What does MOCP mean? (Explanation) | MOCP vs MCA vs FLA vs LRA

Contents

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Key learnings:

MOCP Definition: MOCP, or Maximum Over-Current Protection, ensures electrical equipment operates safely by setting the upper limit for current the equipment can handle.
MCA Importance: Minimum Circuit Ampacity (MCA) is crucial as it dictates the minimum size of wires needed to handle expected electrical loads safely.
FLA Explained: Full Load Ampere (FLA) is the expected current drawn by a device under full operational load, essential for calculating MOCP and MCA.
Calculating MOCP: To ensure safety, MOCP is calculated based on FLA, incorporating additional loads to prevent overheating and damage.
LRA Role: Locked Rotor Ampere (LRA) measures the maximum current during a stalled condition, important for understanding and managing start-up power needs.

What Does MOCP Mean in Electrical Terms?

MOCP (also called MOP) stands for Maximum Over-Current Protection. It is the largest current rating allowed for the fuse or circuit breaker that protects a listed unit, such as a packaged air conditioner. The installer must not exceed the nameplate MOCP. That rating or size of the circuit breaker is a maximum, not a target to “size up to” on every job.

A breaker larger than the nameplate MOCP can leave the unit’s listed protection scheme outside the manufacturer’s listing. Use the nameplate first. Do not invent a larger size.

MOCP sets the largest fuse or breaker the listing allows. MCA, not MOCP, is what you use to pick the conductor size.

So MOCP or MOP = Maximum Overcurrent Protection = the largest listed fuse or breaker rating for that unit.

MCA vs MOCP vs FLA vs LRAs

MCA, MOCP, FLA and LRA appear together on HVAC and similar nameplates. They answer different questions.

MCA

MCA (Minimum Circuit Ampacity) is the smallest ampacity the supply wire or conductor must have for that unit.

MCA names the ampacity the new conductors must meet or exceed, not the current carrying capacity of the conductor or wire already in the wall.

From MCA you pick a conductor whose ampacity, after any required correction, is at least that number.

So MCA = Minimum Circuit Ampacity. That number leads to a minimum wire size; MCA itself is not a wire gauge.

A common HVAC worksheet puts 125% on the largest motor current and then adds other loads. The stored line below multiplies motor FLA and heater current together by 1.25. Some listings put heaters at 100% outside that 1.25. Use the nameplate when it exists. The stored formula is a teaching shortcut.

MCA = 1.25 * (Motor FLA + Heater Current)

MOCP

MOCP is a calculated or listed maximum for the branch-circuit fuse or breaker, not a measured running current.

After the rounding rules, listed MOCP is at least as large as MCA and the two values can match.

MCA sets the minimum wire/conductor ampacity. MOCP sets the maximum fuse or breaker. Follow the nameplate and the adopted electrical code. This page is not a substitute for that code.

NEC 440.22 for many hermetic motor-compressors starts at 175% of rated-load current and allows up to 225% if needed for starting. The stored formula below uses 2.25 (225%). The 2.55 figure in the original sentence does not match that formula or the usual 225% cap. Use 2.25 as written in the equation, then apply the rounding rules and the nameplate.

MOCP = (2.25 * FLA of the Largest Motor) + (Other Motor Loads) + (All other resistive electrical load i.e., Heater Load)

FLA

FLA (full-load amperes, also written Full Load Ampere) is the current motors or other machines draw at rated load. On HVAC nameplates the compressor figure is often RLA, not FLA. For a motor, FLA is the current at rated voltage and rated output.

MCA and MOCP worksheets start from motor FLA or compressor RLA. That is how you size conductors and a fuse, MCB, or breaker. The two rearrangements below (FLA = 0.80 × MCA and FLA = 0.44 × MOCP) only invert the simplified no-heater formulas. They fail when heaters or other loads are present. Do not back-calculate nameplate FLA from MCA or MOCP.

$\begin{align*} FLA = 0.80 * MCA \end{align*}$

and

$\begin{align*} FLA = 0.44 * MOCP \end{align*}$

LRA

LRA stands for Locked Rotor Ampere. It is the current with the rotor stalled. Starting current is in that range. A stored teaching line below uses LRA = 8 × FLA. Many motors sit nearer 5× to 8×. Use the nameplate LRA when it is printed.

$\begin{align*} LRA = 8 * FLA \end{align*}$

LRA is the figure used to estimate starting voltage drop. Motor rules usually want remaining terminal voltage during start to stay above about 80% to 90% of nameplate, so the drop itself is about 10% to 20%, not 80%. A larger drop can stall the motor or cause chatter.

How To Calculate MOCP

If the unit has a nameplate MOCP, that listed value is the largest fuse or breaker you may install. The worksheet below is only for teaching when no nameplate exists. A breaker at MOCP still allows normal start current. It does not cap running current at the MOCP number.

MOCP = (2.25 * FLA of the Largest Motor) + (Other Motor Loads) + (All other resistive electrical load i.e., Heater Load)

NEC 240.6(A) lists standard ampere ratings. The common small sizes start at 15 A, then 20 A, 25 A, 30 A, and so on. 15 A is only the smallest of those everyday sizes; smaller standard ratings also exist, so every circuit is not required to be at least 15 A.

These HVAC worksheets treat two load types on ordinary utilisation-voltage circuits (for example 277 V or 480 V). Those voltages are not high voltage.

Inductive Load i.e., Motor, Compressor, etc.…
Resistive Load i.e., Electric Heaters.

Steps to calculate MOCP

First, find motor FLA or compressor RLA at rated voltage and load. Use the nameplate figure, not a back-calc from MCA.

Second, find simultaneous heater current. That is a resistive load.

After the raw sum, apply a rounding rule. Many HVAC nameplate worksheets round the 225% result up to the next standard size. That listed MOCP is then the largest breaker allowed. The teaching list below rounds down when the raw number is not a multiple of 5. Both appear in the examples. Follow the nameplate when it exists.

If MOCP $\neq$ Multiple of 5 i.e., if the calculated value of MOCP is not an even multiple of 5 then the value of MOCP is rounded down to the nearest standard fuse or circuit breaker size.
If MOCP < MCA i.e., if the calculated value of MOCP is less than the value of MCA then the value of MOCP is taken equal to the MCA and it is rounded up to the nearest standard fuse or circuit breaker size, typically a multiple of 5. Hence the value MOCP is not less than the value of MCA.
If MOCP < 15 A i.e., if the calculated value of MOCP is less than 15 A then it is rounded up to 15 A. This 15 A is the minimum current size or rating of the fuse or circuit breaker permitted by code.

The three examples below follow the stored worksheet and the three rounding rules as written.

Example 1 : Calculate the MOCP value for 3 – phase, 480 V, 10 KW heater load with motor FLA 4.5 A.

Given: 3-phase 480 V, 10 kW heater, motor FLA 4.5 A. The stored working uses 1.73 for √3. Heater current is 12.04 A.

$\begin{align*} \begin{split} for \,\, 3-phase \,\, load \,\, the \,\, current \,\,I = \frac {P}{\sqrt3 * V} \\ Heater\,\,current \,\, (I) = \frac {10000}{\sqrt3 * 480} \\ = \frac {10000}{1.73 * 480} \\ Heater\,\,current \,\, (I) = 12.04 \,\,A \end{split} \end{align*}$

Now,

$\begin{align*} MCA = 1.25 * (Motor\,\,FLA + Heater\,\,Current) \end{align*}$

$\begin{align*} = 1.25 * (4.5 + 12.04) \end{align*}$

$\begin{align*} = 1.25*16.54 \end{align*}$

$\begin{align*} MCA = 20.68 \,\, A \end{align*}$

and

$\begin{align*} MOCP = (2.25 * FLA\,\,of\,\,the\,\,Largest\,\,Motor)+(Other\,\,Motor\,\,Loads)+ \\(All\,\,Heater\,\,Load) \end{align*}$

$\begin{align*} = (2.25 * 4.5) + (0) + (12.04) \end{align*}$

$\begin{align*} = 10.125 + 12.04 \end{align*}$

$\begin{align*} MOCP = 22.17 \,\, A \end{align*}$

22.17 A is not a standard size. Under the stored “round down” rule that becomes 20 A. thus,

MOCP = 20 A (condition 1),

20 A sits below MCA 20.68 A, so the stored Condition 2 raises MOCP to 25 A, the next standard size at or above MCA for this 3-phase load (Condition 2).

(Note that those are typical utilisation voltages, not a universal rule: US 480Y/277 often uses 277 V line-to-neutral and 480 V three-phase. India commonly uses 230 V single-phase and 415 V three-phase).

Example 2: Calculate the MOCP value for 1 – phase, 277 V, 5 KW heater load.

Given: 1-phase 277 V, 5 kW heater, motor FLA 0. Heater current is 18.05 A.

$\begin{align*} \begin{split} for \,\, 1-phase \,\, load \,\, the \,\, current \,\,I = \frac {P}{V} \\ Heater\,\,current \,\, (I) = \frac {5000}{277} \\ Heater\,\,current \,\, (I) = 18.05 \,\,A \end{split} \end{align*}$

Now,

$\begin{align*} MCA = 1.25 * (Motor\,\,FLA + Heater\,\,Current) \end{align*}$

$\begin{align*} = 1.25 * (0 + 18.05) \end{align*}$

$\begin{align*} = 1.25 * 18.05 \end{align*}$

$\begin{align*} MCA = 22.56 \,\, A \end{align*}$

and

$\begin{align*} MOCP = (2.25 * FLA\,\,of\,\,the\,\,Largest\,\,Motor)+(Other\,\,Motor\,\,Loads)+ \\(All\,\,Heater\,\,Load) \end{align*}$

$\begin{align*} = (2.25 * 0) + (0) + (18.05) \end{align*}$

$\begin{align*} MOCP = 18.05 \,\, A \end{align*}$

Raw MOCP 18.05 A is below MCA 22.56 A, so Condition 2 takes the next standard size at or above MCA. MOCP is 25 A for this 1-phase heater load (Condition 2).

Example 3: Calculate the MOCP value for 3 – phase, 480 V, 5 KW heater load.

Given: 3-phase 480 V, 5 kW heater, motor FLA 0. Heater current is 6.02 A.

$\begin{align*} \begin{split} for \,\, 3-phase \,\, load \,\, the \,\, current \,\,I = \frac {P}{\sqrt3 * V} \\ Heater\,\,current \,\, (I) = \frac {5000}{\sqrt3 * 480} \\ = \frac {5000}{1.73 * 480} \\ Heater\,\,current \,\, (I) = 6.02 \,\,A \end{split} \end{align*}$

Now,

$\begin{align*} MCA = 1.25 * (Motor\,\,FLA + Heater\,\,Current) \end{align*}$

$\begin{align*} = 1.25 * (0 + 6.02) \end{align*}$

$\begin{align*} = 1.25 * 6.02 \end{align*}$

$\begin{align*} MCA = 7.53 \,\, A \end{align*}$

and

$\begin{align*} MOCP = (2.25 * FLA\,\,of\,\,the\,\,Largest\,\,Motor)+(Other\,\,Motor\,\,Loads)+ \\(All\,\,Heater\,\,Load) \end{align*}$

$\begin{align*} = (2.25 * 0) + (0) + (6.02) \end{align*}$

$\begin{align*} MOCP = 6.02 \,\, A \end{align*}$

Raw MOCP 6.02 A is below 15 A, so the stored Condition 3 raises it to 15 A, the smallest common standard size in that list (Condition 3).

How To Calculate MCA

Use the nameplate MCA when it is printed. The worksheet below is only a teaching shortcut from FLA and heater current.

Add every load that can run at the same time: compressor, fan motors and heaters. Confirm whether the listing puts heaters inside the 1.25 factor or at 100%.

MCA = 1.25 * (Motor FLA + Heater Current)

The next example computes MCA, not MOCP.

Example: Calculate the MCA value for 3 – phase, 480 V, 12 KW heater load with motor FLA 5 A.

Given: 3-phase 480 V, 12 kW heater, motor FLA 5 A. Heater current in the stored working is 14.45 A.

$\begin{align*} \begin{split} for \,\, 3-phase \,\, load \,\, the \,\, current \,\,I = \frac {P}{\sqrt3 * V} \\ Heater\,\,current \,\, (I) = \frac {12000}{\sqrt3 * 480} \\ = \frac {12000}{1.73 * 480} \\ Heater\,\,current \,\, (I) = 14.45 \,\,A \end{split} \end{align*}$

Now,

$\begin{align*} MCA = 1.25 * (Motor\,\,FLA + Heater\,\,Current) \end{align*}$

$\begin{align*} = 1.25 * (5 + 14.45) \end{align*}$

$\begin{align*} = 1.25 * 19.45 \end{align*}$

$\begin{align*} MCA = 20.7 \,\, A \end{align*}$

The stored shortcut gives MCA = 20.7 A. Size conductors to at least that ampacity after any required correction. This is not a wire gauge by itself.

A sample nameplate is shown below. Read MOCP and MCA from the plate when you have one.

On that stored figure the largest fuse or breaker is 20 A, so listed MOCP is 20 A. Do not install a larger device than that listing.

The same figure shows minimum circuit ampacity 12.2 A, so MCA is 12.2 A. Pick conductors whose ampacity meets or exceeds 12.2 A.

That nameplate also prints fan-motor LRA and FLA. Those are separate from compressor RLA.