recall lecture 9
DESCRIPTION
RECALL LECTURE 9. Introduction to BJT 3 modes of operation Cut-off Active Saturation Active mode operation of NPN. NPN. PNP. I E = I S [ e VBE / VT ]. I E = I S [ e VEB / VT ]. I C = I B. I C = I E. I E = I B ( + 1). = [ / 1 - ]. = [ / + 1 ]. - PowerPoint PPT PresentationTRANSCRIPT
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RECALL LECTURE 9 Introduction to BJT
3 modes of operation Cut-off Active Saturation
Active mode operation of NPN
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IE = IS [ e VBE / VT ]
Based on KCL: IE = IC + IB
IC = IB
IC = IE
= [ / + 1 ]
IE = IB( + 1)
IE = IS [ e VEB / VT]
NPN PNP
= [ / 1 - ]
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DC Analysis of BJT CircuitDC Analysis of BJT Circuit
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DC analysis of BJT BE Loop (EB Loop) – VBE for npn and VEB for pnp CE Loop (EC Loop) - VCE for npn and VEC for pnp When node voltages are known, branch current
equations can be used.
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Assume that the B-E junction is forward biased, so the voltage drop across that junction is the cut-in or turn-on voltage VBE (on).
Common-Emitter CircuitCommon-Emitter Circuit
The figures below is showing a common-emitter circuit with an npn transistor and the dc equivalent circuit.
Unless given , always assume VBE = 0.7V
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Implicitly assuming that VBB > VBE (on), which means that IB > 0. When VBB < VBE (on), the transistor is cut off and IB = 0.
Common-Emitter CircuitCommon-Emitter Circuit The base current: KVL at B-E loop
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Implicitly assuming that the transistor is biased in the forward-active mode.
Common-Emitter CircuitCommon-Emitter Circuit
In the C-E loop of the circuit, we can use: and
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ExamplesBJT DC Analysis
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Example Calculate the base, collector and emitter currents and the C-E voltage for a common-emitter circuit by considering VBB = 4 V, RB = 220kΩ, RC = 2 kΩ, VCC = 10 V, VBE
(on) = 0.7 V and β = 200.
Common-Emitter CircuitCommon-Emitter Circuit
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BJT Circuits at DC = 0.99
KVL at BE loop: 0.7 + IERE – 4 = 0IE = 3.3 / 3.3 = 1 mA
Hence, IC = IE = 0.99 mA
IB = IE – IC = 0.01 mA
KVL at CE loop: ICRC + VCE + IERE – 10 = 0 VCE = 10 – 3.3 – 4.653 = 2.047 V
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Example Find IB, IC, IE and RC such that VEC = ½ VCC for a common-emitter circuit .Consider: VBB = 1.5 V, RB = 580 Ω, VCC = 5 V,
VEB (on) = 0.6 V, β = 100.
Common-Emitter Circuit - PNPCommon-Emitter Circuit - PNP
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IE
EXAMPLEGiven = 75 and VEC = 6V. Find the values of the labelled parameters, RC and IE,