One of the most misunderstood concepts when dealing with transmitters is zero elevation and suppression. Elevation and suppression adjustments are frequently necessary in liquid level measurement when the transmitter cannot be installed on a level with the zero level of the tank. The definition is following, but to understand elevation and suppression, its easiest if you look at it from a mathematical viewpoint, that is described in the second part of this paragraph.

**Zero Elevation** – for an elevated zero range, the amount the measure variable zero is above the lower range value. It may be expressed either in units of measured variables or in percent of span.

**Zero Suppression** – for a suppressed zero range, the amount the measured variable zero is below the lower range value. It may be expressed either in units of the measured variable, or in percent of span.

**Also Read: Zero Suppression & Zero Elevation Examples**

Mathematically you can develop equations that will let you calibrate the transmitter for any given application requiring zero elevation or suppression. It’s important to understand some basic facts about differential pressure transmitters. First, in order to obtain an increasing output, the high side of the transmitter must always be increasing in pressure relative to the low side. Therefore, to achieve a 20 mA output, the net result of all forces on both the high and low sides of the transmitter must be such that the high side is greater than the low side by an amount equal to the calibrated span of the transmitter. Second, the purpose of the calibration biases (elevation and suppression) is to apply a differential pressure to those situations where a 4 mA output is desired at some point other than 0 differential pressure.

Download the Zero Suppression & Zero Elevation Image

We know that for any situation, the sum total of effects on a transmitter should be zero for a 4 mA output.

Therefore at the desired **4 mA condition:**

**H – L+ B = O**

Where:

H = is the pressure (relative to atmosphere) applied at the high side of the transmitter in inches H20

L = is the pressure (relative to atmosphere) applied at the low side of the transmitter in inches H20.

B = is the bias factor, in inches H20, which can be either positive or negative.

At a **20 mA condition:**

**H – L+ B = S**

Where:

S = is the calibrated span of the transmitter, which can never be zero or negative.

So this gives us two equations:

**H – L + B = O at 4 mA**

**H – L + B = S at 20 mA**

With these two equations, it is now possible to determine the calibration for any situation according to the following procedures:

Analyse the specific application to identify all fluid forces on both sides of the transmitter at the point where you want 4 mA Output and the point where you want 20 mA output.

Using the 4 mA condition, solve for B in: H — L+ B = O

Using the 20 mA condition, and the value of B obtained from the above calculation, solve for S: H — L+ B = S

Calibrate the transmitter to a range of: (— B) to (— B + S)

**For example,** Figure below represents a closed tank application where we want a 4 mA output when the level is at the bottom tap and 20 mA output at the top tap.

We also know that the low side leg is always filled with material whose specific gravity is 1.1. If the specific gravity of the fluid in the tank is 1.0, what is the calibration?

**At 4 mA: **

H = 0

L = 100” x 1.1 = 110” H20

So that:

H — L + B = 0

0 —110 + B = 0

B = 110

**At 20 mA: **

H = 100. x 1.0 = 100” H20

L= 100. x 1.1 = 110” H20

So that:

H — L + B = S

100 -110 + 110 = S

S = 100” H20

So our calibration for this application is: (— B) to (— B + S)

So: — 110 to (— 110 + 100)

calibration Range is: — 100 to -10” H20

**Also Read: Pressure Transmitter Uses & Applications**