Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Measuring Power input of DC Motor Control Circuit using PWM

Status
Not open for further replies.

papaisou11

Member level 2
Joined
Aug 2, 2016
Messages
45
Helped
0
Reputation
0
Reaction score
1
Trophy points
8
Activity points
449
Hello,
Greeting to all.

I am having difficulties to measure PWM based motor control circuit power measurement.

Let me explain further. I have a motor that can be controlled by varying DC voltage across the motor or using PWM based signal. I want to make a comparision between How much power in the driver (Including motor obviously) required if I control the motor using DC Voltage vs controlled using PWM Duty cycle.

The PWM frequency is 25Hz.

I could use 100% duty cycle constant in all cases and vary the Motor voltage where the speed is changing. In such case I recorded the Input voltage and current input of the device at different speed sucessfully using normal DC current measurement.

After that, I want to find the point in what duty cycle the motor is getting that speeds where the voltage is at its highest the motor could recive and measure the current and voltage. Since the PWM frequency is low, the voltage and current getting flactuated too much to take proper measurement.

What is the best way to measure the current and the voltage when the PWM duty cycle is 40% or say 65%, 75% and goes on upto 99%.

I have Oscope, True RMS multimeter - METRAVI XB-33CF.

Thanks a lot.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
Hi,

use a suitable low pass filter on both current and voltage. then measure the DC values. Multiply them to get power.
This is wrong. see post#4 (sorry for that)

Klaus
 
Last edited:

papaisou11

Member level 2
Joined
Aug 2, 2016
Messages
45
Helped
0
Reputation
0
Reaction score
1
Trophy points
8
Activity points
449
Thanks a lot. This is what is my plan but not sure. I am working with it and let you know.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
use a suitable low pass filter on both current and voltage. then measure the DC values. Multiply them to get power.
Sorry, this is wrong. Use the DC input voltage (before PWM) and multiply it with the filtered PWM´d current.

Klaus
 

    papaisou11

    Points: 2
    Helpful Answer Positive Rating

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
48,516
Helped
14,271
Reputation
28,805
Reaction score
12,974
Trophy points
1,393
Location
Bochum, Germany
Activity points
280,594
Sorry, this is wrong. Use the DC input voltage (before PWM) and multiply it with the filtered PWM´d current.
Not sure what's filtered PWM'd current in this case. Assume a (lossless) PWM DC motor driver. 12 V input U1, 50% duty cycle, 6V (average) motor voltage U2. Average input current I1 1A , average motor current I2 2A. Respectively motor power is U1*I1 = U2*I2. A real motor driver has switching and conduction losses, respectively U1*I1 > U2*I2.
 

    papaisou11

    Points: 2
    Helpful Answer Positive Rating

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
Not sure what's filtered PWM'd current in this case.
--> Low pass filtered to get the average current.

*****
Average input current I1 1A , average motor current I2 2A
This is what worries me. How do the currents look like?
Some things are clear: (hopefully).
* When switch is ON: input current = motor current
* When switch is OFF: input current = 0, but how does the motor current look like?
To get 1A average input current: 50% of the (OFF) time = 0A, then I need 2A during the other 50% (ON)
Now I know I have 2A of motor current during 50% ON time..... I need still 2A during the other 50% OFF time. This means CCM operation. A big series inductance (motor?) and the use of a free wheeling diode...
Not impossible. But is it realistic with 25Hz switching frequency?

*****
Here are two ways to also measure the power:
* DC bus side (including switching power loss): always DC constant bus voltage multiplied with average current
* load side using both V and I as RMS (excluding switching loss)

In my opinion the DC bus side method is easier to do.

When I´m doing power measurement on a motor (including driver) I multiply the DC (bus) voltage with the average of the current . If the current is PWMd then it needs to be averaged by an extra low pass filter.
Or one measures the current on the other (input) side of the DC bus capacitor ... then the current already should be averaged (by the capacitor).
(This is DC bus side measurement to avoid the (complicated?) RMS calculations.

*********
Example:

I know a motor is not resistive... but just as calculation example with an ideal single transistor PWM and an ohmic resistor.
Input DC voltage: 12V
Load 1 Ohms:

(DC side measurement)
@100% duty cycle: 12V DC; average current; P = 12v * (100% * 12V/1Ohms) = 12V * ( 12A) = 144W
@ 50% duty cycle: still 12V DC; average of current = 12V / 1Ohms x 50% = 6A (as well as load as DC bus); P = U x I = 12V x 6 A = 72W

(load side measurement)
@ 100%: same as above.
@ 50%: But now one can not calculate with average anymore (This was my mistake in post #2), one has to calculate with RMS:
load voltage RMS = 12V x sqrt(50%) = 12V x 0.707 = 8.484V
load current RMS = (12V / 1 Ohms) * sqrt(50%) = 8.484A
Power = V x I = 8.484V x 8.484A = 72W (same as above, because of "ideal" = no loss transistor.

***
Now back to the motor with complex impedance:
With the DC bus method the current may be in two directions in case of power is pushed back to the supply. Here also the averaging method gives correct values.
But with motor side measurement one needs: RMS volt, RMS current ... and additionally phase shift information. One usually doesn´t have. More difficult.

Klaus
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
Hi,

Then you need to know which type of analog meter it is.
* A "moving iron" meter will show the RMS value,
* While most others show average or rectified_average value.

Klaus
 

papaisou11

Member level 2
Joined
Aug 2, 2016
Messages
45
Helped
0
Reputation
0
Reaction score
1
Trophy points
8
Activity points
449
Thanks a lot KlauST.

Such a informative post. I have used a series resitor in the input (0.1 Ohm). The Resistor drop voltage further forwarded to a Low Pass filter that is charged to the mean value. I have taken this voltage and as per ohms law as average current and calculated the power consumption by multiplying the Input voltage. I know it is not accurate but serve the purpose. Is not it?
--- Updated ---

I am measuring total power consumption (Motor+Driver). The driver is a microcontroller based PWM Generetor.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
Hi,

I'm not sure about the circuit... where the resistor is connected...

But do the measured/calculated values make sense?
Or are there still questions?

Klaus
 

    papaisou11

    Points: 2
    Helpful Answer Positive Rating

papaisou11

Member level 2
Joined
Aug 2, 2016
Messages
45
Helped
0
Reputation
0
Reaction score
1
Trophy points
8
Activity points
449
Yes it makes sense. I will further check it. Thanks a lot.

I Think this image would be helpful.

1619252798328.png
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
19,934
Helped
4,369
Reputation
8,747
Reaction score
4,347
Trophy points
1,393
Activity points
131,944
Hi,

If I´m not mistaken then it takes about 10 minutes to settle to 99% (= 1% error). That´s really long time.

If you don´t use a single RC but two RCs you could:
C1 = 1500uF, R2 = 100R, C2 = 150uF, R3= 1k,

Then the 25 Hz ripple is about 0.3% (less error than before) and it settles within 2 seconds. (300 times faster; hopefully my mind calculation is correct)
And because it is lower impedance it will be more accurate.
(I see no drawback)

Klaus
 

papaisou11

Member level 2
Joined
Aug 2, 2016
Messages
45
Helped
0
Reputation
0
Reaction score
1
Trophy points
8
Activity points
449
Umm not 10 minutes but it is taking almost 1-2 minutes. I dont know why it is not that long you assuemed. Maybe because of the Capacitor ESR value or the real scenario. Surely I will divide this into two stages.

Nice help. Thanks a lot.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top