Telemetry: Difference between "motor current" and "battery current"? - HeliFreak
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Old 11-28-2017, 05:16 PM   #1 (permalink)
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Default Telemetry: Difference between "motor current" and "battery current"?

The Kontronik telemetry displays both "motor current" and "battery current". Other ESCs display "plain ole" current. What is the difference between Kontronik "motor current" and "battery current"?
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Old 11-28-2017, 10:01 PM   #2 (permalink)
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Actually very simple, the motor current is current consumed in the motor side to produce the work you demanded from it.

Battery current is what the battery could supply, yes it makes it difficult to understand but on some occasions like high load the motor and battery current can differ due to losses in the motor.
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Old 11-28-2017, 10:12 PM   #3 (permalink)
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In all of my logs, the motor current is higher than the battery current. What does this mean in practical terms? Does this mean that the batteries are not keeping up with the motor demand?
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Old 11-29-2017, 03:17 AM   #4 (permalink)
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In normal load situations the ripple current on the battery side actually makes the current sensing less on that side so all normal.
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Old 12-03-2017, 06:25 AM   #5 (permalink)
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Which number should we be concerned with when compared to the maximum limit of the ESC, e.g for a Kosmik 200A?
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Old 12-03-2017, 06:45 AM   #6 (permalink)
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My understanding of what curmudgeon is saying is the motor is using more amps than the battery is putting out. Would that be like your fuel pump putting out 1 gallon and minute and you burning 1.5 gals a minute in your engine?

Last edited by pgcopter; 12-03-2017 at 06:48 AM.. Reason: adding question
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Old 12-03-2017, 09:09 AM   #7 (permalink)
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What seems to be missed here is that the ESC is in part a switching basically a voltage regulator. In anything but full throttle the ESC steps down (average) voltage.

Assuming output voltage is linear to throttle opening then at 50% throttle if there is 48V on the input there will be 24V (average) on the output. As the power going is must be equal to power going out (ignoring slight efficiency loss to keep things simple) then the current on the output (motor current) would in this case be twice the current of the input (battery current).

This (average voltage step-down) explains why the motor current is usually higher than the battery current. At wide open throttle the two should be the same.

This is no different in principal to your battery charger which outputs 40A (at 25V) but only draws about 10A (at 110V) from the wall socket.
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Old 12-03-2017, 09:16 AM   #8 (permalink)
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Quote:
Originally Posted by Grumpy_Old_Man View Post
What seems to be missed here is that the ESC is in part a switching basically a voltage regulator. In anything but full throttle the ESC steps down (average) voltage.

Assuming output voltage is linear to throttle opening then at 50% throttle if there is 48V on the input there will be 24V (average) on the output. As the power going is must be equal to power going out (ignoring slight efficiency loss to keep things simple) then the current on the output (motor current) would in this case be twice the current of the input (battery current).

This (average voltage step-down) explains why the motor current is usually higher than the battery current. At wide open throttle the two should be the same.
Kind of makes sense but I thought brushless ESCs work by switching the power on and off thousands of times per second and the voltage stays the same. So for a PWM frequency of 8KHz, 50% throttle would be 4000 pulses of power every second.
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Old 12-03-2017, 09:23 AM   #9 (permalink)
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Quote:
Originally Posted by mahbouni View Post
Kind of makes sense but I thought brushless ESCs work by switching the power on and off thousands of times per second and the voltage stays the same. So for a PWM frequency of 8KHz, 50% throttle would be 4000 pulses of power every second.
Yes that's why i said 'average' voltage.

FWIW a PWM of 8KHz has 8000 pulses per second for all throttle opening (except full throttle where all the pulses join up). It's the length (aka 'width') of the pulse that changes with throttle. That's why it's called PWM (Pulse Width Modulation)
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Old 12-03-2017, 09:34 AM   #10 (permalink)
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Quote:
Originally Posted by Grumpy_Old_Man View Post
Yes that's why i said 'average' voltage.

FWIW a PWM of 8KHz has 8000 pulses per second for all throttle opening (except full throttle where all the pulses join up). It's the length (aka 'width') of the pulse that changes with throttle. That's why it's called PWM (Pulse Width Modulation)
Yes it's pulse width modulation but we're dealing with digital equipment that had a clock frequency, so in reality it's 1 second divided by 8,000 equal periods of time, of which 4,000 are 22.2V and 4,000 are 0V. How they are grouped together, e.g 0101010101010101 or 000011110001111 is down to the software.

The bit which I don't understand is how you get average voltage from on and off and how the motor suddenly draws more current every time you turn the power off and on again..

Hmm, a thought just came to me. When you turn on the power it doesn't reach 22.2V instantly and when you turn the power off it doesn't fall to 0 instantly. There's a period of time between during which the voltage rises or falls and is less than full voltage. Perhaps this is where the extra current draw comes from. With long duty cycles it wouldn't be significant but with very short duty cycles it would be more significant.
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Last edited by mahbouni; 12-03-2017 at 10:07 AM..
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Old 12-03-2017, 10:15 AM   #11 (permalink)
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Quote:
Originally Posted by mahbouni View Post

The bit which I don't understand is how you get average voltage from on and off
When a digital signal is switched on/off (i.e., PWM) fast enough what is seen at the output is the average voltage. If the low voltage is 0V and the high voltage is 5V the average voltage is calculated by multiplying the high voltage by the duty cycle, 5V x 0.5 = 2.5V.

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Old 12-03-2017, 10:20 AM   #12 (permalink)
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Originally Posted by Ace Dude View Post
When a digital signal is switched on/off (i.e., PWM) fast enough what is seen at the output is the average voltage. If the low voltage is 0V and the high voltage is 5V the average voltage is calculated by multiplying the high voltage by the duty cycle, 5V x 0.5 = 2.5V.

I think "fast enough" is the key here. I suspect the formula gradually becomes less accurate as the duty cycle increases.
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Old 12-03-2017, 10:46 AM   #13 (permalink)
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Found some interesting explanation here

https://electronics.stackexchange.co...-a-pulse-train
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