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Problems substituting a relay

Bassmo

Jul 14, 2012
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Hi everyone.

I have a pump controller which turns a relay on and off to drive the pump.

Attached is a circuit diagram for the part in question to the best of my examining the actual circuit.

The original relay was a 24v 1200R 40mA one.

The voltage across the relay was about 19v oddly considering there is a 24v zener there and was not *quite* firing the relay properly though obviously it must have at some point. The Zener appears fine. I think that's due to the super simple half wave rectifier.

I thought I could just use a 12v 300R 40mA relay and put a series resistor in line with the relay to make it work off the 19 available volts. However I only get about 6v across the coil at 20mA which clearly is not enough - almost exactly half of what I need!
Not enough resistance in the relay to cause enough of a voltage drop across it to fire it up now that it's only 300R I suspect.

Any ideas how I can make this work easily without reworking the entire power supply (which is also used in other parts of the circuit so I don't want to muck around with it too much in case I wreck something else in the process).

The green LED is a simple power indicator and the other one is a pump on indicator - it comes on with the relay.

Any ideas would be greatly appreciated.

Regards,

-Paul
 

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CDRIVE

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That's an oddball circuit. Why are there two LEDs that will be lit simultaneously? I suggest that you half the value of the resistor.

Edit: I just want to be clear on this. Your relay problem is effected by the ORG LED and resistor in series with the coil. Why is it there when you have another LED (GRN)?
 
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Bassmo

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Whoops

Thanks for your reply CDRIVE.

The green LED next to the 3k3 is simply a power on status LED.

I've originally omitted the circuitry that triggers the pump but it drives a transistor to sink the current to ground as per the updated diagram. The orange LED is supposed to show relay status.

Here is what I perhaps should have submitted.

I've updated my original question to add these obvious oversights :)

-Paul
 

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CDRIVE

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OK, I understand now but it shouldn't be wired like that. The LED should be across the coil with a limiting resistor in series with the LED. The LED shouldn't be in the coil current leg.
 

Bassmo

Jul 14, 2012
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OK, I understand now but it shouldn't be wired like that. The LED should be across the coil with a limiting resistor in series with the LED. The LED shouldn't be in the coil current leg.

Perhaps, though it *was* designed this way by the pump manufacturer and did originally work in this configuration. I'll try changing it in the simulator and see if it makes any difference..
One has to wonder why they did it this way in the first place...
 

Bassmo

Jul 14, 2012
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More info that may help

The original relay had very burnt contacts due to the relay not switching fast enough. The first click always seemed very labored and took a while - I believe the voltage rose too slowly to the relay clicking threshold. It seems to rise as if a capacitor was charging up..
I measured the final voltage across the relay and it was only 19v or so and the relay was rated at 24v so that might also explain it too depending on the relay.

I removed the relay and measured 24v across where the coil would have been. Adding the 1200R relay load seemed to drop the voltage too low to properly operate the relay.

I tested the relay and apart from the contact damage, the coils are fine and the switch time also fine when given the correct voltage.

I figured that there would be no point in replacing the relay as a new one would just get damaged the same way.

So.. I decide to buy the same brand quick action high current relay (space is very limited inside and 16A 250V relays are usually quite biggish :) but instead of 24v 1200R, I get the 12v 300R one and figure I'll just add a series resistor until I get the perfect voltage.

I fiddled with ideal values but got no useful results. As I altered resistances the voltage was always very low (like about 6v). So even with no added resistance, putting a 300R load into the circuit (the new relay) it somehow reduced the voltage across the coil to about half what was needed.

I then put the circuit into a simulator in my iPhone (cool eh!) and ran it and under ideal conditions there is indeed only 19v across the relay coil and it takes a few seconds to fire up. Hmmmm.

Changing the relay also produces about 6v across the relay coil so there must be something I've missed...

So now you're up to date :)
 
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CDRIVE

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The 1uF Cap in your circuit is the primary component that interfaces your 240VAC and limits the current to your board. Are you in the UK? That design would never be approved on this side of the pond. A step down isolation transformer would be mandatory code.

That aside, once the transistor is on the relay contacts should close within a hundred milliseconds or so, not seconds. Contacts will burn if the coil is receiving insufficient current. The transistor MUST be fully on and saturated too.
 

Bassmo

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OK, I understand now but it shouldn't be wired like that. The LED should be across the coil with a limiting resistor in series with the LED. The LED shouldn't be in the coil current leg.

Like this?

(I locked the transistor saturated for the purposes of this discussion)..

The voltage across the coil has now dropped to about 10v (from the previous 19).

hmmm.
 

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Bassmo

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The 1uF Cap in your circuit is the primary component that interfaces your 240VAC and limits the current to your board. Are you in the UK? That design would never be approved on this side of the pond. A step down isolation transformer would be mandatory code.

That aside, once the transistor is on the relay contacts should close within a hundred milliseconds or so, not seconds. Contacts will burn if the coil is receiving insufficient current. The transistor MUST be fully on and saturated too.


I'm not in the UK - New Zealand.

The transistor is on and fully saturated.

It *is* a terrible design and you're quite right about the contacts burning out - that's how it failed the first time I reckon.
 

CDRIVE

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Like this?

(I locked the transistor saturated for the purposes of this discussion)..

The voltage across the coil has now dropped to about 10v (from the previous 19).

hmmm.

Start checking the other components on that board. Pay particular attention to the 1uF and 220uF Cap. Something isn't right here. The base current to saturate that transistor should not load your circuit. What's the value of the base R?

Chris
 
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Bassmo

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Start checking the other components on that board. Pay particular attention to the 1uF and 220uF Cap. Something isn't right here. The base current to saturate that transistor should not load your circuit. What's the value of the base R?

Chris

The base resistor is 1k, but please ignore that whole transistor base circuit as the actual trigger comes from elsewhere on the board. I added that just to turn on the transistor for the simulator. Perhaps to save confusion I should just put a switch there for now :)

But since my simulation shows the same measurements as the real thing, I'm not convinced those caps are to blame though I could replace them anyway to be sure.

Even in the simulation theres quite a bit of ripple everywhere in that circuit.
 

gorgon

Jun 6, 2011
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Hi everyone.

I have a pump controller which turns a relay on and off to drive the pump.

Attached is a circuit diagram for the part in question to the best of my examining the actual circuit.

The original relay was a 24v 1200R 40mA one.

The voltage across the relay was about 19v oddly considering there is a 24v zener there and was not *quite* firing the relay properly though obviously it must have at some point. The Zener appears fine. I think that's due to the super simple half wave rectifier.

I thought I could just use a 12v 300R 40mA relay and put a series resistor in line with the relay to make it work off the 19 available volts. However I only get about 6v across the coil at 20mA which clearly is not enough - almost exactly half of what I need!
Not enough resistance in the relay to cause enough of a voltage drop across it to fire it up now that it's only 300R I suspect.

Any ideas how I can make this work easily without reworking the entire power supply (which is also used in other parts of the circuit so I don't want to muck around with it too much in case I wreck something else in the process).

The green LED is a simple power indicator and the other one is a pump on indicator - it comes on with the relay.

Any ideas would be greatly appreciated.

Regards,

-Paul


If the original relay had a 1200ohm coil, it would be a 20mA relay, corresponding with the current in the LED(minus the 220ohm).

This means that you can't get 40mA out of the circuit, without some changes in the drop capacitor from mains. I would be very sure that you do the right calculation here before changing components.

You can use a 12V relay if you get one with at least 600ohm coil resistance, but I would drop the voltage with a 12V zener in series to protect the circuit and the relay.

These types of drop circuits are designed with a defined current in mind and you can't increase the current, even if you reduce the voltage.

TOK ;)

You should also be aware that if you increase the possible current in the circuit, you'll also need to increase the power capacity of the 24V zener diode, since this will drop the unused current(and power) when the relay is not activated.
 
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CDRIVE

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The original relay was a 24v 1200R 40mA one.
-Paul

If the original relay had a 1200ohm coil, it would be a 20mA relay, corresponding with the current in the LED(minus the 220ohm).

Ha, Good eye gorgon! I never even questioned that statement. It doesn't compute. It's still a crappy power supply design though. ;)
 

CDRIVE

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Paul, I Spiced your original circuit with a 1.2KΩ coil and it's no wonder that the contacts fried. I suspect that even with an exact relay replacement they're going to keep frying. I would love to know what the EE that designed it was thinking, besides cost savings. It's an awful design and not worthy of a commercial product. Please tell me that this isn't a pool pump control!

Chris
 

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KrisBlueNZ

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I don't think it's such a bad design. CDRIVE, I would have liked to hear what specific problems you see with it; you just said that it's "awful". It's a tidy way to obtain a small amount of power from the AC line, the topology is optimal [correction - see edits at end], and component values are appropriate, apart from the zener voltage, although clearly there is not enough safety margin. I don't know which countries it would and wouldn't be legal in, but as long as the control signal comes through an optocoupler and all the live circuitry is properly insulated (ideally, encapsulated), it is not unsafe.

Connecting the LED in series with the relay coil is also a smart move because it avoids the wasted current that would be caused by having two parallel paths, each operating at around 20 mA. I would leave it as designed, rather than putting the LED in parallel with the relay coil.

I agree with gorgon's comments. You simply need more current from the input componentry. A bit more voltage would be helpful too, since the 24V relay will never receive its full specified voltage because (a) the 1N4007 and the LED drop about 3V between them, and (b) there is significant ripple on the derived supply rail, and the zener limits the PEAK voltage, so the average voltage will be lower than 24V by definition.

Changes I would suggest from the original circuit would be:

1. Increase the capacitance of the 1 uF capacitor (make sure you use a 250VAC rated part that is designed for this application, such as a suppression capacitor that's designed to have mains voltage across it continuously). I would try 1.2 uF initially and increase to 1.5 uF if necessary. Your simulation has been accurate so far, so you can probably rely on it to see whether the average derived supply rail voltage is sufficient.

2. Increase the voltage of the zener slightly - to 27V or - perhaps - as much as 33V.

3. Increase the wattage of the zener. I would at least double it from what it is now. Find out the average dissipation using your circuit simulator and add at 30~50% safety margin.

4. Optionally, increase the 220 uF capacitor. You will get little benefit here, and the size of a larger part may be prohibitive. Make sure it's rated for at least 35VDC (actually a 50V capacitor will last longer here) and use a good quality part because it is subjected to significant ripple while the relay is activated.

Edits:
(a) I was thinking a bit more, and actually, with the half-wave approach, increasing the capacitance of the 220 uF capacitor may help significantly, if there's room to do so.

(b) An even better solution would be to use a full-wave rectifier immediately before the zener, in which case you won't need to increase the 1 uF capacitor since you will now be using the energy from positive AND negative half-cycles. Also you won't need to increase the 220 uF capacitor. That would add one component (a bridge rectifier) or four components (individual diodes) and allow you to remove the 1N4007 after the zener, which is another source of voltage loss. Presumably size and/or cost were the reasons this wasn't done originally, but if I had been designing it, I definitely would have done it that way.

This change would increase the power dissipation in the zener, since it would now have the zener voltage across it on both half-cycles instead of just one. You should be able use your simulation software to see the average power dissipation in the zener. You will be able to decrease the 1 uF capacitor, instead of increasing it, and this will reduce the zener dissipation. Try simulating the circuit with a bridge rectifier and 27V zener and look at the average derived supply voltage. Reduce the 1 uF capacitor until the rail drops lower than you'd like, say 22V, and increase it back by around 30% to provide a safety margin. You can then choose the zener's power rating as appropriate. You might want to mount it off the board as it will probably get pretty warm.
 
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CDRIVE

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I disagree with CDRIVE. There's nothing wrong with the design. It's a tidy way to obtain a small amount of power from the AC line, the topology is optimal, and component values are appropriate, although clearly there is not enough safety margin. <snip>

Huh? Think about what you just said. How can there be nothing wrong with this circuit when you're telling Paul that he has to make a page full of modifications to make it work properly? With all do respect, I stand by my opinion. Properly designed circuits don't require re-engineering by the consumer.

Edit: Regarding your Bridge Rectifier suggestion.. How do you propose he do that without an isolation transformer?
 
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Bassmo

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Thanks guys for all your time and thought.

FYI,There are a couple of Variators across the mains - one first thing and one after the relay which switches mains voltage.

The circuit switches a heavy duty sprinkler system pump.

It seems we have a consensus about a couple of points now:

1) the circuit would not exactly win design of the year :) and it's no surprise the relay contacts burnt out.

2) the new relay draws twice the current - I'd overlooked that entirely - I initially thought I'd get away with it since they both use 500mW.

3) subbing the same relay will eventually kill it too.

So...

I need to allow more current in the circuit.

Thanks Gorgon for your suggestions.

I'll try fiddling some values in the simulator with the caps and Zener etc and see
if I can find a working scenario and report my findings..

I really appreciate all your help.. :)

-Paul
 

Bassmo

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Well I tried a couple of things.

Firstly, by increasing that 1u cap to 2.2u, there was enough current available to properly drive the relay instantly - yay for that. About 13v and just over 40mA so thats ok I suppose though I'd rather it was under than over slightly. say 11v and 36mA still works fine.

BUT:

Now there's 40mA flowing through the orange LED leg. Obviously I'd rather avoid having to rewire the whole thing. What is the point in putting a resistor in parallel with the LED? Changing it's value didn't make any difference to the current through the resistor. I reckon the entire current in that circuit was limited only by the 1u cap which was why it was under driving the relay in the first place.

The transistor is there for show in this sub circuit and I bypassed it in my sim so that leg is always on.

How can I keep 40mA going through the relay and not have the same 40mA burning out my LED. A series resistor will reduce the current in the entire leg and the relay won't drive properly.

According to the sim, there's still plenty of room in the zener with 40mA instead of 20mA so I'm thinking I might skip replacing it.

The second circuit is me separating the orange LED leg. That works fine and is a minor adjustment, but why is there still only 12v, 40mA across the relay when theoretically there should be 24v due to the zener. I was expecting to have to add about 220R-300R in series to bring it back but doesn't seem to be the case.

Can anyone verify this or explain it? Seems odd to me...

best regards oh mighty minds :)

And thanks again for all the help!

-Paul
 

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CDRIVE

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These are the results I got with the values shown. They're based on a 12V - 300R coil. I also ran a plot with a 1000uF filter cap not shown. Obviously, ripple was reduced even more.
 

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CDRIVE

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Those plots were based on this circuit.
 

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