MAE 2055: Mechetronics I
Mechanical and Aerospace Engineering
Fall 2013
Homework 8
Irms
Speaker
+
Audio
Amplifier
Vrms
8Ω
-
1) An audio amplifier has the following specifications:
• 8Ω output impedance
• Maximum output power (time-average power) is 50W (when driving an 8Ω speaker)
• Minimum impedance that can be connected to any single output is 6Ω. (A lower impedance load
will damage the amplifier output stage.)
Calculate the values of the rms voltage, (𝑉𝑟𝑚𝑠 )𝑚𝑎𝑥 , and rms current, (𝐼𝑟𝑚𝑠 )𝑚𝑎𝑥 , at the amplifier
output when it is supplying maximum power to an 8Ω speaker.
2) You want to wire the audio amplifier to play music in several rooms at the same time, so you’d like
to connect four speakers to a single amplifier output. Obviously, connecting all four speakers
directly to the amplifier output will exceed the amplifier’s minimum load impedance specification.
Design an impedance matching network, using a transformer, to maximize power transfer to the
speakers and to ensure that the amplifier output is not overloaded.
I1rms
Audio
Amplifier
Z′L
N1:N2
8Ω
I2rms
+
+
V1rms
V2rms
-
-
8Ω
Z′S
8Ω
8Ω
a) Draw a schematic of the equivalent circuit including the audio amplifier, the impedance
matching network, and the four speakers.
b) What is the desired value for the impedance seen by the amplifier output, 𝑍𝐿′ ?
c) What transformer turns ratio, N2:N1, will yield the desired value of 𝑍𝐿′ ? What is the effective
source impedance, 𝑍𝑆′ , seen by the load (the speakers) for this value of the turns ratio?
Comment on (and think about) what this tells you about the role of the transformer in the
circuit.
d) At the maximum amplifier output power level of 50W, what are the values of the rms voltage
and current, (𝑉1𝑟𝑚𝑠 )𝑚𝑎𝑥 and (𝐼1𝑟𝑚𝑠 )𝑚𝑎𝑥 , at the primary winding? Note how these values
compare to those calculated in question 1.
e) What are (𝑉2𝑟𝑚𝑠 )𝑚𝑎𝑥 and (𝐼2𝑟𝑚𝑠 )𝑚𝑎𝑥 at the secondary winding?
f)
How much power is delivered to each speaker?
3) The source voltage in the following circuit is 𝑣𝑠 (𝑡) = 1𝑉𝑐𝑜𝑠(2𝜋 ∙ 10𝑀𝐻𝑧 ∙ 𝑡). Find the current, 𝑖(𝑡).
i(t)
L
1μH
C
vs(t)
R
Answer:
30Ω
𝑖(𝑡) = 9.11𝑚𝐴 ∙ 𝑐𝑜𝑠(2𝜋 ∙ 10𝑀𝐻𝑧 ∙ 𝑡 − 47.2°)
100pF