how to work BJT in saturation in PSPICE

[khaleelrehman]

hey,

what should be the modeling parameters for bjt to make it run in saturation. please help v urgent.

thanks

 

[goldsmith]

Dear khaleelrehman
Hi
It is simple . give a 0.6 volts to the be ( in fact you should give a bit more than this ) , and enough base current , thus it will be at saturation mode as well .
Best Wishes
Goldsmith

 

[WimRFP]

Driving it into saturation is as in real world (so drive hard enough so that Vce becomes alsmost zero so that the BC junction conducts). However how the simulated transistor behaves, depends on the values in the model.

I "met" various standard Silicon transistors (BJT) where the reverse transit time was even less then the forward transit time.

For general purpose transistors, expect reverse transit time to be about 50....100 times the forward transit time.

If this parameter is too small, you will not see the delay time that you normally see in a BJT that goes from saturated to non-saturated condition.

 

[khaleelrehman]

hey gold smith, thanks for your suggestion, i have got a another problem with my circuit, its like my circuit in the figure isnt giving desirable results with the pulse source, i even tried for a simle inverter block but it isnt working satisfactorily, can u tell me whats wrong with this. the parameters i used are given below.
.model Qnpn1 NPN ( LEVEL = 1
+ IS = 1.0E-16 BF = 100 NF = 1.00
+ BR = 1 NR = 1 ISE = 0
+ NE = 1.5 ISC = 0 NC = 2
+ VAF = 74.03 VAR = 10.9081 IKF = 4.36516E-3
+ IKR = 0 RB = 0 RBM = 0
+ IRB = 3.98107E-5 RE = 0 RC = 0
+ CJE = 0 VJE = 0.75 MJE = 0.33
+ FC = 0.5 CJC = 0 VJC = 0.719896
+ MJC = 0.33 CJS = 0 VJS = 0.731664
+ MJS = 0 TF = 0 XTF = 3
+ ITF = 0 VTF = 2.02338 PTF = 0
+ TR = 0 XTB = 1.5 EG = 1.11
+ XTI = 3 TRC1 = 1 )


.model Qbjtp PNP ( LEVEL = 1
+ IS = 1.0E-16 BF = 100 NF = 1.00
+ BR = 1 NR = 1 ISE = 0
+ NE = 1.5 ISC = 0 NC = 2
+ VAF = 74.03 VAR = 10.9081 IKF = 4.36516E-3
+ IKR = 0 RB = 0 RBM = 0
+ IRB = 3.98107E-5 RE = 0 RC = 0
+ CJE = 0 VJE = 0.75 MJE = 0.33
+ FC = 0.5 CJC = 0 VJC = 0.719896
+ MJC = 0.33 CJS = 0 VJS = 0.731664
+ MJS = 0 TF = 0 XTF = 3
+ ITF = 0 VTF = 2.02338 PTF = 0
+ TR = 0 XTB = 1.5 EG = 1.11
+ XTI = 3 TRC1 = 1 )
.MODEL CMOSP PMOS LEVEL=3 PHI=0.600000 TOX=2.1200E-08 XJ=0.200000U
+TPG=-1 VTO=-0.9056 DELTA=1.5200E+00 LD=2.2000E-08 KP=2.9352E-05
+UO=180.2 THETA=1.2480E-01 RSH=1.0470E+02 GAMMA=0.4863
+NSUB=1.8900E+16 NFS=3.46E+12 VMAX=3.7320E+05 ETA=1.6410E-01
+KAPPA=9.6940E+00 CGDO=5.3752E-11 CGSO=5.3752E-11
+CGBO=3.3650E-10 CJ=4.8447E-04 MJ=0.5027 CJSW=1.6457E-10
+MJSW=0.217168 PB=0.850000




.MODEL CMOSN NMOS LEVEL=3 PHI=0.600000 TOX=2.1200E-08 XJ=0.200000U
+TPG=1 VTO=0.7860 DELTA=6.9670E-01 LD=1.6470E-07 KP=9.6379E-05
+UO=591.7 THETA=8.1220E-02 RSH=8.5450E+01 GAMMA=0.5863
+NSUB=2.7470E+16 NFS=1.98E+12 VMAX=1.7330E+05 ETA=4.3680E-02
+KAPPA=1.3960E-01 CGDO=4.0241E-10 CGSO=4.0241E-10
+CGBO=3.6144E-10 CJ=3.8541E-04 MJ=1.1854 CJSW=1.3940E-10
+MJSW=0.125195 PB=0.800000

I will be verythankful for your help GOLDSMITH

 

[khaleelrehman]

this was the inverter circuit i used and the results of it