Question

given a silicon diode circuit, assume $v_b = - 5v$. to plot the output voltage ($v_o$) as functions of the input voltage ($v_i$) over the range $-10leq v_ileq +10v$, we use the analysis:
select one:
a. $v_o = v_i + 4.3$; for $v_igeq - 4.3v$ and $v_o = 0$; for $v_ileq - 4.3v$
b. $v_o = 0$; for $v_igeq - 5.7v$ and $v_o = v_i + 5.7$; for $v_ileq - 5.7v$
c. $v_o = v_i - 5.7$; for $v_igeq 5.7v$ and $v_o = 0$; for $v_ileq 5.7v$
d. $v_o = 0$; for $v_igeq 4.3v$ and $v_o = v_i - 4.3$; for $v_ileq 4.3v$

Explanation:

Step1: Analyze diode conduction

A silicon - diode has a forward - voltage drop of approximately $V_D = 0.7\ V$. When the diode is forward - biased, current flows and there is a voltage drop across it. The voltage across the battery is $V_B=- 5\ V$. For the diode to conduct, the anode (input voltage $V_i$) must be at least $0.7\ V$ higher than the cathode voltage. The cathode voltage is $V_B=-5\ V$, so the diode will conduct when $V_i\geq - 5 + 0.7=-4.3\ V$.

Step2: Calculate output voltage when diode conducts

When the diode conducts ($V_i\geq - 4.3\ V$), we can use Kirchhoff's voltage law. The voltage across the resistor is $V_R = V_i - V_D - V_B$. Since $V_D = 0.7\ V$ and $V_B=-5\ V$, then $V_R=V_i-(0.7)-(-5)=V_i + 4.3$. And the output voltage $V_o = V_R$. When the diode is reverse - biased ($V_i\lt - 4.3\ V$), no current flows through the resistor, and $V_o = 0$.

Answer:

A. $V_O = V_i + 4.3$; for $V_i\geq - 4.3\ V$ and $V_O = 0$; for $V_i\leq - 4.3\ V$