CONFUSED as HELL!!!!!!!!!!

UnclePete

New Member
Can someone plz explain WHY the amp goes up when the voltage goes down?
I really don't understand, because in RC on road cars, there is "classes" according to lipo cells being used.
For example, 6s class, 8s class, open class, etc.
So obviously the higher cell class pulls more power, and are quicker!
Which means they are using MORE voltage. I'm just really confused. Can anyone shed some light on this
 

Derek

Well-Known Member
You will probably get a better explained answer from others on the forum, but I'll take a stab at it.

As you increase your throttle, the RC (car, heli, plane, boat, etc) will need more power and more power to reach top speed. Power is amps. Therefore, the faster you want to go, the more amps you are using to get there. However, to get more amps (more power) you need to suck the voltage from your battery(s). More power from the motor (amps going up) means more voltage being pulled from the battery (voltage going down).

I think that's right. I'm sure someone will come along and either confirm what I've said to correct me. It happens.

- - - Updated - - -

Here is another way that I just thought about it.

An automobile reference...Amps is horsepower of the engine. Voltage is the fuel in your fuel tank. When you stomp the gas pedal to the floor, getting maximum high horsepower (amps) you are using a lot of fuel from your gas tank. You can stomp the pedal down and get high horsepower (amps) but the gas in the tank is running out (voltage).

I probably just "echoed" what I said above, but I'm just trying to help.
 

coolgabsi

Super Mod & DEAL KING!
Power is VxI (voltage x current)


To get the same power lets say 5000W to be delivered to your tires (about 7hp approx.. May be a bit less)

5000= 50V x 100A
5000=25V x 200A
5000= 10V x 500A


And so on... So when voltage comes down... Your system needs more current to make the same power .

So a 12s pack will need to create only 100a burst to give you the power of 7hp on the tires

While a 6s pack will need to create a 200a burst.
:)


Derek:

And power is not amps . Power is watts.. I wanted to clear it up in your analogy... Horsepower is not amps either . It's power :)

Fly safe!! :) have fun!
 
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Derek

Well-Known Member
well now...that's what i get for trying to help, lmao. Sorry about that! I think I'll just get a shower and head to bed now, lol.
 

Lee

Well-Known Member
Here an analogy.
Two cars doing 100mph
A Ferrari (high voltage)
A Ford Focus (low voltage)
Which one is working harder to keep the car doing 100mph?
The lower the Voltage, the more current is needed to get the same performance.
 

pvolcko

Well-Known Member
Higher voltage for the same power, as explained, will require less current. This is easier on power systems, from batteries and wires to ESCs and motors. High current requires larger gauge wire, larger, more expensive electrical components in ESCs, etc. However, higher voltage has some drawbacks. It does require higher cell counts, which can lead to battery sizing and wiring issues. It also has increased risk of arching. Typically needs more windings in motors with lighter gauge wire, which can create stator space problems and heat dissipation problems in higher power applications. All else being equal, voltage oriented motors run higher RPMs to achieve power, whereas higher current motors will not turn as fast but will have more torque in order to deliver mechanical power.

Since higher voltage is higher RPM, generally, it means the bearings have to be higher quality, mechanical dynamic balancing is more important, and overall the machine will need to be very well tuned to achieve and utilize at maximum performance. High current, lower voltage will not need to be as high strung, but the parts will need to be beefier to deal with the torque available which leads to a different balancing act with costs.

As for the car analogy. I might tweak it a bit.

Ferrari = high voltage = light weight = high reving to deliver power = precision engineering in the drivetrain, less challenging on handling
Mustang Shelby = low voltage = massive = still high power but through gobs of lower reving torque = less precision in the drivetrain (but still a challenge due to torque handling requirements), but more challenges on handling
 
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RandyDSok

Well-Known Member
Looking at it mathematically may also help...

watts = volts x amps


To me, this is the simpler way of looking at it.... The watts ( power ) is the result of the voltage times the amperage. Since the V and A are on the same side of the equation, one will always go up when the other goes down to maintain the same power... or visa versa one will go down when the other goes up.
 

UnclePete

New Member
Thanks for everyone's reply.
You guys made it very clear and easy to understand what's going on.
Appreciate the help.
 

Derek

Well-Known Member
I'm sorry that my comment wasn't helpful, but the cool thing is that we both learned something.
 
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