JUMO AQUIS touch S Assembly Instructions Manual page 143

Step 2: Calculate the voltage drop on the individual line segments
Step 3: Calculate the voltage at the respective sensor
Sensor
JUMO digiLine pH/ORP/T
JUMO ecoLine O-DO
JUMO ecoLine NTU
and therefore the supply voltage of the sensor may be lower.
For sensors of type JUMO digiLine pH/ORP/T, the calculation must be per-
formed with the peak values:
I = I + I + I = 20 mA + 20 mA + 20 mA = 60 mA
1 S1 S2 S3
I = I + I = 20 mA + 20 mA = 40 mA
2 S2 S3
I = I = 20 mA
3 S3
For sensors of type ecoLine O-DO/NTU, the highest peak value is used once
and the remaining sensors are considered with their average values. Example
for 1 x O-DO and 2 x NTU:
I = I + I + I = 4 mA + 2 mA + 60 mA = 66 mA
1 S1 S2 S3
I = I + I = 2 mA + 60 mA = 62 mA
2 S2 S3
I = I = 60 mA
3 S3
For further calculation it is assumed that the following sensors are used in the
bus structure shown above:
Sensor 1: JUMO digiLine pH (use peak value)
Sensor 2: ecoLine O-DO (use average value)
Sensor 3: ecoLine NTU (use peak value)
Thus, the following currents result:
I = I + I + I = 20 mA + 4 mA + 60 mA = 84 mA = 0.084 A
1 S1 S2 S3
I = I + I = 4 mA + 60 mA = 64 mA = 0.064 A
2 S2 S3
I = I = 60 mA = 0.06 A
3 S3
The cable lengths of the line segments are 20 m each.
The voltage drop on a line segment is calculated in accordance with the follow-
ing formula:
= ρ × 2 × L / A; where ρ = 1/56 Ωmm
U × I
x
x x
In the above example this means:
= ρ × 2 × L / A = 1/56 Ωmm
U × I
1 1 1
0.177 V
Shown in simplified form:
= 1/56 Ω × 2 × 20 × 0.084 A / 0.34 = 0.177 V
U
1
= 1/56 Ω × 2 × 20 × 0.064 A / 0.34 = 0.135 V
U
2
= 1/56 Ω × 2 × 20 × 0.06 A / 0.34 = 0.126 V
U
3
The value of the supply voltage at the respective sensor is given by the supply
voltage at the point of feed-in minus the sum of all the voltages which drop on
the line segments that are located between the point of feed-in and the sensor.
In the above example this means:
U = U - U = 5.3 V - 0.177 V = 5.123 V ≈ 5.1 V
S1 SV 1
U = U - U - U = 5.3 V - 0.177 V - 0.135 V = 4.988 V ≈ 5.0 V
S2 SV 1 2
U = U - U - U - U
S3 SV 1 2
The required minimum voltage of the sensors is shown in the following table.
Minimum voltage
4.2 V
5 V
5 V
The voltage at sensor 1 (JUMO digiLine pH) is well above the minimum value
2
/m × 2 × 20 m × 0.084 A / 0.34 mm
= 5.3 V - 0.177 V - 0.135 V - 0.126 V = 4.862 V ≈ 4.9 V
3
21 Annex
2
/m and A = 0.34 mm
2
2
=
143