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How to use the electronics package

The problem

A given transistor $T_1$ is used in a transistor amplifier.

The recommended BIAS is:
$U_{\text{CE}}=5\text{V}$	$U_{\text{BE}}=0.6\text{V}$	$I_{\text{C}}=2\text{mA}$	$I_{\text{B}}=10\text{μA}$
H parameters for the BIAS are as follows:
$h_{11}=h_{\text{i}}=2.7\text{kΩ}$	$h_{12}=h_{\text{b}}=1.5·{10}^{-4}$	$h_{21}=h_{\text{f}}=200$	$h_{22}=h_{\text{o}}=18\text{μS}$
Some further constaints are known:
$U_{\text{RE}}=0.6\text{V}$	DC voltage on BIAS stabilization resistor RE
$R_{\text{S}}=3\text{kΩ}$	Signal generator impedance
$R_{\text{A}}=3\text{kΩ}$	Next stage input impedance
$U_{\text{B}}=12\text{V}$	DC power supply

Our task is to find values from the E24 series for $R1$, $R_2$, $R_{\text{C}}$, and $R_{\text{E}}$ to set up the BIAS and to calculate voltage gain $v_u$, amplifiers input impedance $R_{\text{in}}$, and amplifiers output impedance $R_{\text{out}}$.

To have a cut-off frequenzy $f_{\text{c}}=60\text{Hz}$ we choose a maximum attenuation of 1.5 dB for coupling connections.
Capacitors are choosen from the E6 series (20% tolerance).

The solution

Assign symbols to voltages and currents

First we assign symbols to some voltages and currents, either already known or easily to calculate.

For $I_{\text{Q}}$ some books recommend $3\dots 5I_{\text{B}}$, others recommend $5\dots 10I_{\text{B}}$. Larger $I_{\text{Q}}$ results in a more stable circuit, particularly with regard to manufacturing tolerances.
So we use $I_{\text{Q}}=10I_{\text{B}}$ here.

Resistors to set up BIAS

$U_{\text{RC}}=U_{\text{B}}-U_{\text{CE}}-U_{\text{RE}}$	Voltage on $R_{\text{C}}$
$R_{\text{C}}=\frac{U_{\text{RC}}}{I_{\text{C}}}$	Calculated $R_{\text{C}}$ value
$R_1=U_{\text{B}}-U_{\text{BE}}-U_{\text{RE}}$	Voltage on $R_1$
$I_{\text{R1}}=I_{\text{Q}}+I_{\text{B}}=11I_{\text{B}}$	Current through $R_1$
$R_1=\frac{U_{\text{R1}}}{I_{\text{R1}}}$	Calculated $R_1$ value
$U_{\text{R2}}=U_{\text{BE}}+U_{\text{RE}}$	Voltage on $R_2$
$I_{\text{Q}}=10I_{\text{B}}$	Current through $R_2$
$R_2=\frac{U_{\text{R2}}}{I_{\text{Q}}}$	Calculated $R_2$ value

AC equivalent circuit

The AC equivalent circuit shown above can be reduced to the circuit below using

$R_{\text{G}}=R_{\text{S}}||R_1||R_2$	Generator impedance
$R_{\text{L}}=R_{\text{C}}||R_{\text{A}}$	Load impedance / load resistor

Summarized two-gate circuit

Next we want to reduce the two two-gate circuits for transistor and $R_{\text{E}}$ (gray dashed lines) to one large two-gate circuit (red dashed line).
The small two-gate circuits are serial-serial connected, so we have to add the Z parameters of these circuits to find the Z parameters of the resulting two-gate circuit.

$$\begin{array}{ccc}{Z}_{\text{Tr}}& =& \frac{1}{h_{22}}\left(\begin{array}{cc}\det h& h_{12}\\ -h_{21}& 1\end{array}\right)\\ Z_{\text{RE}}& =& \left(\begin{array}{cc}{R}_{\text{E}}& R_{\text{E}}\\ R_{\text{E}}& R_{\text{E}}\end{array}\right)\\ Z& =& Z_{\text{Tr}}+Z_{\text{RE}}\\ v_u& =& \frac{R_{\text{L}}{Z}_{21}}{\det Z+R_{\text{L}}{Z}_{11}}\\ r_{\text{i}}& =& \frac{\det Z+R_{\text{L}}{Z}_{11}}{Z_{22}+R_{\text{L}}}\\ r_{\text{o}}& =& \frac{R_{\text{G}}{Z}_{22}+\det Z}{Z_{11}+R_{\text{G}}}\\ v_i& =& -\frac{Z_{21}}{Z_{22}+R_{\text{L}}}\end{array}$$

Using GNU Octave for calculations

Load “electronics” package.

pkg load electronics;

Enter given values

rs=3e3;     # Signal voltage source impedance
ra=3e3;     # Load resistor
b=200;      # Transistors I_C / I_B relation
ube=0.6;    # BIAS base to emitter voltage
uce=5;      # BIAS collector to emitter voltage
ic=2e-3;    # BIAS collector current
ure=0.6;    # Voltage on feedback resistor R_E
ub=12;      # DC power supply voltage

# h parameters in BIAS
htr = [ 2.7e3 , 1.5e-4 ; 200 , 18e-6 ];

Calculate the resistors

rc = (ub - uce - ure) / ic;
re = ure / (ic * (1 + 1/b));
ib = ic / b;
iq = 10*ib;
r2 = (ube + ure) / iq;
r1 = (ub - ube - ure) / (iq + ib);

Show calculated resistor values

printf("R1 (exact calculation): %g\n", r1);
printf("R2 (exact calculation): %g\n", r2);
printf("RC (exact calculation): %g\n", rc);
printf("RE (exact calculation): %g\n", re);

Find resistor values from E24 series

r1 = E24_value(r1);
r2 = E24_value(r2);
rc = E24_value(rc);
re = E24_value(re);

Show selected values

printf("R1 (E24 series):        %g\n", r1);
printf("R2 (E24 series):        %g\n", r2);
printf("RC (E24 series):        %g\n", rc);
printf("RE (E24 series):        %g\n", re);

Combine parallel resistors

rg = parallel_impedance(rs, r1, r2);
rl = parallel_impedance(rc, ra);

Convert transistors H parameters to Z parameters

ztr = h_params_to_z(htr);

Find Z parameters for R_E

zre = z_params_resistor_vertical(re);

Sum of Z parameters for serial connection of transistor and R_E

z = ztr + zre;

Calculate operational characteristics

[vu, ri, ro, vi] = z_params_op_characteristics(z, rg, rl);

Calculate amplifiers input and output impedance

Rin = parallel_impedance(r1, r2, ri);
Rout = parallel_impedance(ro, rc);

Calculate coupling capacitors

Cin = coupling_capacitor(rs, Rin, 60.0, 1.5);
Cin = E6_value_up( increase_for_tolerance(Cin, 20.0) );

Cout = coupling_capacitor(Rout, ra, 60.0, 1.5);
Cout = E6_value_up( increase_for_tolerance(Cout, 20.0) );

The value returned by coupling_capacitor() is the minimum required to reach the specified maximum attenuation. We want to use capacitors having a tolerance of 20 %, so we have to ensure the capacitor value with tolerance applied does not fall below that calculated value. When selecting an E6 series value we must select a value not less than the calculated value.

Show results

printf("v_u   = %g\n", round_leading_digits(3, vu));
printf("v_i   = %g\n", round_leading_digits(3, vi));
printf("r_i   = %g\n", round_leading_digits(3, ri));
printf("r_o   = %g\n", round_leading_digits(3, ro));
printf("R_in  = %g\n", round_leading_digits(3, Rin));
printf("R_out = %g\n", round_leading_digits(3, Rout));
printf("Two-gate circuit:\n");
printf("v_u    = %g\n", round_leading_digits(3, vu));
printf("v_i    = %g\n", round_leading_digits(3, vi));
printf("r_i    = %g\n", round_leading_digits(3, ri));
printf("r_o    = %g\n", round_leading_digits(3, ro));
printf("Entire amplifier:\n");
printf("R_in   = %g\n", round_leading_digits(3, Rin));
printf("R_out  = %g\n", round_leading_digits(3, Rout));
printf("Coupling capacitors:\n");
printf("CK1    = %g\n", CK1);
printf("CK2    = %g\n", CK2);

Output from the example-with-electronics-package.m file

R1 (exact calculation): 98181.8
R2 (exact calculation): 12000
RC (exact calculation): 3200
RE (exact calculation): 298.507
R1 (E24 series): 100000
R2 (E24 series): 12000
RC (E24 series): 3300
RE (E24 series): 300
Two-gate circuit:
v_u = -4.99
v_i = 193
r_i = 61000
r_o = 988000
Entire amplifier:
R_in = 9110
R_out = 3290
Coupling capacitors:
CK1 = 4.7e-07
CK2 = 1e-06