Adapting
420 ma Signal to 05
vdc Signal Input
Overview
Calculating
Sensor Temperature Range For The Controller
Example
Accuracy
Of The Conversion From Ma To Vdc
Calculating
the Accuracy
Example
Notes
OVERVIEW
We use a 249 Ohm resistor to adapt a 20 ma signal
to a 5 vdc input of
a
Model 5C7371 or Model 5C7385 temperature
controller.
These temperature
controller models are obsolete, but the web page will remain for
reference.
One source for the resistor is
MOUSER
ELECTRONICS.
Most temperature transmitters provide 4 to
20 ma. Use a transmitter
that provides an output that is "temperature linear".
When using a Temperature Transmitter that
outputs 420 ma, here are
the voltages that will be at the sensor input of the temperature
controller...
Maximum voltage at controller's
sensor input connector is 20ma
X 249Ohms = 4.980V
Minimum voltage at controller's sensor
input connector is 4ma X 249Ohms
= 0.996V
CALCULATING SENSOR TEMPERATURE RANGE FOR THE CONTROLLER
If you want to work with celsius,
calculate all temperatures
as celsius values. If you want to work with fahrenheit, calculate all
temperatures
as fahrenheit values.
Variables:
Htt
= High temperature of
the sensors temperature range as provided by the Temperature
Transmitter
(tt) at 20 ma, which is 4.980V.
Ltt
= Low temperature of
the sensors temperature range as provided by the Temperature
Transmitter
(tt) at 4 ma, which is 0.996V.
Htc
= High end of temperature
range defined for the 5V input of the Temperature Controller
(tc)
at 5 vdc.
Ltc
= Low end of temperature
range defined for the 5V input of the Temperature Controller
(tc)
at 0 vdc.
Htc
= Htt
+ (Htt  Ltt)
/ 199.2
Ltc
= Ltt
 (Htt  Ltt)
/ 4
(At the bottom of this page I show how these formulas
were derived.)
EXAMPLE of Calculating
Sensor Temperature Range for the Temperature Controller
Say I have a Temperature
Transmitter providing a temperature
range of 10 to 70 degrees.
Htt
= 70
Ltt
= 10
CALCULATING Htc
Htc
= Htt
+ (Htt  Ltt)
/ 199.2
Htc
= 70 + (70  10) / 199.2
Htc
= 70 + 60 / 199.2
Htc
= 70 + 0.301
Htc
= 70.301
Htc
= 70.3° (round
to a singleprecision number)
CALCULATING Ltc
Ltc
= Ltt
 (Htt  Ltt)
/ 4
Ltc
= 10  (70  10) / 4
Ltc
= 10  60 / 4
Ltc
= 10  15
Ltc
= 5
Ltc
= 5.0° (round
to a singleprecision number)
So even though you will only be reading 10
to 70 degrees (celsius or fahrenheit),
you map this to the controller by telling the controller the LOW TEMP
SCALE
is 5.0 degrees, and the HIGH TEMP SCALE is 70.3 degrees.
Pictured: TEMP SCALE settings
for the control sensor
of a Model 5C7385 Temperature Controller
ACCURACY OF THE CONVERSION FROM MA TO VDC
Accuracy of the temperature reading
is improved by calibrating
the controller/resistor/transmitter/sensor combination to an acceptable
standard. Singlepoint calibration is achieved by using the temperature
controllers OFFSET fields to adjust INPUT1 (controlling sensor) or
INPUT2
to your standard.
To improve the accuracy of the reading
without calibration: select a
highly accurate resistor, a highly accurate transmitter, and a highly
accurate
sensor.
An example of how a 249 Ohm resistor at 3
different "tolerance" values
(10%, 1% and 0.1%) affects accuracy.
 A 10% resistor can be off
by
24.9 Ohms. Price each: Less than 5¢
 A 1% metal film resistor
could
be off by 2.49 Ohms. Price each:
Less than 10¢
 A 0.1% metal film resistor
could be off by 0.249 Ohms. Price each:
Less than $1.25
If you have a 0°100° temperature
range over 05 vdc, the temperature,
due to the resistor alone, can be off by approximately the following
values
when using a 249 Ohm resistor.
A 10% resistor
can be off by 24.9 Ohms
Temperature 
20° 
50° 
75° 
100° 
Transmitter
Output 
4ma 
10ma 
15ma 
20ma 
Voltage
Accuracy as ±vdc 
0.1V 
0.25V 
0.4V 
0.5V 
Temperature
Accuracy as
±Degrees 
2° 
5° 
7.5° 
10° 
A 1% metal film
resistor could be off by 2.49
Ohms
Temperature 
20° 
50° 
75° 
100° 
Transmitter
Output 
4ma 
10ma 
15ma 
20ma 
Voltage
Accuracy as ±vdc 
0.01V 
0.025V 
0.04V 
0.05V 
Temperature
Accuracy as
±Degrees 
0.2° 
0.5° 
0.75° 
1° 
A 0.1% metal film
resistor could be off by
0.249 Ohms
Temperature 
20° 
50° 
75° 
100° 
Transmitter
Output 
4ma 
10ma 
15ma 
20ma 
Voltage
Accuracy as ±vdc 
0.001V 
0.0025V 
0.004V 
0.005V 
Temperature
Accuracy as
±Degrees 
0.02° 
0.05° 
0.075° 
0.1° 
In this example, you could reduce
the possible error by half
if you reduce the temperature span by half. So try not to use a
transmitter
temperature range that is much more that what is needed for the
temperature
controller to maintain the Set Point Temperatures you will be using.
the
smaller the temperature span, the less the resistor can contribute to
inaccuracy.
CALCULATING THE ACCURACY
Variables:
Rt
= Resistor tolerance
of 249 Ohm resistor in decimal. Usually 10% (0.10), 5% (0.05),
1% (0.01), or 0.1% (0.001).
HttAccuracy
= Accuracy
of the high temperature of the sensors temperature range as provided by
the Temperature Transmitter (tt) at 20 ma.
LttAccuracy
= Accuracy
of the low temperature of the sensors temperature range as provided by
the Temperature Transmitter (tt) at 4 ma.
Htt
= High temperature of
the sensors temperature range as provided by the Temperature
Transmitter
(tt) at 20 ma, which is 4.980V.
Ltt
= Low temperature of
the sensors temperature range as provided by the Temperature
Transmitter
(tt) at 4 ma, which is 0.996V.
Htc
= High end of temperature
range defined for the 5V input of the Temperature Controller
(tc)
at 5 vdc.
Ltc
= Low end of temperature
range defined for the 5V input of the Temperature Controller
(tc)
at 0 vdc.
The approximate
accuracy as ±Degrees at Htt
(the highest temperature provided by the transmitter):
HttAccuracy
=~ (Htc
 Ltc)
* Rt
The approximate
accuracy as ±Degrees at Ltt
(the lowest temperature provided by the transmitter):
LttAccuracy
=~ (Htc Ltc)
/ 5 * Rt
(At the bottom of this page I show how these formulas
were derived.)
Between Ltt
and Htt
the accuracy changes linearly as the temperature changes, and so you
can
easily interpolate for the accuracy at any given temperature. For
temperature T
the TAccuracy
is:
TAccuracy
=~ LttAccuracy
+ (T  Ltt)
/ (Htt  Ltt)
* (HttAccuracy  LttAccuracy)
EXAMPLE of Calculating
the Accuracy
Resistor is
a 249 Ohm, 0.1% (Rt)
tolerance, metal film resistor.
The Temperature
Transmitter provides 420 ma representing
20° (Ltt)
to 100° (Htt).
From above
we know that:
Htc
= Htt
+ (Htt  Ltt)
/ 199.2 = 100.4 = approximately 100
Ltc
= Ltt
 (Htt  Ltt)
/ 4 = 0
CALCULATING HttAccuracy
HttAccuracy
=~ (Htc
 Ltc)
* Rt
HttAccuracy
=~ (100  0) * 0.001
HttAccuracy
=~ 100 * 0.001
HttAccuracy
=~ ±0.1° at 100°
CALCULATING LttAccuracy
NOTES
Deriving of Htc
formula.
 Htc
= Htt
+ ((Htt  Ltt)
/ (4.980V  0.996V)) * (5.000V  4.980V)
 Htc
= Htt
+ (Htt  Ltt)
/ 3.984) * 0.020
 Htc
= Htt
+ (Htt  Ltt)
/ 199.2
Deriving Ltc
formula.
 Ltc
= Ltt
 ((Htt  Ltt)
/(20ma  4ma)) * 4ma
 Ltc
= Ltt
 ((Htt  Ltt)
/ 16) * 4
 Ltc
= Ltt
 (Htt  Ltt)
/ 4
Deriving of HttAccuracy
formula. (0.020A = 20ma)
 HttAccuracy
= (Htc
 Ltc)
/ 5V * ((Rt
* 249Ohms) * 0.020A)
 HttAccuracy
= (Htc
 Ltc)
/ 5V * Rt
* 249Ohms * 0.020A
 HttAccuracy
=~ (Htc
 Ltc)
/ 5V * Rt
* 5
 HttAccuracy
=~ (Htc
 Ltc)
* Rt
Deriving LttAccuracy
formula.
(0.004A = 4ma)
 LttAccuracy
= Ltt  (Htc
 Ltc)
/ 5V * ((Rt
* 249Ohms) * 0.004A)
 LttAccuracy
= Ltt  (Htc
 Ltc)
/ 5V * Rt
* 249Ohms * 0.004A
 LttAccuracy
=~ Ltt  (Htc
 Ltc)
/ 5V * Rt
* 1
 LttAccuracy
=~ (Htc
 Ltc)
/ 5 * Rt
Questions?
Contact
McShane
Updated:
Saturday, August 17, 2002
