Finding vT h and RT h using PSpice
The figure shown represents one form of the equivalent circuit for a transistor amplifier. Determine the
open circuit value of v2 and the output resistance RT h of the amplifier.
1
i1 2kΩ
+
vx
5000
2mV
−
3
2
+
−
+
100i1 20kΩ
−
+
vx
−
hk2p43.m4
Figure 1: An equivalent Circuit for a transistor Amplifier
The way to find out Thevenin’s equivalent circuit as seen from the nodes 3 and 0 is to measure the open
oc
circuit voltage voc = v2 = V (3) and the short circuit current isc flowing from 3 to 0 . vT h = voc , RT h = visc
The following PSpice code evaluates the open circuit voltage between 3 and 0 by inserting a very high
resistance across the two nodes and measure the voltage between them. This is voc = vT h .
Prob43c2.cir
*Find Thevenin equivalent circuit for a transistor amplifier
*FIRST V(3)=VOC OPEN CIRCUIT VOLTAGE
V10 1 0 2MV
R12 1 1A 2K
V1A 1A 2 DC 0V; this is needed to measure $i_1$
E20 2 0 3 0 2E-4
F30 3 0 V1A 100
R30 3 0 20K
RL 3 0 1G ; 1 Giga Ohms resistor (almost open circuit)
.END
Run the simulation to get V (3) = −2.4999
The next PSpice code computes the short circuit current between 3 and 0 by inserting a 0 volts voltage
source and a series very small resistance between the two nodes and measure the current flowing through
them. This is isc .
Prob43c2.cir
*Find Thevenin equivalent circuit for a transistor amplifier
* I(V3A)=SHORT CIRCUIT CURRENT
V10 1 0 2MV
R12 1 1A 2K
V1A 1A 2 DC 0V; this is needed to measure $i_1$
E20 2 0 3 0 2E-4
F30 3 0 V1A 100
R30 3 0 20K
V3A 3 3A DC 0
RL 3A 0 1u ; 1 micro Ohms resistor (almost short circuit)
.END
Run the simulation to get I(V 3A) = −1.000E − 04
Next divide the open circuit voltage V (3) obtained from the first simulation by the short circuit current
obtained from the second simulation I(V 3A) to get RT h = −2.4999
−1e−4 = 24.999kΩ.
Use of .TF command
PSPice can also find vT h , RT h using a very high resistance between 3 and 0 and using the .TF command
as follows:
*Find Thevenin equivalent circuit for a transistor amplifier
V10
1
0
2MV
R12
1
1A
2K
V1A
1A
2
DC
0V
E20
2
0
3
0
2E-4
F30
3
0
V1A
100
R30
3
0
20K
RL
3
0
1G
.TF v(3) v10
.END
NODE
VOLTAGE
NODE
VOLTAGE
NODE
VOLTAGE
NODE
VOLTAGE
(
1)
.0020 (
2)-500.0E-06 (
3)
-2.4999 (
1A)-500.0E-06
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V10
-1.250E-06
V1A
1.250E-06
TOTAL POWER DISSIPATION
V(3)/V10 = -1.250E+03
2.50E-09
WATTS
INPUT RESISTANCE AT V10 = 1.600E+03
OUTPUT RESISTANCE AT V(3) = 2.500E+04
From the output of the simulation, V (3) = voc = vT h = −2.4999 volts and the output resistance at V (3)
(open circuit condition) equals RT h = 2.5e + 4 = 25kΩ