Question

In a MOSFET, if the source-to-body voltage \( V_{SB} > 0 \) (for NMOS), the threshold voltage \( V_T \):

• A. Decreases
• B. Increases
• C. Remains constant
• D. Becomes negative

Hint:

Body Effect Rule of Thumb: Reverse-biasing the source-body junction (\( V_{SB} > 0 \) for NMOS) always widens the depletion width, making it harder to turn the device on. Consequently, the threshold voltage magnitude \( V_T \) increases.

Answer 1

The Correct Option is B

Concept: The threshold voltage \( V_T \) of a Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET) is not a static constant. It varies based on the potential difference between the transistor's source terminal and its body (substrate) terminal. This dependency is known as the body effect or substrate bias effect. The mathematical relation governing this behavior is expressed by the body effect equation: \[ V_T = V_{T0} + \gamma \left( \sqrt{|2\phi_F| + V_{SB}} - \sqrt{|2\phi_F|} \right) \] Where:
• \( V_{T0} \) is the threshold voltage when the source is shorted to the body (\( V_{SB} = 0 \)).
• \( \gamma \) is the fabrication-dependent body effect parameter (always positive).
• \( \phi_F \) is the substrate Fermi potential constant.
• \( V_{SB} \) is the source-to-body bias voltage.

Step 1: Examine the physical changes within the NMOS structure when \( V_{SB} > 0 \).
In an NMOS transistor, the substrate body consists of p-type material, and the source/drain regions consist of n-type diffusions. Under normal base operations, the body terminal is typically connected directly to ground (\( 0\text{ V} \)). If a positive voltage bias is applied to the source relative to the body, meaning \( V_{SB} > 0 \), the p-n junction formed between the p-type body substrate and the n-type source channel region becomes reverse-biased.

Step 2: Evaluate the widening of the depletion layer.
Applying this reverse bias extracts holes away from the junction interface and pushes electrons away, which widens the channel depletion region. This expanded depletion region uncovers more fixed negative acceptor ions inside the p-type substrate. These exposed negative charges oppose the formation of an electron inversion layer under the gate oxide.

Step 3: Determine the effect on threshold voltage using the formula.
Because there is an increased volume of negative depletion charge that must be counterbalanced, a larger positive gate-to-source voltage (\( V_{GS} \)) is needed to attract enough free electrons to form the inversion channel. Looking directly at our analytical equation: When \( V_{SB} > 0 \): \[ \sqrt{|2\phi_F| + V_{SB}} > \sqrt{|2\phi_F|} \] This makes the bracketed difference term strictly positive: \[ \left( \sqrt{|2\phi_F| + V_{SB}} - \sqrt{|2\phi_F|} \right) > 0 \] Since \( \gamma > 0 \), the entire secondary correction term is added directly to the baseline threshold: \[ V_T = V_{T0} + \text{positive value} \quad \Rightarrow \quad V_T > V_{T0} \] Thus, the threshold voltage \( V_T \) increases.