recall lecture 10
DESCRIPTION
Recall Lecture 10. DC analysis of BJT BE Loop (EB Loop) – V BE for npn and V EB for pnp CE Loop (EC Loop) - V CE for npn and V EC for pnp. I E. EXAMPLE - PNP. Given = 75 and V EC = 6V. Find the values of the labelled parameters, R C and I E,. DC analysis of BJT - PowerPoint PPT PresentationTRANSCRIPT
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Recall Lecture 10
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
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IE
EXAMPLE - PNP
Given = 75 and VEC = 6V. Find the values of the labelled parameters, RC and IE,
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DC analysis of BJT When node voltages is known, branch current
equations can be used.
R
a b
I
(Va – Vb) / R = I
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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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BJT Circuits at DC
VBE = 0.7VVB – VE = 0.7VVE = 4 – 0.7 = 3.3 V since VB = 4V
= 0.99
IC = IE
IB = IE - IC
VE – 0 = IE
3.3 k
10 - VC = IC
4.7 k
VC = 5.347 VHence, VCE = VC – VE
= 5.347 – 3.3= 2.047 V
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LOAD LINE AND VOLTAGE TRANSFER
CHARACTERISTIC
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Load Line and Voltage Transfer Characteristic can be used to visualize the characteristic of the transistor circuits.
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LOAD LINE
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The input load line is obtained from Kirchhoff’s voltage law equation around the B-E loop, written as follows:
Input Load Line Input Load Line – I– IBB versus V versus VBEBE
Derived using B-E loopDerived using B-E loop
Both the load line and the quiescent base current change as either or both VBB and RB change.
VBE – VBB + IBRB = 0
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The input load line is essentially the same as the load line characteristics for diode circuits.
For example;For example;
IBQ = 15 μA
VBE – 4 + IB(220k) = 0
IB = -VBE + 4
220k 220k
y = mx + c
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Output Load Line Output Load Line – I– ICC versus V versus VCECE
Derived using C-E loopDerived using C-E loop
For the C-E portion of the circuit, the load line is found by writing Kirchhoff’s voltage law around the C-E loop. We obtain:
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For example;For example;
To find the intersection points setting IC = 0,VCE = VCC = 10 V
setting VCE = 0IC = VCC / RC = 5 mA
IC = -VCE + 10
2k 2k
y = mx + c
Q-point is the intersection of the load line with the iC vs vCE curve, corresponding to the
appropriate base current
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Example: Calculation and Load Line
Calculate the characteristics of a circuit containing an emitter resistor and plot the output load line. For the circuit, let VBE (on) = 0.7 V and β = 75.
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Load LineLoad LineUse KVL at B-E loop
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Use KVL at C-E loop – to plot the load line
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IB = 75.1 A
VCE = 12 – 5.63 (1.01)VCE = 6.31 V
VCE = 12 – IC (1.01)IC = - VCE + 12
1.01