# AC amps

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#### mocaquita

##### Active Member
If I have a DC motor rated at 15.4 DC amps powered by a SCR type speed control that converts AC power to DC, how many AC amps would I expect to pull? Would it be the same or more?

Thanks,
DaveB

#### Tin Falcon

##### Well-Known Member
What is your line voltage? what is the motor voltage? watt is 1 joule per second! watt is volts times amps. a 12 volt DC motor rated at 10 amps , hooked to a power supply will pull roughly 1 amp off a 120vac line.
Tin

#### Stan

##### Well-Known Member
A very good question and I have never found the answer. I run several DC motors (both PM and wound field) on SCR type controls. I have the cheap KBIC controls and also very expensive industrial controls. I have tried to measure the AC input current with an Amprobe clamp type meter and the motor draw with both a current shunt and a clamp type meter.

There are so many variables and unknowns that nothing has ever made sense. I have always assumed that the AC reading was realistic but it may not be when only part of the sine wave is being used and the meter is designed to measure sine waves. On the motor input, you are not measuring pure DC, but portions of a 120 hz sine waves Meters designed to measure RMS don't know what to do with a wave form that has peak voltage but only part of the waveform. . There is also the counter EMF that is a big factor in DC motors and that is a function of the motor load which is just another unknown.

DC motors can operate for relatively long periods at way above rated current, so I would base my control requirement on the anticipated load in HP. If this is a 90 volt motor (about 2 HP) and that is the maximum anticipated load, then a 2 HP control would be marginal connected to a 120 volt 20 amp breaker. On the other hand, if you are only anticipating a 1 HP load the 2 HP control on a 15 amp breaker would be fine.

J

#### JorgensenSteam

##### Guest
Standard AC motors generally draw six times rated current when they start, so the circuit breaker for starting across-the-line motors is generally 150-175% of the motor full load current.

Motors operating on VFD's don't surge when they start, so the breaker serving the VFD can be smaller than 150-175%.

The accuracy of AC clamp-on meters varies, and a true RMS meter will be more accurate, but you can get in the ballpark with a non-true RMS clamp on meter.

For a DC motor running off a power supply, or an AC motor running off a VFD, you willl always have the power to the motor, plus the losses of the power supply or VFD, so you need to measure the AC current going into the power supply or VFD using the clamp-on ammeter.

I would guess that power supply or VFD losses many be 10-20% more than a straight motor, but that may vary with the actual power supply or VFD.

Pat J

#### tornitore45

##### Well-Known Member
Let's say that you have a b ridge rectifier feeding the DC motor directly from the 120V line.

DC measures are AVERAGE value, that is the height of a line that divides the waveforms in such a way that the area above the line equal the area below the line ( from line to waveform).

AC measures are usually RMS, that is a the value of a DC dissipating the same power on a resistive load.

DC motors respond thermally to RMS value but mechanically the speed and torque are proportional to the Average Voltage and Average Current respectively.

With sine wave the relationship between the Peak, RMS and Average are well known.

A 100Vrms sine voltage has a peak of 141V and an average value of 90V

The exact ratios are peak = squreroot(2) X rmS = AVERAGE x PIE / 2 this is valid for voltage and current.

Unscrambling the above yield that RMS /Average = 1.11

So if you are running the DC motor full speed absorbing 15.4ADC (Average) the AC current will be the same 15.4A Average but will be 1.11 x 15.4Arms

Disclamer:
If you use the SCR to chop the waveform to reduce speed all this is not valid because the waveform is now a fraction of a sine wave.
Most AC meters are simply responding to the AVERAGE value but are calibrated to display the RMS for a sinewave. In this case this is not important but is a thing to be aware when dealing with difeerent wave forms
The DC motor inductance tends to change the current waveform and things gets more complicated.

#### Ken I

##### Project of the Month Winner!!!
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DaveB - for a simple answer consider the Watts

Assuming your motor is 15.4A @ 90V (guess) - 1386 Watts

Off a 120V AC supply this requires 1386/120 = 11.55A

You must then allow for 90-95% efficiency through the driver (guestimate) and approx 85% power factor (don't ask I'm trying to keep this simple) = 14.3A

Circuit breaker needs to be at least 50% more (for inrush - mentioned in earlier post - depends on whether or not your circuit soft starts or not).

You won't go far wrong with such an estimate.

Ken

#### Stan

##### Well-Known Member
Lots of motor theory! Who has done the measurements with an SCR drive to answer the original question? If you have done them, tell us what you read, especially under different loads and speeds. As I said in my first post, I have tried various ways under various conditions and cannot get any sensible readings. I spent the better part of forty years trouble shooting electronics and that included lots of DC motor circuits. Electric oil well drill rigs use 1000 HP DC motors. Two competing systems are SCR drive and DC drive.

The SCR drive uses three phase AC input with all the control in the SCR circuit while the DC system uses generators with all the control on the generator field. Believe me, it is much easier to troubleshoot the DC system.

#### Troutsqueezer

##### Project of the Month Winner!!!
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Tin has it. Power is the same from the AC line as it is across the load (minus heat loss through components) so it must be the same in the primary as it is in the secondary circuit. You know your secondary current but you don't mention at what voltage that current flows. Multiply that voltage times the current and you have the average power in the secondary. Once you know that you can figure current flow in the primary side easy enough. Divide the power (watts) by 120 and you'll have your current draw in RMS but close to average and close enough for government work.

To get more precise than that by calculation you need to include phase angles and impedance and you really don't want to go there. For current spike calculations an oscilloscope with storage capability will give the answers easy enough.

J

#### JorgensenSteam

##### Guest
You cannot mix and match AC and DC amperes.
You need to think in terms of power flow, since amperes vary with AC/DC, voltage, etc.

It is a mistake to use watts for any electrical calcuation unless you are sure that the power factor can be ignored or perhaps assumed at 0.80. In the electrical power world, ignoring the power factor is a huge, dangerous and often expensive blunder.

Here are some examples:

A DC motor is operating at 12 VDC, and draws 15.4 amps DC.
So the power used by the motor only is 12 * 15.4 A = 184.8 watts.
We can ignore power factor with DC since it does not come into play with DC.

If you have ever installed a surge tank on a water system, you can understand power factor. A surge tank absorbs the energy from sudden pressure changes in the water, and as the water pressure increases, the surge tank absorbs power, and when the water pressure decreases, the surge tank returns the energy to the system.

So for electrical systems, the magnetic field of motors is the surge tank, and when the voltage alternates (in AC electrical systems), the magnetic field of the motor alternately absorbs and releases energy in the form of electrical current. The current is transferred from the power company to the magnetic field of every motor on the first half cylce of every waveform, and on the second half cycle, this energy is transferred back to the power company. This reactive power does not do any actual work such as rotating the motor, so the power company cannot bill you for power for this, but it does cost the power company money to have to send this power to you, and then receive it back every cycle, so the power companies usually have an added fee for low power factor.

So back to the example, we have a DC motor drawing 15.4 amperes at 12 volts DC, with is using 184.8 watts.

The DC power supply is fed from the AC line, lets say for example at 120 volts AC.
So the current on the AC side must take into account the power that the motor is using, plus the power that the power supply is using in the form of waste heat.
If there was no wasted energy, then the current on the 120 volt side would be 184.8 / 120 = 1.54 amperes AC.

Since the power supply is wasting energy, then the current into the power supply on the 120 volt side will be higher, lets say perhaps 10-20% higher, so maybe 1.84 amperes AC.

Large VFD's generally have a high power factor, usually above 90%.

For AC, the equation is V = I * R * (cos theta), where cos theta is the power factor.
For DC, the equation is V = I * R, since the power factor is 1 in DC systems.

The only way to compare apples to apples is to calculate power first from one side or the other, and then go back to current on the other side using the appropriate voltage and power factor if that applies.

The reason the Edison DC systems did not survive is that the DC currents at low voltages are huge compared to the same size motor operating at a higher voltage.
The line losses are very high in high current DC systems since the line losses are current squared times the resistance of the wire.

The main reason for the use of AC systems is so that you can generate power using an alternator without requiring a high wear/ high maintenance commutator to create DC current. The reason for using high voltage AC systems is because as the voltage goes up, the current goes down, as does the wire size (V = I * R * cos theta), so power companies step up the voltage, transmit the power at low losses, and then step the voltage back down.

Doubling the voltage cuts the current in half. Since copper is expensive, and high currents create a lot of line losses just by heating up the wire due to its internal resistance, then it makes perfect sense to operate a motor or other large load at as high a voltage as is practical and safe.

Many residensial motors can be operated at either 120 volts or 240 volts, single phase.
The general rule of thumb in industry is to operate motors less than 1/2 half horsepower at either 120 volts, or 240 volts single phase.
For motors 1/2 hp and above, and where 3-phase is available, then the motor is generally specified to operate at 3-phase, since the smoothness, efficiency and longjevity of a 3-phase motor will quickly offset any added initial cost.

At around 200 hp, medium voltage begins to become more efficient than low voltage. In the power business, low voltage is defined as voltage 600 volts and lower, and medium voltage is above 600 volts but below 69 KV (69,000 volts). Medium voltage circuits are routinely referred to as "high voltage", but from a design standpoint, that refers to a different system.

Most 500 hp motors are operated at 4.16 KV (4,160 volts), although some operate at higher voltages.
The local refinery has a 32,000 hp air compressor motor that operates I think at 13.2 KV, and is fed from a nearby 161,000 volt line via stepdown transformers.

Every appliance in the US is required to have a nameplate on it, and that nameplate will allow you to calcuate the power, and current that the device uses.

You are generally interested in the input voltage and amperes, not the output voltage and amperes, since if you do not size your input circuit breaker and wire large enough, the wire will overheat, and the breaker will nusiance trip.

A general rule of thumb for power factor in AC circuits is 0.80, unless the nameplate states otherwise.

So the answer to the question is "Generally expect much less AC amperes than DC amperes, since the AC voltage is generally much higher than the DC voltage".

The current valves given on the load side are generally meaningless unless you are actually sizing the wire on the secondary side.

Think in terms of power, and assume most of the power is used by the motor, and some is wasted in the power supply and wiring, so more power always has to be input to a device than you will get out of it via the motor.

Hope this helps more than it muddys the water.

Pat J

#### mocaquita

##### Active Member
Thank you all for the responses! Think I'll try this setup on a 20 amp circuit before proceeding further.

Dave

#### Steam4ian

##### Active Member
G'day Dave

Others have said that power in must equal power out plus the losses in the controller.

The dc power is the volts times the current. For the ac input the power is the volts times the current times the power factor. In this case the power factor at full motor speed will be about unit so can be ignored. We have to make some assumptions about the controller but typically the dc current in the motor is the average current whilst the ac power is measured on the RMS currrent. There is a factor of 1.1 between RMS and average current so the ac current is 1.1 times the dc current.

I have just looked up the KB controller manuals and they suggest the following fuse sizes.
Motor fuse 1.5 times the motor current (in your case 1.5 x 15.4 = 23.1 amps)

Line fuse (ac) 1.7 times the motor current (in your case 1.7 x 15.4 = 26.2 amps)

In you case you could probably get away with a 25 amp circuit breaker. You can get circuit breakers with a motor protection curve, called a D curve, which would work for 25 amps.

Regards
Ian

#### mocaquita

##### Active Member
Hi,
As a follow up to my question, I've installed and have been running this 1 1/2 HP 90 volt 15.4 amp DC motor on my mill for about three weeks now. It is on a 20 amp circuit without any problems what so ever. Fully loaded the motor after installation by drilling several large holes in steel. Haven't tripped the breaker yet!

Dave

#### krv3000

##### Well-Known Member
ooo its all a black art