Hi Vasquo. Yes I could do the rotation and manage the routing with another bottom-layer trace but I'm not worried about getting the polarity correct when populating. In the same way you have to make sure the first one is around the right way, it's easy enough to check all of them. (I note the store SS board hasn't worried about this.) But point taken re "best practice".
The ground trace under the relay passes DRC for clearance from the pads. My concern was to ensure good distance (100mil) from the high voltage traces through the relay. I guess I could have simply made the former much thinner but it should be fine if the board house lives up to its DRC.
That's of course fine although I note that the contact rating of the relay in the BoM for the store's board is 10A - the same as for the LY2-0-DC24 here. (Those on lower voltages will probably not be worried about fitting the larger cap.) (The limitation would be governed by copper trace width/thickness: circa 7A.)
One thing I have just noticed: current through the coil models as 38mA versus a coil rated current of 36.9mA. A problem?
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I doubt your conclusion.
A 600VA transformer powering a couple of 200W channels will probably have an average draw of 100W to 200W when playing loud.
For a 110/120Vac supply that amounts to less than 2Aac.
But most Members here do not use 200W+200W amplifiers.
The average current draw playing music is probably more like 500mAac.
The resistance bypass relay in a soft start is not heavily stressed.
At power ON it is open. No spark.
At the end of the delay it shorts across the resistor which may have a few volts or maybe upto 20V across it. On closing there is virtually no spark and therefore virtually no wear.
At power off the relay is still closed. no spark
After the power is off, the relay drops out and goes open. no spark
The current that passes through the switch at the one moment when it actually changes a current is quite low. The transformer has already started.
If you use a 2A fuse then a 2A relay will do. If you have a 5A fuse then a 5A relay will do.
The relay NEVER has to break the transformer primary current. That is when real wear takes place and when there is a risk of damaged contacts.
measure the coil voltage as you reduce the supply to the relay.
It will drop out at about 10% to 20% of it's rated voltage/current.
At this point of incipient drop out the contact pressure will be very low. Don't try to operate a relay way down at 20% of rated voltage.
But you will find that after triggering at rated voltage you can reduce the holding voltage to anywhere between 50% and 100% of the rated voltage and the contacts will perform well.
I typically use 125% to 150% as my triggering voltage and reduce to 50% to 70% as holding voltage.
The high triggering voltage gives very fast closing and probably lots of contact bounce (=bad) with a very long wipe stroke. The low holding voltage significantly reduces power/heating in the coil and slightly reduces the drop out time.
Although the over-voltage triggering will give contacts bounce, it is only shorting across a resistor that now has a low voltage across it. Not much damage can be done during the bouncing event.
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soft starting circuits are meant only for starting.....
nothing to do with normal operation of the amp playing music...
i agree with Andrew, the average user on a 230 volt mains will most likely be running less than an ampere, more than that and you will have your neighbors complaining...
this is also my observation with tube amps....
nothing to do with normal operation of the amp playing music...
i agree with Andrew, the average user on a 230 volt mains will most likely be running less than an ampere, more than that and you will have your neighbors complaining...
this is also my observation with tube amps....
The full primary current runs through the relay at all times. So they do have everything to do with normal operation of the amp playing music. IF one needed more primary current than the relay was capable of passing then they'd have to find another solution, perhaps along the lines of what Mark was suggesting. Now, as to how many people will find a 10A rated relay enough for their needs I'm not sure but I do suspect it's enough for a lot of people's purposes (especially those on 230V).
you mean relay contacts, yes that is true....you can forget about it when the amp is operating normally...
10 amp contacts is more than enough for a circuit that will run 1 amp most of the time...
10 amp contacts is more than enough for a circuit that will run 1 amp most of the time...
I'm not sure I follow you Andrew. Help me through the thinking (and perhaps terminology) here. At power on the voltage potential at the normally open contact is the mains voltage. The voltage potential on the COM contact and normally open contact is lower, building towards mains as the NTC resistance falls with current/temp rise. After the delay, COM connects to NO at mains voltage. Are you saying that because the difference in voltage potential between the contacts when the relay switches is low there's limited wear?
But isn't current effectively 'broken' in that as the relay switches, the path of current changes from (in/out) COM to NO->COM ? NO current goes from zero to full.
Again I'm not following you. Perhaps, again I'm just unfamiliar with the terminology. If you look at my model of the hysteresis-NTC circuit, you will see voltage at and current across the coil of the DPDT relay climbing as the delay progresses (as the charge pump charges?). The relay is triggered two LED drops below the final voltage level. As the LEDs are taken out of the circuit, voltage jumps and so does current. Thereafter, coil current and voltage are constant (ignoring the ripple in the voltage). It is this coil current that I was looking at.
The relay does not turn on the transformer.
The relay does not turn off the transformer.
Some other switch, or you pushing/pulling the plug top does the ON/OFF.
The relay bypasses the resistor that limits transformer start up current.
Forget simulations until you understand what is happening.
What does "bypass" mean?
The relay does not turn off the transformer.
Some other switch, or you pushing/pulling the plug top does the ON/OFF.
The relay bypasses the resistor that limits transformer start up current.
Forget simulations until you understand what is happening.
What does "bypass" mean?
Andrew, I know what bypass means. I know the relay doesn't turn the transformer on or off. It's your communication that baffles me at times.
At switch ON the bypass relay is open. It stays open. It has no effect on the start. The start has no effect on it. There is simply a voltage across the open contacts. That voltage equals the voltage across the current limiting resistor.
No wear, no erosion, nothing.
During the delay interval the transformer starts up and develops a back emf to oppose the voltage being fed in.
That back emf reduces the current flowing into the primary and must also pass through the current limiting resistor.
Due to the current falling to a new lower level, the voltage drop across the resistor also falls to a new lower level. This is the voltage that now appears across the open contacts of the bypass relay. This voltage could be around 1Vac, or upto 20Vac.
This is the voltage that the contacts switch when the relay closes.
You are using a 240Vac rated set of contacts to close on a maximum of 20Vac voltage. No problem.
Now take into consideration that most of the wear and damage to switch contacts happens during opening of the switch contacts. During closing there is almost no wear and no erosion, contacts bounce will create short opening moments and the voltage across the contacts will wear/erode the surfaces, but the switch voltage during these "bounce" induced moments is <20Vac on a 240Vac rated relay. Still no problem.
you said this. Can you now see that what you are thinking about the closing/bypassing operation is wrong?
Confirm you understand this first part and confirm if you require further explanation of your later queries.
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Thanks. So when I said:
you might have said something like:
"More or less, yes. It's the voltage across [i.e. difference in potential?] the contacts that matters. If the timer is set appropriately, by the time the relay activates to connect COM and NO currents have dropped and hence the voltage drop across the resistor has fallen, thereby meaning the voltage 'across the contacts' is low. It's this voltage that the contacts switch when the relay closes." Or is this still incorrect?
you might have said something like:
"More or less, yes. It's the voltage across [i.e. difference in potential?] the contacts that matters. If the timer is set appropriately, by the time the relay activates to connect COM and NO currents have dropped and hence the voltage drop across the resistor has fallen, thereby meaning the voltage 'across the contacts' is low. It's this voltage that the contacts switch when the relay closes." Or is this still incorrect?
That sounds right.
Can you see the relay takes no part in switching to OFF?
Then it follows that the relay does NOT need to be rated for the maximum reactive current that the transformer and any fault it is driving, may pull from the mains.
Can you see the relay takes no part in switching to OFF?
Then it follows that the relay does NOT need to be rated for the maximum reactive current that the transformer and any fault it is driving, may pull from the mains.
Yes, I had understood that already. I do see the relay switching the path of the current (obviously) with significant change in current across the NO contact. I had not fully appreciated that it is the voltage difference that matters which is the question I asked.
But doesn't the current need to remain below the "rated carry current"? In practice these would appear to be the same rating.
Now to my question regarding the sustained current across the relay coil once the relay has closed?
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It is the current that matters. It's the breaking the current that does the damage. The voltage is merely the way to get the size of the current.
No.
If you have a 550VA transformer fed from a 110Vac supply, the nominal current at full load would be 5Aac.
You would fit a T5A mains fuse.
But we don't use our power amplifier transformers like that.
We direct on line start them. We connect them to capacitor input filters.
We run our average loads at way below maximum rating.
Those different conditions all combined result in the start up current being very high, the operating current being very low.
Solve the start up current with a resistive current limiter and the mains fuse only needs to satisfy the operating current requirement.
Expect a 550VA fed domestic power amplifier to run for years on a T5A fuse or less.
The bypass relay could be selected for a similar current of 5A.
A 220VA could use a T2A fuse and a 2A bypass relay.
I often use a current saving resistor in my relay drivers.
I have found that a small capacitor around 10uF to 47uF charged to 1.5times the rated relay coil voltage will trigger the relay quickly and apparently without damage.
I have also found that the relay passes current well when run at substantially below rated coil voltage. I have used 50% to 75% without any problem.
I adopted the more resistive diode back EMF protection after reading a few papers on "speed of release" when using a diode alone.
I use the diode+Zener (plus to plus) with the Zener rated at ~coil voltage.
This apparently speeds up release, but I have not tested for how much.
Say you have a 12V 400r relay.
I'd use a 15V or 18V regulator to charge a 47uF capacitor through a 560r resistor.
The capacitor charges in 5*RC i.e. ~130ms
The capacitor passes a transient current into the relay when the transistor switch closes. This current starts at value defined by V/R (18V/400r in this example) and reduces very rapidly to 18V/[400r+560r].
i.e. it triggers with ~45mA or a bit less due to coil inductance. and runs at ~19mA
It's nominal current is 12/400 = 30mA
The power dissipated in the coil is very much lower than nominal. It runs cool.
It also draws less current from the PSU.
A pair of these relays would consume 60mA and would increase ripple and increase heat load.
Using two 12V in series fed from a 36Vdc supply would consume only 19mA in a typical 50W+50W amplifier.
That is a big saving.
Golly, what's the power dissipated in R3? Dropping R3 from 1 megohm to 0.0075 megohms might increase the dissipation somewhat. It might be prudent to measure R3's power dissipation twice: once, halfway thru the delay period before the relay pulls in; and then again much later, after the current through the zener diode reaches its final, steady state value.
R2 is the 1 mega Ohms resistor. R3 is the 7.5k resistor in the charge pump (as Mark had described it some time ago).
Agree that there seems no point (that I can see) to changing the value of R2.
With C2 set to 1.5uF (mains at 325V peak for this analysis):
The dissipation of R2 over the first second is 49mW. At 750k the dissipation is 65.5mW. (Changing R2's value doesn't make a material change in the delay.)
R3's dissipation over the first second (value 7.5k, R2=1mega) is 56.2mW. Obviously this includes the period of lower dissipation prior to relay pull-in. Post relay pull-in, steady-state dissipation is 71.7mW.
Agree that there seems no point (that I can see) to changing the value of R2.
With C2 set to 1.5uF (mains at 325V peak for this analysis):
The dissipation of R2 over the first second is 49mW. At 750k the dissipation is 65.5mW. (Changing R2's value doesn't make a material change in the delay.)
R3's dissipation over the first second (value 7.5k, R2=1mega) is 56.2mW. Obviously this includes the period of lower dissipation prior to relay pull-in. Post relay pull-in, steady-state dissipation is 71.7mW.