Question 4.11: PSpice/SPICE diode model and analysis The diode circuit show...

PSpice/SPICE diode model and analysis The diode circuit shown in Fig. 4.18 has V_{ S }=10 V , V_{ m }=50 mV \text { at } 1 kHz , R_{ L }=1 k \Omega, \text { and } V_{ T }=25.8 mV . Assume an emission coefficient of n = 1.84.

(a) Use PSpice/SPICE to generate the Q-point and the small-signal parameters and to plot the instantaneous output voltage v_{ O }=v_{ D } .

(b) Compare the results with those of Example 4.9. Assume model parameters of diode D1N4148:

IS=2.682N   CJO=4P   M=.3333   VJ=.5   BV=100   IBV=100U   TT=11.54N

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V_{ S }=10 V , V_{ m }=50 mV , \text { and } R_{ L }=1 k \Omega .

 

(a) From Example 4.7, the Q-point values are V_{ D }=0.7148 V \text { and } I_{ D }=9.284 mA . The diode circuit for PSpice simulation is shown in Fig. 4.23. PSpice simulation gives the following biasing point and small-signal parameters:

 

\begin{array}{lll}\text { ID } & 9.28 E -03 & I_{ D }=9.28 mA \\\text { VD } & 7.18 E -01 & V_{ D }=718 mV \\\text { REQ } & 5.53 E +00 & r_{ d }=5.53 \Omega \\\text { CAP } & 2.10 E -09 & C_{ j }=2.1 nF\end{array}

 

The PSpice plot of the transient response is shown in Fig. 4.24, which gives V_{ D }=718.35 mV and v_{ d (\text { peak })}= v_{ o (\text { peak })}=600.52 \mu V / 2=300.3 \mu V . Thus.

 

v_{ d }=300.3 \times 10^{-6} \sin \omega t \quad \text { and } \quad v_{ d (\text { peak })}=300.3 \mu V

 

(b) Example 4.9 gives V_{ D }=0.7148 V , I_{ D }=9.284 mA , r_{ d }=5.11 \Omega, v_{ d }=254.2 \times 10^{-6} \sin \omega t , and v_{ d (\text { peak })} = 254.2 \mu V , which agree closely with the PSpice results.

Screenshot 2021-10-20 140720
Screenshot 2021-10-20 140720

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