参数资料
型号: ADA4930-2YCPZ-R2
厂商: Analog Devices Inc
文件页数: 15/29页
文件大小: 0K
描述: IC DIFF AMP 1.35GHZ 24-LFCSP
标准包装: 1
放大器类型: 差分
电路数: 2
输出类型: 差分
转换速率: 3400 V/µs
-3db带宽: 1.35GHz
电流 - 输入偏压: 23µA
电压 - 输入偏移: 150µV
电流 - 电源: 34mA
电流 - 输出 / 通道: 30mA
电压 - 电源,单路/双路(±): 3.3V,5V
工作温度: -40°C ~ 105°C
安装类型: 表面贴装
封装/外壳: 24-VFQFN 裸露焊盘,CSP
供应商设备封装: 24-LFCSP-VQ(4x4)
包装: 标准包装
其它名称: ADA4930-2YCPZ-R2DKR
ADA4930-1/ADA4930-2
Rev. A | Page 21 of 28
Terminating a Single-Ended Input in a Single-Supply
Applications
When the application circuit of Figure 50 is powered by a single
supply, the common-mode voltage at the amplifier inputs, VP
and VN, may have to be raised to comply with the specified input
common-mode range. Two methods are available: a dc bias on
the source, as shown in Figure 51, or by connecting resistors RCM
between each input and the supply, as shown on Figure 54.
Input Common-Mode Adjustment with DC Biased Source
To drive a 1.8 V ADC with VCM = 1 V, a 3.3 V single supply
minimizes the power dissipation of the ADA4930-1/ADA4930-2.
The application circuit of Figure 50 on a 3.3 V single supply with a
dc bias added to the source is shown in Figure 51.
ADA4930
RL VOUT, dm
1.990V p-p
3.3V
RS
50
RG1
142
VP
VN
RG2
142
RF2
301
RF1
301
VOCM
VS
2V p-p
VDC
RT
64.2
50
09
20
9-
1
51
Figure 51. Single-Supply, Terminated Single-Ended-to-Differential System with G = 1
To determine the minimum required dc bias, the following steps
must be taken:
1.
Convert the terminated inputs to their Thevenin equivalents,
as shown in the Figure 52 circuit.
ADA4930
RL VOUT, dm
1.99V p-p
3.3V
VON
VOP
RTH
28.11
RG1
142
VP
VN
RG2
142
RF2
301
RF1
301
VOCM
VTH
1.124V p-p
VDC-TH
0
92
09
-15
9
RTH
28.11
Figure 52. Thevenin Equivalent of Single-Supply Application Circuit
2. Write a nodal equation for VP or VN.
()
TH
DC
TH
ON
TH
DC
TH
P
V
+
=
28.11
142
301
OP
TH
DC
N
V
28.11
142
301
+
=
Recognize that while the ADA4930-1/ADA4930-2 is in its
linear operating region, VP and VN are equal. Therefore,
both equations in Step 2 give equal results.
3.
To comply with the minimum specified input common-mode
voltage of 0.3 V at VS = 3.3 V, set the minimum value of VP
and VN to 0.3 V.
4.
Recognize that VP and VN are at their minimum values when
VOP and VS are at their minimum (and therefore VON is at its
maximum).
Let
VP min = VN min = 0.3 V, VOCM = VCM = 1 V, VTH min = VTH/2
VONmax = VOCM + VOUT,dm/4 and VOPmin = VOCM VOUT,dm/4
Substitute conditions into the nodal equation for VP and solve
for VDC-TH.
0.3 = 1.124/2 + VDC-TH + 0.361 × (1 + 1.99/4 = 1.124/2 – VDC-TH)
0.3 + 0.562 0.361 0.18 0.203 = 0.639 VDC-TH
VDC-TH = 0.186 V
Or
Substitute conditions into the nodal equation for VN and
solve for VDC-TH.
0.3 = VDC-TH + 0.361 × (1 1.99/4 VDC-TH)
0.3 – 0.361 + 0.18 = 0.639 × VDC-TH
VDC-TH = 0.186 V
5.
Converting VDC-TH from its Thevenin equivalent results in
V
0.33
0.186
=
×
+
=
TH
S
DC
R
V
The final application circuit is shown in Figure 53. The
additional dc bias of 0.33 V at the inputs ensures that the
minimum input common-mode requirements are met when
the source signal is bipolar with a 2 V p-p amplitude and
VOCM is at 1 V.
3.3V
ADA4930
RL VOUT, dm
1.990V p-p
RS
50
RG1
142
RG2
142
RF2
301
RF1
301
VOCM
VS
2V p-p
RT
64.2
09
20
9-
1
60
VP
VN
50
VDC
0.33V
Figure 53. Single-Supply Application Circuit with DC Source Bias
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