CN0337 Datasheet, PDF (3/8 Pages) Analog Devices – EVALUATION AND DESIGN SUPPORT

Circuit Note
CN-0337
RTD LINE INPUT
(Pt100)
1
r₁
1
R10
1kΩ
U1C
1/4
AD8608
V_R
+0.1V
R_X = R₀ + ∆R ²
2
r₂
3
r₃
3
0ºC TO 300°C
100Ω TO 212.05Ω
U1D
1/4
AD8608
R5
+3.3V
2kΩ
R1′
95Ω
R6′
1.9kΩ
U1A
AD¹8^/6⁴ 08
V_OUT
0.1V TO 2.4V
R3
R4
GND
1kΩ
39.2kΩ
Figure 2. RTD Signal Conditioning Circuit Using a Three-Wire Connection
Circuit Design
The circuit shown in Figure 2 converts the RTD resistance
change from 100 Ω to 212.05 Ω to an output voltage change of
0.1 V to 2.4 V, which is compatible with the ADC input range.
In addition, the circuit removes the errors associated with the
wiring resistances r1 and r2.
The transfer function of the circuit in Figure 2 is obtained using
the superposition principle:
V _OUT 
V _{R ( r1}
R1'
 R_X
 r₂

R1'
)
R6 '
R5  R6
'

1

R4
R3

(1)

V_R
R1'
(
r
2

R1')
R4
R3
where:
RX = R0 + ΔR
R1′ = R1||R2 =R0, R6′ = R6||R12
r1 = r2, and neglects the voltage drop across r3.
Expand Equation 1, set the term containing r1 to zero, and solve
for R6′:
R6 ' R5 ^R4/R3
(2)
2  R4/R3
Meeting the criteria in Equation 2 removes the error due to the
lead resistances, r1 = r2, (r3 is not taken into account because it is
connected to the high impedance input of U1D).
Substituting Equation 2 into Equation 1, obtain the transfer
function:
V _OUT
 ^VR
2 R₀
 ^R4 R
R3
(3)
Equation 3 shows that the lead wire resistance is fully
compensated provided Equation 2 is met. The gain is set to the
desired value by adjusting the ratio of R4/R3.
Calculation of the Gain, Output Offset, and Resistor
Values and Tolerances.
For temperature range of 0°C to 300^oC, the RTD Pt100 resistance
range is 100 Ω to 212.05 Ω, and the input resistance change, ΔR,
for the circuit in Figure 2 is 0 Ω to 112.05 Ω. Therefore, the gain
of the circuit from Equation 3 is:
Gain  ^{V R}  ^R4  ^{V OUT}  ^{2.4 V  0.1 V}  20.53 mA ⁽⁴⁾
2 R₀ R3 R 112.05 Ω  0 Ω
Assuming that the current through the sensor is equal to 1 mA
and R0 = 100 Ω, the required reference voltage VR is:
V _R  100   1 mA  0.1 V ^.
Then, Equation 4 is solved for R4/R3:
R4 _{ 2 } 100 Ω _{ 20.53 mA  41.06}
R3
0.1 V
Choose R3 = 1 kΩ, then R4 = 41 kΩ.
Choosing a standard value of 2 kΩ for Resistor R5, Resistor R6′
can be calculated from Equation 2.
R6 ' R5 ^R4/R3  2 kΩ  ^41.06  1.907 kΩ
2  R4/R3
2  41.06
An easy way to ensure Equation 2 is met is to use the following
relationships:
R5 = 2R3, R6′ = R5||R4, as shown in Figure 1.
If this condition is met, R1′ = R0 = 100 Ω at 0°C, and VOUT = 0 V.
The output offset of the circuit must now be set to 0.1 V. An
easy way to shift the output is to make the resistor R1′ slightly
less than R0. Note that this affects the gain proportionally. The
output offset of 0.1 V is approximately 4.35% of the total span of
2.3 V, therefore the ratio R1′/R0 must be less than 0.9565. To
keep the high output level equal to 2.4 V, the ratio R4/R3 can be
proportionally corrected. For example, R4 = 0.9565 × 41.06 ×
R3 = 39.27 kΩ. Using standard resistors values as shown in
Figure 1, the circuit gives a good approximation to the required
gain and the output offset. Resistor R1′ is formed by connecting
Resistor R2 = 1.91 kΩ in parallel with resistor R1 = 100 Ω.
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