Important Solved Problems of Power Amplifiers and DC/DC Converters

SOLVED PROBLEMS

Problem 5.1

Design a class B power amplifier to deliver an average power of 20 W to an 8 Ω load. V_CC is selected to be 5V greater than the peak output voltage. Determine the supply voltage required, peak current drawn from each supply, total supply power, power conversion efficiency. Also determine the maximum power that each transistor must be able to dissipate safely.

Given:

P_L = 20 W

R_L = 8 Ω

Solution:

V_CC should be selected 5V greater than peak output voltage, V_o.

So, V_CC = 18 + 5 = 23 V

Peak current drawn from each supply

Power conversion efficiency, η = 

Maximum power dissipation in each transistor

Problem 5.2

In class AB amplifier, V_CC = 15 V, R_L = 100 Ω, and the output is sinusoidal with a maximum amplitude of 10V. I_S = 10^-13A, β = 50. Assume biasing diodes have one-third the junction area of the output devices. Find the value of I_Bias that guarantees a minimum of 1 mA through the diodes at all times. Determine the quiescent current and quiescent power dissipation in the output transistors (V_o = 0). Also find V_BB for V_o = 0, +10V and -10V.

Given:

V_CC = 15 V

R_L = 100Ω

V_max = 10 V

I_S = 10_-13 A

β = 50

Solution:

Maximum current i_Lmax = V_max/R = 10/100 = 0.1 A = 100 mA

To maintain a minimum current of 1 mA, I_Bias is selected as I_Bias = 3 mA

It is given that Biasing diodes have 1/3 junction area of output devices.

So, quiescent current is 9 mA.

Quiscent power dissipation, P = 2 x 15 x 9 = 270 mW

For V_o = 0,

Base current I_B through Q_N = 9/51 = 0.18 mA

So, current flowing through diodes is 3 - 0.18 = 2.82 mA

At V_o = + 10 V,

Current through diodes will decrease to 1 mA.

Substitute in V_BB equation,

V_BB ≈ 1.21 V

For V_o = -10 V,

Q_N will be conducting a very small current, so I_B will be negligibly small. Hence I_Bias of 3mA flow through the diodes.

Substitute I_B in V_BB equation

V_BB ≈ 1.26 V

Problem 5.3

A BJT is specified to have a maximum power dissipation of 2 W at ambient temperature of 25°C and a maximum junction temperature of 150°C. Find

(a) Thermal resistance θ

(b) Maximum power that can be safely dissipated at an ambient temperature of 50°C

(c) Junction temperature if the device is operating at T_A = 25°C and dissipating 1W.

Given:

Maximum power dissipation = 2 W

Ambient Temperature T_A = 25°C,

Maximum junction temperature = 150°C

Solution:

Problem 5.4

In Figure R₁ = 10 K Ω, R₂ = 2.2 KΩ, R_C = 3.6 KΩ, R_E = 1.1 KΩ V_CC = +10V. Find the dc power drawn from the supply by the amplifier.

Solution:

DC voltage across R₂ is

DC voltage across R₂,

DC emitter current, I_E = V_E/R_E

Total DC current drawn from the supply is

I_T = I_C + I₁ = 1mA + 0.82 mA = 1.8 mA

DC Power drawn from the supply is

Problem 5.5

A power amplifier operated from 12 V battery gives an output of 2W. Find the maximum collector current in the circuit.

Solution:

Problem 5.6

In CE amplifier shown in Figure, Assume β_dc = β_ac = 100.

1. Determine dc Q-point

2. Determine voltage gain and power gain

3. Signal power in load RL

4. Efficiency

Solution:

Problem 5.7

A boost regulator has an input voltage of V_in = 5 V. The average output voltage, V_a = 15 V and the average load current, I_a = 0.5 A. The switching frequency is 25 KHZ. If L = 150 μH and C = 220 μF. Determine (a) duty cycle, k (b) ripple current of inductor, ∆I (c)Peak current of inductor, I₂, (d) ripple voltage of filter capacitor, V_C

Given:

V_in = 5V, V_a = 15 V, f = 25 KHZ, L=150 μH. C = 220 μF

Solution:

(a) Duty Cycle

Problem 5.8

The buck regulator has an input voltage of V_in = 12 V. The required average output voltage is V_a = 5V and the peak-to-peak output voltage is 20 mV. The switching frequency is 25 KHZ. If the peak-to-peak ripple current of inductor is limited to 0.8 A. Determine (a) duty cycle, K (b) Filter inductance, L (c) filter capacitor C.

Given:

V_in ≈ 12 V, ∆V_C = 20 mV, ∆I = 0.8 A, f = 25 KHZ, V_a = 5V

Solution:

(a) Duty Cycle

(b) Filter Inductance, L

(c) Filter Capacitor, C

 

Problem 5.9

A buck-boost regulator has an input voltage of V_in = 12 V. The switching frequency is 25 KHZ. If L = 150 μH, C = 220μF. Duty cycle K = 0.25. The average load current I_a = 1.25 A. Determine (a) average output voltage, (b)Peak-to-peak output voltage ripple ∆V_C (c) Peak-to-peak ripple current of inductor ∆I, (d) peak current of the transistor I_P.

Given

V_in = 12 V, K = 0.25, I_a = 1.25 A, f = 25 KHZ, L = 150 μH, C = 220 μF

Solution:

(a) Average Output Voltage

(b) Peak-to-Peak Output Voltage Ripple

(c) Peak-to-Peak Ripple Current of Inductor, ∆I

(d)Peak-peak current of the transistor I_P

Peak-peak current of transistor

Problem 5.10

An ideal class B complementary symmetry push pull amplifier operates with V_CC = 12 V and R_L = 5 Ω, if the input is sinusoidal. Calculate (a) maximum power output (b) Maximum power dissipation (c) power dissipation in each transistor (d) conversion efficiency.

Given:

V_CC = 12 V

R_L = 5 Ω

Solution:

(b) Maximum Power Output

(b) Maximum Power Dissipation

(c) Power Dissipation in each Transistor

(d) Efficiency

Problem 5.11

The maximum collector dissipation of a transistor used in class A amplifier is 10 W. The collector efficiency of the circuit is 32%. Calculate the ac power output.

Given:

η = 32% = 32/100 = 0.32

P_d.c = 10 W

To Find:

ac power output, P_o(ac)

Solution:

Problem 5.12

A single transistor is operating as an ideal class B amplifier with 1 KΩ load. A dc meter in the collector circuit reads 10 mA. How much power is delivered to the load?

Given:

R_L = 1ΚΩ= 1 x 10³ Ω

I_C = 10 mA = 10 x 10^-3 A

To Find:

Power delivered to the load

Solution:

Power delivered to the load,