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CD4066BE 参数 Datasheet PDF下载

CD4066BE图片预览
型号: CD4066BE
PDF下载: 下载PDF文件 查看货源
内容描述: CMOS四路双向开关 [CMOS QUAD BILATERAL SWITCH]
分类和应用: 复用器开关复用器或开关信号电路光电二极管输出元件
文件页数/大小: 20 页 / 545 K
品牌: TI [ TEXAS INSTRUMENTS ]
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CD4066B  
CMOS QUAD BILATERAL SWITCH  
SCHS051D – NOVEMBER 1998 – REVISED SEPTEMBER 2003  
electrical characteristics  
LIMITS AT INDICATED TEMPERATURES  
PARAMETER  
TEST CONDITIONS  
25°C  
UNIT  
V
(V)  
V
DD  
IN  
125°C  
–55°C –40°C  
85°C  
(V)  
TYP  
0.01  
0.01  
0.01  
0.02  
MAX  
0.25  
0.5  
1
0, 5  
5
0.25  
0.5  
1
0.25  
0.5  
1
7.5  
15  
7.5  
15  
0, 10  
0, 15  
0, 20  
10  
15  
20  
Quiescent device  
current  
I
µA  
DD  
30  
30  
5
5
150  
150  
5
Signal Inputs (V ) and Outputs (V  
)
os  
is  
V
R
= V  
,
DD  
C
L
5
800  
310  
200  
850  
330  
210  
1200  
500  
1300  
550  
470 1050  
= 10 kreturned  
VDD VSS  
On-state resistance  
(max)  
10  
15  
180  
125  
400  
240  
r
on  
to  
V
,
2
= V  
to V  
DD  
300  
320  
is  
SS  
5
15  
10  
5
On-state resistance  
difference between  
any two switches  
r  
R
= 10 kΩ, V = V  
DD  
10  
15  
on  
L
C
V
V
R
= V  
= 5 V, V = –5 V,  
SS  
C
DD  
= 5 V (sine wave centered on 0 V),  
Total harmonic  
distortion  
THD  
0.4  
40  
%
is(p-p)  
= 10 kΩ, f = 1-kHz sine wave  
is  
L
–3-dB cutoff  
frequency  
(switch on)  
V
C
= V  
= 5 V, V  
DD SS  
= –5 V, V = 5 V  
is(p-p)  
MHz  
MHz  
µA  
(sine wave centered on 0 V), R = 1 kΩ  
L
–50-dB feedthrough  
V
C
= V  
= –5 V, V = 5 V  
is(p-p)  
SS  
1
frequency (switch off) (sine wave centered on 0 V), R = 1 kΩ  
L
Input/output leakage  
current (switch off)  
(max)  
V
and  
= 0 V, V = 18 V, V = 0 V;  
is os  
C
–5  
10  
I
is  
18  
0.1  
0.1  
1
1
0.1  
V
C
= 0 V, V = 0 V, V = 18 V  
is  
os  
V (A) = V  
= 5 V,  
= –5 V,  
C
DD  
SS  
–50-dB crosstalk  
frequency  
V (B) = V  
C
8
MHz  
ns  
V (A) = 5 V , 50-source,  
is  
L
p-p  
R
= 1 kΩ  
R
= 200 kΩ, V = V  
,
5
20  
10  
7
40  
20  
15  
L
C
DD  
Propagation delay  
(signal input to  
signal output)  
V
V
= GND, C = 50 pF,  
SS  
is  
L
= 10 V  
10  
15  
t
pd  
(square wave centered on 5 V),  
t , t = 20 ns  
r f  
C
C
C
Input capacitance  
Output capacitance  
Feedthrough  
V
DD  
V
DD  
V
DD  
= 5 V, V = V  
= –5 V  
= –5 V  
= –5 V  
8
8
pF  
pF  
pF  
is  
C
SS  
SS  
SS  
= 5 V, V = V  
C
os  
ios  
= 5 V, V = V  
C
0.5  
4
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