SME
Measurements & Instrumentation · Problem type
A 4–20 mA pressure transmitter
Use this for “A transmitter spans 200 to 1000 kPa on 4–20 mA. What current at 350 kPa, and what pressure gives 12.8 mA?”
Chart
I(350) = 7
This is an example. Type your own values to start a new problem.
I (mA)
=
4
mA
+
16
mA
×
P (kPa)
−
Pₘᵢₙ (kPa)
Pₘₐₓ (kPa)
−
Pₘᵢₙ (kPa)
Formulas
I
=
4
+
16
×
P
− Pₘᵢₙ
Pₘₐₓ − Pₘᵢₙ
7
=
4
+
16
×
350 − 200
1,000 − 200
Assumptions
•
The transmitter is linear: equal steps in pressure give equal steps in current.
•
The low end sends 4 mA, so a dead loop (0 mA) shows as a fault, not as Pₘᵢₙ.
•
Pₘᵢₙ and Pₘₐₓ are the range set on the transmitter, in gauge pressure.
Step-by-step
WE KNOW
P = 350 kPa, Pₘᵢₙ = 200 kPa, Pₘₐₓ = 1,000 kPa
FIND
loop current (I)
Step 1 · Find loop current (I)
Use I = 4 + 16 × (P − Pₘᵢₙ) ÷ (Pₘₐₓ − Pₘᵢₙ)
The share of the span the pressure has reached, times the 16 mA span, on top of the 4 mA zero.
I
=
4
+
16
×
350 − 200
1,000 − 200
I
=
4
+
16
×
150
800
I
=
4
+
2,400
800
I
=
4
+
3
I
=
7
mA
Check
Put the numbers back in. Both sides should match.
7
=
4
+
16
×
350 − 200
1,000 − 200
✓
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