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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

- DC analysis of BJT
- When node voltages is known, branch current equations can be used.

(Va – Vb) / R = I

R

b

a

I

BJT Circuits at DC

= 0.99

KVL at BE loop: 0.7 + IERE – 4 = 0

IE = 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

BJT Circuits at DC

= 0.99

VC = 5.347 V

Hence,

VCE = VC – VE

= 5.347 – 3.3

= 2.047 V

IC = IE

IB = IE - IC

VBE = 0.7V

VB – VE = 0.7V

VE = 4 – 0.7 = 3.3 V since VB = 4V

VE – 0 = IE

10 - VC = IC

3.3 k

4.7 k

- Load Lineand Voltage Transfer Characteristic can be used to visualize the characteristic of the transistor circuits.

Input Load Line – IB versus VBE

Derived using B-E loop

- The input load line is obtained from Kirchhoff’s voltage law equation around the B-E loop, written as follows:

VBE – VBB + IBRB = 0

- Both the load line and the quiescent base current change as either or both VBBand RB change.

IB = -VBE + 4

220k

220k

For example;

- The input load line is essentially the same as the load line characteristics for diode circuits.

VBE – 4 + IB(220k)= 0

y = mx + c

- IBQ = 15 μA

Output Load Line – IC versus VCE

Derived 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:

IC = -VCE + 10

2k

2k

For example;

y = mx + c

- To find the intersection points setting IC= 0,
VCE = VCC = 10 V

- setting VCE = 0
IC= VCC / RC = 5 mA

Q-point is the intersection of the load line with the iCvsvCE curve, corresponding to the appropriate base current

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.

Use KVL at B-E loop

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