BJT as an Amplifier
❑ A transistor must be dc-biased in order to operate
as an Amplifier – Current Amplifier, Voltage
amplifier, Power Amplifier
❑ A dc operating point (Q-point – quiescent point)
must be set first
❑ An ac signal is then applied to the input to cause
an amplified output
❑ When a transistor is biased, certain current and
voltage conditions are established (IB, IC and VCE)
Department of EE 1
,DC Load Line
IC (mA)
IB5
IB4
IB3
IB2
IB1
VCE
0 (V)
Department of EE 2
,DC Load Line
RC
RB 200Ω IC
10kΩ
VCE
IB VC C
+ +
VBB Βdc = 100
10V
3.7V - -
Output voltage equation VCC = ICRC + VCE
This is a straight line equation
Department of EE 3
, DC Load Line
❑ When IC = 0 i.e. IB = 0
VCE = VCC
i.e. (VCE, IC) = (VCC, 0) = (10V, 0A)
❑ When VCE = 0
ICRC = VCC
i.e. IC = VCC/RC = 10V/200Ω = 50mA
(VCE, IC) = (0, 50mA)
Department of EE 4
❑ A transistor must be dc-biased in order to operate
as an Amplifier – Current Amplifier, Voltage
amplifier, Power Amplifier
❑ A dc operating point (Q-point – quiescent point)
must be set first
❑ An ac signal is then applied to the input to cause
an amplified output
❑ When a transistor is biased, certain current and
voltage conditions are established (IB, IC and VCE)
Department of EE 1
,DC Load Line
IC (mA)
IB5
IB4
IB3
IB2
IB1
VCE
0 (V)
Department of EE 2
,DC Load Line
RC
RB 200Ω IC
10kΩ
VCE
IB VC C
+ +
VBB Βdc = 100
10V
3.7V - -
Output voltage equation VCC = ICRC + VCE
This is a straight line equation
Department of EE 3
, DC Load Line
❑ When IC = 0 i.e. IB = 0
VCE = VCC
i.e. (VCE, IC) = (VCC, 0) = (10V, 0A)
❑ When VCE = 0
ICRC = VCC
i.e. IC = VCC/RC = 10V/200Ω = 50mA
(VCE, IC) = (0, 50mA)
Department of EE 4