# Class-B output stage

Thread Starter

#### Rumination

Joined Mar 25, 2016
74

The figure shows a class B output stage with two complementary output transistors Q3 and Q4 and two complementary driver transistors Q1 and Q2. The transistors are connected in Darlington connection for sufficiently high current gain from input to output at the output stage.
R1 is inserted to increase the quiescent current in Q1 and Q2, and increases the speed of Q3 and Q4 when the transistors to turn off (Class B). RL is the load output stage to drive. Vbb1 and Vbb2 creates a voltage between the bases of the two symmetrical halves amplifier so that the quiescent current IQ in the output transistors can be controlled, it applies to VBB = Vbb1 + Vbb2 and Vbb1 = Vbb2. Rest current is the current flowing through the two output transistors collectors when RL is disconnected and Vout = 0V
For transistors: Is-q1 = Is-q2 = 5 * 10^-14 A. and Is-q3 = Is-q4 = 2 * 10^-12 A.

a) Calculate the maximum power dissipation in the output transistors PQ3 + PQ4.

b) Calculate Vbb1 = Vbb2 so that the quiescent current in the output transistors are I_Q-X, and R1 as the quiescent current of the driver transistors (I_R1) will be equal to I_R1-X.
β is ∞ for all transistors.

c) Determine the maximum efficiency of the amplifier when the maximum output voltage is limited by the following (Vneg + 4V <= Vout <= V POS-4V).

----------------
The given values are:

Vpos = 24 V.
Vneg = 24 V.
RL = 7 Ω.
R2 = 768 Ω.
R3 = 768 Ω.
I_R1-X = 27 mA.
I_Q-X = 66 mA.
Vout-X = 16 V.

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
I have calculated the first question:

a) P_max in Q3 and Q4:
Vout = (2/ π) * Vcc = 0,64 * Vcc
P_max = (Vcc^2) / ((π^2) * RL) = (24^2) / ((π^2) * 7) = 8,34 W.

c) Can I determine the maximum efficiency of the amplifier with this table:

Can you guys help me with the rest? The deadline is tomorrow night.

#### Jony130

Joined Feb 17, 2009
5,127
For part B you must use a Shockley equation and solve for Vbe3+Vbe4 at Ic = 66mA and next solve for Vbe1 + Vbe2 for Ic = 27mA.
Then by help of a Ohm's law you can find R1 and Vbb1 + Vbb2 from KVL ( Kirchhoff's voltage law , II Kirchhoff's law ).

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
For part B you must use a Shockley equation and solve for Vbe3+Vbe4 at Ic = 66mA and next solve for Vbe1 + Vbe2 for Ic = 27mA.
Then by help of a Ohm's law you can find R1 and Vbb1 + Vbb2 from KVL ( Kirchhoff's voltage law , II Kirchhoff's law ).
I haven't heard about the shockley equation.

#### Jony130

Joined Feb 17, 2009
5,127
You have never seen this equation?

Ic = Is * e^(Vbe/Vt)

$$Ic = Is * e^{\frac{Vbe}{Vt}}$$

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
You have never seen this equation?

Ic = Is * e^(Vbe/Vt)

$$Ic = Is * e^{\frac{Vbe}{Vt}}$$
Never But now I have. I know Ic, but what is Is.Vbe is 0,7and Vt is 24 V

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
Okay. The thermal voltage is 26 mV. So I have to solve for Vbe:

66 = (2*10^-12) e^(Vbe/26)

27 = (5*10^-14) e^(Vbe/26)

When I have the two Vbe, what to do next.

#### Jony130

Joined Feb 17, 2009
5,127
When I have the two Vbe, what to do next.
Solve for R1 using Ohm's law and next Vbb1 and Vbb2 with help of a II Kirchhoff's law.

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
Solve for R1 using Ohm's law and next Vbb1 and Vbb2 with help of a II Kirchhoff's law.
I found:

66 = (2*10^-12) e^(Vbe/26)
Vbe = 809,33

27 = (5*10^-14) e^(Vbe/26)
Vbe = 882

Using Ohm's law to find R1:

R = U / I = (809,33 + 882) / 27 mA = 62,64 Ω

#### Jony130

Joined Feb 17, 2009
5,127
66 = (2*10^-12) e^(Vbe/26)
Vbe = 809,33

27 = (5*10^-14) e^(Vbe/26)
Vbe = 882
Totally wrong, next time try to track the units and prefixes more carefully.
Using Ohm's law to find R1:

R = U / I = (809,33 + 882) / 27 mA = 62,64 Ω
This is also wrong. But why do you add Vbe3 and Vbe1 to get the voltage drop across R1 resistor?

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
66 mA = (2*10^-9 mA) * e^(Vbe/26 mV)

66 mA /(2*10^-9) mA = e^(Vbe/26 mV)

3,3*10^10 = e^(Vbe/26 mV)

Better?

#### Jony130

Joined Feb 17, 2009
5,127
Definitely much better now.

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
How can I isolate Vbe?

3,3*10^10 = e^(Vbe/26 mV)

I get it to Vbe = 629,7

#### Jony130

Joined Feb 17, 2009
5,127
Good, but what is the unit of this result ?? As for Vbe we can use maths and we can rearrange this formula Ic = Is * e^(Vbe/Vt) to get Vbe directly.
Vbe = Vt * In(Ic/Is)
So for Ic = 66mA and Is = 2pA we have Vbe = 26mV * In(66mA/2pA) = 629.714mV

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
Yes. And the other Vbe, I get:

Vbe = 702,386 mV.

Using Ohm's law to find R1:

R = U / I = (U) / 27 mA =

Is my voltage 24?

#### Jony130

Joined Feb 17, 2009
5,127
Yes. And the other Vbe, I get:

Vbe = 702,386 mV.
OK

Using Ohm's law to find R1:

R = U / I = (U) / 27 mA =

Is my voltage 24?
Why 24V ?
What is the voltage at the top end of a R1 resistor ? What is the voltage at the bottom end of a R1 resistor ?

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
OK

Why 24V ?
What is the voltage at the top end of a R1 resistor ? What is the voltage at the bottom end of a R1 resistor ?
I'm thinking it's Vpos = 24 V. and Vneg = 24 V.

#### Jony130

Joined Feb 17, 2009
5,127
I'm thinking it's Vpos = 24 V. and Vneg = 24 V.
If so, the voltage across R1 is equal to:
VR1 = 24V - (-24V) = 48V which is of course not true. What about Q1 and Q2 Vce voltage is this voltage equal to zero ?
Also notice that R1 is connected between Q3 and Q4 bases ? Do you see this ?

Thread Starter

#### Rumination

Joined Mar 25, 2016
74
I'm sorry, I don't get it.

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