Adapting 4-20 ma Signal to 0-5 vdc Signal Input

OVERVIEW

We use a 249 Ohm resistor to adapt a 20 ma signal to a 5 vdc input of a Model 5C7-371 or Model 5C7-385 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 4-20 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 single-precision 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 single-precision 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 5C7-385 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. Single-point 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 0-5 vdc, the temperature, due to the resistor alone, can be off by approximately the following values when using a 249 Ohm resistor.

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).

Htt-Accuracy = Accuracy of the high temperature of the sensors temperature range as provided by the Temperature Transmitter (tt) at 20 ma.

Ltt-Accuracy = 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):

Htt-Accuracy =~ (Htc - Ltc)  * Rt

The approximate accuracy as ±Degrees at Ltt (the lowest temperature provided by the transmitter):

Ltt-Accuracy =~  (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 =~ Ltt-Accuracy +  (T - Ltt) / (Htt - Ltt) * (Htt-Accuracy - Ltt-Accuracy)

 

 

EXAMPLE of Calculating the Accuracy

 

Resistor is a 249 Ohm, 0.1% (Rt) tolerance, metal film resistor.

The Temperature Transmitter provides 4-20 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 Htt-Accuracy

Htt-Accuracy =~ (Htc - Ltc)  * Rt

Htt-Accuracy =~ (100 - 0)  * 0.001

Htt-Accuracy =~ 100  * 0.001

Htt-Accuracy =~ ±0.1° at 100°

CALCULATING Ltt-Accuracy

• Ltt-Accuracy =~  (Htc - Ltc)  / 5 * Rt
• Ltt-Accuracy =~  (100 - 0)  / 5 * 0.001
• Ltt-Accuracy=~  100  / 5 * 0.001
• Ltt-Accuracy =~  20 * 0.001
• Ltt-Accuracy =~  ±0.02° at 20°


Accuracy at any given temperature...
 

NOTES

Deriving of  Htc formula.

1.   Htc = Htt + ((Htt - Ltt) / (4.980V - 0.996V)) * (5.000V - 4.980V)
2.   Htc = Htt + (Htt - Ltt) / 3.984) * 0.020
3.   Htc = Htt + (Htt - Ltt) / 199.2

1.   Ltc = Ltt - ((Htt - Ltt) /(20ma - 4ma)) * 4ma
2.   Ltc = Ltt - ((Htt - Ltt) / 16) * 4
3.   Ltc = Ltt - (Htt - Ltt) / 4

1.   Htt-Accuracy = (Htc - Ltc) / 5V * ((Rt * 249Ohms) * 0.020A) 
2.   Htt-Accuracy = (Htc - Ltc) / 5V * Rt * 249Ohms * 0.020A
3.   Htt-Accuracy =~ (Htc - Ltc) / 5V * Rt * 5
4.   Htt-Accuracy =~ (Htc - Ltc)  * Rt

1.   Ltt-Accuracy = Ltt - (Htc - Ltc) / 5V * ((Rt * 249Ohms) * 0.004A)
2.   Ltt-Accuracy = Ltt - (Htc - Ltc) / 5V * Rt * 249Ohms * 0.004A
3.   Ltt-Accuracy =~ Ltt - (Htc - Ltc) / 5V * Rt * 1
4.   Ltt-Accuracy =~  (Htc - Ltc)  / 5 * Rt

Updated: Saturday, August 17, 2002