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What Capacitor to use in this half wave rectifier circuit to encode signal?

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silas greenback

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Hi So, I have designed a circuit to send two switching signals down one wire, Basically i have a heating controller that can control both the central heating and hot water, The Cable its on is only 4 core including Earth, so a live and negative, a switch return out and insulated earth.

Because i can not easily replace the cable, i can currently only use the central heating portion of the control, because i need an extra core for the hot water

I have come up with this circuit
When SW1 only the positive part of the part rectified signal is sent and closes RLY1, when SW2 is closed only the negative part of the signal is sent and RLY2 is closed, when both switches are closed, the full wave is sent, so both relays close

Each Relay needs a capacitor to hold it closed during the half of the wave thats missing, i have built this as a test circuit running on 12 volts 50hz and 1000uF is ideal with only .7 volt ripple

I now would like to make this circuit work at 240 volts ac 50hz, the relay coils use 4.5mA and are 28k ohms, does anyone know how to work out the value of the capacitor for 240 volts?
 

silas greenback

New Member
Thanks for taking the time Nigel!

Humm I hand not noticed that when i scaled up from 12dc the relay became 240ac, at the controller end, there's no room for a step down transformer or i would just use a 12 volt system that i currently have working on the bench

maybe a triac instead of a relay?
 

dr pepper

Well-Known Member
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A 'proper' ac rated relay usually has a magnetic shunt on the solenoid, if you try it you might find one of these will hold on with only half wave ac.
The fact you mention negative and positive parts of the cycle means your using diodes or similar to do this, by doing that you are rectifying the ac into dc, if so you'll need dc relays.
 
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ronsimpson

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1) The diode idea is cleaver.
2) Now you have DC on the relays not AC.

Please send information on the relay. Any numbers?
My guess is the relay say 110VAC for the coil.

Have you tried your circuit? I think the coil will be very hot.

I looked up a relay that I use.
120VAC coil 2250 ohms 2 watts
240VAC coil 9110 ohms 2 watts
48VDC coil 1800 ohms 1.3 watts

So using AC it takes 2 watts to work. If you were to put 220 on a 110 relay coil you will use 8 watts. (hot coil)
Using DC on a AC coil. You will have about 5 to 6 watts of heat. This is because the coil was made for AC not DC.
I found that a 48VDC coil as close to the same amount of resistance in the coil. My guess is that the 110AC coil will work fine with 48VDC. I have used AC relays with DC but I reduced the supply down to where the relay worked with about the same watts in the coil.

I will think about this and come back 8 hours later. Hopefully some one will have an idea.
 
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silas greenback

New Member
Thanks Very much Ron!

I have the circuit working with 12 volts, so a 12 volt ac supply, 12dc relays, and 1000uF Smoothing capacitors (Sized by trail and error as I had a few to try)

In order to just put this Circuit into action I will have to find space in the wall for the step down transformer (240 ==> 12) its a solid wall, the task is not impossible, but if it can be avoided, I would like to

But as you and a few other have pointed out, at first I didn't give consideration for that fact a 240 volt relay will nearly always be AC due to the voltage invovled

It looks like i might be digging a hole in the wall!

Thanks for looking and taking the time to write!
 

spec

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Hi SG,

Welcome to ETO. Which county are you at? If you put it next to 'Location' on your user page, it will display in the box at the left of your posts; that will help us to answer your questions.

Like Ron says, a clever approach you have devised for driving two relays with one wire.

I am a bit confused by your figures though. You say that the relays function with half wave rectified 12V AC at 50Hz (I know that because I know you are from the UK). But then you say that the relays have a resistance of 28K and operate with a current of 4.5mA. That implies a voltage of 126V.

You ask how you calculate the capacitors:

The formula you need to know is Q=CV=IT
Where
Q= charge in coulombs
C= capacitance in Farads
V= voltage in Volts
I= current in Amps
T= time

The time is simple: 50Hz has a period of 20mS. With half wave rectification you have to wait a full period (to a first approximation) before the capacitor gets topped up again. So the discharge time of the capacitor is roughly 20mS. Thus T in the above formula is 20mS.

I, you have stated is 4.5mA

C, you have stated is 1uF

So you need to work out what V (the ripple voltage is or voltage droop between peaks of the AC supply) is.

Extracting and transposing the above formula, you get V = IT/C

Thus, V = (0.0045A * 0.050S) /0.000,001F = 45V

So the capacitors are not doing much.:)

I would remove the capacitors. Then put a suitable sized resistor of a suitable power rating in series with the input to the circuit (between live and the two switch wipers). Start with 56K and work down until the relays start operating. Then reduce the resistor by a further 20%.

Then go and have a pint of beer.:happy:

spec
 
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silas greenback

New Member
Hi SG,

Welcome to ETO. Which county are you at? If you put it next to 'Location' on your user page, it will display in the box at the left of your posts; that will help us to answer your questions.

Like Ron says, a clever approach you have devised for driving two relays with one wire.

I am a bit confused by your figures though. You say that the relays function with half wave rectified 12V AC at 50Hz (I know that because I know you are from the UK). But then you say that the relays have a resistance of 28K and operate with a current of 4.5mA. That implies a voltage of 126V.

Hi thanks for the reply and compliment, so the numbers were for a 240 relay (the one i don't know the capacitance for)

the the 12 volt system i have made i have used this I have used 1000uF caps on this and it works very nicely so far

in the 240 volt system i was proposing, i was thinking of this and it was that that gave the nubers of 28k 4.5mA
 
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spec

Well-Known Member
Most Helpful Member
So i didn't really find any, I'm based in the Uk so mostly used Rapid Electronics,
Oxford... nice, apart from the traffic and tourists, like me.:D

Many makers like, DigiKey, but there is also Farnell (Element 14), RS, Mouser, and CPC (part of Element 14 group).

spec
 
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silas greenback

New Member
Oxford... nice, apart from the traffic and tourists, like me.:D

Many makers like DigiKey, but there is also Farnell (Element 14), RS, Mouser, and CPC (part of Element 14 group).

spec
Haha too true, its also a bit artsy, so not many hardware/electrical shops, best you can do is Maplins, and I don't need a radio controlled car ;-) I was going to say helicopter, but who doesn't need a radio controlled helicopter
 

MaxHeadRoom78

Well-Known Member
So i didn't really find any, I'm based in the Uk so mostly used Rapid Electronics, perhaps i should used google more ;-)

Thanks
There are more listed in the industrial sites than the small control relays, although I have used the octal and plug in types with 240vdc coil, P&B/Tyco/Idec etc.
(My home town, BTW, got all my initial training there).
Max.
 

ronsimpson

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Most Helpful Member
Some one check my work!
1) I tried to determine what the coil on a 230VAC relay looks like. Using data provided in the last link in post #9.
At dc the resistance is 28k ohms. At 60hz 230V the coil draws 4.5mA.
That gives me the idea the coil looks like 28k ohms resistor and a 100h coil in series. If true the coil should operate at 230 VAC 50/60hz and at 120VDC. This is close to what some of my relays act like.
2) Using LT spice:
L2+R2 is a 230VAC relay drawing 4.5mA. Working the way it should
L1+R1 is a second relay running on DC. Drawing 4.5mA. Creating the same amount of magnetic pull.
D1 is 1N4006 or 4007.
C1 is 2.2uF 250V cap.
R3 is 9.1k 2 watt resistor. 1 watt might work.
The voltage on C1 (and the relay) is 120VDC. With some ripple.
You can change R3 to get different amounts of current in the relay.
upload_2016-10-28_21-49-46.png
Please comment.
----edit----
Note the relay looks like a resistor and inductor together.
 

spec

Well-Known Member
Most Helpful Member
Only a small point Ron: the mains is 50Hz in the UK.

Can't you eliminate the two capacitors? I would have thought that the relays would still work OK with half a sine wave.

spec
 

ronsimpson

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Can't you eliminate the two capacitors?........half a sine wave
I thought so but the spec (data sheet) shows a release time of <10mS. And that is too fast for 50hz half wave. (OK borderline)
The relay probably will chatter with 1/2 wave power.
The difference of 50 or 60hz is very little.
 

crutschow

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Ron I think your circuit in post #14 is a good way to go.
Note the TS mentioned 240Vac at 50Hz not 220Vac at 60Hz, so you may have to adjust the component values slightly.
 

ronsimpson

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240Vac at 50Hz not 220Vac at 60Hz
By moving to DC the frequency makes no difference.
240/220 So make the resistor 10% bigger. or maybe 10k ohms.
By adding the resistor where it is we get away from grabbing the very top of the sign wave. We sag down to any value we want.
AND
We want the same current (AC or DC). You can put a meter on the coil and measure the current.
 

spec

Well-Known Member
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I thought so but the spec (data sheet) shows a release time of <10mS. And that is too fast for 50hz half wave. (OK borderline)
The relay probably will chatter with 1/2 wave power.
Nice one Ron.:)
The difference of 50 or 60hz is very little.
:D I was just being pedantic.

By the way, R3 can be common to both relays if you connect R3 between live and the two switches.

spec
 
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spec

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UPDATE of 29_10_29. As pointed out by Les in post #22 this circuit will only allow one relay to be energized at a time. If both switches are made there will be an effective short circuit across the mains supply.

Fewer components:

2016_10_29_ISS1_ETO_RELAY_SWITCHING_ONE_WIRE_ Ver3.jpg
 
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