frequency vs voltage vs current draw

Fred Raud

Member
Join Date
Jul 2004
Location
San Antonio
Posts
213
tonight i was called on an inverter that kept faulting on overcurrent (F18,,TBwoods Etrac inverter 4001),,what was happening was after the inverter is a starter,,and the operator basically operated the starter,,

for a little background,,yesterday the inverter blew up, due to a bad motor,,so was replaced,,and not all of its parameters were reset,,

now theres a conveyor next to it and for all purposes in this thread can be considered its twin,,same 56c type motor,,same 1hp,,same 480v,,(supposed to be same gearbox,but as we are about to find out looks to not be the case),,so i used that one as a guideline to set the parameters of the one that blew being we had no back up or hard copy of that inverter,,

so,,all to gether,,set it to run at 17hz,,and most of the regular parameters set the same,,and fired up fine,,but ran really really slow,,so increased the frequency to 60hz to get it to the same speed as its 'twin',,figuring that somewhere along the ways the gear box got switched out with one we had and they used theinverter to match them up,,(no sprockets or chains to change ratio),,

figured all was ok,,but noticed that when we stopped the starter,,the inverter wasnt 'notified' so when we would push start,,the inverter would fault on overcurrent F18,,looking through the book and comparing the inverters parameters i noticed parameter 82(actually that was pointed out by my boss) was set at 0,,and should be set at 1 so as the inverter would ramp it up as it should,,well,since today rolled around,,i figured on their break id change the parameter,,and did such,,and it still had the same problem,,the only other parameter that i noticed different was 68(fault reset times,,basically the number of times within a 10 minute period the inverter would restart on its own if faulted,,i set it to 3 as its twin was set),

while they were on break,,i got curious and set the inverter to 17hz as its twin is,,started it,,and then started the starter,,and no fault,,so,,i put it back to 60hz,,and on starting would fault on overcurrent,,this is where my question lies,,

at 17hz,,thats roughly 25% of the frequency used to drive the conveyor with the problem,,so i am assuming the pwm is putting out a voltage of around 25% as well,,so around 120vac??and that would put the current draw 4x as much???if so,,wouldnt go to say that the motor running at 17hz shoujld be the one pulling the fault??or is my logic all messed up here?now,,i did do a current draw check on the motor while it was at 60hz,,was around .80amps,,,and i regretfully didnt do it at 17hz while i was at work tonight,,

also,,all things being the same,,frequency,,conveyor drive sprockets,,(they sit side by side,,and feed the same spiral cooler),,motor speed,,motor voltage,,motor hp,,inverter type,,and the only unknown is the gearbox(no data plate and didnt want to pull the motor to manually figure gear ratio),,is it safe to assume that the gear box ratio is the reason one inverter is set at 17hz,,and the other at 60hz to run at the same speed,,

and,,anybody familiar with etracs,,know if there is a way to have the inverter not have this problem when started from the starter without wiring it to the contactor or the start switch?i was told parameter 82 would do it,,(id define the parameter for those not familiar with etracs but dont have the book infront of me,,),

i think that about covers it for now,,

Fred Raud
 
Fred,

What a weird setup. If I understand you correctly you have the VSD set to a fixed speed, and then you close a contactor between the VSD output and the motor. Is this really what is being done? If so then it is totally the wrong way to be applying a VSD.

There is no need for the contactor. Get rid of it. The VSD needs to start each time from a slow speed and ramp the output frequency to the desired running speed. The ramp rate needs to be set so that the conveyor accelerates fast enough for the application, but the motor is always operating close to it's normal synchronous slip speed. ie with VSD operation the motor speed must ALWAYS be within 2-5% of the VSD's output frequency. In that case the current the motor draws will never likely exceed 120% FLA.

What you are doing at present is treating the VSD as a DOL starter that is attempting to draw 600-700% FLA from the VSD every time it starts. Unless the VSD is grossly oversized it will not supply that current and trip exactly as you descibe.
 
im aware of that,,and thats why its going to change,,i was mainly curious on how themath works out for the current draw,,

also,,since you brought it up,,if i pull the starter out from between the motor and vfd,,what do i use for overload protection(aside from the inverter,,i just blew one up yesterday,,thats an expensive form of overload protection),

hmmm,,have the starter before the vfd,,?

Fred Raud

ps,,i noticed you said VSD,,i assume its the same as my VFD,,if so,,what does the S stand for?
 
Yep, get rid of the contactor between the VFD and the motor. (We tend to call them Variable Speed Drives in this part of the world). Almost all VFD's will have a perfectly good electronic overload built in and it is normal to use that. (If you want to keep the external one you have then it can be left in circuit, but this is usually only done on multi-motor installations where one VFD is powering two or more motors and each motor needs individual protection.)

Putting the starter before the drive will not help...in fact it will make matters worse. If we imagine the conveyor is a constant torque load, then the power it consumes will increase linearly with motor speed. ie HP= Speed * Torque, so although the load current into the motor will be more or less constant through the whole speed range, but on the supply side it is a different story.

On the supply side the Voltage is constant, and if the Power supplied is to increase with Speed, then the Current will have to increase with speed, until it reaches FLA at 60Hz, exactly as if it was a DOL. This means that if you put the overload on the supply side of the VFD the current it will see is not constant, and it will fail to protect the motor at all speeds less than 60Hz.

Another reason for not putting the starter upstream of the VFD, is that the supply section of all VFD's (the rectifier, DC bus and inrush current limiting) will usually have a limited number of "starts per hour" that it can tolerate. Don't confuse this with the output section of the VFD...that can acel/decel a motor as often as you like because it rarely runs at more than about 120% FLA, usually less.
 
that all makes sense to me,,now im still wondering on why the fault at 60hz and not at 17hz being it SHOULD draw more current at 17hz

Fred Raud
 
I think in fact if you measure the motor current is will be fairly constant through the whole speed range. It won't be exactly constant, but if the conveyor load (torque) is constant with speed, then so will the motor current.

If you measure the current on the supply side of the VFD you will see the current rises dramatically with speed. It's worth doing on the other twin VFD just to see what is happening.

What I suspect caused your overcurrent is that either the VFD is at the reference speed (17 or 60Hz) BEFORE the contactor is pulled in (which is creating a DOL 600% current draw) OR the contactor is pulling in when the VFD is at zero speed, but the acel ramp rate is too fast, and the VFD output is rising in frequency much faster than the motor can develop enough torque to keep up. When this happens the slip between the VFD and the motor exceeds about 3-5%, then the motor current will rise dramatically over 150% FLA and the VFD will fault on overload.
 
thats just it,,the vfd is turned on,,then the starter is then used as a switch to just turn the motor on and off,,not alloowing the vfd to do its job and ramp,,i know its set up wrong,,im just trying to find out why at 17hz i get no current overload,,and at 60hz i do

Fred Raud
 
Fred..Just a guess here but...

Lets say the motor darws 60amps at 60hz the same motor would draw around 17 amps at 17 hz..(I know its not quite like that but...) Across the line starting at 17 hrz woul give you about 51 amps...(17x3) Now at 60hz you would have an inrush of 180amps..(60x3) That would be enough...

As i said just a guess but it makes scense to me..

as for the overloads..i wouldnt use them. i would use the drive to start/stop and take care of overload..
 
im just trying to find out why at 17hz i get no current overload,,and at 60hz i do

I guess there is a wrong set up for your max. reference frequency,
maybe you check the parameter of it. If you exceded with your set point it will trigger the over current fault.

One thing to consider is your ramp-up time & ramp-down time parameter, this also trigger the over current fault.


Hope this will help!!!
 
Phillip's analysis is right on target.

I think Darren's point about inrush may also be pertinent - after all, we use reduced voltage starters to cut inrush, don't we.

There are a couple of other points. First, did you make sure the new VFD is rated for constant torque loads, and not just picked to match horsepower?

Second, even with that naughty contactor in the mix, it will probably take longer to acellerate to a "speed" of 60 Hz than 17 Hz, although the final conveyor speed is the same. The over-current trip is based on time as well as Amps, so this might cause the trips.
 
Maj. Toxido said:
im just trying to find out why at 17hz i get no current overload,,and at 60hz i do

I guess there is a wrong set up for your max. reference frequency,
maybe you check the parameter of it. If you exceded with your set point it will trigger the over current fault.

One thing to consider is your ramp-up time & ramp-down time parameter, this also trigger the over current fault.


Hope this will help!!!
the ramp up and ramp down times dont matter being the inverter is already running,,which is the cause of the problem,,being it doesnt actually ramp up,,the inverter is already running so the starter is acting like a knife switch,,,,lets just say,,"hot connect" to the motor,,in normal operation the inverter would be turned on and control ramp up to get to speed,,when i force the inverter to do so,,it works well with no overcurrent,,the way i test that is by turning off the breaker to the inverter,,push the start button which pulls in the starter,,,,then turn on the breaker to the inverter and it ramps up in the 3 seconds it should,,and all is well,,

Lets say the motor darws 60amps at 60hz the same motor would draw around 17 amps at 17 hz..
the whole point of me posting is due to my thinking that the current draw is inverse proportional to the voltage applied,,so if at 60hz,,drawing 60 amps,,at 17hz,,it should draw in the area of 240 amps,,i did measure 120vac at the inverter under load at 17hz(close enough to 120vac) and 480vac at 60 hz,,

if that is true,,im back to my original question,,shouldnt the higher current being drawn at 17hz be overloading the inverter?instead of the 60hz?,,or maybe the motor speed is the factor letting the inverter handle the 'hot start' of the 17hz and not the 60hz?

Fred Raud
 
you are looking at it backwards..The whole pupose of the VFD is to reduce current at lower frequencys. I understand where you are coming from thou..couse a 120v motor drawing 10 amps will draw 5 amps at 240.. It just dosnt work the same way with an inverter..


Just reread your post..dont forget you are also changing the frequency to the motor if you were to just change the voltage and try to keep the same torque then yes current would go up becouse you would be stalling the motor..
 
Fred Raud said:
the whole point of me posting is due to my thinking that the current draw is inverse proportional to the voltage applied,,so if at 60hz,,drawing 60 amps,,at 17hz,,it should draw in the area of 240 amps,,i did measure 120vac at the inverter under load at 17hz(close enough to 120vac) and 480vac at 60 hz,,

It doesn't work that way. A conveyor is typically a constant torque load because it takes a fixed force to pull the load uphill.

A drive drops voltage at reduced speed because a more or less constant ratio of Volts to Hz is generally best for the motor operation.

In a motor, even at reduced speed, Voltage is a function of current. So, if your load is indeed constant torque, it will require a constant current at steady state, regardless of speed and/or voltage.
 
My first point--Fred--your use of an inverter would make an excellent classroom example of how to do as many things wrong as possible. Please take Philip's recommendations seriously and change the system.

Second point--regarding the overcurrent fault. You need to think of this as an across-the-line start. One is at 460V 60Hz and the other is at 460v times 17/60 and 17Hz which works out to 130v 17Hz. If you look in your Electric Engineering Handbook pocket edition and at the page showing motor inrush currents, you will see that a NEMA Design B motor (most likely what you have) will have an inrush current of around 6-8 times nameplate Full Load Amps. This assumes that you have a motor designed for 60Hz starting at nameplate voltage and 60Hz, and the motor is standing still--0 Hz. That's a difference of 60Hz in applied voltage and motor rotation frequency.

Now, you take this same motor and start it on a 130V 17Hz supply. The voltage is reduced to only one quarter of 460V and the difference in frequency between the applied voltage and the motor rotation is only 17Hz. So the motor is seeing only 25% voltage (which by itself would reduce the inrush to 25% of what it was at 460V) and only 25% of the difference frequency (actually this is called slip speed). Is it any wonder that the resulting inrush current is much lower?

Fred, you mentioned that you knew that lower voltage on a motor results in higher current. That's true for a motor that's running and loaded but, at start, a whole different set of conditions apply as you can see above.

Hope this helps you.
 

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