DC Analysis of a NPN BJT using Voltage Divider Bias
FIGURE 1: NPN Voltage Divider Bias Circuit
GIVEN:
A
VCC := 12V
RB2 := 30kΩ
β := 150
VBE := 0.7V
RB1 := 60kΩ
RE := 6.8kΩ
A
RC := 2.7kΩ
FIND: All node voltages and branch currents for the amplifier.
To do this we can proceed using node equations and find the solution for the
node voltages and branch currents. This is a valid approach yet there is a much
easier way by using Thevenin's analysis of the circuit. To Theveize the circuit we
first find the Thevenin voltage at the base of the NPN.
VTH := VCC ⋅ 


 RB2 + RB1 
RB2
VTH = 4 V
Next, we find the Thevenin resistance at the base of the NPN:
RTH :=
 1 + 1 
 RB1 RB2 


−1
RTH = 20 kΩ
Next. we redraw the NPN circuit using the Thevenin voltage and resistance and
inserting the VBE voltage and the load of RE in the base-emitter loop. We have
IB entering the base and IE = (β+1) x IB sank into RE. To determine IB, we simply
multiply RE by (β+1) resulting in the second circuit in Figure 2.
FIGURE 2: Thevein's Equivalent Circuit for the Base-Emitter Cicuit
Solving for IB, IE and IC:
IB :=
VTH − VBE
RTH + ( β + 1 ) ⋅ RE
IB = 3.152 µA
IE := ( β + 1 ) ⋅ IB
IE = 476.022 µA
IC := β ⋅ IB
IC = 472.87 µA
Solving for Node Voltages:
VE := IE⋅ RE
VE = 3.237 V
VB := VE + VBE
VB = 3.937 V
VR1 := VCC − VB
VR1 = 8.063 V
VC := VCC − IC⋅ RC
VC = 10.723 V
Solving for the Remaing branch currents:
IB1 :=
VR1
RB1
IB1 = 134.384 µA
I2 :=
VB
RB2
I2 = 131.232 µA
CE Amplifier with voltage divider bias Small-Signal Analysis
Continuing with the small-signal analysis of the amplifier
Figure 3: Common-Emitter Amplifier with Voltage Divider Bias
Using the resistance values from the DC analysis and the corresponding DC bias point
information found we now perform a small-signal analysis on the circuit as shown in Figure 3.
The coupling capacitors CC1 and CC2 are used to pass the signal component while blocking DC
thereby maintaing the DC bias point. The bypass capacitor CB is used to place the emitter at
small-signal ground. These capacitances will be considered ideal where they have no impedance
on the signal and infinite impednace for DC. To perforn the small-signal analysis the small-signal
equivalent circuit is next presented in Figure 4.
Figure 4: Small-Signal Equivalent Circuit for the CE Amplifier
vs := 10mV
Given:
f := 1kHz
RL := 3.3kΩ
RS := 50Ω
VA := 150V
Calculation of ro:
ro :=
VA
ro = 317.212 kΩ
IC
Calculation of small-signal base resistance RB:
RB :=
 1 + 1 


 RB1 RB2 
−1
RB = 20 kΩ
Calculation of emitters incremental resistance (Temp = 25'C) :
VT := 25.8651mV
re :=
VT
re = 54.336 Ω
IE
π:
Calculation of rπ
rπ := ( β + 1 ) ⋅ re
rπ = 8.205 kΩ
Calculation of transconductance gm:
α :=
β
gm :=
β+1
α
gm = 0.018 S
re
Calcularion of Input and Output resristances:
Rin :=
 1 + 1


 RB rπ 
−1
1
1
Rout := 
+

 RC ro 
Rin = 5.818 kΩ
−1
Rout = 2.677 kΩ
Calculation of small-signal circuit gains:
Av1 :=
Rin
Av1 = 0.991
RS + Rin
1
1 
Av2 := −gm⋅ 
+

 Rout RL 
AV := Av1⋅ Av2
V
Av1 : Gain from vs to vb vb/vs
V
−1
Av2 = −27.022
V
AV = −26.792
V
Av2: Gain from the base to the collector vc/vs
V
V
AV: Total gain vc/vs
Calculation of base and collector signal voltages:
vb := Av1⋅ vs
vb = 9.915 mV
vc := AV ⋅ vs
vc = 0.268 V
t := 0 , 0.01ms .. 1ms
v_s( t ) := vs ⋅ sin( 2 ⋅ π ⋅ f⋅ t )
vo( t ) := AV ⋅ vs ⋅ sin( 2 ⋅ π ⋅ f⋅ t )
vo(t)
0.2
0.13
vo( t)
0
v_s ( t)
0.07
0.2
0
2 .10
4
4 .10
6 .10
4
4
8 .10
t
Time (s)
Figure 4: Amplifier's Output Signal across RL
4
0.27