Battery Questions...Voltage/Amperage/Cells...

[Scott M. Foster]

I just ordered a Traxxas Rustler today. Got (2) 1900mAh - 6 cell
batteries. Question: How much faster will going to a 7 cell 1900mAh
battery bring me? How much run time will I loose? Is going to a 7
cell battery better for me or should I be setting my sights on a
different motor to get more speed? Which motor is a good choice
(price/performance)? I'm very new to this hobbie and want to start
with good information...

Scott

[Justin Mahn]

Adding a cell will not shorten your run-time. Ideally the car would be
16.67% faster, but due to drag and friction, will probably be 10-15% faster.
It will definately be noticeably faster. Requiring 7 cells for all your
packs will probably be more expensive than getting a cheep mod motor to
begin with. Be sure to get a motor that your ESC (I hope you got the ESC
version) can handle. Less turns on the motor requires a more robust ESC.
Your manual will let you know the minimum turns you can install.

--
Justin Mahn
remove Panties to reply


"Scott M. Foster" <scott_m_foster@yahoo.com> wrote in message
news:533ddfdc.0306201444.40dae0ac@posting.google.c om...
> I just ordered a Traxxas Rustler today. Got (2) 1900mAh - 6 cell
> batteries. Question: How much faster will going to a 7 cell 1900mAh
> battery bring me? How much run time will I loose? Is going to a 7
> cell battery better for me or should I be setting my sights on a
> different motor to get more speed? Which motor is a good choice
> (price/performance)? I'm very new to this hobbie and want to start
> with good information...
>
> Scott

[Rick Russell]

In article <vf7ue3fq14s4f0@corp.supernews.com>,
Justin Mahn <gabrilPanties@tcainternet.com> wrote:
> Adding a cell will not shorten your run-time.

It will shorten it. Increasing voltage also increases current draw
(I=V/R).

> Requiring 7 cells for all your
> packs will probably be more expensive than getting a cheep mod motor to
> begin with.

Right. A decent motor will probably be more efficient than the
Traxxas Stinger, resulting in more speed and runtime.

Rick R.

[Jim Schlatter]

"Justin Mahn" <gabrilPanties@tcainternet.com> wrote in
news:vf7ue3fq14s4f0@corp.supernews.com:

> Adding a cell will not shorten your run-time. Ideally the car would
> be 16.67% faster, but due to drag and friction, will probably be
> 10-15% faster. It will definately be noticeably faster. Requiring 7
> cells for all your packs will probably be more expensive than getting
> a cheep mod motor to begin with. Be sure to get a motor that your ESC
> (I hope you got the ESC version) can handle. Less turns on the motor
> requires a more robust ESC. Your manual will let you know the minimum
> turns you can install.
>

Scott
I just upgraded to a Trinity Speed Gem 2 19T double wound motor. Do a
google search for my results that I posted a week or so ago.

For the Rustler, I found that the 18T pinion and 87T spur gear worked
best. With the original 84T spur gear, the batteries ran down a bit
faster, the motor and batteries were hotter, and it didn't go any faster.
I think the new spur gear was ~$3 from Tower. If you decide on the Speed
Gem, you will also need power leads. I got the kind that clip on the
motor spade lugs, but wish I would have gotten solder ones instead. The
clips fell off twice during my first race.



--
================================================== ====
This just in:
- Energizer Bunny arrested, charged with battery.
================================================== ====

[John]

In article <bd20m0$egm$1@joe.rice.edu>, rickr@is.rice.edu says...
> In article <vf7ue3fq14s4f0@corp.supernews.com>,
> Justin Mahn <gabrilPanties@tcainternet.com> wrote:
> > Adding a cell will not shorten your run-time.
>
> It will shorten it. Increasing voltage also increases current draw
> (I=V/R).

I love it when people quote formulas without stating the right
assumptions or just blindly apply it to anything and everything.

You're *wrong*, it will not "increase" the current draw by virtue of
Ohm's law(V=I*R). Ohm's law works for purely RESISTIVE devices(i.e.
resistors).

The motor (the load) appears as an *inductor* in series with a small
resistance, hence you cannot simply apply Ohm's law here!

With RC car setups, the motor(and driving situation) will set the load
current. The load current is what controls the engine torque, while the
applied voltage controls the engine RPM. When the motor is heavily
loaded (i.e. car/truck is going up a hill), the motor will be under load
and draw more current vs moving on a flat piece of road.

ESCs pulse-width modulate the battery voltage, effectively controlling
the engine RPM. Under full-throttle, the ESC (as best as it can) allows
the battery voltage to be fully applied to the motor.

With a 7-cell setup, since the added cell is in series with the pack, it
will increase the net pack voltage and the car can potentially move
faster (vs. 6-cell).

Adding another cell in series will gain him a faster vehicle but will
not increase the total capacity of the battery pack (the number of Amp-
Hours).

So here's the bottom line: The run-time will be slightly lower because
the added cell will add weight but not capacity(Amp-Hour). The vehicle
will now have to move this extra weight around, resulting in a slightly
increased current draw(but not because of "V=IR") vs. a 6-cell setup.

And yes, I'm a little late in the thread, but that's because I've been
out of town for sometime.

Thanks,
TR.

[Rick Russell]

In article <MPG.1973c45731e7556298968b@news.surfcity.net>,
John <ngp1011@yahoo.com> wrote:
> I love it when people quote formulas without stating the right
> assumptions or just blindly apply it to anything and everything.

Well, I wasn't applying it blindly. At full throttle, the motor will
settle into an equilibrium state that is essentially equivalent to a
resistive load.

> You're *wrong*, it will not "increase" the current draw by virtue of
> Ohm's law(V=I*R). Ohm's law works for purely RESISTIVE devices(i.e.
> resistors).

Ohm's law is still valid, and applies to anything with resistance.
Inductors subject to an oscillating voltage (that is, motor armatures)
have resistance. That's what inductors do.

> The motor (the load) appears as an *inductor* in series with a small
> resistance, hence you cannot simply apply Ohm's law here!

Sure you can. For AC circuits, the resistance of an inductor is
related to the frequency of oscillation. Motors are AC devices, since
the current to the inductors constantly being switched back and forth.

I would have to guess that the inductive load in a motor is extremely
complex, since the magnetic fields of the inductor are constantly
twiddled by the permanent magnets. But it's still an inductive load in
an AC circuit.

> With RC car setups, the motor(and driving situation) will set the load
> current. The load current is what controls the engine torque, while the
> applied voltage controls the engine RPM. When the motor is heavily
> loaded (i.e. car/truck is going up a hill), the motor will be under load
> and draw more current vs moving on a flat piece of road.

Agreed. But if you have two identical motors under identical driving
conditions, the motor with the higher voltage will draw more current.

Think about the physics of the motor for a second. If the motor is
going faster, and the car is moving faster... doesn't the motor have
to produce more torque to keep the car moving? It must produce more
torque, to overcome the additional drag. There is your additional
current draw.

> ESCs pulse-width modulate the battery voltage, effectively controlling
> the engine RPM. Under full-throttle, the ESC (as best as it can) allows
> the battery voltage to be fully applied to the motor.

Yes, but that's only relevant at less than full throttle. For full
throttle, the ESC can be considered a closed switch.

> With a 7-cell setup, since the added cell is in series with the pack, it
> will increase the net pack voltage and the car can potentially move
> faster (vs. 6-cell).

Yep.

> Adding another cell in series will gain him a faster vehicle but will
> not increase the total capacity of the battery pack (the number of Amp-
> Hours).

Yep.

> So here's the bottom line: The run-time will be slightly lower because
> the added cell will add weight but not capacity(Amp-Hour). The vehicle
> will now have to move this extra weight around, resulting in a slightly
> increased current draw(but not because of "V=IR") vs. a 6-cell setup.

If all other things remain equal, the current draw increases with
voltage, I'm afraid. To prove it to yourself, do the math on a simple
inductor in an AC circuit.

Rick R.

[John]

In article <begm7v$n7o$1@joe.rice.edu>, rickr@is.rice.edu says...
> In article <MPG.1973c45731e7556298968b@news.surfcity.net>,
> John <ngp1011@> wrote:
> > I love it when people quote formulas without stating the right
> > assumptions or just blindly apply it to anything and everything.
>
> Well, I wasn't applying it blindly. At full throttle, the motor will
> settle into an equilibrium state that is essentially equivalent to a
> resistive load.

No, this is NOT even near correct.

Just for the sake of reference, the only part about this discussion that
it's even under question is in the instance where the output from the 7-
cell pack via the ESC is above the 7.2V from the 6-cell pack. In all
other instances, the 7-cell pack (with ESC's VOUT <=7.2V) looks like a
6-cell pack to the motor.

*Even* at equilibrium, the motor will produce a back emf that's slightly
less than the applied voltage...show me a resistor which does that.

But don't take it from me: I'll quote from: "Electrical Machines, Drives
and Power Systems" 4th Ed, by Theodore Wildi, Chapter 5, Direct-Current
Motors.

Page 96

"In the case of a motor[DC], the induced voltage E0 is called the
counter-electromotive force because its polarity always acts *against*
the source voltage Eg.

He goes on to state:

"The net voltage acting in the armature circuit is (Eg-E0) [where Eg is
applied and Eo is the back-emf]...As the speed increases, the counter-
emf E0 increases, with the result that the value of (Eg-E0) diminishes.
It follows that the armature current I drops progressively as the speed
increases".

This doesn't match the operation of a resistor.

Next paragraph:

"Although the armature current decreases, the motor continues to
accelerate until it reaches a definite, maximum speed. At no-load this
speed produces a counter-emf E0 slightly less than the source voltage
Eg...thus when a motor runs no-load, the counter-emf must be slightly
less than Es, so as to enable a small current to flow, sufficient to
produce the require torque"

So again, even under steady-state conditions your applications of Ohm's
law was incorrect.

There are only a few ways you can use Ohm's law here:

1. Winding loss in the motor. The windings of the motor are made with
wire which are non-ideal, they have resistance(s) and the power loss due
to this is:

I^2*R(winding).

2. You could state that the MAXIMUM current the motor could ever draw is
I=(Eg-E0)/R(winding). This simply means if the motor was put under
maximum load (I.e. clamped in place and unable to move, RPM is at 0),
and Eg (applied voltage) didn't buckle, the load current would shoot to
( Eg - 0(motor is not spinng) ) / R(winding) = Eg/R(winding).

As soon as the motor is let go (no-load or light-load), the RPM would
build-up and thus the back-emf would build-up and the current would
drop.

> > You're *wrong*, it will not "increase" the current draw by virtue of
> > Ohm's law(V=I*R). Ohm's law works for purely RESISTIVE devices(i.e.
> > resistors).
>
> Ohm's law is still valid, and applies to anything with resistance.
> Inductors subject to an oscillating voltage (that is, motor armatures)
> have resistance. That's what inductors do.

The "Ohm's" law part applies to the purely resistive portion of the
circuit, not it's entire operation(which is what you tried to boil it
down to). See the winding losses stated above.

> > The motor (the load) appears as an *inductor* in series with a small
> > resistance, hence you cannot simply apply Ohm's law here!
>
> Sure you can.

Again, you cannot apply Ohm's law in the way you did, the inductive part
of the motor never goes away, it never "becomes" a resistor under steady
state operation, as t --> inf, etc.

Blindly stating the current will be higher because the voltage is higher
is wrong, motors have never behaved like resistors.

> For AC circuits, the resistance of an inductor is
> related to the frequency of oscillation. Motors are AC devices, since
> the current to the inductors constantly being switched back and forth.

What you said is a mishmash of terminology which doesn't directly apply.

AC signals are anything time varying(I.e. not constant for all time,
like DC). The voltage applied to the motor could be constant (DC) if the
ESC is fully on or fully off, or the voltage applied could be AC
(chopped by the ESC). Thus, the current through the motor would more
than likely be AC, though under full-throttle, perfect load, "steady-
state", it could very well be DC.

The motors we are dealing with (RC car motors) are intended to work on
"DC" voltages (though technically we're Pulse-Width Modulating it, but
*ideally* the PWM results in an average Vout that = DutyCycle*Vin),
calling it an "AC" device is not correct(and it's meaningless since you
really haven't defined AC).

And with respect to the *impedance*, yes, for any non-purely resistive
circuit (I.e. a circuit with inductance and capacitance) the impedance
of the circuit or element(s) is frequency dependant, but that's not
"resistance", that's *impedance*.

If we look at the frequency response of the motor with respect to the
square-wave voltages being applied to it, you will find that the motor's
inherent inductance combined with the output capacitance from the ESC
combine to become a 2-pole low-pass filter, removing many harmonics from
the square-wave, resulting in close to the average output voltage
(Vin*Duty_Cycle) being applied to the motor. The additional capacitors
added to to the can suppress high-frequency noise.

The ESC and motor combine to basically build a buck switching power
supply, if we want to get that detailed in the analysis.

> I would have to guess that the inductive load in a motor is extremely
> complex, since the magnetic fields of the inductor are constantly
> twiddled by the permanent magnets. But it's still an inductive load in
> an AC circuit.

Actually it's not too complex, it's easily analyzed and boiled down by
the book.

> > With RC car setups, the motor(and driving situation) will set the load
> > current. The load current is what controls the engine torque, while the
> > applied voltage controls the engine RPM. When the motor is heavily
> > loaded (i.e. car/truck is going up a hill), the motor will be under load
> > and draw more current vs moving on a flat piece of road.
>
> Agreed. But if you have two identical motors under identical driving
> conditions, the motor with the higher voltage will draw more current.

Again, you are wrong. You are again neglecting the fundamental way in
which a motor behaves. I will not provide the derivation, but the book
states: "...the torque produced by the motor is directly proportional to
the armature current":

From the book:

T = (Z*theta*I)/(2pi)

where:

T= torque in Nm
Z= number of conductors on the armature
theta = effective flux per pole [Wb, given by 60*E0/(Z)

Substituting for theta and solving for I that becomes:

I = Torque*pi
-----------
30*E0

Which shows the current will be *inversely* proportional to the applied
voltage(by a factor of 30).

Intuitively, this makes sense:

Assume we have a load that draws 10W regardless of input voltage(and
we'll assume DC voltages and current here). Now say we provide 10V to
that load, this means the load will draw 1A.

Now assume we provide 20V to this load, since it still only wants 10W,
it will only end up needing .5A of current.

Similairly, the RC car with the 7-cell pack will, for the exact same
load conditions(hence same power needs), draws less current than the 6-
cell pack, assuming full power conversion by the ESC. Even when the
output voltage of the ESC goes above 7.2V (the 6-cell limit), the load
the car places on the motor hasn't really changed, thus there is no
reason why *automatically* the load current should increase...it should
actually decrease.

This is the same argument that motivates us to use higher voltages for
in our car (I.e. 42V systems vs. 12V systems). The same load requires
less current at higher voltages (assuming we can convert efficiently,
which we can with switching power supplies, which ESC almost are), thus
the I^2*R losses are less.

An example(DC voltages and currents here):

Car's automotive system provides 12V and the headlamps need 12V and draw
2A of current = 24 W

The wiring losses in the car for the 12V conversion system is = (2A)^2
*R. Arbitrarily, let's make the wiring resistance(from the 12V system to
the headlamps) equal to 1 Ohm, resulting in a wiring loss of 4W.

Now assume we moved the car to a 42V system and assume the headlights
still need 12V/2A.

Assume we have a power converter between the 42V car electronics and the
headlights, and assume this power converter is 95% eff (which is
realizable today) and gives the headlamps the 12V/2A it needs. What
happens?

With 95% eff, the headlamp "circuit" will need 25.26W and will draw
(42V)/(25.26W) = 1.66A from main power.

The wiring losses from the 42V electronic system to the headlamp circuit
is now 2.76W, assuming the same length and type of wire as before.

This is 69% of the power we were previously drawing, or a savings of
31%, simply by moving to a higher voltage.

> Think about the physics of the motor for a second. If the motor is
> going faster, and the car is moving faster... doesn't the motor have
> to produce more torque to keep the car moving? It must produce more
> torque, to overcome the additional drag. There is your additional
> current draw.

The motor will always be under load to keep the car moving, but that is
the case regardless of the number of cells. And as I stated, the extra
cell will add extra weight, thus needing slightly more energy, but not
in the way you're claiming. Again, look at the equation that shows
output current is inversely proportional to the applied voltage.

Your argument is also not self-consistent. If the car is increasing in
speed(moving "faster"), then it is not in steady-state. Steady-state
would occur when the velocity of the car has stabalized. Ramping up to
that steady-state point might require more energy given the extra cell
weight, but I think I've covered that before.

> > ESCs pulse-width modulate the battery voltage, effectively controlling
> > the engine RPM. Under full-throttle, the ESC (as best as it can) allows
> > the battery voltage to be fully applied to the motor.
>
> Yes, but that's only relevant at less than full throttle. For full
> throttle, the ESC can be considered a closed switch.

Full throttle and no-throttle are the two extremes, and full throttle is
really what we need to look at, which is why I mentioned what an ESC
actually does. Technically, you can get the same output voltage from the
two packs under different "operating" conditions:

6-cell, duty cycle 83.3% PWM = ~6V(AVG)
7-cell, duty cycle 71.5% PWM = ~6V (AVG)

So when considering the load current, how long the pack will last, etc,
you have to either talk about WHAT output voltage or you have to talk
about full throttle in both cases. This is what the disclaimer at the
top was getting at.

The 7-cell pack has a higher Watt-Hour rating (Joules) than the 6-cell
pack, "period". If the 7-cell pack were being run below 7.2V, and we
assume the ESC is efficient, then the run-times would actually be longer
compared with the 6-cell pack, given the extra energy the cell provides.
Even at full-throttle, past 7.2V, there is nothing inherently in the 7-
cell setup, aside from the small weight of the extra cell, causing the
run-time to be less.

Rick, do you mind if I ask what your technical qualifications are? I'm
an Electrical Engineer with notable experience designing switching power
supplies, motor controls and other power conversion related circuits.

Thanks,
TR.

[Mike]

Thanks for the education!


"John" <ngp1011@yahoo.com> wrote in message
news:MPG.1974868c3a362cdb98968f@news.surfcity.net. ..
>
> In article <begm7v$n7o$1@joe.rice.edu>, rickr@is.rice.edu says...
> > In article <MPG.1973c45731e7556298968b@news.surfcity.net>,
> > John <ngp1011@> wrote:
> > > I love it when people quote formulas without stating the right
> > > assumptions or just blindly apply it to anything and everything.
> >
> > Well, I wasn't applying it blindly. At full throttle, the motor will
> > settle into an equilibrium state that is essentially equivalent to a
> > resistive load.
>
> No, this is NOT even near correct.
>
> Just for the sake of reference, the only part about this discussion that
> it's even under question is in the instance where the output from the 7-
> cell pack via the ESC is above the 7.2V from the 6-cell pack. In all
> other instances, the 7-cell pack (with ESC's VOUT <=7.2V) looks like a
> 6-cell pack to the motor.
>
> *Even* at equilibrium, the motor will produce a back emf that's slightly
> less than the applied voltage...show me a resistor which does that.
>
> But don't take it from me: I'll quote from: "Electrical Machines, Drives
> and Power Systems" 4th Ed, by Theodore Wildi, Chapter 5, Direct-Current
> Motors.
>
> Page 96
>
> "In the case of a motor[DC], the induced voltage E0 is called the
> counter-electromotive force because its polarity always acts *against*
> the source voltage Eg.
>
> He goes on to state:
>
> "The net voltage acting in the armature circuit is (Eg-E0) [where Eg is
> applied and Eo is the back-emf]...As the speed increases, the counter-
> emf E0 increases, with the result that the value of (Eg-E0) diminishes.
> It follows that the armature current I drops progressively as the speed
> increases".
>
> This doesn't match the operation of a resistor.
>
> Next paragraph:
>
> "Although the armature current decreases, the motor continues to
> accelerate until it reaches a definite, maximum speed. At no-load this
> speed produces a counter-emf E0 slightly less than the source voltage
> Eg...thus when a motor runs no-load, the counter-emf must be slightly
> less than Es, so as to enable a small current to flow, sufficient to
> produce the require torque"
>
> So again, even under steady-state conditions your applications of Ohm's
> law was incorrect.
>
> There are only a few ways you can use Ohm's law here:
>
> 1. Winding loss in the motor. The windings of the motor are made with
> wire which are non-ideal, they have resistance(s) and the power loss due
> to this is:
>
> I^2*R(winding).
>
> 2. You could state that the MAXIMUM current the motor could ever draw is
> I=(Eg-E0)/R(winding). This simply means if the motor was put under
> maximum load (I.e. clamped in place and unable to move, RPM is at 0),
> and Eg (applied voltage) didn't buckle, the load current would shoot to
> ( Eg - 0(motor is not spinng) ) / R(winding) = Eg/R(winding).
>
> As soon as the motor is let go (no-load or light-load), the RPM would
> build-up and thus the back-emf would build-up and the current would
> drop.
>
> > > You're *wrong*, it will not "increase" the current draw by virtue of
> > > Ohm's law(V=I*R). Ohm's law works for purely RESISTIVE devices(i.e.
> > > resistors).
> >
> > Ohm's law is still valid, and applies to anything with resistance.
> > Inductors subject to an oscillating voltage (that is, motor armatures)
> > have resistance. That's what inductors do.
>
> The "Ohm's" law part applies to the purely resistive portion of the
> circuit, not it's entire operation(which is what you tried to boil it
> down to). See the winding losses stated above.
>
> > > The motor (the load) appears as an *inductor* in series with a small
> > > resistance, hence you cannot simply apply Ohm's law here!
> >
> > Sure you can.
>
> Again, you cannot apply Ohm's law in the way you did, the inductive part
> of the motor never goes away, it never "becomes" a resistor under steady
> state operation, as t --> inf, etc.
>
> Blindly stating the current will be higher because the voltage is higher
> is wrong, motors have never behaved like resistors.
>
> > For AC circuits, the resistance of an inductor is
> > related to the frequency of oscillation. Motors are AC devices, since
> > the current to the inductors constantly being switched back and forth.
>
> What you said is a mishmash of terminology which doesn't directly apply.
>
> AC signals are anything time varying(I.e. not constant for all time,
> like DC). The voltage applied to the motor could be constant (DC) if the
> ESC is fully on or fully off, or the voltage applied could be AC
> (chopped by the ESC). Thus, the current through the motor would more
> than likely be AC, though under full-throttle, perfect load, "steady-
> state", it could very well be DC.
>
> The motors we are dealing with (RC car motors) are intended to work on
> "DC" voltages (though technically we're Pulse-Width Modulating it, but
> *ideally* the PWM results in an average Vout that = DutyCycle*Vin),
> calling it an "AC" device is not correct(and it's meaningless since you
> really haven't defined AC).
>
> And with respect to the *impedance*, yes, for any non-purely resistive
> circuit (I.e. a circuit with inductance and capacitance) the impedance
> of the circuit or element(s) is frequency dependant, but that's not
> "resistance", that's *impedance*.
>
> If we look at the frequency response of the motor with respect to the
> square-wave voltages being applied to it, you will find that the motor's
> inherent inductance combined with the output capacitance from the ESC
> combine to become a 2-pole low-pass filter, removing many harmonics from
> the square-wave, resulting in close to the average output voltage
> (Vin*Duty_Cycle) being applied to the motor. The additional capacitors
> added to to the can suppress high-frequency noise.
>
> The ESC and motor combine to basically build a buck switching power
> supply, if we want to get that detailed in the analysis.
>
> > I would have to guess that the inductive load in a motor is extremely
> > complex, since the magnetic fields of the inductor are constantly
> > twiddled by the permanent magnets. But it's still an inductive load in
> > an AC circuit.
>
> Actually it's not too complex, it's easily analyzed and boiled down by
> the book.
>
> > > With RC car setups, the motor(and driving situation) will set the load
> > > current. The load current is what controls the engine torque, while
the
> > > applied voltage controls the engine RPM. When the motor is heavily
> > > loaded (i.e. car/truck is going up a hill), the motor will be under
load
> > > and draw more current vs moving on a flat piece of road.
> >
> > Agreed. But if you have two identical motors under identical driving
> > conditions, the motor with the higher voltage will draw more current.
>
> Again, you are wrong. You are again neglecting the fundamental way in
> which a motor behaves. I will not provide the derivation, but the book
> states: "...the torque produced by the motor is directly proportional to
> the armature current":
>
> From the book:
>
> T = (Z*theta*I)/(2pi)
>
> where:
>
> T= torque in Nm
> Z= number of conductors on the armature
> theta = effective flux per pole [Wb, given by 60*E0/(Z)
>
> Substituting for theta and solving for I that becomes:
>
> I = Torque*pi
> -----------
> 30*E0
>
> Which shows the current will be *inversely* proportional to the applied
> voltage(by a factor of 30).
>
> Intuitively, this makes sense:
>
> Assume we have a load that draws 10W regardless of input voltage(and
> we'll assume DC voltages and current here). Now say we provide 10V to
> that load, this means the load will draw 1A.
>
> Now assume we provide 20V to this load, since it still only wants 10W,
> it will only end up needing .5A of current.
>
> Similairly, the RC car with the 7-cell pack will, for the exact same
> load conditions(hence same power needs), draws less current than the 6-
> cell pack, assuming full power conversion by the ESC. Even when the
> output voltage of the ESC goes above 7.2V (the 6-cell limit), the load
> the car places on the motor hasn't really changed, thus there is no
> reason why *automatically* the load current should increase...it should
> actually decrease.
>
> This is the same argument that motivates us to use higher voltages for
> in our car (I.e. 42V systems vs. 12V systems). The same load requires
> less current at higher voltages (assuming we can convert efficiently,
> which we can with switching power supplies, which ESC almost are), thus
> the I^2*R losses are less.
>
> An example(DC voltages and currents here):
>
> Car's automotive system provides 12V and the headlamps need 12V and draw
> 2A of current = 24 W
>
> The wiring losses in the car for the 12V conversion system is = (2A)^2
> *R. Arbitrarily, let's make the wiring resistance(from the 12V system to
> the headlamps) equal to 1 Ohm, resulting in a wiring loss of 4W.
>
> Now assume we moved the car to a 42V system and assume the headlights
> still need 12V/2A.
>
> Assume we have a power converter between the 42V car electronics and the
> headlights, and assume this power converter is 95% eff (which is
> realizable today) and gives the headlamps the 12V/2A it needs. What
> happens?
>
> With 95% eff, the headlamp "circuit" will need 25.26W and will draw
> (42V)/(25.26W) = 1.66A from main power.
>
> The wiring losses from the 42V electronic system to the headlamp circuit
> is now 2.76W, assuming the same length and type of wire as before.
>
> This is 69% of the power we were previously drawing, or a savings of
> 31%, simply by moving to a higher voltage.
>
> > Think about the physics of the motor for a second. If the motor is
> > going faster, and the car is moving faster... doesn't the motor have
> > to produce more torque to keep the car moving? It must produce more
> > torque, to overcome the additional drag. There is your additional
> > current draw.
>
> The motor will always be under load to keep the car moving, but that is
> the case regardless of the number of cells. And as I stated, the extra
> cell will add extra weight, thus needing slightly more energy, but not
> in the way you're claiming. Again, look at the equation that shows
> output current is inversely proportional to the applied voltage.
>
> Your argument is also not self-consistent. If the car is increasing in
> speed(moving "faster"), then it is not in steady-state. Steady-state
> would occur when the velocity of the car has stabalized. Ramping up to
> that steady-state point might require more energy given the extra cell
> weight, but I think I've covered that before.
>
> > > ESCs pulse-width modulate the battery voltage, effectively controlling
> > > the engine RPM. Under full-throttle, the ESC (as best as it can)
allows
> > > the battery voltage to be fully applied to the motor.
> >
> > Yes, but that's only relevant at less than full throttle. For full
> > throttle, the ESC can be considered a closed switch.
>
> Full throttle and no-throttle are the two extremes, and full throttle is
> really what we need to look at, which is why I mentioned what an ESC
> actually does. Technically, you can get the same output voltage from the
> two packs under different "operating" conditions:
>
> 6-cell, duty cycle 83.3% PWM = ~6V(AVG)
> 7-cell, duty cycle 71.5% PWM = ~6V (AVG)
>
> So when considering the load current, how long the pack will last, etc,
> you have to either talk about WHAT output voltage or you have to talk
> about full throttle in both cases. This is what the disclaimer at the
> top was getting at.
>
> The 7-cell pack has a higher Watt-Hour rating (Joules) than the 6-cell
> pack, "period". If the 7-cell pack were being run below 7.2V, and we
> assume the ESC is efficient, then the run-times would actually be longer
> compared with the 6-cell pack, given the extra energy the cell provides.
> Even at full-throttle, past 7.2V, there is nothing inherently in the 7-
> cell setup, aside from the small weight of the extra cell, causing the
> run-time to be less.
>
> Rick, do you mind if I ask what your technical qualifications are? I'm
> an Electrical Engineer with notable experience designing switching power
> supplies, motor controls and other power conversion related circuits.
>
> Thanks,
> TR.
>

[SD]

OK, and now for my contribution:

At equal ESC duty cycles (throttle positions) the 7 cell battery will be
supplying MORE current than the 6 cell battery. If throttle position is
varied to produce equal power output from the motor, then the 7 cell battery
will supplying LESS current than the 6 cell battery.

In the full throttle condition, increasing voltage will increase current to
the motor at EVERY operating point. It's not a buck converter at full
throttle.

-SD (BSEE)

[John]

In article <-7KcnevDIcUB3ISiXTWJhQ@comcast.com>, user@network.net
says...
> OK, and now for my contribution:
>
> At equal ESC duty cycles (throttle positions) the 7 cell battery will be
> supplying MORE current than the 6 cell battery. If throttle position is
> varied to produce equal power output from the motor, then the 7 cell battery
> will supplying LESS current than the 6 cell battery.
>

Actually, this was one of the cruxes of the argument. The fact of the
matter is that this RC car setup essentially a series circuit, with the
battery being the source of the power, the ESC chopping *the voltage*
and the motor being the load. All three in series, which means the
current that all three see will be the same (and time varying).

The ESC is effectively a buck converter REGARDLESS of duty cycle! If you
look at equations for a buck converter (Vout = Duty_Cycle * Vin), you'll
notice there is no statement of current. Iout in a buck/boost converter
is whatever the load draws.

Inductance in the motor and output capacitance will supply current in
the "off-time" of the ESC (ton+toff = 1 period for the PWM, I'm talking
toff here). Since the motor has mechanical inertia, it also serves to
further low-pass filter the applied voltage(which is already low-pass
filtered by the output capacitance and inductance).

> In the full throttle condition, increasing voltage will increase current to
> the motor at EVERY operating point. It's not a buck converter at full
> throttle.

At full throttle, the ESC is just a resistor (effectively), meaning the
full battrey voltage is applied to the motor. It means that Vout is just
a small resistive drop away from Vin. If we run a buck converter at 100%
duty cycle, we get the same thing(Vout is a small IR drop from Vin) so
being at full-throttle doesn't make our ESC any less of a buck
converter.

The load (in this case, the motor) **sets** the load current. By
applying a larger voltage (higher duty cycle) the motor RPM goes up. If
you have the motor running with no load, full voltage applied (highest
RPM), the current draw will be at some mininum(proportional to the
friction torque of the motor).

If you suddenly clamp that motor or put a heavy load on it, the current
draw will shoot up. The applied voltage to that motor will still be the
same as it was BEFORE you put the load on the motor, but because the
load has increased, the motor *requires* more current to exert a torque.

In a previous e-mail I typed equations for a DC motor from a book on
motors(third-party, you can get the book and look up the equations for
yourself if you want).

The equations showed that the motor RPM is controlled by the voltage and
the *current draw* is *proportional* to the motor *torque output*. That
means that if the motor is not exerting much torque (if the motor is NOT
driving a heavy load) it will draw less current than if a load is
attached REGARDLESS of the "fixed" applied voltage.

Again, having Vout=Vin doesn't change the motor actually setting the
load current. Again, we are *not* dealing with a simple resistive load.
The motor is a "complex" load, analyzing it, how it draws current and
what voltage does to it requires knowledge of it's construction
(windings), mechanics and magnetics (how much flux, etc).

Rick tried to use the same simply analysis as you are trying(Ohm's law)
stating that because Vout is higher, Iout is automatically higher. A
motor doesn't follow Ohm's law (aside from the resistive losses in the
windings), Iout is a function of motor torque and vice-versa.

One final note: I'm glancing at a data sheet for a DC motor and I have a
"Torque Constant": 1.2 oz-in/Amp for this particular motor . That means
every 1.2 oz-in of torque will require a current draw of 1A. If I need 6
oz-in of torque for a load, it will need 5A of current. As you can see,
current drawn is proportional to the load / required torque, not a
function of voltage.

I am also given a "velocity constant" of 1,123 RPM/Volt. It means if I
want to spin the motor at 3600 RPM, I'll need an output voltage of 3.2V
to get my desired RPM.

It's an informal proof (I went through the formal one with the equations
from the books), but you can see the that load current depends on output
torque and output RPM depends on an applied voltage.

With the above motor, if I needed an output torque of 6 oz-in (5A
current draw) and 3600 RPM (3.2V applied voltage), assuming RMS currents
& voltages, I would need P = V(rms) * I(rms) = 16W of power.

All of this may seem counter-intuitive, but it's true. If you don't
believe me, hook up a motor to a voltage source, put a resistor in
series(to measure the load current), use an oscilliscope and SEE how the
current changes with load (vs. time). You'll start to see everything I
am saying is true.

John.

[John]

Hi there,

In article <PuCdnXM-JYipSYeiXTWJiA@comcast.com>, user@network.net
says...
>

> That's true for Ton but the motor has current flowing though it during at
> least part of Toff and the battery doesn't.

I thought that was implied given the ESC operation, but it's true
regardless.

>
> > The ESC is effectively a buck converter REGARDLESS of duty cycle!
>
> At 100% duty cycle no current flows though the free wheeling diode, there's
> no switching going on. It's been reduced to a FET resistor at that point.

Right, it's not bucking at 100%, I didn't quite follow when you said
it's not a buck converter at 100%. I now realize you just wanted to
simplify it down to resistive losses in the FETs at 100%, which is a
perfectly sane simplification.


> > If you
> > look at equations for a buck converter (Vout = Duty_Cycle * Vin), you'll
> > notice there is no statement of current. Iout in a buck/boost converter
> > is whatever the load draws.
>
> There's no statement of current because you didn't state it. Iout/Iin =
> Vin/Vout = 1/D (page 167)

Page 167 of what book? That equation would only be true for a resistor
as the load. A buck converter doesn't FORCE a current to flow unless the
control-loop is configured to do that (but ESCs are effectively open-
loop and most switching power supplies are configured to stabalize
output voltage, not output current).

You could have a buck converter, with zero load and the output voltage
would still be Vout = Vin*Duty , with I0ut = 0 (aside from the small
currents flowing due to the ESR of the caps). Just as you could have a
linear power supply with no load, thus IOut there is close to 0(we have
some quiescent current, but I think you know what I mean).

Given how the load sets the current, I don't really follow your page 167
equation, it must be for some very specific operating conditions/load.

Note: I'm disregarding the ripple current which will be factor of
switching frequency, inductance, etc. The ripple current will (ideally)
be a triangluar variation riding on top of the IOut "DC" value, but it
doesn't even matter in our discussion.


> Why does page 332 of my book say armature current of a DC motor is:
> Ia = (Vt - Eg) / Ra? There is a relationship between voltage and
> current, even if it isn't seen in the motor constants.

You didn't state verbatim what your book shows and technically my book
has that equation...but that defines the MAXIMUM current that could
flow, i.e. you hold the armature in place while Vt is applied, what will
Ia be. That Ia is the no-load current, not loaded.

I'm sure as you leafed further along you'd see how the author relates
the current to torque and the RPM to voltage. Current then becomes a
function of torque. Note, back-EMF is also involved in the equation
which is a function of RPM, thus the current draw is not simply a factor
of the applied voltage.

> In real loads, increasing RPM requires the motor to provide more torque.
> Show me an RC application where increasing RPM will not result in increased
> torque. I think Rick said this before, more voltage means more RPM, more
> RPM means more torque, more torque means more current. Thus more voltage
> means more current.

I'm not saying there won't be a transient where the current draw
increases, that will of course be the case as you try to increase the
RPM with some fixed load. To gain those RPM it will indeed need more
torque to do that.

But I say when the car reaches steady-state (with full throttle
applied), the car will reach some "terminal" velocity. The question is,
with this steady-state velocity, will the load current be higher than at
a previous 6-cell steady-state operating point? I do concede it might be
slightly higher due to the drag and friction Rick mentions, but our
battery pack has more capacity, so I don't think the run-time will be
compromised by much (1-3 mins maybe).

What happens actually in steady-state depends heavily on the motor. My
own profiling showed non-linearities in particular DC motors I am
working with(at the moment).

Note: If we are talking energy here, yes the car is consuming more
energy at a faster speed since Ke = .5 m*V^2. But that's a side issue
given that a 7-cell pack also HAS more energy than the 6-cell pack and
we know faster speed == more kinetic energy == more electrical energy.

> Is that the whole point of all this?
> That an RC car will draw more current from a 7 cell battery than a 6 cell battery?

The original question is what will happen to the run-time if a user went
from a 6-cell setup, to a 7-cell.

I reason, due to non-linearities of the motor, (Kv/Kt are not really
constants, there is a particualr point at which the motor is most
efficient) all of which must be taken into consideration it's difficult
to provide a pure yes or no, only a maybe.

My gut tells me the run-time won't be affected too heavily unless the
motor is particularly inefficient at those high RPMs (which could
possibly be the case), but that isn't the same reason you or Rick feels
the run-time will be shorter.

Thanks,
John.

[Rick Russell]

In article <MPG.198582ff7d78817b9896aa@news.surfcity.net>,
John <ngp1011@yahoo.com> wrote:
> But I say when the car reaches steady-state (with full throttle
> applied), the car will reach some "terminal" velocity. The question
> is, with this steady-state velocity, will the load current be higher
> than at a previous 6-cell steady-state operating point? I do concede
> it might be slightly higher due to the drag and friction Rick
> mentions, but our battery pack has more capacity, so I don't think
> the run-time will be compromised by much (1-3 mins maybe).

1-3 minutes might be a lot if the 6-cell runtime isn't very much.

In any case, I suggest that the "drag and friction" are not
slight. The drag and friction are what determine the torque on the
motor, in a very direct way. And torque determines current.

I readily agree that drag and friction are non-linear, but they
definitely go up with increasing velocity.

Rick R.

[SD]

"John" <ngp1011@yahoo.com> wrote in message
news:MPG.198582ff7d78817b9896aa@news.surfcity.net. ..
> Hi there,
>
> In article <PuCdnXM-JYipSYeiXTWJiA@comcast.com>, user@network.net
> says...
> >
>
> > That's true for Ton but the motor has current flowing though it during
at
> > least part of Toff and the battery doesn't.
>
> I thought that was implied given the ESC operation, but it's true
> regardless.

I mentioned it because you said:
"...all three in series, which means the current that all three see will be
the same (and time varying)."

> > > The ESC is effectively a buck converter REGARDLESS of duty cycle!
> >
> > At 100% duty cycle no current flows though the free wheeling diode,
there's
> > no switching going on. It's been reduced to a FET resistor at that
point.
>
> Right, it's not bucking at 100%, I didn't quite follow when you said
> it's not a buck converter at 100%. I now realize you just wanted to
> simplify it down to resistive losses in the FETs at 100%, which is a
> perfectly sane simplification.
>
>
> > > If you
> > > look at equations for a buck converter (Vout = Duty_Cycle * Vin),
you'll
> > > notice there is no statement of current. Iout in a buck/boost
converter
> > > is whatever the load draws.
> >
> > There's no statement of current because you didn't state it. Iout/Iin =
> > Vin/Vout = 1/D (page 167)
>
> Page 167 of what book? That equation would only be true for a resistor
> as the load. A buck converter doesn't FORCE a current to flow unless the
> control-loop is configured to do that (but ESCs are effectively open-
> loop and most switching power supplies are configured to stabalize
> output voltage, not output current).

Power Electronics 2nd ed. Mohan, Underland, Robbins.
Iout/Iin = Vin/Vout = 1/D does not imply forcing of anything.

> You could have a buck converter, with zero load and the output voltage
> would still be Vout = Vin*Duty , with I0ut = 0 (aside from the small
> currents flowing due to the ESR of the caps). Just as you could have a
> linear power supply with no load, thus IOut there is close to 0(we have
> some quiescent current, but I think you know what I mean).
>
> Given how the load sets the current, I don't really follow your page 167
> equation, it must be for some very specific operating conditions/load.

It's not. It's a general equation where power losses are neglected.

> Note: I'm disregarding the ripple current which will be factor of
> switching frequency, inductance, etc. The ripple current will (ideally)
> be a triangluar variation riding on top of the IOut "DC" value, but it
> doesn't even matter in our discussion.
>
>
> > Why does page 332 of my book say armature current of a DC motor is:
> > Ia = (Vt - Eg) / Ra? There is a relationship between voltage and
> > current, even if it isn't seen in the motor constants.
>
> You didn't state verbatim what your book shows and technically my book
> has that equation...but that defines the MAXIMUM current that could
> flow, i.e. you hold the armature in place while Vt is applied, what will
> Ia be. That Ia is the no-load current, not loaded.

I've got two books with that equation. The other is Electrical Machines and
Transformers 2ed. George McPherson, Robert D. Laramore. The equation is for
steady state DC motors with constant field flux. By the way, if your book
calls the locked rotor current the no load current, then you need to burn
it.

If you hold your armature still, the Eg term drops out. Ia = Vt / Ra


> I'm sure as you leafed further along you'd see how the author relates
> the current to torque and the RPM to voltage. Current then becomes a
> function of torque. Note, back-EMF is also involved in the equation
> which is a function of RPM, thus the current draw is not simply a factor
> of the applied voltage.

Are you now admiting current draw is affected by voltage? FWIW I never said
current was only a function of voltage.

> > In real loads, increasing RPM requires the motor to provide more torque.
> > Show me an RC application where increasing RPM will not result in
increased
> > torque. I think Rick said this before, more voltage means more RPM,
more
> > RPM means more torque, more torque means more current. Thus more
voltage
> > means more current.
>
> I'm not saying there won't be a transient where the current draw
> increases, that will of course be the case as you try to increase the
> RPM with some fixed load. To gain those RPM it will indeed need more
> torque to do that.
>
> But I say when the car reaches steady-state (with full throttle
> applied), the car will reach some "terminal" velocity. The question is,
> with this steady-state velocity, will the load current be higher than at
> a previous 6-cell steady-state operating point? I do concede it might be
> slightly higher due to the drag and friction Rick mentions, but our
> battery pack has more capacity, so I don't think the run-time will be
> compromised by much (1-3 mins maybe).

It will be more than slightly higher, at least 20%. And adding cells in
series does not change the capacity (mAh).

> What happens actually in steady-state depends heavily on the motor. My
> own profiling showed non-linearities in particular DC motors I am
> working with(at the moment).

What is it that you do exactly? Do you have any RC cars?

> Note: If we are talking energy here, yes the car is consuming more
> energy at a faster speed since Ke = .5 m*V^2. But that's a side issue
> given that a 7-cell pack also HAS more energy than the 6-cell pack and
> we know faster speed == more kinetic energy == more electrical energy.

Ke = .5 m*V^2 tells us nothing about the rate of energy consumption.

> > Is that the whole point of all this?
> > That an RC car will draw more current from a 7 cell battery than a 6
cell battery?
>
> The original question is what will happen to the run-time if a user went
> from a 6-cell setup, to a 7-cell.
>
> I reason, due to non-linearities of the motor, (Kv/Kt are not really
> constants, there is a particualr point at which the motor is most
> efficient) all of which must be taken into consideration it's difficult
> to provide a pure yes or no, only a maybe.
>
> My gut tells me the run-time won't be affected too heavily unless the
> motor is particularly inefficient at those high RPMs (which could
> possibly be the case), but that isn't the same reason you or Rick feels
> the run-time will be shorter.

Even with an ideal motor that held perfectly to it's Kv and Kt constants,
you'd still see a noticeable reduction in run time if the car ran full
throttle the whole time.

-SD

[John]

In article <PuCdnXM-JYipSYeiXTWJiA@comcast.com>, user@network.net
says...

Ok, a quick follow up...

> There's no statement of current because you didn't state it. Iout/Iin =
> Vin/Vout = 1/D (page 167)

I figured out what you or your book is trying to say. It is doing the
same as your motor book was, giving an *upper bound* on the output
current and output voltage. In this case your book is using PowerIn &
PowerOut as the relation.

Your equation reduces to: Vin * Iin = Vout * Iout.

which is the same as saying

Power_In = Power_Out. (in reality we know it will be Power_Out <=
Power_In, but that's another issue).

Your equation simply means we cannot create power, that the output power
(Vout*Iout) is equal to (in the case of 100% effieciency) Vin*Iin.

My point however is that you have *NO* equation that tells you what IOut
is, *purely* as a function of Vout(you don't have one because it doesn't
exist). If I have a buck converter with 12Vin, 5V out, what is Iout? You
can't state what Iout is unless there's a load(that you know) or you're
told what Iin is(again implying a load setting Iout). You can tell me
that Iin will be less than Iout using your above relation, but that
really buys you nothing.

This was the same point I was trying to make when you were saying more V
= more I for the ESC. Given a resitor increasing output voltage will
certainly mean the output current goes up, but in reality the *load sets
the current*, no difference with an ESC and a motor, except the motor
doesn't behave as a resistor, meaning the current is time varying and
non-linear. Just because you shove V up for the motor doesn't
automatically mean I goes up(and stays up), unless you are dealing with
a resistor...

Iin is irrelevant, we know given fixed Vin, Vout and Iout, that Iin will
be proportionally less than Iout. If that weren't the case there would
be no point to using switching power supplies, they wouldn't be
efficient.

I just wanted to clear up that the above relation isn't anything new (or
helpful for our discussion).

John.