LinSolve

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BJT Circuit Analysis and Design: DC Analysis of a Two Stage BJT Amplifier

Introduction: DC voltages have to be applied to a bipolar junction transistor in order to "bias" the device so that it operates in the linear region. One such circuit for a two stage amplifier is given below. This worksheet can be used to analyze such a circuit.

Statement: Analyze the circuit below to determine the voltages at all nodes and the currents through all branches.

Parameters: Here are the known values of resistors and voltages.

R.B1 100 kΩ * :

R.B2 51 kΩ * :

R.C2 2700 Ω * :

1 V * 1 kg m 2^* A s 3^* / * ≡

V.BE 0.7 V * :

V.CC 15 V * :

R.E2 2200 Ω * :

β 100 :

R.C1 5100 Ω * :

R.E1 3300 Ω * :

Solution: We begin by calculating some intermediate values.

R.B R.B1 R.B2 * R.B1 R.B2 + / : kΩ 33.775

V.B R.B2 R.B1 R.B2 + / V.CC * : 5.066 V *

V.CE1 5 V * :

V.BC2 4 V * :

I.C2 0.001 A * :

I.B2 0.00001 A * :

I.E2 0.001 A * :

V.EC2 5 V * :

I.C1 0.001 A * :

I.B1 0.00001 A * :

I.E1 0.001 A * :

V.B - I.B1 R.B * + V.BE + I.E1 R.E1 * + 0 ≡

V.BC2 V.BE + V.EC2 ≡

V.CC - I.E2 R.E2 * + V.EC2 + I.C2 R.C2 * + 0 ≡

I.E1 I.C1 I.B1 + ≡

V.CE1 I.E1 R.E1 * + V.BC2 I.C2 R.C2 * + ≡

I.C1 β I.B1 * ≡

I.C1 I.B2 - (R.C1 * I.E2 R.E2 * V.BE + ≡

I.C2 β I.B2 * ≡

I.E2 I.C2 I.B2 + ≡

LinSolve LinSolve

Löst ein lineares Gleichungssystem, gegeben als Vektor von Gleichungen, nach allen
unbekannten Variablen auf. Für symbolische Ergebnisse definieren Sie bekannte Größen
mit eigenen Maßeinheiten. Solves a linear system, given as a vector of equations, for any unknown variable. 
Use custom units for parametric (symbolic) solutions. sys# LinSolve subtract right hand sides of equations n# sys# length : i# 1 n# range Nsys# i# el sys# i# el num2str ≡ -(strrep) concat str2num : sys# i# el num2str trace 21 line for Compute Jacobi matrix and it's determinant u# Nsys# Unknowns : J# Nsys# u# Jacob : d# J# det : compute right hand side of linear system i# 1 n# range RHS# i# el Nsys# i# el - J# i# j# el u# j# el * (j# 1 n# sum + : for multiply inverse jacobi matrix by rhs i# 1 n# range r# i# el J# j# i# alg d# / RHS# j# el * (j# 1 n# sum : for r# 121 line :

RHS J u * G - :

J 1 -^RHS *

Residual(sys): Berechnet den Residuenvektor des Gleichungssystems sys. Die Lösung muss den 
Unbekannten schon zugewiesen sein. Geeignet zur Prüfung von Lösungsverfahren mit eingebetteter
Ergebniszuweisung. Residual(sys): Computes the residual vector of system of equations sys. All variables must
be assigned. Suitable to check solutions with embedded assignment. sys# Residual subtract right hand sides of equations Nsys# sys# num2str ≡ - strrep str2num : 21 line :

Residual(sys,L): Berechnet den Residuenvektor des Gleichungssystems sys für die Lösung L, die
den Unbekannten nicht vorab zugewiesen wurde. Dies erfolgt innerhalb der Funktion. Geeignet zur 
Prüfung von Lösungsverfahren ohne eingebettete Ergebniszuweisung. Residual(sys,L): Computes the residual vector of system of equations sys for a given solution L. 
The solution is assigned to the unknowns and the residual is evaluated. Suitable to check 
solutions without embedded assignment. sys# L# Residual subtract right hand sides of equations n# sys# length : u# sys# Unknowns : Assign solution to unknowns i# 1 n# range u# i# el num2str : L# i# el num2str concat str2num 11 line for Nsys# sys# num2str ≡ - strrep str2num : 61 line :

A V.B - I.B1 R.B * + V.BE + I.E1 R.E1 * + 0 ≡ V.CC - I.E2 R.E2 * + V.EC2 + I.C2 R.C2 * + 0 ≡ V.CE1 I.E1 R.E1 * + V.BC2 I.C2 R.C2 * + ≡ I.C1 I.B2 - (R.C1 * I.E2 R.E2 * V.BE + ≡ I.C2 β I.B2 * ≡ V.BC2 V.BE + V.EC2 ≡ I.E1 I.C1 I.B1 + ≡ I.E2 I.C2 I.B2 + ≡ I.C1 β I.B1 * ≡ 91 mat :

A LinSolve 1.1895 105 -^* A * 2.3609 105 -^* A * 0.0012 A * 0.0024 A * 0.0012 A * 0.0024 A * 2.6798 kg m 2^* A s 3^* / * 5.0896 kg m 2^* A s 3^* / * 3.3798 kg m 2^* A s 3^* / * 91 mat

Lexicographic sol'ns A Unknowns I.B1 I.B2 I.C1 I.C2 I.E1 I.E2 V.BC2 V.CE1 V.EC2 91 mat

1.1895 105 -^* A * 2.3609 105 -^* A * 0.0012 A * 0.0024 A * 0.0012 A * 0.0024 A * 2.6798 V * 5.0896 V * 3.3798 V * 91 mat

Volt [V]... is the SI derived unit for electric potential

. . . the difference of potential that would drive

one ampere [A] of current against one ohm [Ω] resistance.

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V = potential_energy/charge = N.m/C = kg.m²/A.s³

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1 kg m 2^* A s 3^* / * V 1

1 kg m 2^* A 2^s 3^* / * Ω