From: Mark on

>
> Approximately 10% losses occur in the transformer, bridge rectifier and
> darlington. This could be reduced by using a higher voltage secondary and a
> MOSFET when the losses could be 5% or lower. The control electronics side
> is isolated by the transformer so you are working with low voltage DC.
>
>
OK, I can see this working well for nearly off and nearly full on, but
when you are at say 1/2 power, doesn't the darlington on the secondary
side dissipate a lot of power?

It seems to me it is the equivalent of putting a resistor in series
with the load, its just that the resistor value is transformed by the
turns ratio and the rectifier makes it DC instead of AC, but you are
still basically putting a resistor (not an inductor) in series with
the load...

no?

Mark
From: George Herold on
On Jul 13, 8:53 am, Mark <makol...(a)yahoo.com> wrote:
> > Approximately 10% losses occur in the transformer, bridge rectifier and
> > darlington. This could be reduced by using a higher voltage secondary and a
> > MOSFET when the losses could be 5% or lower. The control electronics side
> > is isolated by the transformer so you are working with low voltage DC.
>
> OK, I can see this working well for nearly off and nearly full on, but
> when you are at say 1/2 power, doesn't the darlington on the secondary
> side dissipate a lot of power?
>
> It seems to me it is the equivalent of putting a resistor in series
> with the load, its just that the resistor value is transformed by the
> turns ratio and the rectifier makes it DC instead of AC, but you are
> still basically putting a resistor (not an inductor) in series with
> the load...
>
> no?
>
> Mark

Yeah, I think you are right the 'beauty' from my point of view is that
you are now regulating a lower voltage DC signal.

George H.
From: ontherails on
>OK, I can see this working well for nearly off and nearly full on, but
>when you are at say 1/2 power, doesn't the darlington on the secondary
>side dissipate a lot of power?
>
>It seems to me it is the equivalent of putting a resistor in series
>with the load, its just that the resistor value is transformed by the
>turns ratio and the rectifier makes it DC instead of AC, but you are
>still basically putting a resistor (not an inductor) in series with
>the load...
>
>no?
>
>Mark
>

From a previous post this arrangement doesn't replace triacs with their
associated RFI etc because triacs work as pulse width modulators (PWM) and
dissipate very little power. The darlington will have a maximum dissipation
(at 1/2 power) of about 1/4 of the power rating of the motor it is driving
so it will need a heat sink.

What this circuit does, as you say Mark, is to create a variable resistor.
However, it isn't quite as simple as that as because as the resistance
increases at some point the inductance of the secondary will take over. So
considering the operation more carefully I think this circuit can be more
accurately represented as a variable resistor in parallel with the large
inductance of the primary winding. The power factor will therefore be
variable between 1 (low resistance - full power) and zero (entirely
inductive - off). A shaded pole motor will be an inductance with a series
resistance and will have a fixed power factor.

To be honest I've never really given the theory of operation too much
thought as it just worked from the first time I tried it. I might have a go
at producing a phasor diagram in Excel.
Peter
greenpaper usualsymbolhere btinternet.com

---------------------------------------
Posted through http://www.Electronics-Related.com
From: ontherails on
>OK, I can see this working well for nearly off and nearly full on, but
>when you are at say 1/2 power, doesn't the darlington on the secondary
>side dissipate a lot of power?
>
>It seems to me it is the equivalent of putting a resistor in series
>with the load, its just that the resistor value is transformed by the
>turns ratio and the rectifier makes it DC instead of AC, but you are
>still basically putting a resistor (not an inductor) in series with
>the load...
>
>no?
>
>Mark
>

From a previous post this arrangement doesn't replace triacs with their
associated RFI etc because triacs work as pulse width modulators (PWM) and
dissipate very little power. The darlington will have a maximum dissipation
(at 1/2 power) of about 1/4 of the power rating of the motor it is driving
so it will need a heat sink.

What this circuit does, as you say Mark, is to create a variable resistor.
However, it isn't quite as simple as that as because as the resistance
increases at some point the inductance of the secondary will take over. So
considering the operation more carefully I think this circuit can be more
accurately represented as a variable resistor in parallel with the large
inductance of the primary winding. The power factor will therefore be
variable between 1 (low resistance - full power) and zero (entirely
inductive - off). A shaded pole motor will be an inductance with a series
resistance and will have a fixed power factor.

To be honest I've never really given the theory of operation too much
thought as it just worked from the first time I tried it. I might have a go
at producing a phasor diagram in Excel.
Peter
greenpaper usualsymbolhere btinternet.com

---------------------------------------
Posted through http://www.Electronics-Related.com
From: Michael on
On Jul 3, 9:01 am, "JosephKK"<quiettechb...(a)yahoo.com> wrote:
> On Thu, 1 Jul 2010 06:18:49 -0700 (PDT), George Herold
>
>
>
> <gher...(a)teachspin.com> wrote:
> >On Jul 1, 7:06 am, "JosephKK"<quiettechb...(a)yahoo.com> wrote:
> >> On Tue, 29 Jun 2010 17:00:03 -0500, "ontherails"
>
> >> <peter.elbro(a)n_o_s_p_a_m.btinternet.com> wrote:
>
> >> >>Does anyone have experience with speed control of a small shaded pole
> >> >>motor about 25W, with a ghastly PF of about 0.36 ?.
>
> >> >> This drives the water circulation pump in an experimental setup much too
> >> >>hard, and the churning
> >> >>measurably heats the water.
>
> >> >>I thought a variable frequency inverter might be nice, but wondered how
> >> >>sinusoidal it would need to be. I doubt that a simple phase shift let-it-
> >> >>slip-more controller, would be good enough. Works for fans up to a point.
>
> >> >>Rather than suck it and see, I thought I'd politely probe the brains of
> >> >>the experts first.
>
> >> >>All the best
> >> >>Ian Macmillan
>
> >> >Use my variable inductance controller. Carry out this simple experiment to
> >> >see how it works. Get a 100VA transformer, something like 240V to 40V.. Put
> >> >the 240V windings in series with the motor and connect to the mains -
> >> >nothing happens. Short the 40V windings - motor runs at nearly full speed
> >> >but you are still working with AC.
>
> >> >Put a bridge rectifier on the 40V windings and short the DC output. Motor
> >> >runs at nearly the full speed it did before but now you are working with
> >> >DC. Put a transistor (I normally use a darlington - TIP121 for a 40V
> >> >secondary and add simple control electronics. You can normally get up to
> >> >about 90% full speed.
>
> >> >Approximately 10% losses occur in the transformer, bridge rectifier and
> >> >darlington. This could be reduced by using a higher voltage secondary and a
> >> >MOSFET when the losses could be 5% or lower. The control electronics side
> >> >is isolated by the transformer so you are working with low voltage DC..
>
> >> >Have fun experimenting but beware of the 240V on the transformer.
> >> >Peter
>
> >> Gosh that is only about 40 years old.  Over 75 if you count tubes.- Hide quoted text -
>
> >> - Show quoted text -
>
> >You make it sound like an old design is not as good as something more
> >recent.  I'd never heard of this so it is 'new' to me.
>
> >George H.
>
> It is not so much as saying old is good, but that i find it disingenuous
> to represent it as a new circuit / discovery.


What was this circuit called 40-75 years ago?

I Googled around for "motor control transformer short secondary" but
didn't come up with anything relevant.

Thanks!

Michael