Topic: Analog Circuits
What will be the voltage reading of DC Voltmeter placed across the terminals of the Diode in the circuit below, having the following periodical input signal $V_i(t)$? (Assume cut-in voltage of the Diode = 0 V Forward resistance of the Diode = 2 $\Omega$)
- A. 1.25 V
- B. 2.5 V
- C. 0 V
- D. 0.1 V
Answer: B. 2.5 V
Explanation: The transformer has a 1:2 turns ratio, so the secondary voltage $V_s(t)$ is a triangle wave with peak voltage $V_{s(peak)} = 2 \times 5.2\text{V} = 10.4\text{V}$. During the positive half cycle, the diode conducts. The voltage across the diode is $V_d = V_s(t) \times \frac{R_d}{R_d + R_L} = V_s(t) \times \frac{2}{52} = V_s(t) / 26$. During the negative half cycle, the diode is reverse-biased (acts as an open circuit), so the entire secondary voltage drops across it, $V_d = V_s(t)$. The average voltage over a full cycle $V_{avg}$ is the area under the curve divided by the period. The positive half cycle area scales by $1/26$, while the negative half cycle area remains intact. Since the raw triangle areas are $\pm 5.2\pi$, the average voltage is $\frac{1}{2\pi} [\frac{5.2\pi}{26} - 5.2\pi] = \frac{1}{2\pi} [0.2\pi - 5.2\pi] = -2.5$ V. The voltmeter measures $V_{cathode} - V_{anode} = -V_d$, so it displays $+2.5$ V.