Can it work ?

It won't work for an AC-coil relay. A series resistor or capacitor or inductor (another relay coil!) can work.

John

Since you don't know the simple basics, don't screw around with mains voltages. Your project will fail and you may kill yourself or someone else.

that's most likely to destroy the coil, even on 115VAC using a diode could be bad.

use a step-down transformer, (if you have a 230V transformer with a centre-tapped primary you could use it as an autotransformer: put the relay coil in parallel with half the primary.

eg: this one:

or this one: (stocks limited)

The inductance of the coil is going to rise as the relay closes so solutions involving only series dropping elements could be tricky to get working correctly.

--- Maybe not...

I've got a 120V 60Hz relay, a P&B KA11AY with a DC resistance of 2310 ohms which I hooked up like this in order to determine its impedance: (View in Courier)

E1 E2

R1 is a power resistor decade box, and initially I adjusted it for zero ohms and raised E1 to 120V to get the relay to latch. Then I increased the resistance and the VARIAC's output voltage until I had 240V out of T1 and 120V across the coil.

With the relay latched, the same current through R1 and L1, (since they're in series) and the voltages across each of them equal, the resistance of R1 had to be equal to the impedance of the coil.

We know that since:

Z = sqrt (Xl² + R²)

we can get the reactance by rearranging and solving:

Xl = sqrt (Z² - R²)

= sqrt (7000² - 2310²)

~ 6608 ohms.

To get the inductance of the coil, we rearrange

Xl = 2pi f L

and solve for L:

Xl 6608R L = ------- = ------------- = 17.53 henrys 2pi f 6.28 * 60Hz

YOW!!!

Seems absurd, so let's attack the problem in a different way...

E1 E2

Here, we measure the current in the secondary and use:

E2 Z = ---- I

to determine the impedance of the coil and then, as before, get the reactance and the inductance of the coil.

Just for grins, I started with E1 at 20V and then calculated the data for 10 volt increments up to 120V to see how much the inductance varied as current through the coil changed, and what happened around the swithing point.

Here it is:

E2 It Z Xl L VOLTS mA OHMS OHMS HENRYS

--------------------------------------- 20 4.8 4167 3468 9.2 30 6.9 4348 3684 9.77 40 9.1 4395 3739 9.92 50 11.5 4348 3684 9.77 60 13.7 4380 3721 9.87 70 15.2 4602 3980 10.56

No. You are ignoring the fact that the impedance of the inductance of the coil is at 90 degrees to the resistance of the coil and R1.

Lets start with the simpler case where the resistance of the relay is zero. In this case then the impedance of the coil would be purely inductive and at 90 degrees. Let us assume that this impedance is equal to R1 and at 90 degrees. Then the load on the secondary of your transformer is sqrt(R*82 + Xl**2) = 1.414 * R1. (Please note that this is not 2 * R1 even though the two impedances are the same. This is due to the 90 phase shift which forces us to do vector sums.) The current through R1 and the relay is 240 / 1.41.4 * R1 and the voltage across either R1 or the relay is 169.7 volts.

Your relay has non zero resistance (you said 2310 ohms). This means that the phase angle for its impedance will not be 90 degrees. However unless there is no inductance in the coil, its phase angle will not be zero.

Now you have stated that both E1 = 240 volts and E2 = 120 volts. This can only happen when the impedance of the relay has zero phase, i.e. it is purely resistive. Yet your other measurements indicate that the resistance of the relay is much less than 7000 ohms so you probably have a significant reactance.

This is contradictory!!

I suggest that you verify your measurements of the voltages, currents, and resistances.

Dan

Xl**2)

R1

My previous statements are wrong. It is possible to have 120 volts across the relay and 240 volts across the resistor/relay combination. However since the relay impedance has a non zero phase angle, the voltage across the resistance will not be 120 volts. Instead it will be some higher value. Likewise the magnitude of the resistance will be higher than the magnitude of the impedance of the relay.

Check the voltage across the resistance when you have 120 volts across the relay and 240 volts across the resistor/relay combination. You should see more than

:Dan Coby wrote: :> John Fields wrote: :>> On 4 Sep 2009 12:22:01 GMT, Jasen Betts wrote: :>>

:>>> On 2009-09-03, mowhoong wrote: :>>>> I would like to use a diode to act as a step down tranformer to drive :>>>> a small ac 115 volts relay from the 230 volts ac. :>>>> I measured the resistance of the coil was 2470 ohm. I saw this method :>>>> used in soldering iron to redue temperature ,but relay appliction :>>>> unheard. Does any one has any idea on this ? :>>>> Thanks in advance. :>>> that's most likely to destroy the coil, even on 115VAC using a diode :>>> could be bad. :>>>

:>>> use a step-down transformer, (if you have a 230V transformer with a :>>> centre-tapped primary you could use it as an autotransformer: put :>>> the relay coil in parallel with half the primary. :>>>

:>>> eg: this one: :>>>

:>>>

:>>> or this one: (stocks limited) :>>>

:>>>

:>>>

:>>>

:>>> The inductance of the coil is going to rise as the relay closes so :>>> solutions involving only series dropping elements could be tricky to get :>>> working correctly. :>>

:>> --- :>> Maybe not... :>>

:>> I've got a 120V 60Hz relay, a P&B KA11AY with a DC resistance of 2310 :>> ohms which I hooked up like this in order to determine its impedance: :>> (View in Courier) :>>

:>> E1 E2 120VAC>----+ :>> 240VAC 120VAC :>> | T1 / R1 / :>> [VARIAC]> | P||S |L1 :>> | R||E [2310] :>> | I||C | :>> 120VAC>----+---------+ +----------+ :>>

:>> R1 is a power resistor decade box, and initially I adjusted it for zero :>> ohms and raised E1 to 120V to get the relay to latch. Then I increased :>> the resistance and the VARIAC's output voltage until I had 240V out of :>> T1 and 120V across the coil. :> :> No. You are ignoring the fact that the impedance of the inductance of the :> coil is at 90 degrees to the resistance of the coil and R1. :> :> Lets start with the simpler case where the resistance of the relay is zero. :> In this case then the impedance of the coil would be purely inductive and :> at 90 degrees. Let us assume that this impedance is equal to R1 and at 90 :> degrees. Then the load on the secondary of your transformer is sqrt(R*82 + Xl**2) :> = 1.414 * R1. (Please note that this is not 2 * R1 even though the two :> impedances are the same. This is due to the 90 phase shift which forces :> us to do vector sums.) The current through R1 and the relay is 240 / 1.41.4

• R1 :> and the voltage across either R1 or the relay is 169.7 volts. :> :> Your relay has non zero resistance (you said 2310 ohms). This means that the :> phase angle for its impedance will not be 90 degrees. However unless there :> is no inductance in the coil, its phase angle will not be zero. :> :> Now you have stated that both E1 = 240 volts and E2 = 120 volts. This can :> only happen when the impedance of the relay has zero phase, i.e. it is purely :> resistive. Yet your other measurements indicate that the resistance of the :> relay is much less than 7000 ohms so you probably have a significant :> reactance. :> :> This is contradictory!! : :My previous statements are wrong. It is possible to have 120 volts across the :relay and 240 volts across the resistor/relay combination. However since the :relay impedance has a non zero phase angle, the voltage across the resistance :will not be 120 volts. Instead it will be some higher value. Likewise the :magnitude of the resistance will be higher than the magnitude of the impedance :of the relay. : :> I suggest that you verify your measurements of the voltages, currents, and :> resistances. : :Check the voltage across the resistance when you have 120 volts across the relay :and 240 volts across the resistor/relay combination. You should see more than :120 volts.

Perhaps this patent application might help explain the method.

An afterthought: I wonder how you can patent a method of applying standard mathematical equations and derivations?

Xl**2)

• R1