Designing a two stage push-pull class B audio amplifier

Question

I am designing a two stage push-pull class B audio amplifier with s 2N4401 used as my pte-amplifier stage and TIP31C & TIP32C as my push-pull stage where my RL is a 10W speaker.

I am trying to set both my pre-amplifier stage and push-pull stage to be 30V VPP or at least push-pull, but I am unable to obtain those results no matter how many times I calculate. Also, I am still not sure about my diodes (1N4001) since I might replace them with a 10k potentiometer.

These are my calculations done via MATLAB:

%% First Stage 
R1= 50; R2= 20E3; R3= 1.45E3; R4= 196; R5= 11.5;
B= 300;
C= 10E-6;
F= 6E3;
Vs= 0.7;
Vcc= 30;
XC1= (1./(2*pi*F*C));

R6= 1E3; R7= 1E3; RL= 8;
B2= 50;
Vd1=0.7;
Vd2=Vd1;

%DC Analysis
VB= Vcc*(R3/(R2+R3));
VE= VB-0.7;
IE= (VE/R4);
re= (25E-3/IE);
Rin= ((1/R2)+(1/R3)+(1/(B*(R5+re))))^-1;
Rb= ((1/R2)+(1/R3))^-1;
Ieq=(VE)/((Rb/(B+1))+R5);
Vce= Vcc-(Ieq*(R4+R5));

%AC Analysis
Gain= (R4/(R5+re));
Vout= Vs*(Rin./realsqrt((Rin+R1)^2+(XC1).^2))*Gain;
Gaindb= 20*log10(Vout/Vs);

FC= (1/(2*pi*(R3)*C));
%% Second Stage

%DC Analysis
It=(Vcc-Vd1-Vd2)/(R6+R7); %if I have diodes in the middle
%It=(V2-(-V3))/(R7+R8+R10); %if I have resistor in the middle
Vb1=Vcc-(It*R6);
Vb2=Vb1-(Vd1+Vd2);
%Vb2=-Vb1; %no diodes
Ve1=Vb1-0.7;
Ve2=Ve1;
Vceq1=Vcc/2;
Vceq2=Vceq1;

%AC Analysis
Rinpp=((1/R6)+(1/R7)+(1/(B2*(RL))))^-1;
Vout2=Vout*(Rinpp/(Rinpp+R4));

Updated My circuit:

Answer 1

If you really want to get this 10W you need a drive stage between the output stage and voltage amplifier stage (Q3).

You could try to use this topology:

^{simulate this circuit – Schematic created using CircuitLab}

And ballpark calculations will look like this:

Assuming \$P = 10W\$ for \$R_L = 8Ω\$

This means that \$V_{L_{max}} = \sqrt{2*P*R_L} = \sqrt{2* 10W * 8Ω} ≈ 12.7V\:\: [peak]\$

And the load current:

\$ I_{L_{max}} = \frac{V_{L_{max}}}{R_L} = 1.6A \:\:[peak] \$

For good thermal stability of an output stage, I decided to choose:

\$R8 = R9 = \frac{V_{BE}}{ I_{L_{max}}} ≈ \frac{0.8V}{1.6A} ≈ 0.47Ω \:\:[2W]\$

The output stage quiescent current (\$I_Q\$) can be set via a pot (R12) connected across VBE multiplier \$Q_8\$. And this current we need to be set on the lab bench.

\$I_Q = (2...5)\% I_{L_{max}} ≈ 50mA\$

Now if we assume \$\beta_7 = 20\$ for \$Q_6\$ and \$Q_7\$ and \$\beta_4 = 100\$ for \$Q_4\$ and \$Q_5\$ we can select \$I_{C3}\$ quiescent current.

\$ I_{C3} ≥ (5..20) * \frac{ I_{L_{max}}}{\beta_7 * \beta_4} = (5..20)* \frac{1.6A}{20 * 100}≈ 5mA\$

Due to negative feedback action, we know that the DC output voltage will be close to 0V. Thus, we can select \$R_{10}\$ and \$R_{11}\$.

\$ R_{10}+R_{11} ≈ \frac{Vcc}{I_{C3}} = \frac{15V}{5mA} = 3kΩ\$

\$R_{10} = R_{11} = \frac{3kΩ}{2} = 1.5kΩ\$

The input stage quiescent current we can find in a similar fashion, knowing Q3 DC current.

\$I_{C1} = 10 * \frac{I_{C3}}{\beta_{min}} = 10 * \frac{5mA}{100} = 0.5mA\$

\$R_2 = \frac{V_{BE3}}{I_{C1}} = \frac{0.7V}{0.5mA} ≈ 1.3kΩ\$

\$R_1 = \frac{Vcc - Vbe}{2*I_{C1}} = \frac{15V - 0.6V}{1mA} ≈ 13kΩ\$

Notice that \$R_3\$ determines the amplifier input resistance. And to reduce the "DC offset" \$R_5\$ should be equal to \$R_3\$

\$R_3 = R_5 = 10kΩ\$

\$R_4\$ will set the amplifier voltage gain.

\$R_4 = \frac{10kΩ}{10} ≈ 1kΩ\$ (gain equals to 10)

Input capacitor and C2 w set the low-frequency corner frequency

\$C_2 = \frac{0.16}{R_4 * F_C} = \frac{0.16}{1kΩ * 10Hz} ≈ 22μF\$

\$C1 = \frac{0.16}{R_3*F_C} = \frac{0.16}{10kΩ*10Hz} ≈ 2.2μF\$

\$C_C = 10pF\$ but we select the correct size of this capacitor by testing it on the test bench.

The bootstrapping capacitor: Bootstrap in power amplifier

\$C_3 = \frac{0.16}{R_{10}||R_{11} * F_C} ≈ 22μF \$

The average power dissipation in the output stage transistor (in Q6 and Q7) is equal to:

\$P_{D} ≈ 0.1*\frac{V_{CC}^2}{RL} = 0.1*\frac{15V^2}{8Ω} ≈ 2.8W\$ plus a DC current looses \$I_Q*V_{CC} = 50mA*15V = 0.75W\$

And the peak power dissipation \$P = \frac{15V^2}{4*8Ω} = 7W \$

The power in \$Q_4\$ and \$Q_5\$ is \$\frac{P_D}{\beta_7}\$. And the power dissipation in \$ Q_3\$ is \$5mA * 15V = 0.075W\$

And \$Q_8\$ (Vbe multiplier transistor) should be thermally connected with \$Q_4\$.

Now it is time to test the circuit and tweak the component values if needed. To meet the requirements.