Problems substituting a relay

[CDRIVE]

OK, here's a solution that should satisfy both Peak and RMS values.

Provide two text boxes. One labeled RMS, the other labeled Peak. Entering data in the RMS box will convert and display it in the Peak box and visa versa. Of course, the plots will remain peak values.

Chris

 

[Bassmo]

The Results are in...

The results are in!

But it isn't all good.

I had to get the cap in from England and they lost it the first time so I had to order it twice and wait for two international deliveries. Grrr.

Anyway... it finally arrived.

I used the 24v correct original type of relay and swapped in the 2.2uF cap, upped the zener to a 28v 5W, also increased the wattage of the 8.2 zener to be safe. The relay has exactly 24v on it and switches perfectly now and the circuit performs well. Yay!

BUT...

We calculated the zener should be 1.3-2W. I used a 5W one to be safe but it still gets too hot to touch after about 10-20 seconds, even when the relay is on using the load, and the 100R resistor gets warm too but I didn't feel it before so I don't know how hot it was meant to get. I left plenty of room around the big zener for ventilation.

Any ideas where the error is? I can take some measurements if anyone needs them..

 

[KrisBlueNZ]

The zener is going to get pretty warm, since it's dissipating up to 1.3 watts. The zener is rated for 5 watts but this doesn't mean that it won't get hot. Power is converted into heat which is radiated by the surface area of the component, so if 1.3W of power is being dissipated, a 5W zener will get as hot as a 2W zener if they're in the same package with no heatsinking.

Reading the data sheet for 5W zeners, I see that they rely on the heatsinking effect of the PCB copper. The length of the leads between the device body and the PCB affects how hot the body gets.

You can check that the zener current is reasonable. You'll need two diodes (e.g. 1N4001) and a multimeter with a 200 mA range or thereabouts. Lift the anode of the zener out of the board and solder a wire into the hole it came out of. Connect the first diode directly between the two pads that the zener was connected to, with the same orientation as the zener. This diode will take the "forward current" for the zener, when the input voltage is negative.

Then connect the anode of your second new diode to the anode of the zener, and connect your milliammeter with positive to the cathode of that second diode, and negative to the wire from the PCB at the "zener anode" position. Don't touch anything while you make the measurement! The mean current should be less than 50 mA. Multiply it by the zener voltage to get the mean power dissipation. It should be less than 1.3 watts.

A smallish plastic package like the one the 5W zener is supplied in will have a pretty low thermal conductivity to ambient, because of its small surface area. Perhaps around 150 degrees Celsius per watt. So without the heatsinking effect of the PCB copper, at 1.3W dissipation its temperature could rise to 200 degrees Celsius. So keep the leads short.

Exactly how hot does it get? Have a look at Steve's finger calibration scale in https://www.electronicspoint.com/roland-keyboard-works-1-min-then-dies-t233730.html#post1388706.

I'm not aware of any heatsinks available for this type of package. But you'll probably need to do something if you want to avoid damaging the PCB in the long term.

 

[CocaCola]

I'm not aware of any heatsinks available for this type of package. But you'll probably need to do something if you want to avoid damaging the PCB in the long term.

You can likely take a small section of copper or brass tubing (hobby store) the fits tight around the diode and slide that over the diode with some thermal grease, and ideally have thermal pads on the PC board that you could solder the copper/brass sleeve to... Or even solder more brass/copper to the outside of the tube, even solder two tubes in a 8 pattern or whatever for a DIY heat sink... Might not be pretty but it will help sink the heat...

 

[CDRIVE]

JB Weld is handy stuff. They make different blends of this product, one of which contains aluminum particulates. The stuff is easy to work with and is high temp tolerant. If you have room in your enclosure you can use it to bond an aluminum heat sink to the Zener. If the enclosure has a metal case you can increase the heats sink's effectiveness by making contact between the sink and the case, using thermal compound and pressure fit.

Chris

 

[Bassmo]

Thanks for the tips guys. First I'll determine whether it's actually an issue or not or whether I'm just over reacting to a normal heat dissipation. Though I'm worried that it all goes inside the pump enclosure which is sealed so is both air and watertight. No ventilation there. Just heat absorption by being next to a pump full of cold water so might not be as bad as it sounds. There are signs of previous overheating though as plastic has warped in places.

I've done a "finger test" and when the relay is on, the 100R resistor gets way hotter than the zener - or at least more quickly - after about 12 seconds I had to take my finger off the resistor (it's a 2W resistor by the looks of it).

Question: Can I just measure the voltage drop across the 100R resistor which is in series with the entire circuit to see if I'm overdoing it now with the increased current and need a higher wattage one?

It's mains AC going through it, so what's the best way to measure number of Watts it's passing?

 

[KrisBlueNZ]

Measure the AC voltage across it, and use P = V^2 / R. The voltage and current in that resistor is roughly a sinewave, so that method should be accurate to about 10% or so.

 

[Bassmo]

Hmmm.

Well the current in the 100R resistor is P = (13.22VAC^2) / 98R5 = 1.77W and 134mA.

Total current required by circuit is:

7.23mA for the green LED leg
11.6mA for the 8.2v zener leg
20mA for the relay leg
= total of about 40mA

The 27v zener is absorbing all the rest of the unneeded 87mA or so.

Oddly, changing the 100R resistor for a 200R one (in a sad attempt to halve the current available to the circuit) made no difference to the current through the resistor which remained at 116mA in the sim. Perhaps a series resistor in line with the 27v zener will reduce the current similar to the 8.2v zener leg?

Why does increasing that 100R resistor make no difference to the current draw of the circuit?
How can I reduce the current down to about 50mA? This should resolve everything...

-Paul

 

[CDRIVE]

The primary conduit for this current is the 2.2uF cap.

Chris

 

[Bassmo]

So possibly I can reduce it a bit to keep the heat down.

However, can anyone explain why increasing a resistor in series with the entire circuit doesn't change the current flow in the entire circuit? Surely just popping in a larger value ought to keep the current down?

 

[KrisBlueNZ]

The 100 ohm resistor doesn't limit the current flowing in the circuit - it's limited by the reactance of the 2.2 uF capacitor.

These values you've given, are they measured on the actual unit or are they from a simulation?

I've re-run the simulation. Please check that my schematic matches yours. Don't worry about the zener voltage being slightly different. It makes very little difference to the zener current. I've assumed the relay is NOT energised, since this means more power dissipation in the zener.

The red trace is the zener current. When it's above the 0mA line, the zener is conducting in the forward direction and the voltage across it is only ~0.7V so the power dissipated is negligible. It's only when current flows in the other direction, shown below the 0mA line, that significant power is dissipated.

I've filled in a couple of these excursions to show you where power is dissipated, and how much. I also did a higher resolution capture and calculated the area inside the wave as a fraction of the total cycle time and the peak current, and it works out to about 23%. The wave peaks around 185 mA, and 23% of that is 43 mA. So the mean zener power dissipation is about 1.16 watts in this simulation (assuming a 27V zener).

Can you tell me WHERE your simulation, or your actual circuit, differs from that simulation?

In my simulation, the RMS current in the 100 ohm resistor is about 155 mA which is about 2.4 watts dissipation! You can reduce the resistor to, say, 47 ohms to halve this dissipation (P = I^2 R, and I is mainly determined by the 2.2 uF capacitor). Another way to reduce the dissipation in the 100 ohm resistor would be to use a bridge rectifier, because the existing design wastes half the incoming current, but that's a major change so let's not go there.

Do you have access to an oscilloscope? An isolating transformer?

 

[Bassmo]

The 100R resistor was measured in the actual real live circuit while the relay was on:
P = (13.22vAC ^ 2) / 98R5 = 1.77W and 134mA.

My sim shows 116mA. Probably due to the 20% variance on the 2.2u. Also note the resistance of the part in the circuit is 98R5 not quite 100R.

The ideal load values I estimated from my sim by "measuring" current flow through each leg:
7.23mA for the green LED leg measured through the 3k3.
11.6mA for the 8.2v zener leg measured through the 1k5.
20mA for the relay leg.
87mA through the zener with relay leg on so I suppose that means 87+20=107mA when off.

I'm guessing you used 99999R to "turn off" the relay leg.

So apart from the zener voltage and the diode across the relay coil your circuit sim appears to match mine. Though I have used 240v AC on my sim which I am hoping is RMS. I'd better double check that...

Unfortunately I do not have an oscilloscope - just a DMM. Happy to make some real world measurements if you like..

I don't have an isolating transformer but have been testing the circuit via an RCD.

What is the purpose of the 100R resistor and what are the ramifications on reducing it?

I still can't grasp why that resistor doesn't limit the current. The 2.2u cap allows the current in in the first place, I get that; but surely at the end of the day, the impedance of the 100R would limit the overall current.. How can it not?

-P

 

[CDRIVE]

I know this is going to sound weird, possibly impossible to you but if you doubled the value of R1 to 200R there would be barely any current reduction yet it would dissipate about twice the wattage. As I said and Kris confirmed, C1 is the primary current limiter in this circuit.

I tunneled back through this thread and found this. C1 remained 1uF but I increased the Zener to 28V. I also included a second plot with C2 increased to 500uF but the gain in performance wasn't significant.

https://www.electronicspoint.com/problems-substituting-relay-t250057p2.html#post1479120

https://www.electronicspoint.com/problems-substituting-relay-t250057p2.html#post1479114

If you don't want to go back to 1uF, .. replacing C1 with 1.5uF would definitely reduce the PD in R1 and DZ1.

Chris

 

[KrisBlueNZ]

The 100R resistor was measured in the actual real live circuit while the relay was on:
P = (13.22vAC ^ 2) / 98R5 = 1.77W and 134mA.

I get about 220 mA peak (about 155 mA RMS assuming it was a proper sinewave) with my simulation.

My sim shows 116mA. Probably due to the 20% variance on the 2.2u. Also note the resistance of the part in the circuit is 98R5 not quite 100R.

Is that 116 mA RMS? It's quite a lot lower than my simulation shows. I wonder why.

The ideal load values I estimated from my sim by "measuring" current flow through each leg:
7.23mA for the green LED leg measured through the 3k3.
11.6mA for the 8.2v zener leg measured through the 1k5.
20mA for the relay leg.

Those figures all sound fine. I wasn't questioning them.

87mA through the zener with relay leg on so I suppose that means 87+20=107mA when off.

How did you measure the zener current with your simulator? Did you follow the method I used to calculate the mean zener current from the current waveform and convert that into a mean power dissipation figure? Can you do the same calculation with your simulation?

I'm guessing you used 99999R to "turn off" the relay leg.

Right.

So apart from the zener voltage and the diode across the relay coil your circuit sim appears to match mine. Though I have used 240v AC on my sim which I am hoping is RMS. I'd better double check that...

Good idea, check it!

Unfortunately I do not have an oscilloscope - just a DMM. Happy to make some real world measurements if you like. I don't have an isolating transformer but have been testing the circuit via an RCD.

I guess we should get the simulations to match up first...

What is the purpose of the 100R resistor and what are the ramifications on reducing it?

It's there to absorb spikes on the mains supply, and to pop if the capacitor fails. Both of those functions will be compromised if you reduce it, but you may not have much choice.

If the capacitor fails, the resistor will see most of the applied 230VAC across it. If it's 100 ohms, it will limit the current to 2.3A RMS, and will dissipate V^2 / R watts, which is 530 watts! It will go up in a puff of smoke. (It should be a fusible resistor, so it shouldn't catch fire in the process). If it's 47 ohms, it will limit the current to 4.9A RMS, and will dissipate 1125 watts, so it should fuse more quickly, but it exposes the other components to a higher current before it fuses.

I still can't grasp why that resistor doesn't limit the current. The 2.2u cap allows the current in in the first place, I get that; but surely at the end of the day, the impedance of the 100R would limit the overall current.. How can it not?

It does, but it's in series with a much higher "resistance" - the reactance of the 2.2 uF capacitor, which is calculated from Xc = 1 / (2 pi f C) which is 1 / (2 x 3.14 x 50 x 2.2e-6) which is 1447 ohms. The capacitor's reactance dominates and is the main current limiting factor.

 

[Bassmo]

Well I *do* still have the original 1u cap so it's a pretty short test to see what happens in a live test. Can't do any harm anyway.

 

[Bassmo]

I get about 220 mA peak (about 155 mA RMS assuming it was a proper sinewave) with my simulation.

Is that 116 mA RMS? It's quite a lot lower than my simulation shows. I wonder why.

It's supposed to be RMS I believe. If it were p-p it would be worse.

How did you measure the zener current with your simulator? Did you follow the method I used to calculate the mean zener current from the current waveform and convert that into a mean power dissipation figure? Can you do the same calculation with your simulation?

My sim will show the current through any component so I just selected it and read off the numbers...

I'll do it the way you suggested also to see if the figures agree.

It does, but it's in series with a much higher "resistance" - the reactance of the 2.2 uF capacitor, which is calculated from Xc = 1 / (2 pi f C) which is 1 / (2 x 3.14 x 50 x 2.2e-6) which is 1447 ohms. The capacitor's reactance dominates and is the main current limiting factor.

Ahhhh.. That's what I was grasping for! Cheers!

 

[CDRIVE]

Well I *do* still have the original 1u cap so it's a pretty short test to see what happens in a live test. Can't do any harm anyway.

No, it can't. Give it a shot.

Chris