# Audio Amplifier Computation

Please compute and choose the values of all components of the Audio Amplifier in the figure as to obtain a power of 6W.

The computations and explanations for each component follow.

Start by choosing the speaker standard load resistance $R_L =8 ohm$.

Compute voltage swing on load resistance (speaker resistance) from maximum power

$P = V^2/R_L$

$V =sqrt{P*R_L} = sqrt{6*4} =4.9 V$

Compute Value of Capacitor $C1$ such as it has minimum reactance ($X_c= RL/10$) at minimum audio frequency ($F=50 Hz$). The capacitor is in series with the speaker so its reactance (which is highest at low frequencies) need to not count.  This is why its reactance needs to be $Xc = RL/10$

$Xc =1/(2pi FC) = 0.8 ohm$

$C1 = 1/(2pi F*0.4) =1/(2pi*50*0.4) = 3.98 miliF = 4000 mu F$

Choose voltage supply by ad adding the Collector-Emitter voltage drop ($1-2.5 V$) to the speaker voltage swing.   $Vcc = V +V(CE) = 4.9 + 2 = 6.9 =7 V$ Take negative voltage supply symmetric   -Vcc =-7 V
Compute the maximum current through transistors from maximum power  $P = U*I$ so that $I = P/U =6/7 = 0.85 A$

Choose transistors to have $2*V=12 V$ minimum $Vce$ and $I_C = 0.85 A$ minimum. These transistors are BD235 (npn) and BD 236 (pnp) ($Vce= 45 V$, $Ic =2A$, $P=25 W$)

Bias the base of the transistors with 2 diodes in forward conduction in series with 2 resistors.

The voltage on each of the open diodes is $0.7 V$ ($=Vbe$ of transistors), the voltage on each of the remaining 2 resistors is  $V(R) =Vcc-0.7 V =7-0.7=6.3 V$

Choose the current through the divider 10 times the base current of transistors.

$I(divider) =I(base)*10 = (Ic/beta)*10 =(0.85/100)*10 =0.085 A =85 mA$

The idea here is that the current that goes into the base does not count on the total current through the divider. (You can increase the current through the divider but in this case the power of the resistances will increase).

Chose diodes as to have a forward minimum current of $I=85 mA$ and forward minimum voltage $2*Vcc =14 V.$ The diodes are $D1=D2 =1N4001 – 1N4007$  ($1A$ and $100-1000 V$)

Compute $R1=R4 = V(R)/I(divider) =6.3/0.085 = 74 ohm = 75 ohm$ (standard)

Compute power dissipated by the resistors $R1=R4$

$P1=P4 =I^2*R =0.085^2*75 =0.54 W$

(The resistors are just exactly 0.5 Watt -“standard”)

Observation: Additionally you can modify the circuit like this:

Choose Darlington transistsors: $BD675-677-679$ and $BD676-678-680$

For Darlington transistors $beta =100*100 =10000$

$I_{base} =Ic/10000 =0.85/10^4 =85 mu A$

$I_{divider} = 10*I_{base} =850 mu A =0.85 mA$

(This time the current in the divider is very small)

Bias each base with 2 diodes in series (there will be a total of 4 diodes).

There are 2 diodes for each transistors because they are Darlignton and now $V_{be} =2*0.7 V =1.4 V$. Choose diodes $D1,D2, D3, D4$ as to have $I_{min}=850 mu A$. The diodes are $1N4148$ ($100 mA$).

Recompute $R1=R4 =(7-1.4)/0.85 m =6.59 Kohm$  (You can increase a bit the current in the divider by taking $R1=R4 =4.7-5 Kohm$ Now the power dissipated by the resistors is very small)

Now, the load of the capacitor $C2$ is the divider $R1,R4(D1,D2)$  $R_{load} = R_{divider}/2 =150/2 =75 ohm$

Choose $C2$ to have reactance $X_{C2}$, 10 times smaller than $R_{load}=75 ohm$ at $F_{min} =50 Hz$ (again, the capacitor is in series with $R_{load}$ so that its reactance need to not count).

$1/(2pi*F*C2) = 37.5$

$C2 =1/(2pi*50*75) = 42 mu F$

Choose $R0$ for the minimum out impedance of Op Amp ($R0$ is just for protection of Op Amp not to go into short circuit) $R0 = 75 Ohm$ for 741 OpAmp Choose common OpAmp: 741 . Power supply of 741 will be the same as of transistors: $+/-6 V$. Choose standard input impedance $Rin = 10 Kohm$.

It means $R4 = 10 KOhm$ since $(In-)$ of the op amp is virtual ground. Compute R5 from voltage gain G:

$G = V_{out}/V_{in} = 7$   $R_5/R_4 =G$  $R_5 = 70 Kohm$ (for $Gain =7$)

Observations: What counts in practice to obtain the power desired is the supply voltage.  This is somehow find by trials (varying with about 1 V its value).  The values of the capacitors $C14$ and $C2$ is always better to be taken as high as possible. For example instead of $4000 mu F$ is a good idea to increase it to $6000 mu F$. Instead of $40 mu F$ is a good practice to increase it up to $100 mu F$. The capacitors are both electrolytic with $+$ terminal connected to voltage source and $-$ terminal to the load