john c. polasek

The ads for hand drills push higher and higher voltages at higher and
higher prices but I fail to see the logic. What can be done at 24V can
be done at 9.6, just use up the battery space and size the motor
windings properly.
High volt batteries have many more cells and many solder joints. A
24v battery has 20 cells and 21 joints; a 9.6V has 9 joints (for Nicad
at 1.2V/cell). Further more there are than many more metal walls to go
bad.
High voltage only has an advantage when there are long runs of wire or
the load cannot be adjusted, but there are none that I can see here.
Are the manufacturers putting something over on what they perceive to
be stupid customers?

John Polasek

dave

sure, its just like marketing vacuum cleaners where the higher the current
on the motor the better it is.

rp

Higher voltage drills have more cells, thus providing longer use between
chargings. That is, more cells means more total stored energy.
The greater power of the higher voltage drills is a result of the fact
that if you load the lower voltage pack to the same extent, then the
effiiciency will be greatly reduced. This in turn is due to the physics
of batteries; higher load equals lower efficiency. The reason for this
is that the battery pack has an internal resistance. For maximum
efficiency of energy transfer the external resistance should be much
higher than the internal battery resistance. Thus you basically want to
keep the ratio above a certain level, regardless of voltage, that is the
ratio targets a minimum desireable efficiency. Higher voltage at the
same efficiency means greater torque, hp, and length of use.


Richard Perry

john c. polasek

Poppycock. The intrinsic energy storage of Nicads joules/volume is a
constant.

And a 24V needs 20 cells whose probability of failing is twice that
for a 12v battery. I've had a number of such tools and when I throw
them away it's always because the $20 or $45 battery has flunked, and
it's always 1 cell (how could it be 2?).

If I have a 24v motor with 100 turns carrying 1 amp, I can duplicate
what's going on with a 12 volt battery.

Just take 50 of the turns and solder them in parallel with the other
50 and the 12v will deliver exactly the same flux and power,

The 24v power is 24^2/R equal to 12^2/(R/4). The original winding
resistance is cut by 4 in parlleling the 2 50 turn portions.

No there's more to this subterfuge.

John Polasek

dave

nothing fancy, its just 'bigger is better'.

pi

It is true that the energy storage in Joules/volume is constant. But
there is another factor to consider:

Lets suppose two drill handtools. If they really were resistive this
would give (both tools are of the same power, lighly loaded 48 Watt):
Tool1: 12V, so I = 48/12 = 4 A, R seems to be 12/4 = 3 Ohm
Tool2: 24V, so I = 48/24 = 2 A, R seems to be 24/2 = 12 Ohm

BUT: a motor does not act as resistor. In fact the resistance of the
windings is very low. For example 0.5 Ohm (measure it if you do not
believe this). So whats happening that makes it look like a resistor?
The answer is that the motor is a the same time a generator, dynamo.
When you do not load the motor heavily, the motor generates an EMF
(electrical feedback). A DC motor with magnets can actually be used as
generator. This feedback voltage depends on the rotation-velocity.

So the 12V drill at 48 Watt load generates 12V-4A*0.5 Ohm (resistive
losses) = 12-2 = 10 Volt. 2 Volts are lost in heat!! When you load it
heavily where this drill almost stops (so no generatorfunction), the
current is 12V/0.5 Ohm = 24 A. This gives 24A*12V = 288 Watt max peak
power (Try it, but don't do this too long at home, burns out the
drill). Remember now that the torque = current * windings.

The 24V drill at 48 Watt has because it has twice the windings also
twice the internal resistance, so 0.5*2 = 1 Ohm. In our example at a
48 Watt load the gerator delivers 24 - 2A*1 Ohm = 24 - 2 = 22 Volt.
Notice that at light load the losses are the same. But now we have to
drill a concrete wall where the drill almost comes to a stop. The
generator voltage is near to zero. The current is 24V / 1 Ohm = 24 A.
But the power is 24V*24A = 576 Watt!!!!!. Twice the power!The current
is the same as in the 12V model, but twice the windings, so twice the
torque (Note that the real technician is not interested in Watts, but
wants the torque at heavy load).

So twice the power (I know, torque) at full load where you are in fact
using the device a bit out of spec. When you load it not too heavily,
the battery will last the same time in both devices as the same power
(I know, Joules) is used. Only at heavy load the higher voltage model
gets an advantage. But only when they have the same quality of motor
(it is a bit tempting for the manufacterer to use a somewhat cheaper
motor in the 48V devices, degrades performance not so much as in the
12V device). And of course the same gearbox.

Do not try this at home, or make at least a video of it and post that
to give us a good laugh.

Pieter
email): pieter@hoeben.com
(yes I have a good spamfilter)
http://www.hoeben-electronics.com

rp

He was suggesting splitting the 24v motors windings in two, wiring them
in parallel and sending 12v to each. This would provide, as he says,
exactly the same performance with 12v as with 24v.

The advantage of the 24 over the 12 is that the 24 *usually* has twice
as many cells, thus twice the staying power. If OTOH both packs were
identical except for the parallel wiring in the 12v version, he is
correct, a motor could easily be manufactured to equal the 24v version
by simply paralleling the winding. OTOH, there is still this advantage
of the higher voltage version, that the wire size, and any other control
components, can be halved in ampere rating with the 24v version, thus
saving a few dimes on every unit, while also eliminating IR losses in
the delivery circuit. So IOW, by Polasek's reasoning, we shouldn't need
high voltage cross country transmission lines, the same power can be
delivered to the homes at a much lower voltage. Sure, and even if this
were magically correct what would the advantage of a lower voltage drill
be? He was just pissed because his battery packs keep crapping out on
him. He probably doesn't know that you aren't supposed to over-torque
the drill or battery damage can occur

hvacrmedic

pi

When very, very, very thick cables are used, this is true...


As you can see in my estimation of powers of the drill, a factor 10
overload may be possible. However, good batteries can handle this for
a shorter time period.

Here is why the batteries wear out so quickly: most people want them
to load fast. The batteries can handle short discharge peaks. There is
recovery time between the drilling (or whatever you do) for the
batteries. But when you charge them fast, most equipment sends a
current through them as high as possible, at a high temperature.

Good batteries that can be charged/discharged 1000 times can be
reduced to 50 times this way (wearing out means here a 50% or worse
capacity reduction)! There several tests done in this area. THose
batteries will wear out quickly. Only special batteries (as used in
airplane models etc) can handle it somewhat better.

So all the manufacterer has to do is add a switch on the charger:
fast-slow. But they do now want to pay for that. And if the batteries
wear out, you have to buy new ones....so no switch.

Pieter

john c. polasek

Not so fast my friend. In my 12V drill with your .5ohms, I choose to
put 2 1 ohm windings in parallel, making .5 ohms.
Now I can make a 24V drill by putting the same two windings in series
so the winding resistance is now 2 ohms not 1 ohm.
So the volts per winding is 12 in both cases and the motor does not
know the difference, 12 or 24.


Not at all. It's 24V/2ohms and the same power.


John Polasek

john c. polasek

Beg to differ. A 24 volt nicad has 20 cells:
UUUUUUUUUUUUUUUUUUUU
That's 20 cells and 38 walls (vs 10 & 18) that not only take up
space, but have to be soldered together. The probability of failure is
twice as high. There is in fact a loss in "staying power" just due to
the extra walls.


Forget wire size. The only advantage of lower current is in the
(admittedly flakey) battery contacts and switch surfaces.


I think you mean by YOUR reasoning (or whatever process you use to
come up with for example "twice the staying power").


John Polasek
John Polasek
http://www.dualspace.net