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XTR101AG 参数 Datasheet PDF下载

XTR101AG图片预览
型号: XTR101AG
PDF下载: 下载PDF文件 查看货源
内容描述: 高精度,低漂移的4-20mA两线制变送器 [Precision, Low Drift 4-20mA TWO-WIRE TRANSMITTER]
分类和应用:
文件页数/大小: 15 页 / 197 K
品牌: BB [ BURR-BROWN CORPORATION ]
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DETAILED ERROR ANALYSIS  
The ideal output current is  
A. AT THE LOWER RANGE VALUE (T = +25°C).  
σO = IOS RTO = ±6µA  
iO IDEAL = 4mA + K eIN  
VCC  
σI = VOSI + (IB1 R + IOS1 R4) +  
PSRR  
K is the span (gain) term, (0.016+ (40/RS))  
(3)  
(e1 + e2)/2 – 5V  
In the XTR101 there are three major components of error:  
+
CMRR  
1. σO = errors associated with the output stage.  
2. σS = errors associated with span adjustment.  
3. σI = errors associated with the input stage.  
R = RT 25°C – R4 = 109.4 – 109 0  
VCC = (24 X 0.005) + 4mA (250+ 100) + 0.6V  
= 120mV + 1400mV + 600mV  
= 2120mV  
The transfer function including these errors is  
iO ACTUAL = (4mA + σO) + K (1 + σS)(eIN + σI)  
(4)  
When this expression is expanded, second order terms  
(σS σ1) dropped, and terms collected, the result is  
e1 = (2mA X 2.5k) + (1mA X 109) = 5.109V  
e2 = (2mA X 2.5k) + (1mA X 109.4)  
= 5.1094V  
iO ACTUAL = (4mA + σO) + K eIN + KσI + KσS eIN  
(5)  
(e1 + e2)/2 – 5 = 0.1092V  
PSRR= 3.16 X 105 for 110dB  
CMRR = 31.6 X 103 for 90dB  
The error in the output current is iO ACTUAL – iO IDEAL and  
can be found by subtracting equations (5) and (3).  
iO ERROR = σO + Kσ1 + KσS eIN  
(6)  
σ1 = 30µV + (150nA X 0 + 20nA X 109)  
This is a general error expression. The composition of each  
component of error depends on the circuitry inside the  
XTR101 and the particular circuit in which it is applied. The  
circuit of Figure 9 will be used to illustrate the principles.  
2120mV  
3.16 X 105  
0.1092V  
3.16 X 103  
+
+
(10)  
= 30µV + 2.18µV + 6.7µV + 3.46µV  
= 42.34µV  
1. σO = IOS RTO  
2. σS = εNONLINEARITY + εSPAN  
(7)  
(8)  
(9)  
σS = εNONLIN + εSPAN  
VCC  
= 0.0001 + 0 (assumes trim of RS)  
IO error = σO + K σI + K σS eIN  
3. σI = VOSI + (IB1 + R4 – IB2 RT) +  
PSRR  
(e1 + e2)/2 – 5V  
+
40  
RS  
40  
K = 0.016 +  
= 0.016 +  
= 0.340  
CMRR  
123.3Ω  
The term in parentheses may be written in terms of offset  
current and resistor mismatches as IB1 R + IOS' R4.  
eIN = e2 – V4 = IREF1 RT 25°C – IREF2 R4  
since RT 25°C = R4,  
VOSI* = input offset voltage  
IB1*, IB2* = input bias current  
IOSI* = input offset current  
eIN = (IREF1 – IREF2) R4 = 0.4µA X 109Ω  
= 43.6µV  
I
OS RTO* = output offset current error  
R = RT – R4 = mismatch in resistor  
VCC = change supply voltage between  
pins 7 and 8 away from 24V nominal  
PSRR* = power supply rejection ratio  
Since the maximum mismatch of the current references is  
0.04% of 1mA = 0.4µA,  
IO error = 6µA + (0.34 X 42.34µV) + (0.34  
X
0.0001 X 43.6µV) = 6µA + 14.40µA + 0.0015µA  
= 20.40µA  
CMRR* = common-mode rejection ratio  
εNONLIN* = span nonlinearity  
20.40µA  
εSPAN* = span equation error. Untrimmed error  
= 5% max. May be trimmed to zero.  
% error =  
X 100%  
16mA  
Items marked with an asterisk (*) can be found in the  
Electrical Specifications.  
0.13% of span at lower range value.  
B. AT THE UPPER RANGE VALUE (T = +150°C).  
R = RT 150°C – R4 = 156.4 – 109.4 = 47Ω  
VCC = (24 X 0.005) + 20mA (250+ 100) +  
0.6V = 7720mV  
EXAMPLE 3  
The circuit in Figure 9 with the XTR101BG specifications  
and the following conditions: RT = 109.4at 25°C, RT =  
156.4at 150°C, IO = 4mA at 25°C, IO = 20mA at 150°C,  
RS = 123.3, R4 = 109, RL = 250, RLINE = 100, VDI  
=
e1 = 5.109V  
0.6V, VPS = 24V ±0.5%. Determine the % error at the upper  
and lower range values.  
e2 = (2mA X 2.5k) + (1mA X 156.4) = 5.156V  
(e1 + e2)/2 – 5V = 0.1325V  
®
XTR101  
14  
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