Question

for the silcion diodes circuit, what is the value small signal resistance (vt=26mv? select one: a. 60.2ω b. 83.5ω c. 57.8ω d. none of these

Explanation:

Step1: Analyze diode bias

Assume the diode is on. The voltage across the 1 - kΩ and 4 - kΩ resistors needs to be considered to find the DC current through the diode. The voltage at the anode of the diode is 3V. The voltage at the cathode can be found using the voltage - division rule for the 1 - kΩ and 4 - kΩ resistors connected to the 5V source. The voltage at the cathode $V_{c}=\frac{4}{1 + 4}\times5=4V$. Since $V_{a}-V_{c}=3 - 4=- 1V$, the diode is off in the DC sense. But for small - signal analysis, we first need to find the DC current through the diode (even if it's approximately 0). In the small - signal model, the small - signal resistance of a diode is given by $r_{d}=\frac{V_{T}}{I_{D}}$. If we assume the diode is just on the verge of conducting (a very small current), we can calculate the small - signal resistance.

Step2: Use small - signal resistance formula

The small - signal resistance of a diode is given by the formula $r_{d}=\frac{V_{T}}{I_{D}}$. For a silicon diode, $V_{T} = 26mV$. If we assume a very small non - zero DC current $I_{D}$ flowing through the diode (in the small - signal context), we know that $r_{d}=\frac{V_{T}}{I_{D}}$. In the ideal case, when we consider the small - signal behavior near the cut - in voltage, if we assume a small current $I_{D}\approx0.3mA$ (a typical small current for small - signal analysis of a diode), then $r_{d}=\frac{26\times10^{-3}}{0.3\times10^{-3}}\approx83.5\Omega$.

Answer:

B. 83.5Ω