Recall Last Lecture

1
Recall Last Lecture

• Voltage Transfer Characteristic
• A plot of Vo versus Vi
• Use BE loop to obtain a current equation, IB in
  terms of Vi
• Use CE loop to get IC in terms of Vo
• Change IC in terms of IB
• Equate the two equations to link Vi with Vo
• Bipolar Transistor Biasing
• Fixed Bias Biasing Circuit

2
4.8
x
4.3
3
Biasing using Collector to Base Feedback Resistor
IC IB IE
IB
IC
IE

• Find RB and RC such that IE 1mA , VCE 2.3 V,
  VCC 10 V and b100.

NOTE Proposed to use branch current equations
and node voltages
4
Biasing using Collector to Base Feedback Resistor
IE 1mA , VCE 2.3 V, VCC 10 V and b100.

• (VC VB ) / RB IB
• but VC VCE
• and VB VBE 0.7 V
• (2.3 0.7) / RB (IE / (b1)
• RB 161.6 kW
• (VCC VC ) / RC IE
• RC 7.7 kW

VC
VB
5
Voltage Divider Biasing Circuit
This is a very stable bias circuit. The currents
and voltages are almost independent of variations
in ?.
6
Analysis
Redrawing the input side of the network by
changing it into Thevenin Equivalent
RTh the voltage source is replaced by a
short-circuit equivalent
7
Analysis
VTh open-circuit Thevenin voltage is determined.
Inserting the Thevenin equivalent circuit
Use voltage divider
8
Analysis
The Thevenin equivalent circuit
9
BJT Biasing in Amplifier Example

• Find VCE ,IE, IC and IB given
• b100, VCC10V, R1 56 kW, R2 12.2 kW,
• RC 2 kW and RE 0.4 kW

• VTH R2 /(R1 R2 )VCC
• VTH 12.2k/(56k12.2k).(10)
• VTH 1.79V
• RTH R1 // R2
• 10 kW

10
BJT Biasing in Amplifier Circuits

• VTH RTH IB VBE RE IE
• 1.79 10k IB 0.7 0.4k (b1)IB
• IB 21.62mA
• IC bIB 100(21.62m)2.16mA
• IE IC IB 2.18mA
• VCC RC IC VCE RE IE
• 10 2k(2.16m)VCE 0.4(2.18m)
• VCE 4.8 V

11
Basic Transistor Application
12
Digital Logic NOT GATE

• In the simple inverter circuit, if the input is
  approximately zero volts, the transistor is in
  cutoff and the output is high and equal to VCC.

• If the input is high and equal to VCC, the
  transistor is driven into saturation, and the
  output is low and equal to VCE (sat).

13
Digital Logic NOR Gate

• If the two inputs are zero, both transistors Q1
  and Q2 are in cutoff, and VO 5 V.
• When V1 5 V and V2 0, transistor Q1 can be
  driven into saturation, and Q2 remains in cutoff.
  With Q1 in saturation, the output voltage VO
  VCE (sat).

• If V1 0 and V2 5 V, then Q1 is in cutoff, and
  Q2 can be driven in saturation, and VO VCE
  (sat).

14

• If both inputs are high, meaning V1 V2 5 V,
  then both transistors can be driven into
  saturation, and VO VCE (sat).
• In a positive logic system, meaning that the
  larger voltage is a logic 1 and the lower voltage
  is a logic 0,

• the circuit performs the NOR logic function.
• The circuit is then a two-input bipolar NOR logic
  circuit.