參數(shù)資料
型號(hào): THS6062D
廠商: Texas Instruments, Inc.
英文描述: LOW-NOISE ADSL DUAL DIFFERENTIAL RECEIVER
中文描述: 低噪聲寬帶雙差分接收機(jī)
文件頁(yè)數(shù): 22/31頁(yè)
文件大?。?/td> 540K
代理商: THS6062D
THS6062
LOW-NOISE ADSL DUAL DIFFERENTIAL RECEIVER
SLOS228B – JANUARY 1999 – REVISED JUNE 1999
22
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
APPLICATION INFORMATION
optimizing frequency response
Internal frequency compensation of the THS6062 was selected to provide very wide bandwidth performance
and still maintain a very low noise floor. In order to meet these performance requirements, the THS6062 must
have a minimum gain of 2 (–1). Because everything is referred to the noninverting terminal of an operational
amplifier, the noise gain in a G = –1 configuration is the same as in a G = 2 configuration.
One of the keys to maintaining a smooth frequency response, and hence, a stable pulse response, is to pay
particular attention to the inverting terminal. Any stray capacitance at this node causes peaking in the frequency
response (see Figure 42 and Figure 43). There are two things that can be done to help minimize this effect. The
first is to simply remove any ground planes under the inverting terminal of the amplifier. This also includes the
trace that connects to this terminal. Additionally, the length of this trace should be minimized. The capacitance
at this node causes a lag in the voltage being fed back due to the charging and discharging of the stray
capacitance. If this lag becomes too long, the amplifier will not be able to correctly keep the noninverting terminal
voltage at the same potential as the inverting terminal’s voltage. Peaking and possibly oscillations will then
occur.
Figure 42
_
+
150
VO
VI
50
CIN–
300
7
6
2
0
100 k
1 M
10 M
O
8
9
f – Frequency – Hz
OUTPUT AMPLITUDE
vs
FREQUENCY
10
100 M
500 M
5
4
3
1
CIN– = 10 pF
No CIN–
(Stray C Only)
VCC =
±
15 V
Gain = 2
RF = 300
RL = 150
VO(PP) = 0.4 V
300
Figure 43
_
+
360
150
VO
VI
56
CIN–
360
1
0
–4
–6
100 k
1 M
10 M
O
2
3
f – Frequency – Hz
OUTPUT AMPLITUDE
vs
FREQUENCY
4
100 M
500 M
–1
–2
–3
–5
CIN–= 10 pF
No CIN–
(Stray C Only)
VCC =
±
15 V
Gain = –1
RF = 360
RL = 150
VO(PP) = 0.4 V
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