In cutoff mode, a transistor is essentially acting as an open switch, meaning it is not conducting any current between its collector and emitter terminals. When a transistor is in cutoff mode, there is no significant flow of current through it, and its output voltage (V_CE) is at its highest level, typically close to the power supply voltage (V_CC) for an NPN transistor or close to ground (0V) for a PNP transistor.
To understand the concept of the voltage drop across a transistor in cutoff mode using Ohm's Law, we need to focus on the voltage across the collector and emitter terminals, denoted as V_CE.
Ohm's Law states that the voltage (V) across a resistor is equal to the product of its resistance (R) and the current (I) passing through it:
V = I * R
In the case of the transistor in cutoff mode, we can model it as an open switch, which means the current flowing through it (I_CE) is essentially zero. Since there is no current flowing through the transistor, we can consider the equivalent resistance (R_CE) to be infinite.
Now, let's apply Ohm's Law to the transistor in cutoff mode:
V_CE = I_CE * R_CE
Since I_CE is almost zero and R_CE is infinite, the voltage drop across the transistor (V_CE) is negligible. Therefore, in cutoff mode, the voltage drop across the transistor is essentially zero, and the output voltage (V_CE) is nearly equal to the supply voltage (V_CC) for an NPN transistor or 0V for a PNP transistor.
In summary, when a transistor is in cutoff mode, it acts as an open switch, and according to Ohm's Law, with negligible current flowing through it, the voltage drop across the transistor (V_CE) is almost zero.