PMDC Motor Wiring and Fuse Question

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
A 600W PMDC motor, 12V to 48V,
wiring length is 10' from 12V battery

The suggested fuse size is supposed to be 70% of the 600W, which would mean a 35A fuse, correct?

For less than a 3% loss, does the wire sizing need to be for 50A or the 35A fuse?
 
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MaxHeadRoom

Joined Jul 18, 2013
19,006
The wiring needs to be sized for maximum motor current, normally the recommended protection is rated at 1.5x the maximum.
For 70%, it would need to be time delay fusing, even then it could cause nuisance blowing.
Max.
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
Not sure I understand. Let me see if I understand this correctly.

Since a permanent magnet motor generates power as it turns, some of the power being supplied has to overcome that. So, a 600W PMDC motor has a maximum power output of between 65-70% of the 600W/50A 'rating' of the windings. Therefore, the maximum (delayed) fuse size needs to be 70% of the 50A, or 35A. Yes?

Then to feed the motor, for less than a 3% loss for 10', do I need to size it for 35A or 50A?

I am also looking at a 1700W PMDC, so if it works for one scenario, it should work for the other, .....except that the latter will likely need much bigger wiring and fuse.

In other words, the windings are 1700W/140A. 70% of 140A = roughly 100A for a fuse. Is the wire size 4 AWG for 100A fuse, or '0' AWG for the 140A windings?
 

MaxHeadRoom

Joined Jul 18, 2013
19,006
If no motor controller that allows for soft start, then the current can be very high at switch on , governed by whatever the resistance of the armature is, i.e. there is no generated BEMF at that point.
The normal practice is to fuse motor circuits at least 1.5x the FLA of the motor, you cannot rely on motor generated voltage to limit the current as the fuse/CB etc is intended for fault conditions, overloading and stalling where the BEMF is either greatly reduced or non-existant.
Any fusing lower than maximum rated current can cause nuisance blowing.
Max.
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
Oh my. I sure wish I understood what you wrote. Could you dummy it down for me? I particularly need a translation of the BEMF, FLA, and CB sentence.

I've adapted a tarp motor to use to pull crab and shrimp pots. It is a 600W PMDC that I fused with a 40A delay self-resetting, on 10' of 8ga AWG wire, with a 12V 250A switch relay from the power winch on a truck (the price was right, and it allows me to easily reverse the motor). The plug is a 200A 3 pole, slightly oversized, but I already had it. It works well, but with two friends with bigger motors having issues that I am not, I would like to understand why.

(Moreover, how I could beef it up if I happened to overeat the motor one day if I snagged something on the bottom. I've been thinking about going to a 900W with the same 60:1 gearing.)

These two friends can't do a fraction of what I can with my 600W. One has a 1700W, and the other a 1450W, both with 90:1 gearing, use 80A fuses on 6ga AWG wiring, and what looks like a 200A switch. Both are overheating, stalling, and gradually losing power. (I understand that overheating will destroy a PMDC motor, but need to know why they are overheating.)

I believe they need a bigger line feed, but I don't know whether it should be rated on the fuse size or the motor wattage. 1700W @ 12V works out to roughly 140A, and 70% is 100A. My belief is that the wire needs to be matched to the wattage of the motor, not the fuse.

But even with only an 80A fuse, the fuses are not blowing when their motors are overheating. Why? Please, keep it simple.
 
Really numbing it. The motor is an inductor and per rules of the Universe the current in an inductor CANNOT change instantaneously.

The voltage across an inductor is given by:


So, when any motor starts up, the power supply sees a very low pure resistance of the winding resistance. If you cannot supply that surge, the motor may not start.
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
Really numbing it. The motor is an inductor and per rules of the Universe the current in an inductor CANNOT change instantaneously.

The voltage across an inductor is given by:


So, when any motor starts up, the power supply sees a very low pure resistance of the winding resistance. If you cannot supply that surge, the motor may not start.
I get that, ...I think, ...just not sure that answers my sizing question. The motors all run, so the power supply is definitely ample for that. But given the size difference, theirs bog down, where mine doesn't.
 

TeeKay6

Joined Apr 20, 2019
381
I get that, ...I think, ...just not sure that answers my sizing question. The motors all run, so the power supply is definitely ample for that. But given the size difference, theirs bog down, where mine doesn't.
@Captain Crunchie
BEMF="back EMF" of a running motor; BEMF=0V if armature is not turning.
FLA="full load amperage"; the rated motor load
CB="circuit breaker"
When the motor is not turning, as when you have just applied power, it has no BEMF and current is limited only by the resistance of the motor winding (and any external wire you add); this is called the stall current. This current will always be substantially larger than the rated running current. Also, if the situation permits the motor to be reversed while running, the current surge lasts longer because (1) the BEMF now adds (rather than subtracts) current as the motor is slowed down and (2) the motor must still come to a full stop and then accelerate in the opposite direction.

For safe operation, the wires to the motor must be able to handle the stall current continuously until a protective device (e.g. a circuit breaker) can react; this is because some mechanical failure may block the armature from turning.
 

MaxHeadRoom

Joined Jul 18, 2013
19,006
@Captain Crunchie
BEMF="back EMF" of a running motor; BEMF=0V if armature is not turning.
FLA="full load amperage"; the rated motor load
CB="circuit breaker"
When the motor is not turning, as when you have just applied power, it has no BEMF and current is limited only by the resistance of the motor winding (and any external wire you add); this is called the stall current. This current will always be substantially larger than the rated running current. Also, if the situation permits the motor to be reversed while running, the current surge lasts longer because (1) the BEMF now adds (rather than subtracts) current as the motor is slowed down and (2) the motor must still come to a full stop and then accelerate in the opposite direction.

For safe operation, the wires to the motor must be able to handle the stall current continuously until a protective device (e.g. a circuit breaker) can react; this is because some mechanical failure may block the armature from turning.
What he said!:p
Max.
 

MaxHeadRoom

Joined Jul 18, 2013
19,006
These two friends can't do a fraction of what I can with my 600W. One has a 1700W, and the other a 1450W, both with 90:1 gearing, use 80A fuses on 6ga AWG wiring, and what looks like a 200A switch. Both are overheating, stalling, and gradually losing power. (I understand that overheating will destroy a PMDC motor, but need to know why they are overheating.)

I believe they need a bigger line feed, but I don't know whether it should be rated on the fuse size or the motor wattage. 1700W @ 12V works out to roughly 140A, and 70% is 100A. My belief is that the wire needs to be matched to the wattage of the motor, not the fuse.
But even with only an 80A fuse, the fuses are not blowing when their motors are overheating. Why? Please, keep it simple.
Depends on the type of fuse they, time delayed etc, motors are rated for continuous current level, they also have a peak current area, which should only be allowed to occur for a very brief time, otherwise overheating and/or destruction can occur.
Motor drives are generally equipped with setting levels to accommodate the limits of the motor, torque/current limit etc.
Max.
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
For safe operation, the wires to the motor must be able to handle the stall current continuously until a protective device (e.g. a circuit breaker) can react; this is because some mechanical failure may block the armature from turning.
Thanks for that explanation. Now that I know what the terms mean, I understand what you are saying, ....except ....

Then what you are saying here is that the wires need to be bigger than the fuse so that the fuse blows long before the wires melt, right? So, theoretically, the back EMF needs to be added to the input current, which means the wires need to be rated for the 1700W, and not the fuse. Makes sense....
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
Depends on the type of fuse they, time delayed etc, motors are rated for continuous current level, they also have a peak current area, which should only be allowed to occur for a very brief time, otherwise overheating and/or destruction can occur.
Motor drives are generally equipped with setting levels to accommodate the limits of the motor, torque/current limit etc.
Max.
The problem, Max, is that their motors are not performing like they should. They are stalling, yet my little 600W is doing what theirs aren't. Mine barely gets to any kind of warm, though I have had to assist with pulling briefly when I heard the motor slow/groan.

So, let's just deal with the permanent magnet motors. I read that voltage controls speed, and amperage handles the torque. What happens to the 1700Ws when a load is applied? Does the voltage/speed drop to allow the amperage to increase?

What regulates the speed of a PMDC motor? My 600W is 3600rpm.

Does 12V fed into a PMDC motor turn slower than the same motor at 48V (understanding that the W = VA is consistent).

And is the power curve of the peak efficiency of both voltages the same?

I do have one more question I think I already know the answer to, but I'd like your take on it. Since only 65-70% the wattage rating on a PMDC motor is actually doing the work, is putting an horsepower rating as based on that wattage erroneous?

In other words, if a 1700W PMDC motor can only use 65-70% of the fed wattage to do the work, using the calculation of 745.6W/HP, how can they rate this motor at 2.3HP?
 
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MaxHeadRoom

Joined Jul 18, 2013
19,006
Yes, simply RPM is dependent on voltage and torque on current, with no load if the 12v is fed to a 48v motor, the motor will generate the 12v BEMF at an RPM directly proportional to whatever the RPM is at the motor maximum ((48v).
IOW if the rpm at 48v is 2000, at 12v it will be 500rpm.
When a load is applied, the BEMF becomes proportionately lower than the applied voltage, hence the current will increase.
I prefer Torque rating, rather than HP, a typical DC motor has maximum torque at zero rpm, with a torque curve that is fairly flat over the rpm range, sloping slight downward as rpm increases.
Max.
 

TeeKay6

Joined Apr 20, 2019
381
Thanks for that explanation. Now that I know what the terms mean, I understand what you are saying, ....except ....

Then what you are saying here is that the wires need to be bigger than the fuse so that the fuse blows long before the wires melt, right? So, theoretically, the back EMF needs to be added to the input current, which means the wires need to be rated for the 1700W, and not the fuse. Makes sense....
@Captain Crunchie

Response to Post#11:

Certainly the fuse must melt long before the wires get warm; otherwise, why bother with a fuse? Just let the wires melt...and start a fire to burn the house, boat, or whatever. :eek: And since wires melt rather slowly compared to a fuse (that is designed to melt quickly), the motor may also overheat while we wait for the wires to melt.

Note that BEMF is a voltage, not a current. How much current is created by a BEMF depends on the load across which the BEMF appears. A low power motor can have the same BEMF as a high power motor, but for the high power motor that BEMF will likely create higher currents (for one reason among others, the wires are larger and lower resistance).

As far as fuse sizing is concerned, you can ignore BEMF--it's current, not voltage that matters to a fuse. For fuse sizing what you need to be concerned with is (a) what is the largest motor current that will exist under normal operation of the motor with its load and (b) what is the expected motor current in an abnormal, failure mode against which you wish to protect. Normal operation includes start-up from 0 rpm and, optionally, includes effects of motor direction reversal. Generally this normal current will be the "running current under load" and its value will be greater than the running current under no load (indirectly due to BEMF)--but always less than the motor stall current when the rotor/armature is stopped/stalled in a fault condition. Most motors will tolerate a short period of stall but motor temperature will begin to rise due to the high stall current. However, a stall condition is generally not considered a normal operating mode (as always, there are exceptions). So, the fuse must tolerate current due to normal operation on a continuous basis but ought to tolerate a stall current for only a very brief time (depending on motor and application, that time could be milliseconds or many minutes). For a prolonged stall condition, the fuse must "blow" prior to any damage to the motor or connecting wires or switches. Thus, usually, the fuse rating must be a bit higher than the normal running current and be able to tolerate the start-up current, even though the start-up current will normally drop rapidly to the running current as the motor gains speed. A time delay in the fuse (all fuses have some time delay, but some "slow blow" fuses are designed to show long delays) allows the fuse to ignore the brief, but high, start-up current. However, if that high current persists long enough (depending on the specific motor & application) the fuse must blow. Note that the high start-up current is about equal to the stall current (because the motor is not moving in either case). So, summarizing, the proper fuse will not blow due to a brief start-up current but will blow if that current persists (it is then a stall current). The connecting wires should not get hot under any current condition other than a direct short across the motor. Finally, note that wires get hot due to their having resistance (power=resistance times current times current); thinner wires have more resistance than thicker wires, and hot wires mean that power is being lost in the wires rather than, as desired, in the motor.
 

TeeKay6

Joined Apr 20, 2019
381
The problem, Max, is that their motors are not performing like they should. They are stalling, yet my little 600W is doing what theirs aren't. Mine barely gets to any kind of warm, though I have had to assist with pulling briefly when I heard the motor slow/groan.

So, let's just deal with the permanent magnet motors. I read that voltage controls speed, and amperage handles the torque. What happens to the 1700Ws when a load is applied? Does the voltage/speed drop to allow the amperage to increase?

What regulates the speed of a PMDC motor? My 600W is 3600rpm.

Does 12V fed into a PMDC motor turn slower than the same motor at 48V (understanding that the W = VA is consistent).

And is the power curve of the peak efficiency of both voltages the same?

I do have one more question I think I already know the answer to, but I'd like your take on it. Since only 65-70% the wattage rating on a PMDC motor is actually doing the work, is putting an horsepower rating as based on that wattage erroneous?

In other words, if a 1700W PMDC motor can only use 65-70% of the fed wattage to do the work, using the calculation of 745.6W/HP, how can they rate this motor at 2.3HP?
@Captain Crunchie

In response to your Post#12:
Motor power ratings are generally based on how much power can be dissipated in the motor without overheating it or causing other damage to the motor (e.g. weakening the permanent magnet if the motor uses such). The rating is not necessarily closely related to how much external work the motor can do. As previously noted, there are always exceptions. Motors range in size from a grain of rice to as large as a tractor-trailer truck. Motors are designed for widely different applications. Some motors are designed to accept stall; most motors are not. Some motors are designed to have extremely fast acceleration while for others acceleration is a minor issue. Etc.

As for the 2.3Hp rating: 1700W / 745.6 = 2.28Hp (pretty close to 2.3Hp). As I noted above, the rating is based on how much power you can shove into the motor without damage to the motor. It is not rated on the actual work that the motor can do to some external load.
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
As for the 2.3Hp rating: 1700W / 745.6 = 2.28Hp (pretty close to 2.3Hp). As I noted above, the rating is based on how much power you can shove into the motor without damage to the motor. It is not rated on the actual work that the motor can do to some external load.
So, my question is, ....HP is either the energy used to move 550lbs moving 1', or 745.6W, the motor isn't putting out the former, and since the fuse is limiting a 140A motor to 80A, how can it be rated as a 2.3HP motor? Because if you were to allow 140A into it, it would fry. It seems misleading and erroneous.
 

TeeKay6

Joined Apr 20, 2019
381
So, my question is, ....HP is either the energy used to move 550lbs moving 1', or 745.6W, the motor isn't putting out the former, and since the fuse is limiting a 140A motor to 80A, how can it be rated as a 2.3HP motor? It seems misleading and erroneous.
A flashlight battery remains rated at 1.5V, even when it is dead. A cable with a rated breaking load of 1000 pounds, still has the same rating after it has broken; the rating applies only to the unbroken cable. The motor rating relates to when it is driven as hard as the manufacturer specified; that rating is power input, not power output. If you drive the motor with less power, you will certainly get less mechanical power out--but that does not change the rating.

Have I answered your question?
 

Thread Starter

Captain Crunchie

Joined Apr 4, 2019
11
A flashlight battery remains rated at 1.5V, even when it is dead. A cable with a rated breaking load of 1000 pounds, still has the same rating after it has broken; the rating applies only to the unbroken cable. The motor rating relates to when it is driven as hard as the manufacturer specified; that rating is power input, not power output. If you drive the motor with less power, you will certainly get less mechanical power out--but that does not change the rating.

Have I answered your question?
No, and please indulge me here. I get that a rating will stick whether it works or is broken. But is there any time that that 1700W motor would have 1700W surging through its windings? Then, to add a restriction to that feed seems even more erroneous, because to have to protect that winding by limiting the feed makes the rating moot.

Moreover, HP is a rating of its ability to do work.

Using the formula (V * I * Eff)/746=HP, determine your electric motor’s horsepower. Multiply the voltage, the current, and the efficiency, then divide the result by 746. For example, the horsepower of a 230v motor pulling 4 amps and having 82% efficiency would equal 1 horsepower.

In the above example, the HP rating is based partly on the motor's efficiency. But in the 1700W motor, its efficiency doesn't seem to be considered.

Or are you saying that the 1700W rating is actually 65-70% of the core capability?
 

TeeKay6

Joined Apr 20, 2019
381
No, and please indulge me here. I get that a rating will stick whether it works or is broken. But is there any time that that 1700W motor would have 1700W surging through its windings? Then, to add a restriction to that feed seems even more erroneous, because to have to protect that winding by limiting the feed makes the rating moot.

Moreover, HP is a rating of its ability to do work.

Using the formula (V * I * Eff)/746=HP, determine your electric motor’s horsepower. Multiply the voltage, the current, and the efficiency, then divide the result by 746. For example, the horsepower of a 230v motor pulling 4 amps and having 82% efficiency would equal 1 horsepower.

In the above example, the HP rating is based partly on the motor's efficiency. But in the 1700W motor, its efficiency doesn't seem to be considered.

Or are you saying that the 1700W rating is actually 65-70% off the core capability?
@Captain Crunchie
Since I do not have the motor specification, I cannot answer as to what it says. Otherwise, if the manufacturer says that the motor is rated for 1700W, then it is customary--for motors--to convert that wattage to Hp via the formula you know. If the manufacturer says that the motor is rated for 2.3Hp, then it is customary--for motors--to convert that Hp rating to wattage via the formula you know. If you wish to know how much external work the motor can do, you must get that info from the manufacturer, possibly as an efficiency %. Multiply the motor input power (e.g. 2.3Hp) by the efficiency to get the available output power. The efficiency may not vary linearly with drive power; only the manufacturer (or datasheet) can give you that info. I do not know from where you got the 65-70% value and I have no knowledge as to whether it is relevant to this discussion.

As to your question "is there any time that that 1700W motor would have 1700W surging through its windings?", I must refer you to the motor's datasheet or manufacturer. It would be usual to rate a motor at its max power allowed output condition, assuming sufficient drive (electric input power) and appropriate load. It is my belief that if the manufacturer claims the motor is rated at 2.3Hp, then that input power (electrical) is what is required to achieve the maximum mechanical output power that the motor is rated to deliver. Due to inefficiencies, that output power will certainly be less than the input electrical power.

BTW, to make it more complicated. Motors are not resistive loads; there is a "power factor" involved due to the phase shift between voltage and current in the inductive & resistive windings. The example you cited (230VAC, 4A) ignores that factor.

Your question: "...to have to protect that winding by limiting the feed makes the rating moot." I do not understand why limiting the power to the rating poses any limitation. The protection is against a stall condition; that is not the condition at which motor power is specified. The data sheet should show the electrical drive and at what rpm and what external mechanical load (i.e. torque) ( OR the equivalent mechanical power) the motor is operating to meet its spec.
 
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captain crunchie said:
In the above example, the HP rating is based partly on the motor's efficiency. But in the 1700W motor, its efficiency doesn't seem to be considered.[.quote]

have you ever looked at vacuum cleaner motors:

https://www.intervacdesign.com/f4/The-Truth-About-Vacuum-Horsepower.htm

Audio Amplifiers?

https://www.audioholics.com/audio-amplifier/amplifier-power-ratings

it's mostly garbage and designed to sell.

Friends had a store that sold car audio and did consumer electronics repairs.

one guy comes in and asks:

How many Watts is this.
How many Watts is that.
That's all he cared about.

Now try to size a generator for your house.
 
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