# PWM relationship to current

T

#### Troll

This question has been on my mind for some time. The best way I can try to explain this is by giving you a senerio:
When I apply 50% duty cycle of the PWM to a motor that has a load against it, the initial current draw is high, then comes back down to a steady state level. In this case the high current spike is at 10Amps, and the Voltage setting is at 12VDC. How can you get such a high current draw if the motor has terminal resistance of 1.5 Ohms? Using Ohm's law: V=IR, so I=V/R, the current can be no greater than 8 amps (12V/1.5Ohms), but I see 10 amps. Also, it should be alot lower since I only applied 50% duty cycle. Doesn't 50% duty cycle mean 50% of 12VDC, which is 6VDC? Am I thinking about this wrong?

T

#### Thomas Hergenhahn

On January 15, 2004, Troll wrote:
> This question has been on my mind for some time. The best way I can try to explain this is by giving you a senerio:
> When I apply 50% duty cycle of the PWM to a motor that has a load against it, the initial current draw is high, then comes back down to a steady state level. In this case the high current spike is at 10Amps, and the Voltage setting is at 12VDC. How can you get such a high current draw if the motor has terminal resistance of 1.5 Ohms? Using Ohm's law: V=IR, so I=V/R, the current can be no greater than 8 amps (12V/1.5Ohms), but I see 10 amps. <

You are right in principle. Three things may influence the resistance you measure:

1st, at 1.5 Ohms theresistance of your instruments cables may be greater than neglectable. Short them and subtract any resistance you measure this way.

2nd, the contact to the motors armature is through brushes and resistance may vary depending on surface and pressure. Try to turn the motor a bit and repeat the measurement.

3rd, depending on position the brushes may contact two neighboured segments od the commutator, thus shortening paert of the winding.

Next, when you measure the current on a PWM signal, your instrument has to deal with the waveform of the current. Depending on frequency and motor inductance, it is likely to be somewhere between rectangular (low freq, no inductance) and DC with an overlayed triangular AC signal (high freq, high inductance). Particularly in the first case digital instruments will show wrong values.

> Also, it should be alot lower since I only applied 50% duty cycle. Doesn't 50% duty cycle mean 50% of 12VDC, which is 6VDC? Am I thinking about this wrong? <

Generally, the current should be lower than what is determined by the resistance. The motor, while spinning, generates a voltage counteracting the external voltage. The current should be proportional to the force necessary to make the motor spin at a rate so the voltages cancel out each other. In real motors, this current has to go to the resistance of wires and other parts, causing a voltage drop. The resulting rotational speed is so that generated voltage + voltage drop are equal to the external voltage. If the motor is blocked, generated voltage is zero, so external voltage=voltage drop and current is determined by resistance only.

R

#### Robert Scott

Theoretically you are right. A pure R-L (resistance-inductance) circuit cannot have 10 Amps under the conditions you describe. So I can only guess that your measurement is in error, or there is some capacitance in parallel with the motor circuit. That could briefly take more than 8 Amps when the capacitor is charging.

Robert Scott
Real-Time Specialties
Embedded Systems Consulting

H

#### Henry

Please look up (lock rotor amp) and (back emf.) in AC motors. This will explain the condition. LRA = the motor act like a shorted transformer. Back EMF =in run mode the motor act like a generator.

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#### Matthew Hyatt

I am working with an application which uses a 24vdc brushless motor and we use a off the shelf PWM controller. At full power to the motoer, the signal is 100% Duty Cycle to the motor, which draws about 4 amps.

WIth a PWM of 50% the signal is approx. a 20KHz pulse train and the current is approx. 40 to 50% of the nominal mentioned above. The pulse width varies to deliver the same relative power to the motor at the set speed (pulse width modulation). The longer the pulse the more current or power, the shorter, the less power. I have observed that at low speed settings the pulses are short in duration and have a greater time between pulses and at higher speeds the pulses are wider and closer together until the pulse is a constant on state.

If you're measuring current use a true RMS meter. Better yet measure the pulse duty cycle and do a correlation between speed and current or pulse width and current.

In the case of start up, it takes more current to get the motor turning than it does to keep it turning. Initial start current is usually fairly high even in AC motors - Locked rotor condition as mentioned in another posting.
By the way you will always have 12vdc to the motor, just a varing pulse width at the set speed.

Regards,

M Hyatt

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#### Meir Saggie

Hi,

First order guess - either or both your measurements (resistance, current) have errors. I would suspect more on the resistance because:

1. If you measured it with a regular ohm-meter, you could easily get an error of 0.2 or so ohms from bad contacts, meter zero, etc.

2. Likewise, the brushes might show a few extra tenths of an ohm when measured with an ohm-meter - versus when the real motor current flows.

And, if the load rotated in the opposite direction before you applied the voltage, you have an internal generator in the wrong direction in the motor (also called active braking condition).

Meir

A

#### Anonymous

Thomas Hergenhahn - I have a question on your last comment. Are you saying that 50%PWM from a 12VDC supply does not equal 6VDC, but still 12VDC? Does this mean that if your resistance is 1 Ohm, you can still get 12Amps at 50%PWM?

So if this is the case, then you cannot limit your current based of the %PWM. IS this correct?

B

#### Bouchard, James \[CPCCA\]

That is my understanding of PWM. The average value of voltage will be 50% Since motors are not a pure resistance the current may not rise to 100% but the average will be lower that the full value. It is sort of like the old single cylinder motors that fired only as the fly wheel slowed down a bit to keep it going. The higher the load the more often they fired. They always fired with full power but the result was an average power level that met the load requirement.

James Bouchard