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STA575 数据表(PDF) 13 Page - STMicroelectronics

部件名 STA575
功能描述  100100W STEREO BASH POWER AMPLIFIER
PDF  20 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

STA575 数据表(HTML) 13 Page - STMicroelectronics

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STA575
APPLICATION HINTS (refer to fig. 6)
PREAMPLIFIER AND COMPRESSOR
In the test circuit showed in figure 6, R1/R3 (or R2/R4) ratio fix the gain of the preamplifier.
If the input signal is very low, is possible to increase the gain fixing the product Vin
∗G = cost.
In that case is possible to increase G decreasing R1,2 from 10KΩ until 2KΩ without relevant effetcs on the cir-
cuitbehavior and remaining in the operating range Iin_max = Vin_max/R1(2),<1mA.
So it is possible to increase the preamplifier gain until 25.
If no compression is present (equivalnt compressor Gm=0), the effects are:
– The output voltage offset increase
– The SNR decrease
The following table shows these variations:
R3(4) = 50KΩ and all the other external components are the same
Attenuation = 0 dB
If the compression is active the circuit behaviour is the same.
It”s also possible to eliminate the compressor. In this case the ATT_REL (1,2) pin must be connected to gnd.
STBY-MUTE CIRCUIT
In the suggested application circuit (figure 6), the resistor for Standby/Mute function (R13) is connected between
the Standby/Mute switches and 5V Supply.
It is possible to connect the resistor to another Supply Voltage level VL, but in that case also the resistor value
(R13,14) must be changed according to the following formula (fixing VSTBY/MUTE = 2.5V and R15 = 10KΩ):
HEADROOM
In the suggested application circuit the supply voltage to obtain 75W (Power Output) on 8
Ω (Rload)
is:
It is also possible to increase the system’s efficiency forcing the headroom to follow the output signal (variable
drop insteadof a constant drop).
In that case:
R1,2
VIN MAX
G
VOFFSET
EN
10K
8V
5
15mV
10
µV
5K
4V
10
30mV
13
µV
2K
1.6V
25
75mV
20
µV
R
13
4V
L
10
()KΩ
=
R
14
4V
L
10
+
()KΩ
=
V
suppl y
∆VI
LM AX
,
R
DS on
+
=
V
sup ply
∆VI
L V
() R
DS on
+
=



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