Question

[2] Consider the following circuit given below. Use $V_T = kT/q = 26 mV$ as necessary. $I_{in}$ $R_1$ $V_{in}$ + $V_{out}$ $D_1$ [a] Using the ideal diode model (with $V_{D,on} = 0 V$), plot $V_{out}$ (y-axis) vs. $V_{in}$ (x-axis). Include all four quadrants. [b] Using the constant-voltage diode model (with $V_{D,on} = 0.8V$), plot $V_{out}$ (y-axis) vs. $V_{in}$ (x-axis). Include all four quadrants. [c] Using the pn-junction-based exponential model [$I_D = I_s exp(V_D/V_T)$], find $R_1$ and $I_s$ when the circuit shows that $V_{in} = 1 V \rightarrow I_{in} = 0.2 mA$ and $V_{in} = 2V \rightarrow I_{in} = 0.5 mA$. [d] Now, $V_{in}(t) = 2 + (0.001)sin(2\pi ft)$ [V]. Calculate $V_{out}(t)$ using a small-signal analysis. Use the results from [c] and the pn-junction-based exponential model.

          [2] Consider the following circuit given below. Use $V_T = kT/q = 26 mV$ as necessary.

$I_{in}$ $R_1$

$V_{in}$ + $V_{out}$
$D_1$

[a] Using the ideal diode model (with $V_{D,on} = 0 V$), plot $V_{out}$ (y-axis) vs. $V_{in}$ (x-axis). Include all four
quadrants.
[b] Using the constant-voltage diode model (with $V_{D,on} = 0.8V$), plot $V_{out}$ (y-axis) vs. $V_{in}$ (x-axis). Include
all four quadrants.
[c] Using the pn-junction-based exponential model [$I_D = I_s exp(V_D/V_T)$], find $R_1$ and $I_s$ when the circuit
shows that $V_{in} = 1 V \rightarrow I_{in} = 0.2 mA$ and $V_{in} = 2V \rightarrow I_{in} = 0.5 mA$.
[d] Now, $V_{in}(t) = 2 + (0.001)sin(2\pi ft)$ [V]. Calculate $V_{out}(t)$ using a small-signal analysis. Use the results
from [c] and the pn-junction-based exponential model.

[2] Consider the following circuit given below. Use VT = kT/q = 26 mV as necessary.

Iin R1

Vin + Vout
D1

[a] Using the ideal diode model (with VD,on = 0 V), plot Vout (y-axis) vs. Vin (x-axis). Include all four
quadrants.
[b] Using the constant-voltage diode model (with VD,on = 0.8V), plot Vout (y-axis) vs. Vin (x-axis). Include
all four quadrants.
[c] Using the pn-junction-based exponential model [ID = Is exp(VD/VT)], find R1 and Is when the circuit
shows that Vin = 1 V → Iin = 0.2 mA and Vin = 2V → Iin = 0.5 mA.
[d] Now, Vin(t) = 2 + (0.001)sin(2π ft) [V]. Calculate Vout(t) using a small-signal analysis. Use the results
from [c] and the pn-junction-based exponential model.

Added by Nicholas S.

Question

Please give Ace some feedback