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TDA2005 数据表(PDF) 9 Page - STMicroelectronics

部件名 TDA2005
功能描述  20 W bridge/stereo amplifier for car radio
PDF  25 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

TDA2005 数据表(HTML) 9 Page - STMicroelectronics

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TDA2005
Electrical specifications
Doc ID 1451 Rev 6
9/25
2.3.2
Bridge amplifier design
The following considerations can be useful when designing a bridge amplifier.
Where:
VCE sat = output transistors saturation voltage
VS = allowable supply voltage
RL = load impedance
Voltage and current swings are twice for a bridge amplifier in comparison with single ended
amplifier.
In other words, with the same RL the bridge configuration can deliver an output power that is
four times the output power of a single ended amplifier, while, with the same max output
current the bridge configuration can deliver an output power that is four times the output
power of a single ended amplifier, while, with the same max output current the bridge
configuration can deliver an output power that is twice the output power of a single ended
amplifier.
Core must be taken when selecting VS and RL in order to avoid an output peak current
above the absolute maximum rating.
From the expression for IOmax, assuming VS = 14.4 V and VCE sat = 2 V, the minimum load
that can be driven by TDA2005 in bridge configuration is:
The voltage gain of the bridge configuration is given by (see Figure 36):
Table 5.
Bridge amplifier design
Parameter
Single ended
Bridge
Vo max
Peak output voltage (before clipping)
Io max
Peak Output current (before clipping)
Po max
RMS output power (before clipping)
1
2
--- V
s
2V
CEsat

V
s
2V
CEsat
1
2
---
V
s
2V
CEsat
R
L
-----------------------------------
V
s
2V
CEsat
R
L
-----------------------------------
1
4
---
V
s
2V
CEsat

2
2R
L
-------------------------------------------
V
s
2V
CEsat

2
2R
L
-------------------------------------------
R
Lmin
V
s
2V
CEsat
I
Omax
------------------------------------
14.4
4
3.5
---------------------
2.97
==
=
G
v
V
0
V
1
-------
1
R
1
R
2 R4
R
2
R
4
+
---------------------



--------------------------
R
3
R
4
-------
++
==



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