The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: "Jonathan Bromley"
Subject: Re: Hand Mixer
Date: Tue, 7 Jan 2003 10:14:56 -0000
References: <firstname.lastname@example.org> <3E19A589.email@example.com> <firstname.lastname@example.org> <email@example.com>
Reply-To: "Jonathan Bromley"
NNTP-Posting-Date: Tue, 7 Jan 2003 10:14:57 +0000 (UTC)
X-Newsreader: Microsoft Outlook Express 5.50.4807.1700
"Samatha" wrote in message
> Sorry for not being very clear in my question. Firstly I was talking
> about the Kitchen Appliance - hand mixer. I wanted to know the
> mechanism as to what makes it work. I know that there is a motor
> inside it and know the basic use of the motor. But I wanted to the
> details of how the rotation of the motor depends of voltage/watts
> etc..Anything someone could post about the working of a handmixer
> would be appreciated.
Don't you just hate it when people post simple questions with
impossibly difficult answers?
There's a real problem with asking "how does X work", because
there are so many different possible answers. It would be
correct, but probably unhelpful, to say "plug it in, switch it
on, and insert the beaters into the stuff you want mixed".
It would also be true to say "there's an electric motor that
turns the beaters via some shafts and gears". And in one
sense it's that easy, because everything described in that
sentence - electric motor, gears... - is standard, well-
established technology that you can buy off-the-shelf. So
if you want to make a mixer, you don't need to work out from
scratch how to build an electric motor - you go buy one. But
in the nineteenth century, people wrote huge textbooks about
gears - how to cut the teeth the right shape so they were
as efficient as possible, how to make them run quietly,
and so on. It's not easy, but we *think* it's easy because
the standard recipes for making gears are so well known.
And then there's the question of how an electric motor works.
Once again it's not easy. A few things you might find
helpful to know, though... I've used lots of technical terms
here, but you should be able to plug them into a search
engine to find out more.
* In many small electric appliances, there is a so-called
"universal motor". This means it's a series-wound motor. It
has coils of wire to provide the magnetic field in the fixed
part of the motor, more coils of wire on the rotating part,
and a commutator and brushes to change the direction of
current in the rotating part so that it keeps turning.
Universal motors of this type are a cheap and nasty way
to get a motor working correctly on alternating current,
which otherwise is a bit more tricky.
* Most hand mixers I've come across (and I use lots - I have
an unfortunate fondness for overly rich puddings, which
often start life as egg yolks and sugar beaten together)
have some kind of speed control with three or four settings,
rather than a continuously variable control. This kind of
speed control commonly works by switching-in different
coils on the fixed part of the motor, modifying the
motor's characteristics so it goes faster or slower.
But there are plenty of other schemes that could be used,
including electronic control for continuously variable speed.
* howstuffworks.com is a great website for "obvious"
technology questions like how an electric motor works.
* If someone tells you that electric motors are simple to
understand, don't believe them. Most people I talk with
about motors have a really hazy idea about what's going
on, and much of it is quite subtle.
However, I can at least try to give you a clue about your
> how the rotation of the motor depends of voltage/watts
You probably know the so-called "Ohm's law" (terrible name,
it's not a law at all!) relating the voltage and current in
a resistor (V=I.R). You probably also know that if an
electrical circuit delivers a current I to something, with
a voltage V across that something's terminals, then the
electrical circuit is pushing an amount of power P=V.I into
that "something". (You can't blindly apply these rules
when alternating current is involved, but we'll skip that
for now). One more thing you need to know: Electric
motors and electric generators are pretty much the same
thing. If you spin a motor mechanically, it will generate
a voltage across its terminals. That voltage is larger
if the motor spins faster.
Now let's think about a motor that's spinning under power.
There is current flowing in the motor's windings, and
those windings have some resistance (not much though!)
so you get a voltage drop V(drop) = I.R We also have
some voltage developed by the generator effect in the
motor, related to the speed. This voltage is usually
called the "back EMF" so let's label it V(back).
These two voltages V(back) and V(drop) are effectively
in series so they add together, and the voltage appearing
across the motor's terminals is therefore
V(motor) = V(back) + V(drop) = V(back) + I.R
But hang on a moment... The voltage across the motor's
terminals is exactly the voltage put there by the power
outlet, battery or whatever - it's the supply voltage!
So we can see that the supply voltage is shared between
V(back) and V(drop).
The next useful step is to think about how much power is
being pushed around the place. The power we are taking
from the supply is [V(supply).I] but we already decided
that V(supply)=V(motor) so...
(power consumed) = V(motor).I = [V(back) + I.R].I
= V(back).I + I.I.R
(sorry, have to write I.I instead of I-squared!)
The I.I.R part is simply the power that's wasted in
warming the motor winding resistance. It's lost.
That's why motor makers try to make R as small as
The V(back).I part is the electrical power that's
being converted into mechanical power by the motor.
Some of that mechanical power is of course wasted
in frictional losses, but most of it goes into
stirring your mixture.
Now let's suppose the mixture thickens a bit (eggs
and sugar starting to get nice and fluffy and stiff).
The increased drag will of course slow the rotation
a bit, and therefore V(back) will fall - maybe
proportionally to the speed, maybe not, but it will
fall anyhow. But your power source keeps V(motor)
close to constant. Since V(motor) = V(back) + I.R
that means I.R must increase to make up for the
reduction in V(back). R is constant - the resistance
of the motor windings - so I increases. Hence, a
motor forced to go slower will consume more current
from a given voltage supply. Increased current in
the motor windings leads to stronger magnetic
fields in the motor, and hence more torque. So the
motor is trying to pull the speed back up again.
If R is fairly small, a small reduction in V(back)
will give rise to a large increase in current -
that's what we want, because we don't want the
motor slowing down too much when it meets an
I hope this gives you some kind of mental picture
of what's going on so that you can make good use
of other resources.
Don't ask me about the gears part - I'm only an
electronics guy, I don't understand mechanical stuff :-)
Jonathan Bromley, Consultant
DOULOS - Developing Design Know-how
VHDL * Verilog * SystemC * Perl * Tcl/Tk * Verification * Project Services
Doulos Ltd. Church Hatch, 22 Market Place, Ringwood, Hampshire, BH24 1AW, UK
Tel: +44 (0)1425 471223 mail: firstname.lastname@example.org
Fax: +44 (0)1425 471573 Web: http://www.doulos.com
The contents of this message may contain personal views which
are not the views of Doulos Ltd., unless specifically stated.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup