80amp ESC for 1/10

Old 10-31-2014, 08:46 AM
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Default 80amp ESC for 1/10

What's the best way to calculate how many amps your ESC should have? I always just made sure my ESC had a higher amp rating than my motors but I'm sure there is a better explanation
Reason being I need to get a new ESC for my novak 13.5T motor but I can't find how many amps it draws. I bought it as a system so I never really worried about the draw with this system. I'm thinking it probably pulls around 80 amps (of course there are other variables that come into play). So I'm not sure if I should just get a 120amp esc or if that would be overkill.

I'm using a 4200mah 100C battery so 4.2x100C=420 so I'll definitely have enough juice for the motor/esc.

Any help would be appreciated.

Thanks!

Last edited by skeasor; 10-31-2014 at 09:34 AM.
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Old 11-01-2014, 07:56 AM
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For 13.5t even 80 amp esc more than enough. A took a look at viper 6.5t motor's specs. And the power is generated by this motor is 440watts. This means 440watts/7.4volts= 59,46amps. I don't think even most powerful 13.5 motor pulls more than 40 amps. Look at specs of 13.5 reedy or viper motors. Divide the found value to the 7.4 then, you got avarage current that your motor can pull.
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Old 11-01-2014, 09:05 AM
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Thanks! I'm going plug 8.4v into the formula instead of 7.4v

Last edited by skeasor; 11-01-2014 at 09:16 AM.
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Old 11-01-2014, 11:41 AM
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Originally Posted by skeasor
Thanks! I'm going plug 8.4v into the formula instead of 7.4v
it is better not to plug it. Because, when you apply load (push throttle) to the motor, voltage of the each cell drops from 4.2v to the assumed 3.7v (nominal voltage)
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Old 11-01-2014, 02:39 PM
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Originally Posted by murat61
For 13.5t even 80 amp esc more than enough. A took a look at viper 6.5t motor's specs. And the power is generated by this motor is 440watts. This means 440watts/7.4volts= 59,46amps. I don't think even most powerful 13.5 motor pulls more than 40 amps. Look at specs of 13.5 reedy or viper motors. Divide the found value to the 7.4 then, you got avarage current that your motor can pull.
Not quite true. Depends how the motor was rated. It could be much more if the motor was tested with no-load, as a lot of them are.

Originally Posted by murat61
it is better not to plug it. Because, when you apply load (push throttle) to the motor, voltage of the each cell drops from 4.2v to the assumed 3.7v (nominal voltage)
Plugging 8.4 in will help you a bit because the voltage starts out at this level. The higher the voltage, the lower the amperage when there is no change in load.
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Old 11-01-2014, 03:56 PM
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Originally Posted by Hoese37
Not quite true. Depends how the motor was rated. It could be much more if the motor was tested with no-load, as a lot of them are.



Plugging 8.4 in will help you a bit because the voltage starts out at this level. The higher the voltage, the lower the amperage when there is no change in load.
Ok, at the very begining voltage is 8.4v for lipos. It quickly drops faster initial load period. Then, the change in voltage slows down. I added a picture of voltage drop of lipo on constant load. Also, I am wondering that as voltage drops, does not current increase to suppy same power. So load on esc will increase. I would take into lower voltage values to make sure i choose right one. For, motor choice i did not take test conditions into consideration. Current draw could change drastically. But again, 80 amp esc for 13.5t motor much more than enough.
Attached Thumbnails 80amp ESC for 1/10-10390081_775907772441704_3546507592144732943_n.jpg  
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Old 11-01-2014, 04:24 PM
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That's why I said only a bit. Youd be really surprised how much amperage a 13.5 can pull. 17.5 2wd buggy class here looking at datalogs we see spikes up to 120A, and constants of 90-100A
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Old 11-02-2014, 04:37 AM
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Motors rating in wattage is what they can handle with full load continuously (nominal). Obviously, we will not have this continues load when we drive the car, since we coast, break and accelerate. During peak conditions like under full acceleration of your car, the ESC will see a much higher current (amps) during these conditions (up to 5 times the nominal) and ESC are designed to handle these currents for short periods (bursts). And yes, you can narrow down the size of an ESC with a simple calculation based on the nominal wattage and voltage. Lower voltage means higher current (power = voltage x current). I would use 7,2V, since I never go below 3,6V per cell. So, for a Novak 13.5T with 195W nominal power, the max amps continuously would be 27A. So you could getaway with a smaller ESC, e.g. 40A.

Another aspect is the capacity and C rating of the battery. The capacity is given in mAh, thus a 5000 mAh battery could theoretically supply 5000mA (5A) for 1 hour. With a rating of let's say 40C, it should be able to supply up to 40 x 5A = 200A. High C ratings is never needed, and I will actually make a statement that you would probably be more consistent with lower C rating (20-40C). The picture of the voltage drop given by "murat61" is a test of different batteries with a 98 Amps continuously load. It's a good indication of how voltage drops over time for Lipo batteries. The purple curve (5000mAh battery) is most likely drained 100%, but a good thumb rule is not to use more than 80% of your capacity.
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Old 11-02-2014, 06:47 AM
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Originally Posted by cokemaster
The purple curve (5000mAh battery) is most likely drained 100%, but a good thumb rule is not to use more than 80% of your capacity.
Yes, when i learn that using lipo between 4.1v(max) and 3.5v(min) increases lipos life up to 1000 cycles, i was amazed
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Old 11-02-2014, 08:49 AM
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Originally Posted by Hoese37
Not quite true. Depends how the motor was rated. It could be much more if the motor was tested with no-load, as a lot of them are.



Plugging 8.4 in will help you a bit because the voltage starts out at this level. The higher the voltage, the lower the amperage when there is no change in load.
Originally Posted by cokemaster
Motors rating in wattage is what they can handle with full load continuously (nominal). Obviously, we will not have this continues load when we drive the car, since we coast, break and accelerate. During peak conditions like under full acceleration of your car, the ESC will see a much higher current (amps) during these conditions (up to 5 times the nominal) and ESC are designed to handle these currents for short periods (bursts). And yes, you can narrow down the size of an ESC with a simple calculation based on the nominal wattage and voltage. Lower voltage means higher current (power = voltage x current). I would use 7,2V, since I never go below 3,6V per cell. So, for a Novak 13.5T with 195W nominal power, the max amps continuously would be 27A. So you could getaway with a smaller ESC, e.g. 40A.

Another aspect is the capacity and C rating of the battery. The capacity is given in mAh, thus a 5000 mAh battery could theoretically supply 5000mA (5A) for 1 hour. With a rating of let's say 40C, it should be able to supply up to 40 x 5A = 200A. High C ratings is never needed, and I will actually make a statement that you would probably be more consistent with lower C rating (20-40C). The picture of the voltage drop given by "murat61" is a test of different batteries with a 98 Amps continuously load. It's a good indication of how voltage drops over time for Lipo batteries. The purple curve (5000mAh battery) is most likely drained 100%, but a good thumb rule is not to use more than 80% of your capacity.
Increased voltage on the same load will drive more current, not less. A given motor will produce more power at higher voltage. Put a 3S lipo on the same motor, current and power will increase and run time will decrease, compared to a 2S pack.

The high voltage-low current concept requires a change in load/motor to work properly, to maintain a constant power comparison. Adjusting gearing on the same motor can help some, but this generally requires a motor change in practice to realize the full benefit of higher voltage.


+ on needing to consider peak loading, not just nominal dyno ratings, to size the ESC and other amp requirements. In practice the vehicle weight and gearing, even driving style, impacts this to some degree.
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Old 11-02-2014, 12:46 PM
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Originally Posted by Dave H
Increased voltage on the same load will drive more current, not less. A given motor will produce more power at higher voltage. Put a 3S lipo on the same motor, current and power will increase and run time will decrease, compared to a 2S pack.

The high voltage-low current concept requires a change in load/motor to work properly, to maintain a constant power comparison. Adjusting gearing on the same motor can help some, but this generally requires a motor change in practice to realize the full benefit of higher voltage.


+ on needing to consider peak loading, not just nominal dyno ratings, to size the ESC and other amp requirements. In practice the vehicle weight and gearing, even driving style, impacts this to some degree.
Higher voltage will always result in lower current (amps) for the same load.
Lower voltage will always result in higher current (amps) for the same load.
Further, if your theory with higher/lower voltage would affect your available power, what would happen with your car when the battery voltage drop from 8.4v to 7.2v, you would have less power? No, the esc/motor still have the same power rating (wattage), so to be able to keep up with the demand from the esc/motor the current from your battery has to increase. Remember that all system is rated for a certain voltage 1s, 2s, 3s....6s. So, normally you cannot just increase voltage since it will create to much heat and you will fry it. Traxxas and other RTR cars where you can swap between e.g. 2s and 3s is designed in such way that you will have max rating and efficiency with 3s. The difference in power is actually not that much. What happens is that your motor will have a higher rpm, but you will have less torque. So, yes in most cases you would have to change your gearing due to this. Your last statement is true, but that is why you typically have dedicated esc for 1/10, 1/8, SC, etc. so you do not have think about sizing your esc based on amp/voltage ratings.
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Old 11-02-2014, 06:01 PM
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Originally Posted by cokemaster
Higher voltage will always result in lower current (amps) for the same load.
Lower voltage will always result in higher current (amps) for the same load.
Further, if your theory with higher/lower voltage would affect your available power, what would happen with your car when the battery voltage drop from 8.4v to 7.2v, you would have less power? No, the esc/motor still have the same power rating (wattage), so to be able to keep up with the demand from the esc/motor the current from your battery has to increase. Remember that all system is rated for a certain voltage 1s, 2s, 3s....6s. So, normally you cannot just increase voltage since it will create to much heat and you will fry it. Traxxas and other RTR cars where you can swap between e.g. 2s and 3s is designed in such way that you will have max rating and efficiency with 3s. The difference in power is actually not that much. What happens is that your motor will have a higher rpm, but you will have less torque. So, yes in most cases you would have to change your gearing due to this. Your last statement is true, but that is why you typically have dedicated esc for 1/10, 1/8, SC, etc. so you do not have think about sizing your esc based on amp/voltage ratings.
The first couple statements are true for the same power level, not for the same load. Hardly my claim, nor a mere theory. Credit traditionally goes to Georg Ohm, for which Ohm's Law is named after. One relevant derivation:

Power = voltage squared / resistance

Note the same load implies the resistance is the same, thus power is proportional to the square of voltage, meaning it changes rather quickly with voltage changes. So yes, as the battery voltage drops off from 8.4 to 7.2V during the run the power drops off. This may not be so obvious with many of today’s rather sufficiently powered RCs, lower powered examples make it far easier to detect or notice. Granted this is only one aspect, an electrical motor is a little more complex than simple resistance, there are secondary considerations, but the voltage is nearly always a dominant consideration.

The published motor power ratings are based on a specific set of conditions, not a definitive measure of the power output in all conditions. For example this motor data sheet from Novak, footnote 3 clearly specifies the voltage used to generate the power ratings. Member John Stranahan has an interesting motor dyno thread in the on road section, I can’t recall but it may include some voltage comparison data, I’ll try and take a look when I have the time.

The heat in the motor and ESC is largely a result of current-resistance losses, aka Joule’s first law. The reason the 3S example mentioned runs hotter than the 2S is because the current is higher, driven by the higher voltage. (Thus the max rating on 3S mentioned, it changes from 2S doesn’t it?)


From a practical perspective, there are many examples in RC past and present that demonstrate or are a result of the change in power with voltage changes:

- Not so obvious with good lipos which have a comparatively flat discharge voltage and typically limited discharge amount due to the high capacities and comparatively short race heats, the drop off in power during a run was far more obvious with the nickel based batteries of old. Was especially obvious with the 1200-1400mAh cells. Of course power loss due to heating up over the run is also a factor, both then and now.

- Electric 1/8 off roaders, where much of this discussion typically occurs, generally don’t use the same motor for different voltage setups. With 4S a 1900kv or so motor is common, but those running 6S tend to run 1400kv or so motors (of course there are exceptions for both). If voltage did not impact power why change motors (load)?

- 1/12 scale on road now uses 1 cell lipos, as 2S was determined to be too much power. Going back a ways, some used to run 6 nickel cells on large outdoor tracks, while 4 cell was used on the smaller indoor tracks. The difference in power was rather dramatic.

- Back in the nickel cell era, it was not unheard of to take a cell out to reduce power on slippery off road tracks. A couple examples, if I remember correctly it was the 2005 Worlds where some drivers went to 5 cells instead of 6 to make the cars easier to handle. Going back further, Gil Losi Jr. won a big race, I think in Japan, by switching to 6 cells instead of the 7 allowed at the time in modified classes. As I remember it this win was credited by some as being a significant part of the success of the launch of the original Losi, the JRX2, perhaps even part of the decision to proceed with releasing the car, undoubtedly a big gamble at the time for Pops Losi.

- All of the effort and importance the power limited stock class guys put into having the best, lowest internal resistance, highest voltage under load, max voltage charged, batteries. Been the case for something like 30 years or better. Hardly a placebo effect, there really is a difference.
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Old 11-03-2014, 06:58 AM
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Lots of good information on here. Thanks for the help guys!
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Old 11-03-2014, 11:54 AM
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I start looking into this in more detail, but it raises more questions than answers. Just for clarification; the difference in this discussion is what your system can deliver without permanent damage (i.e. temperature or current getting to high so you get irreversible damage) and what is required to move a car from A to B in a certain time frame. For the same condition (weight, time) it will take the same energy regardless of what system you have (2S or 3S) to move the car from A to B. This energy is equivalent to what the “car” load your system with in wattage. So for the same load you require the same wattage (power) from your system. So derived from P = U x I (power = voltage x current). This is a simplification, but more about that later. Consequently, when voltage increases the current decreases and vice versa for constant power demand.

The Ohms law is only true for ohmic system like resistor and wires which is not the case for a brushless motor. If you look up a motor equation you will see that you have to take into account reactive forces like EMF, etc. But if you could apply Ohms law your statement would be true. However, I have stated from the beginning that you cannot increase voltage above your rating without risking permanent damage. If your system is rated for 3S you can run 3S and lower, A 3S system does not run hotter than a 2S if you e.g. use a Traxxas VXL system. Yeah, it will if you do not change your gearing, but that it’s like running any wrong pinion in your 2S car.

Obviously, by lowering your voltage you will have less available power, and for the old system using NiCd or NiMh batteries with a typical cell voltage of 1.2V that would discharge to around 1.0V would give you a total loss of 1.2V for a six cell pack. That is equal voltage drop for a lipo battery (8.4V -7.2V), so why do we not see this “power loss” from start to end for a brushless system, like we did for the “old” system? It got me thinking. An ESC is a PMW device (Pulse Width Modulation) and it’s basically using its duty cycle to control the average voltage the motor sees. The longer the cycle, the higher the average voltage which again increase the RPM of your motor. What if the ESC modulates so you never get above 7.2V regardless you battery voltage (8,4V -7,2V). Thus the car will be 100% consistent during your race. Also, that make sense that Novak rate their motors with 7.2V and not 8.4V. But this is just my theory.

- Electric 1/8 off roaders, where much of this discussion typically occurs, generally don’t use the same motor for different voltage setups. With 4S a 1900kv or so motor is common, but those running 6S tend to run 1400kv or so motors (of course there are exceptions for both). If voltage did not impact power why change motors (load)?


I never stated that voltage did not impact power (well it was not my intention anyway)

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Old 11-03-2014, 05:51 PM
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Originally Posted by cokemaster
I start looking into this in more detail, but it raises more questions than answers. Just for clarification; the difference in this discussion is what your system can deliver without permanent damage (i.e. temperature or current getting to high so you get irreversible damage) and what is required to move a car from A to B in a certain time frame. For the same condition (weight, time) it will take the same energy regardless of what system you have (2S or 3S) to move the car from A to B. This energy is equivalent to what the “car” load your system with in wattage. So for the same load you require the same wattage (power) from your system. So derived from P = U x I (power = voltage x current). This is a simplification, but more about that later. Consequently, when voltage increases the current decreases and vice versa for constant power demand.
It is not a constant power situation with a given motor, the 13.5 originally asked about, as you acknowledge a couple times later in your post. But for discussion, achieving the equal power, and equal A-B times, with significantly different voltages likely requires a different motor to accomplish the current change.

The Ohms law is only true for ohmic system like resistor and wires which is not the case for a brushless motor. If you look up a motor equation you will see that you have to take into account reactive forces like EMF, etc. But if you could apply Ohms law your statement would be true. However, I have stated from the beginning that you cannot increase voltage above your rating without risking permanent damage. If your system is rated for 3S you can run 3S and lower, A 3S system does not run hotter than a 2S if you e.g. use a Traxxas VXL system. Yeah, it will if you do not change your gearing, but that it’s like running any wrong pinion in your 2S car.
The original question is about sizing the ESC, which needs to handle the peak amps involved. Peak amps occur at stall where the more complex aspects of a motor are not terribly significant, and it is mostly resistance. But even at higher RPM where back EMF comes into play, effectively creating complex impedance and reducing current, Ohm's Law still works, just requires complex math to work out and it varies with RPM. Trends remain the same.

Obviously there are other considerations and limits to the system as mentioned.

Obviously, by lowering your voltage you will have less available power,
YES!
and for the old system using NiCd or NiMh batteries with a typical cell voltage of 1.2V that would discharge to around 1.0V would give you a total loss of 1.2V for a six cell pack. That is equal voltage drop for a lipo battery (8.4V -7.2V), so why do we not see this “power loss” from start to end for a brushless system, like we did for the “old” system?
Much of it is due to the greater capacities. With lipo at least in racing cells are frequently not drained very far with a 13.5 in typical race heats. Wasn't always the case with lower capacity nickel cells combined with less efficient brushed motors which used more. A situation where say only 30% of the pack is used compared to 80 or 100% makes a big difference all by itself, regardless of cell chemistry.

To be fair fresh really good nimh cells can come close in discharge flatness, but in practice are very hard to keep that way.

It got me thinking. An ESC is a PMW device (Pulse Width Modulation) and it’s basically using its duty cycle to control the average voltage the motor sees. The longer the cycle, the higher the average voltage which again increase the RPM of your motor. What if the ESC modulates so you never get above 7.2V regardless you battery voltage (8,4V -7,2V). Thus the car will be 100% consistent during your race. Also, that make sense that Novak rate their motors with 7.2V and not 8.4V. But this is just my theory.
Pure hypothetical, that's not the way ESC's work. Although some would like that capability. The 7.2V used for the motor data is likely a remnant of nimh days.



I never stated that voltage did not impact power (well it was not my intention anyway). I don’t know much about 4S versus 6S, but lower KV gives you higher rating (watts) so would it not be a normal thing to increase your voltage so you can decrease your current which give you less voltage drop and lower temperature (less current give you less temperature rise in a motor).
Point remains it takes a motor/load change to accomplish. Perhaps not in the way you seem to be stating, as all else equal a lower KV motor has a lower watt capability at a given voltage. Reference the Novak chart linked earlier.

Last edited by Dave H; 11-03-2014 at 06:02 PM.
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