I've blown many fuses lately, so I'm Thinking of building a soft start circuit, but I don't know much about relays yet, this being my first project using one. I got a free 1200 watt 20-0-20 transformer for my 5 channel surround sound gainclone amp, and I bought This hoping it might work, but I'm sure it won't after looking at the little plastic relays.
So what would be a good relay for this? I'm kinda shooting in the dark here, but will one of these for each 20V output work (the High Current General Purpose Power Relays at the bottom)?
I'm looking for a price efficient unit. I could also use a little info in English (i.e. easy to understand, I'm stupid) on relays, but I think I get the jist of how they work.
Do I want a 30 amp relay for 12vdc (separate 12V trafo to turn it on), I think thats cutting it too close, tho.
Thanks
So what would be a good relay for this? I'm kinda shooting in the dark here, but will one of these for each 20V output work (the High Current General Purpose Power Relays at the bottom)?
I'm looking for a price efficient unit. I could also use a little info in English (i.e. easy to understand, I'm stupid) on relays, but I think I get the jist of how they work.
Do I want a 30 amp relay for 12vdc (separate 12V trafo to turn it on), I think thats cutting it too close, tho.
Thanks
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The relay duty is not particularly onerous.
The relay is open when the transformer starts up and stays open for hundreds of ms. No damage, no wear, no erosion.
The transformer is drawing current through the resistive element when the relay does close. There is an increase in current at that moment when the transformer is connected direct in line. I think any relay that is rated for 250Vac and Irelay > Iprimary would survive a long time.
The biggest wear/erosion will come when the relay opens on shutting down. This should be addressed with a switch snubber to minimise the arc across the opening contacts of the main power off switch. By the time that the relay contacts open there should be little or no current flowing.
Single pole single throw is sufficient to bypass the resistive element. SPST 250Vac 10Aac (normally open = NO) would do in my opinion.
The coil voltage is chosen to suit the voltage you are using as "control".
The relay is open when the transformer starts up and stays open for hundreds of ms. No damage, no wear, no erosion.
The transformer is drawing current through the resistive element when the relay does close. There is an increase in current at that moment when the transformer is connected direct in line. I think any relay that is rated for 250Vac and Irelay > Iprimary would survive a long time.
The biggest wear/erosion will come when the relay opens on shutting down. This should be addressed with a switch snubber to minimise the arc across the opening contacts of the main power off switch. By the time that the relay contacts open there should be little or no current flowing.
Single pole single throw is sufficient to bypass the resistive element. SPST 250Vac 10Aac (normally open = NO) would do in my opinion.
The coil voltage is chosen to suit the voltage you are using as "control".
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Just a side remark: even when the transformer was for free I would not use it for a gainclone. Your electricity bill will sum up very quickly, and the efficiency of your amp will be very little. I'm not that much into transformers, but I would guess that a 1200VA unit will draw at least 50W idle power. I would rather spend the money and buy a much much smaller unit.
Andrew, I'm confused.
According to Soft-Start Circuit For Power Amps, "The relay contacts must be rated for the full mains voltage, and at least the full power current of the amplifier. The use of a relay with 10A contact rating is strongly recommended".
He is basing this on a 500VA trafo with 230V mains, I have a 1200 watt and 120V mains. Also on the same website, he describes how to select the correct resistance, but states they should handle "twice" the expected Amperage, (I=VA rating of transformer/mains)*200%. I realize this is for a different method, but such an inrush of current I'd like to be safe.
In my case, I=VA rating of transformer/mains is as follows:
I=(1200/120)*200%
So, expected inrush of current could be as high as 20 Amps.
Again, this is based on the resistor method, which I don't plan to use. But I wanna be safe.
So is a 20 Amp relay overkill, safe, or....any ideas? I know you know alot more on this topic than I do so I'd love to hear your input.
I have tossed this idea back and forth, but decided to go with my 1200 watt for the following reasons:
Quoted from another website:
"On paper to get full power you should have about 150VA per LM3886" (@ 44VCT per the example), "but in practice it has been proven that as low as 90VA or so has worked..."
If that is the case, a 750 watt trafo should work for a 5 channel amp @ 44VCT. Mine measures 40VCT, so around 700 watts should suffice. Giving it a little headroom, 800 watt should be plenty. But it seems even that would draw a decent amount of power @ idle from my mains, plus the cost of finding and purchasing such a trafo would be in the neighborhood of around $100 AND I would still need a pretty hefty soft start circuit to be safe. I think I'll opt for what I have right now, and I'll still have plenty of headroom for additional amps (7 channels in the future, lol). If I do notice the idle watts used is on the high side, I can still downgrade the trafo and I will have an overkill soft start.
According to Soft-Start Circuit For Power Amps, "The relay contacts must be rated for the full mains voltage, and at least the full power current of the amplifier. The use of a relay with 10A contact rating is strongly recommended".
He is basing this on a 500VA trafo with 230V mains, I have a 1200 watt and 120V mains. Also on the same website, he describes how to select the correct resistance, but states they should handle "twice" the expected Amperage, (I=VA rating of transformer/mains)*200%. I realize this is for a different method, but such an inrush of current I'd like to be safe.
In my case, I=VA rating of transformer/mains is as follows:
I=(1200/120)*200%
So, expected inrush of current could be as high as 20 Amps.
Again, this is based on the resistor method, which I don't plan to use. But I wanna be safe.
So is a 20 Amp relay overkill, safe, or....any ideas? I know you know alot more on this topic than I do so I'd love to hear your input.
I have tossed this idea back and forth, but decided to go with my 1200 watt for the following reasons:
Quoted from another website:
"On paper to get full power you should have about 150VA per LM3886" (@ 44VCT per the example), "but in practice it has been proven that as low as 90VA or so has worked..."
If that is the case, a 750 watt trafo should work for a 5 channel amp @ 44VCT. Mine measures 40VCT, so around 700 watts should suffice. Giving it a little headroom, 800 watt should be plenty. But it seems even that would draw a decent amount of power @ idle from my mains, plus the cost of finding and purchasing such a trafo would be in the neighborhood of around $100 AND I would still need a pretty hefty soft start circuit to be safe. I think I'll opt for what I have right now, and I'll still have plenty of headroom for additional amps (7 channels in the future, lol). If I do notice the idle watts used is on the high side, I can still downgrade the trafo and I will have an overkill soft start.
I missed the point that it was a 5-ch. Makes it (a bit
more reasonable for 5x75 watts rms output power (@4 ohms, would that be right?). I go the other way though and have good results. My (class D) stereo amp drives 250 w rms into each (8 ohm) speaker and is supplied with an SMPS capable of 100w continuous (600w short time peak). Plus large buffer capacitors. I don't manage to exceed 50°C at any heat sink no matter how loud I "listen".Have a look at Soft-Start Circuit For Power Amps
Basically the circuit puts a resistor in series with the mains input at power on, after a short delay the relay shorts across the resistor so that the full mains is applied to the transformer.
The relay must be rated for the current that the transformer is expected to draw. Too small and the relay will fail, if it fails the resistor(s) will overheat and fail.
Another option could be using a triac based soft start circuit. More complex but has the advantage that it does not contain bulky resistive parts with high dissipation. I have briefly looked at http://www.atmel.com/dyn/resources/prod_documents/doc4712.pdf, and the typical application diagram looks just like what you are searching for.
indeed you are.
we give you the explanation and we give you the answers and you are confused.
Do not be confused when working with mains electricity. Your life depends on you understanding exactly what you are doing. More research and more questions.
Now back to my answers.
I said 250Vac for the relay. That is higher than any of the domestic voltages available anywhere in the world which range from 100Vac to 220Vac
I hope there is no confusion in there.
I said 10Aac. This equals the maximum rated current for your 1200VA 120Vac transformer. I hope there is no confusion with this bit.
Now to soft start and why we need it for inductive loads (eg. motors and transformers).
At start up the inductive load has no current flowing and no flux in the magnetic field (this is a simplification but does not significantly alter what follows). If you were to measure the resistance of the primary and obtain an estimate of the mains supply impedance and add on the resistance of your internal household circuit to feed the transformer you will find a total resistance of <<10r.
Apply 120Vac to say 5r and the AC current could be as high as 120/5 ~ 24Aac.
But at the instant of start up the voltage available at the socket may be anywhere between -170Vpk and +170Vpk with reference to Neutral.
The worst case peak transient current at start up is ~170/5 (substitute your mains circuit resistance) i.e. <=34Apk
I have done this calculation for smaller and much smaller transformer and find that worst case instantaneous peak start up currents can approach 100Apk.
I use soft starts to reduce this potential worst case current.
Add a resistive device in series with that 5r household resistance. say 30r.
the worst case start up current is now 170/[5+30] <=5Apk
While this current is flowing the flux in the magnetic circuit builds up and the inductive load starts to develop an inductive impedance that adds to the 35r seen by the mains. The current starts falling towards the running current of the transformer, (this occurs in the first few cycles of the mains supply).
During this time or very soon after, the relay clicks over to bypass the 30r. The 5r + transformer impedance now resists the mains voltage at the moment of contacts closing. If the transformer has no load or very little load then that inductive impedance is likely to be around 1500ohms for this fairly large transformer. The current will increase slightly from the before relay bypass to after relay bypass. But it will never reach 10Aac, when the transformer is lightly loaded.
If the transformer is heavily loaded (eg. the smoothing caps are discharged) then we are in a completely different ballgame.
I cannot estimate the increased current that will flow after the relay bypasses, but I suspect it will not exceed 10Aac for long, if at all.
The damage to the relay contacts is not when they close.
Does this get rid of the confusion?
Now to switch off, when damage/wear can occur to the relay contacts.
The bypass relay is closed.
The mains switch is closed.
The transformer is feeding the lightly loaded amplifier (caps fully charged).
The mains switch opens and is susceptible to damage and wear. This mains switch MUST be RATED for the duty of switching OFF the transformer.
After the power has been switched off there may be an arc across the opening contacts, but this should not last longer than half a cycle of the mains frequency. As the relay driving circuit loses it's hold in voltage the relay drops out. Provided the relay drops out AFTER the arc has extinguished the relay contacts cannot suffer damage/wear. They will simply wipe themselves as they part.
Is this bit clear?
I will repeat my safety message.
Never guess when working with mains electricity. Understand what you propose or get professional help if you have doubts.
To ALL,
this is the first time I have gone through this version of the extended discussion.
Can anyone see any errors that must be corrected, or even a better way of explaining why and how the soft start works?
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In my opinion, phase-control circuits used in such applications should be avoided, because they generate noise.
Spectrum generated is very large, these frequencies are radio noise.
I missed the point that it was a 5-ch. Makes it (a bit
Hi tatus,
could you please describe your SPMS for the class D amp?
Thanks,
max
Agree, on the other hand this will be the case during the first few seconds after turn on. After that period, it would be wise to short and also inhibit that circuit. Reverse procedure at turn off. This way I would expect that the relais has no heavy duty in either direction. Plus the circuit will be much smaller compared to resistive soft-start.
See http://www.diyaudio.com/forums/power-supplies/179703-problem-hypex-smps400-67-oem-v4.html. Runs like a charm, except for that startup issue.
Andrew, I agree especially to your safety message.
caxxxxxx, please judge by yourself: are you experienced enough in understanding electronics to work with dangerous mains voltage? Are you aware of the design principles, guidelines and regulations applicable for this type of work? This is not to sound arrogant, its just the necessary diligence that people like AndrewT must show in this kind of forum.
Regarding the electrical explanations, I agree to everything except the considerations at start-up. The transformer's primary is an inductance and it will behave as one. That means applying a constant voltage to it causes the current to ramp linearily and, provided the inductance is large, slowly. As the winding has a high quality factor (xl / r is large), the inductive term will always be dominant. In other words, just connecting the transformer to the mains, no matter at which phase, does not directly cause excessive inrush current.
What can actually happen at transformer startup (and what causes the fuse to blow) is core saturation.
(Remember: when you plot the flux curve of an inductor at steady-state operation at a sinusodial voltage, it lags the primary voltage by 90°. I am thinking of an unloaded transformer here.) The flux at the instance of connecting the transformer to mains is zero (neglecting remanence). Consider the mains voltage also starts with zero (zero phase). During normal operation, the instantaneous flux would have its lower extreme at this point of time. This means the flux has the wrong bias in our situation, and will reach a peak value of twice the peak value during steady-state. It's a bit difficulty for me to explain, I found this nice picture that shows it and also shows the process of moving into steady-state: http://www.mathworks.com/help/toolbox/physmod/powersys/ug/sess7swsn.gif . Normally transformers are constructed with enough core material to work with the steady-state peak flux, plus some margin. Going up twice as high will sature the core. In this situation, the core cannot magnetize to a higher level, which in turn renders it without effect. That means that the primary effectively becomes an air core, with much much less inductance, allowing that high current increase that blows the fuse. The funny point about this is: connecting the transfomer at mains voltage zero crossing is actually the worst situation. An interesting experiment could be using a 220V transformer at 110V. That one should never saturate and not require a startup circuit.
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I am not sure it happens this way.
When the transformer is off and you first apply voltage there is no flux in the core. It does not behave as an iron cored inductor on that first half cycle of mains power.
Could you try again at explaining it and maybe you can convince me of the true start up condition.
Lol, I am in no way an expert in anything, lol. But I think I use enough common sense to not have electrocuted myself yet with electronics in my limited use with it and mains wiring (keeps fingers crossed, knock on wood). I have thrown more than 1 breaker, but I trust myself enough that I lean or step away while powering it up first time. I'm safe, not crazy....fire extinguishers are close at hand, lol!!
Actually I have, in fact, rewired my first house and taken building trades course back in high school, which included installing and wiring an electrical panel with circuit breakers and ground wires and the wire(s) they hooked up to the meter, etc. I'm no expert, but common sense has served me well so far.
Anyway back on topic I think I see how this circuit works. Taking things step by step (give me time, I'm a slow learner but greatly appreciate the help), The Snubber's got me a little confused. Do I have to use a calculation or is there a standard that will work??? I think I can read about it for days and still get lost in the math. I'm gonna finish reading up on them, and I'm sure there will be a lot more questions to follow.
. I'm safe, not crazy....fire extinguishers are close at hand, lol!!Actually I have, in fact, rewired my first house and taken building trades course back in high school, which included installing and wiring an electrical panel with circuit breakers and ground wires and the wire(s) they hooked up to the meter, etc. I'm no expert, but common sense has served me well so far.
Anyway back on topic I think I see how this circuit works. Taking things step by step (give me time, I'm a slow learner but greatly appreciate the help), The Snubber's got me a little confused. Do I have to use a calculation or is there a standard that will work??? I think I can read about it for days and still get lost in the math. I'm gonna finish reading up on them, and I'm sure there will be a lot more questions to follow.
When you wind a conductor around a magnetizable core, you end up with an inductor with a specific inductance (ideally, neglecting all the dirt effects). Its inductance does not depend on the instantaneous magnetic flux in the core. It is there and will stay there - as long as you do not saturate the core. In saturation condition, the construct will behave like a wound conductor without the core, resulting in significant less inductivity provided the core's permeability is significant.
The primary winding of a transformer will behave in the same way when the transformer has no load at the secondary. In your mind, you can remove the secondary winding (it is open in this scenario) and what you get is a cored inductor. That will always behave like an inductor unless saturated. Here is a good excurse on saturation: How Transformers, Chokes and Inductors Work, and Properties of Magnetics : CWS Coil Winding Specialist, manufacturer of transformers, inductors, coils and chokes
Of course the situation is different when a (in the case of having a rectifier followed by a capacitor) initally short-circuiting load is present. In this case, excessive inrush current is caused by shorting the secondary (with the empty capacitor). As you say correctly, in this situation only the transformer's leakage inductance and ohmic/core losses limit that inrush current.
I hope I could convince you. If not - please explain me why an inductor with a core shall not behave like an inductor with a core when flux=0?
Lol, I am in no way an expert in anything, lol. But I think I use enough common sense to not have electrocuted myself yet with electronics in my limited use with it and mains wiring (keeps fingers crossed, knock on wood). I have thrown more than 1 breaker, but I trust myself enough that I lean or step away while powering it up first time
I believe you in that, but that was not my point. It was about knowing all the relevant design principles and regulations of working with mains voltage. Do you know how to handle PE, do you know the minimum creepage and clearance distances, do you know the difference between creepage and clearance, can you distinguish the difference between functional, single and reinforced insulationn, do you know the impact on the transformer and the way you have to mechanically construct the cabling inside your case? You should not feel uncomfortable with any of these questions before continuing. Imagine that you finish your project and start using it. But something gets loose, shorts L with your nice aluminium case, and your kid touches it. Or: the subcircuit that is continuously connected to mains voltage starts overheating at night because of a design mistake and causes fire in your house. Your insurance company will not pay, provided you survive it.
The idea behind a snubber is connecting a capacitor across the contacts. This way the change of voltage cannot change rapidly. In other words, at the moment of time the relay contacts open, the capacitor will be empty and keep the voltage across the contacts zero (for a very short time, the more capacitance the longer). The capacitor will take the load current and get charged by it. That would be nice, if we would not have to close the contacts as well. The worst case here is that the mains voltage is just at its maximum, and due to this the capacitor is fully charged. If we now close the contacts the stored energy will instantaneously discharge into the contacts and can destroy them. For this reason, we also need a resistor in series with the capacitor that limits the current in this sitation to the maximum that the relay contacts can handle.
Summarizing:
1. capacitance as large as possible and reasonable, depending on load current.
2. resistance as small as possible, enough that the maximum switch current is not exceeded.
Condition 1 has no upper limit. Reasonable value is 100nF per 1A load current. For 10A load current you end up with a 1uF capacitor.
Condition 2: voltage = 250*1.41 = 350V, current = 10A (assuming 230V mains voltage, and that the relay is rated for 10A) ---> R=350/10 = 35 Ohms. Power rating depends on capacitor selection, as the resistor must be capable of dissipating the entire energy stored in the capacitor. The maximum stored energy in the capacitor is Q=1/2 * C * V^2 = 1/2 * 1uF * 350^2 = 0.06 joules. Assuming that the switch is not operated quicker than once per second, we get a rough picture of the required power rating of that resistor: P = Q/t = 0.06 / 1 = 0.06 W. So a standard 0.25 W resistor should be enough here.
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