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Re: (ET) field weakening potentiometer with a Curtis 1204



David

Well, it looks like we may have to agree to disagree here, but let me try one last example. 

Again, using your example for 7/8 strength in in the field.  (a 12.5% reduction in field voltage)

>Finally, how about 7/8 strength, or 1.75a?  R = 36 / 1.75 = 20.57 ohms.
>Series resistor = 20.57 - 18 = 2.57 ohms.  Resistor voltage = 4.5v and field
>voltage = 31.5v.  Series resistor power dissipation at 7/8 field is exactly
>the same as it was when we were at 1/8 field,  7.9 Watts.

I propose using an 80 ohm rheostat for the field control, as that seems to be the total used in some of the tractors.  Lets use 25 watts for our example.  So, an 80 ohm, 25 watt rheostat, replaces the relay controlled field weakening circuit.

The basic problem is this.  When you only use 2.57 ohms of this 80 ohm rheostat, you only have a 2.57/80 of the 25 watt resistor in use.  You are only using a small portion of the total area available to dissipate the heat.  So you are doing the field weakening with the 25 watt rheostat with a portion that can only dissipate ~.8 watts. There is nothing wrong with any of your math.  But a rheostat power rating  must be down-rated in proportion to the amount of the resistance that is in use.  You have not taken this into account. 

Robbie Laird






On Wed, Jun 21, 2017 at 5:48 PM, David Roden <etpost drmm net> wrote:
On 21 Jun 2017 at 11:32, Robert Laird wrote:

>  If 1/30th of the resistor is 7.9 watts, then the entire resistor has
> to be around 237 watts.

I don't understand this at all.  Where on earth are you getting 237 watts?
Nothing like that is being dissipated anywhere in the example field circuit.
No current is passing through the other 29/30 of the resistor.  It might as
well not be there.

Even with no series resistance, the motor field itself was dissipating only
72 watts.  And as you add resistance, the total power dissipated in the
circuit will fall.  So there's no way you can get to 237 watts.  (Actually,
there IS a way to get 237 watts on the field: raise the battery voltage to
65 volts and connect it right to the field. I don't recommend trying that,
however.)

> But when you first put the rheostat into the circuit, (the first
> click, or first turn on winding), it pretty much has to carry nearly
> the entire current of the field, since the resistance is still quite
> low at that point.

No.  First, that one turn is NOT carrying the full field current.  That's
the whole purpose of the series resistance, to REDUCE the current in the
field circuit.  As I said above, by Ohm's law, if the total circuit
resistance increases, the circuit current falls.  And remember, in a series
circuit, the current is the same everywhere.

More importantly, on that "first click," the added series resistance is a
small fraction of the total circuit resistance.  Therefore the voltage drop
across that resistance is a small fraction of the system voltage.  Since P =
E * I (power = voltage times current), power dissipation in the added series
resistance is a fraction of the full-field power.

Maybe you're still thinking of the series resistance as if it were the
entire load across the power supply (battery).  It's not.  It and the motor
field IN SERIES form the load.  That changes everything.

Look at the calculations in my previous post again.  Run them for yourself.
I think you'll see that there's no need to panic.  :-)

And if you don't see that, fine.  For your own ET, just use whatever power
rating your intuition suggests is necessary for the field rheostat.  You
can't harm anything except your wallet by absurdly oversizing the field
control.


David Roden - Akron, Ohio, USA

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