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Transcript of audiomusicengpart1-lecture_slides-W6_L2_slidesA.pdf
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7/27/2019 audiomusicengpart1-lecture_slides-W6_L2_slidesA.pdf
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Audio Circuits
Week 6 - Lecture 2Mark Bocko & Dave Anderson
Topics: Guitar amplifier schematic Tone control Gain and distortion Speaker cabinet design
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Full Schematic
2
TL072TL072
Tone Gain Volume
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Tone control circuit
3
inout-
+
5.6 k 50 k
10 nF
(1-g)RfgRf
Cf
-
+
Zf
Zin
G =Z
f
Zin
0 g 1
Zf= (1 g)R
f+
1
jCf
gRf
1
jCf
+ gRf
G =1
Rin
(1 g)Rf+
gRf
1 + jgRfC
f
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Frequency response versus tonepot setting
4
in out-+
5.6 k 50 k
10 nF
g
0.0
0.40.2
0.8
0.6
1.0
G =1
Rin(1 g)R
f+
gRf
1 + jgRfCf
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Gain control - distortion
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Diode
I
+ V -
I = I0
eV V
T 1( )
VT 0.026 V
V/VT
-0.5
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
I
Vin 0
Rin
=
0 Vout
Rf
+ I0e
(0Vout
) VT 1
( )
KCL: i1 = i2 + i3
Vin= R
inI
0e
Vout
VT 1
V
out
Rf
Vin-
+
RinVout
i1
i3
Rfi2
Cant solve analytically for Vout butwe can use a graphical technique
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Graphical solution for Vout vs. Vin
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Vin= R
inI
0e
Vout
VT 1
V
out
Rf
Find Vin as a function of
Vout but then plot it asVout vs. Vin
But dont forget that thisis an inverting amplifier
configuration!
Thats why we plotted themagnitude of the transfer
function.
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Making the transfer function symmetrical
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Vout
Vin
D1 conducts for Vin > 0,(Vout will be negative)
D2 conducts for Vin < 0,(Vout will be positive)
Vin-
+
Rin
Vout
Rf
D1
D2
D2 forward
biased
D1 forwardbiased
Initial slope (Gain)is determined by
Rf/Rin
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Speaker Cabinet Design
Guitar amplifiers normallyhave an open back so thecabinet has little acoustical
effect
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Bass cabinet design toobtain better bass
response from theamplifier the cabinetshould be properly sized.
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Thiele-Small parameters for project speaker
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Following Robs Week 5 Lecture 1 example
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fs = 40 HzQts = 0.424VAS = 26.93 liters
Choose Qcb = 1 so Qcb/Qts = 2.36
Vas
Vbox
=
Qcb
Qts
2
1V
as
Vbox
= 2.36( )2
1 = 4.57
Vbox
=
VAS
4.57=
26.93
4.57= 5.89 liters = 5890 cm
3
Using the ratio, 0.618 : 1 : 1.618% %
we find approximately% % %11 cm x 18 cm x 29 cm
11 cm
18 cm
29 cm
