, $V_{CE}$ (which is $V_o$), becomes very small and nearly constant (typically $\sim 0.2$V).
This region, where the transistor is fully "on", is called the saturation region.
Region II:
This is the transition region between cutoff and saturation.
In this region, a small change in the input voltage $V_i$ causes a large change in the output voltage $V_o$.
The collector current $I_C$ is proportional to the base current $I_B$ ($I_C = \beta I_B$).
The transistor is operating as an amplifier in this region. This is called the active region.
The question asks for the active, saturation, and cutoff regions, respectively.
Based on our analysis:
Active region = Region II Saturation region = Region III Cutoff region = Region I Therefore, the correct sequence is II, III, and I.