Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a...
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Transcript of Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a...
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
Biasing using Collector to Base Feedback Biasing using Collector to Base Feedback ResistorResistor
Find RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and =100.
IC
IE
IB
IC + IB = IE
NOTE: Proposed to use branch current equations and node voltages
Biasing using Collector to Base Feedback Resistor
• (VC – VB ) / RB= IB
• but VC = VCE
• and VB = VBE = 0.7 V
• (2.3 – 0.7) / RB = (IE / (+1)
• RB = 161.6 k
• (VCC – VC ) / RC = IE
• RC = 7.7 k
IE = 1mA , VCE = 2.3 V, VCC = 10 V and =100.
VC
VB
This is a very stable bias circuit.The currents and voltages are almost independent of variations in .
Voltage Divider Biasing CircuitVoltage Divider Biasing Circuit
Redrawing the input side of the network by changing it into Thevenin Equivalent
RTh: the voltage source is replaced by a short-circuit equivalent
AnalysisAnalysis
VTh: open-circuit Thevenin voltage is determined.
Inserting the Thevenin equivalent circuit
AnalysisAnalysis
VTH
Use voltage divider
VTH
BJT Biasing in Amplifier ExampleFind VCE ,IE, IC and IB given
=100, VCC=10V, R1 = 56 k, R2 = 12.2 k
RC = 2 kandRE = 0.4 k
VTH= R2 /(R1 + R2 )VCC
VTH = 12.2k/(56k+12.2k).(10)
VTH = 1.79V
RTH = R1 // R2
= 10 k
BJT Biasing in Amplifier Circuits
VTH = RTH IB + VBE + RE IE
1.79 = 10k IB + 0.7 + 0.4k (+1)IB
IB = 21.62A
IC = IB = 100(21.62)=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
Digital Logic – NOT GATEDigital 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).
Digital Logic – NOR GateDigital 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).
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.