Can it work ?

[mowhoong]

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.

Regards

 

[John Larkin]

On Thu, 3 Sep 2009 04:15:02 -0700 (PDT), mowhoong
<mowhoong@hotmail.com> 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.

Regards

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

John

 

[David Eather]

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.

Regards

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.

 

[Jasen Betts]

On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD

or this one: (stocks limited)
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434



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.

 

[John Fields]

On 4 Sep 2009 12:22:01 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD

or this one: (stocks limited)
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434



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]<----+ +--[7000]--+
| 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.

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 (Xl˛ + R˛)


we can get the reactance by rearranging and solving:

Xl = sqrt (Z˛ - R˛)

= sqrt (7000˛ - 2310˛)

~ 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!!!

17 henrys???

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



E1 E2
120VAC>----+ 240VAC 120VAC
| T1 / R1 /
[VARIAC]<----+ +--[AMMETER]---[7000R]--+
| P||S |L1
| R||E [2310]
| I||C |
120VAC>----+---------+ +-----------------------+

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
<-- SWITCHING POINT
80 12.8 6250 5807 15.4
90 14.7 6122 5669 15.04
100 16.4 6097 5642 14.97
110 18.2 6043 5584 14.81
120 20.1 5970 5504 14.6

Quite a bit different from the 17.5 henrys found earlier.

Checking into it, it turns out that rounding E2 and It has a greater
than expected influence on Z, Xl, and L.

More on that later, in another post...

From the data above, it appears that the average impedance before
switching is about 4379 ohms and, after switching, 6096 ohms. That's
about a 39% increase in impedance, which means that until the relay
switches in, the current in the circuit will be higher than expected.
That shouldn't be a problem, though, because the switching time is on
the order of tens of milliseconds and we're talking about melting the
copper wire in the relay coil or the nichrome or manganin or whatever
the resistor is wound with, which ain't gonna happen.

Once the relay has closed, with 20.1 mA in the circuit and 120V being
dropped across R1 it'll dissipate about:


P = IE = 120V * 0.0201A ~ 2.4 watts


A 5 watt resistor would work, and Mouser has a nice Vishay/Dale
CW0057K000JE12 (7k +/-5%, 5W, Mouser P/N 71-CW5-7K-E3) in stock for
US$ 0.97, so that would be a good choice.


Next installment:

Will more precise measurements cause the results in both methods
described previously to converge?


Installment 3:

What happens if we replace R1 with a capacitor?


Installment 4:

What happens if we replace R1 with an inductor?

JF

 

[Dan Coby]

John Fields wrote:

On 4 Sep 2009 12:22:01 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD

or this one: (stocks limited)
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434



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]<----+ +--[7000]--+
| 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!!

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


Dan

 

[Dan Coby]

Dan Coby wrote:

John Fields wrote:
On 4 Sep 2009 12:22:01 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD


or this one: (stocks limited)
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434




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]<----+ +--[7000]--+
| 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.

 

[Ross Herbert]

On Mon, 07 Sep 2009 20:41:16 -0700, Dan Coby <adcoby@earthlink.net> wrote:

an Coby wrote:
:> John Fields wrote:
:>> On 4 Sep 2009 12:22:01 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:
:>>
:>>> On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
:>>>
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD
:>>>
:>>>
:>>> or this one: (stocks limited)
:>>>
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434
:>>>
:>>>
:>>>
:>>>
:>>> 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]<----+ +--[7000]--+
:>> | 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
f 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.
http://www.freepatentsonline.com/6842014.pdf

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

 

[John Fields]

On Mon, 07 Sep 2009 20:41:16 -0700, Dan Coby <adcoby@earthlink.net>
wrote:

Dan Coby wrote:
John Fields wrote:
On 4 Sep 2009 12:22:01 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-09-03, mowhoong <mowhoong@hotmail.com> 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:
http://jaycar.co.nz/productView.asp?ID=MT2030&keywords=transformer&form=KEYWORD


or this one: (stocks limited)
http://nz.farnell.com/dagnall-electronics/d2034/transformer-6va-2x-12v/dp/899434




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]<----+ +--[7000]--+
| 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.

---
Yes.
---

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.

---
Yes, you're right.

I mistakenly lumped the coil's reactance in with the coil's resistance
and forgot to include the effects of the reactance on the external
resistance. Thanks for the reality check.
---

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

---
For the circuit that works:

E1 E2
120VAC>----+ 240.7 VAC 120.06 VAC
| T1 / R1 /
[VARIAC]<----+ +--[AMMETER]---[6944R]--+
| | | <--I--> |L1
| | | +----|----+
| P||S | [2310R] |
| R||E | | |
| I||C | [L] |
| | | +----|----+
| | | |
120VAC>----+---------+ +-----------------------+



I did, and in this case, to get the impedance of the coil we say:

E2
Z = ----
I

Then, to get the reactance:


Xl = sqrt (Z˛ - R˛),

where R is the resistance of the coil and, finally, the inductance:

Xl
L = -------
2Pi f


So here's the new table:

.. E2 It Z Xl L
.. VOLTS mA OHMS OHMS HENRYS
..---------------------------------------
.. 0
.. 10.16 2.5 4064 3344 8.87
.. 20.15 4.91 4103 3392 8.99
.. 30.42 7.25 4196 3503 9.29
.. 40.48 9.50 4261 3581 9.50
.. 50.44 11.71 4307 3636 9.64
.. 59.96 13.84 4332 3665 9.72
.. 70.28 15.93 4412 3760 9.97
.. <-- SWITCHING POINT
.. 79.72 12.79 6233 5789 15.35
.. 90.52 14.67 6170 5722 15.18
.. 99.73 16.24 6141 5690 15.09
.. 110.42 18.19 6070 5614 14.89
.. 120.91 19.99 6048 5590 14.83
---

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.

---
Indeed. It's 138V, and with 20 mA through it when there's 120V across
the relay, it'll dissipate:


P = IE = 138V * 0.02A = 2.76 watts,


so the earlier recommended 7000 ohm 5 watt resistor will still work.