0
\$\begingroup\$

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:

enter image description here

\$\endgroup\$
23
  • \$\begingroup\$ There is something called a vbe-multiplier, which is commonly found instead of D1 and D2. Would that be unacceptable? Also, you are hitting on 15 W output, which is bordering on non-trivial. (Heat will certainly now be an issue.) I have read an earlier question (maybe two) but I'm not sure what this is about. Part of a process of education? Or? And your 1st stage is almost never used in cases like this. In fact, the one you show is to be specially discouraged. Are you open to considering differing approaches? Or locked into that one? \$\endgroup\$
    – jonk
    Jun 5 at 18:11
  • \$\begingroup\$ You say you are using a 10 W speaker but are demanding +/- 15 V at the output of your driver, which is closer to 15 W (given the 8 Ohm load I see.) You may want a bigger speaker or a smaller specification that fits the speaker you have. And again, is this an exercise for simulation? Or do you plan on trying your hand at building this? And after all this, you aren't really asking a question so much as "any help would be appreciated." (Kind of like throwing mud at the wall to see what sticks there.) You may want to take learning about these things in steps. \$\endgroup\$
    – jonk
    Jun 5 at 18:23
  • \$\begingroup\$ To get 10W across 8-ohm resistor you need a peak voltage to be equal \$V_L = \sqrt{2*10W*8\Omega} \approx 12.7 Vpeak\$. This means you need a 25.4V peak-to-peak. Thus, your supply voltage is too low. What about the load current? ILpeak= 12.7V/8Ω = 1.6A. The \$\beta\$ of TIP31 will be at best around 50. Thus, the base current needed is equal to 32mA. So, for sure R6 = 2kΩ will be unable to provide such a large current. \$\endgroup\$
    – G36
    Jun 5 at 18:36
  • \$\begingroup\$ Also, your first stage will be unable to drive such a demanding load. And no one uses this topology to drive the push-pull stage. We are using the "current driven technic" to drive the push-pull stage, not the voltage. See the example here: ecircuitcenter.com/Circuits_Audio_Amp/Advanced_Amplifier/… \$\endgroup\$
    – G36
    Jun 5 at 18:39
  • \$\begingroup\$ ecircuitcenter.com/Circuits_Audio_Amp/Basic_Amplifier/… \$\endgroup\$
    – G36
    Jun 5 at 19:06

1 Answer 1

2
\$\begingroup\$

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:

schematic

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.

\$\endgroup\$
11
  • \$\begingroup\$ just a small question about why Sziklai transistor was used and not Darlington I am still not that well versed about them. Also, where does the pre-amplifier stage stop, and does the amplifier stage start? \$\endgroup\$ Jun 6 at 16:21
  • \$\begingroup\$ The main difference between Sziklai and the Darlington is that the base-emitter voltage drop (between the input and the output ) of the Darlington will be 2Vbe compared to a single Vbe drop in the Szikail pair. This circuit is a 3 stage amplifier. The input stage (Q1, Q2), the voltage gain stage (Q3), and the output stage (current gain stage) Sziklai pair. ecircuitcenter.com/Circuits_Audio_Amp/Basic_Amplifier/… \$\endgroup\$
    – G36
    Jun 6 at 16:41
  • \$\begingroup\$ is this correct? Circuit, since I am getting distorted sine wave \$\endgroup\$ Jun 6 at 17:07
  • \$\begingroup\$ Q4/Q5 chance to BD139/BD140 and increase Cc to 47pF. \$\endgroup\$
    – G36
    Jun 6 at 17:30
  • \$\begingroup\$ it's fixed now, and I replaced R12 with 10 potentiometers Circuit, but in case BD139/BD140 is not available what should I substitute it with? and does this consider a class b amplifier, since it looks like Class AB? sorry for the weird question \$\endgroup\$ Jun 6 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.