Honeywell's published documentation for their R182C switching relay
contains the following warning:
IMPORTANT: The transformer [120VAC primary, 24VAC secondary] on the R182C may overheat when used with a series 20 thermostat if the total resistance of the thermostat circuit exceeds 2.5 ohms. If the measured resistance of the thermostat (including thermostat wire and thermostat contact resistance) exceeds 2.5 ohms, add a 100 ohm, 10 watt resistor between the W and R terminals.
I don't understand that -- if the total resistance is, say, 10 ohms, clearly the current drawn will be *lower* than it would be at 2.5 ohms, or at 1 ohm.
So how is the transformer in danger of overheating at a *lower* current? Seems to me that the greatest danger of overheating would occur with a circuit resistance of near zero ohms, i.e. a dead short across the transformer secondaries. Someone please explain this to me.
[Please note that I *do* understand the purpose of the 100 ohm 10W shunt resistor, in reducing the total resistance of the connected load. What I
*don't* understand is how a *lower* resistance avoids overheating the transformer.]
Doug Miller (alphageek at milmac dot com)
That makes sense, if it's an alternating current relay, if the relay wouldn't pull in due to the line resistance so the inductance of the relay would stay low, so current through it would stay high so it'd all heat up.
I'm guessing the resistor does something to boost the power from the transformer
It makes no sense, but Homer has an idea there. Is there a power contactor type of relay? Thingy where a laminated core has to pull in to move the contacts? If so, there's the possibility that the contactor's coil could overheat when it doesn't pull in.
And there's a somewhat hinky type of low voltage transformer/relay that was popular in the '60's. I don't remember much about how it worked, but it was a sort of AC relay with extra winding on it. The winding acted as a transformer secondary and you were able to run a low voltage, isolated control pair out to a switch - the average homeowner could do it legally without an electrician. You short the control wires and it activates the mains circuit.
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Sounds counter intuitave doesn't it? Might be a mistake. Maybe me it works somethiing like this, the load acts as a break, so imageing drinking a a car with the brakes on a bit, that would make the engine do more work? Or something like that? I do notice, transformers supplied with small devices alike modems or speakers etc tend to be hot even when the device is in active (drawing no current).
And I think you need to be clear aout what is over heeating, the transformer or the load, with a low resistance is is likly that the thermostate would over heat, but you are concerned about the transformer.
This is a sort of guess, but say you are putting Xammount of energy in, then X amount of energy mush also go out (sounds reasonable?) OK so the output energy can only be disapaited in the load and the transformer. Now if the load is ssmaller (higher resistance) then only place the rest of the eneeergy can go is into the transformer itself, so the lower the load (the higher the thermostate resistance) then the more energy must be disapaited in the tranformer itself. I don't need to specify (or know) why this happens, its just basics maths.
E(in) = E(transformer) + E(thermostat)
So with a higher resistance in the thermostate it will use less energy (V*V/R) of the rest of the input energy must go into the transformer itself.
I don't know if that is right but it seems a line of thought which would make sense?
It would follow that with no load *all* the input energy must be disapaited in the transformer. Sounds reasonable?
Anotheer car analogy, imagine the transformer is the clutch of a car. Now if you drove that car against an imovable object then the clutch would start to slip and burn loads of energy (and probably burn out) however with a low load (level road freewheeling) hardly any energy would would be spent iin the clutch (transformer).
I assume a similar thing happens in a transformer.
In short you are putting energy in and it has to go somewhere, so if its not going into the load it must go into the transformer, the mechanism of that process is academic but I will hazard a guess at it another time.
Yours Sincerely Lord Turkey of Norwich(bird flu free zone ;O))
Looks as if the transformer needs a minimum load. This may be caused by a false or cheap contruction of the transformer itself. The transformer may get saturated driving light loads. On the other hand the real reason for this remark could be caused by other problems with won't sound better to the customer (i.e. overvoltage problems etc.)
My reading of the doc is that the instruction in the panel when using type 20 thermostats is WRONG.
While it is a little difficult to make out the text for the center wiring arrangement in Fig. 3 it seems to refer to the type 20 thermostat. Note that this arrangement shows a solid wire jumper between W and R which would apply for the maximum length of wire for any particular gauge of wire shown in table 1. Since the transformer is supplying operating current to its own contactor coil via the long cable run and the thermostat contact the maximum resistance of the cable run can not be allowed to exceed a certain value, otherwise the contactor coil will not operate reliably.
What the instruction mean to say is that when the loop resistance of the wiring connecting the thermostat is LESS THAN 2.5 OHMS, then replace the solid wire jumper between W and R with a 100 ohm 10W resistor.