Soft-Start Circuit For Power Amps

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=20

Elliott Sound Products

Project = 39

Soft-Start Circuit For Power Amps
Rod = Elliott=20 (ESP), Updated 18 April 2006=20 =20

Shar= e |

=20

=20 Please Note: PCBs are available for the latest = revision of=20 this project. Click the PCB image for details.

WARNING: This circuit = requires=20 experience with mains wiring. Do not attempt construction unless=20 experienced and capable. Death or serious injury may result from = incorrect=20 wiring.

Updates ...
PCBs are available for a somewhat modified = version of the=20 soft-start project. Rather than the MOSFET switch, the PCB version uses = a cheap=20 opamp, and provides power and soft start switching. Full details are = available=20 when you purchase the PCB, but the schematic and a brief description is = shown below.=20

The delay time for all circuits shown has been revised. The optimum = is around=20 100ms - sufficient for around 5 full cycles at 50Hz, or 6 cycles at = 60Hz. It is=20 also quite alright to run the transformer to around 200-500% of full = load=20 current at start-up, and the formulae have been revised for up to 200%. = Without=20 the soft-start, inrush current can be so high as to be limited only by = wiring=20 resistance - well in excess of 50A is not at all uncommon for average = sized 230V=20 transformers.

Thermistors - Important !
Using thermistors rather than = resistors is a=20 common question, and while there are many caveats they will generally = work well.=20 Unfortunately, it can be very difficult for the novice (and = not-so-novice) to=20 determine the proper value and size, and manufacturers don't help much. = The=20 specification format from one maker rarely matches that of another, and = making=20 direct comparisons is rarely easy. Some quote a maximum current, others = a rating=20 in Joules, and some include almost nothing except the nominal resistance = at 25=B0C=20 and the dimensions - hardly helpful.=20

Many people like the idea of using NTC (negative temperature = coefficient)=20 thermistors for inrush limiting, with a common claim being that no = additional=20 circuitry is needed. In a word, DON'T. This is possibly = controversial,=20 because they are used by a number of major manufacturers so must be = alright - or=20 so it might seem. If used in a switched system as described here, they = are safe=20 enough, but I have personally seen (yes, with my very own eyes) NTC = thermistors=20 explode mightily if there is a fault. Resistors can also fail, but the = failure=20 is (usually) contained - there are exceptions of course. In general, NTC = thermistors are designed for very high peak current, but as noted = earlier, you=20 will see many different ways to describe the same thing, with almost no=20 commonality between makers.=20

If the relay fails to operate because you didn't listen to me and = used the=20 amp's supply, the thermistor will (in theory) become a low resistance = due to the=20 current flow and the fuse will blow. However, if current is too high due = to a=20 major fault, the thermistor may explode before the fuse has a chance. = I'm unsure=20 why some people insist that the thermistor is somehow "better" than = resistors -=20 it isn't, and in some cases may even be a less robust solution. As noted = below,=20 a resistor (or thermistor) value of about 50 ohms (230V) or 25 ohms = (120V) is a=20 pretty good overall compromise, and works perfectly with transformers up = to=20 about 500VA. The resistance should be reduced for higher power = transformers.=20

If a thermistor is used, it needs to be sized appropriately. While = some small=20 thermistors may appear quite satisfactory, they will often be incapable = of=20 handling the maximum peak current. I suggest that you read the article = on inrush=20 protection circuits for more information. A suitably rated = thermistor can be=20 used in any version of this project (including the PCB based unit shown = in=20 Figure 6).=20

Under no circumstances will I ever suggest a thermistor without a = bypass=20 relay for power amplifiers, because their standby or low power current = is=20 generally insufficient to get the thermistor hot enough to reduce the = resistance=20 to a sensible value. You may therefore get power supply voltage = modulation, with=20 the thermistor constantly thermally cycling.=20

If there is enough continuous current (Class-A amplifier for = example),=20 the surface temperature of any fully functioning thermistor is typically = well=20 over 100=B0C, so I consider bypassing mandatory to prevent excess = unwanted heat. A=20 bypass circuit also means that the thermistor is ready to protect = against inrush=20 current immediately after power is turned off. Without the bypass, you = may have=20 to wait 90 seconds or more before it has cooled.

Introduction=20

When your monster (or not so monster) power amplifier is switched on, = the=20 initial current drawn from the mains is many times that even at full = power.=20 There are two main reasons for this, as follows ...=20

Transformers will draw a very heavy current at switch on, until = the=20 magnetic flux has stabilised. The effect is worst when power is = applied as=20 the AC voltage passes through zero, and is minimised if power is = applied at=20 the peak of the AC waveform. This is exactly the opposite to what you = might=20 expect
At power on, the filter capacitors are completely discharged, and = act as a=20 short circuit for a brief (but possibly destructive) period

These phenomena are well known to manufacturers of very high power = amps used=20 in PA and industrial applications, but 'soft start' circuits are not = commonly=20 used in consumer equipment. Anyone who has a large power amp - = especially one=20 that uses a toroidal transformer - will have noticed a momentary dimming = of the=20 lights when the amp is powered up. The current drawn is so high that = other=20 equipment is affected.

This high inrush current (as it is known) is stressful on many = components in=20 your amp, especially ...

Fuses - these must be slow-blow, or nuisance fuse blowing will be=20 common
Transformer - the massive current stresses the windings = mechanically and=20 electrically. It is not uncommon to hear a diminishing mechanical = buzz as the=20 chassis and transformer react to the magnetic stress
Bridge rectifier - this must handle an initial current way beyond = the=20 normal, because it is forced to charge empty filter capacitors - = these look=20 like a short circuit until a respectable voltage has been reached
Capacitors - the inrush current is many times the ripple current = rating of=20 the caps, and stresses the internal electrical connections

It should come as no surprise to learn that a significant number of = amplifier=20 failures (especially PSU related faults) occur at power on (unless the = operator=20 does something foolish). This is exactly the same problem that causes = your=20 lights at home to 'blow' as you turn on the light switch. You rarely see = a light=20 bulb fail while you are quietly sitting there reading, it almost always = happens=20 at the moment that power is applied. It is exactly the same with power=20 amplifiers.

The circuit presented here is designed to limit inrush current to a = safe=20 value, which I have selected as 200% of the full load capacity of the = power=20 transformer. Please be aware that there are important safety issues with = this=20 design (as with all such circuits) - neglect these at your peril. Up to = 500% of=20 full power is quite alright, and the decision as to which value to use = is up to=20 you. The transformer manufacturer may have some specific = recommendations.

NOTE: Do not attempt this project if you = are=20 unwilling to experiment - the relay operation must be 100% = reliable, your=20 mains wiring must be to an excellent standard, and some metalwork = may be=20 needed. There is nothing trivial about this circuit (or any other = circuit=20 designed for the same purpose), despite its apparent=20 simplicity.

Description=20

Although the soft start circuit can be added to any sized = transformer, the=20 winding resistance of 300VA and smaller transformers is generally = sufficient to=20 prevent a massive surge current. Use of a soft start circuit is = definitely=20 recommended for 500VA and larger transformers.

As an example, a 500VA transformer is fairly typical of many high = power=20 domestic systems. Assuming an ideal load (which the rectifier is not, = but that=20 is another story), the current drawn from the mains at full power is ... =

I =3D VA / V (1) Where VA is the VA = rating of the=20 transformer, and V is the mains voltage used

Since I live in a 230V supply country I will use this for my = calculations,=20 but they are easy for anyone to do. Using equation 1, we will get the = following=20 full power current rating from the mains (neglecting the transformer = winding=20 resistance) ...

I =3D 500 / 230 =3D 2A (close enough)

At a limit of 200% of full power current, this is 4A AC. The = resistance is=20 easily calculated using Ohm's law ...

R =3D V / I (2)

so from this will get ...

R =3D 230 / 4 =3D 60 Ohms (close enough)

Not really a standard value, but 3 x 180 Ohm 5W resistors in parallel = will do=20 just fine, giving a combined resistance of exactly 60 Ohms. A single 56 = Ohm=20 resistor could be used, but the power rating of over 900W = (instantaneous) is a=20 little daunting. We don't need anything like that for normal use, but be = aware=20 that this will be the dissipation under certain fault conditions.

To determine the power rating for the ballast resistor, which is 200% = of the=20 transformer power rating at full power ...=20

P =3D V=B2 / R (3)

For this resistance, this would seem to indicate that a 930W resistor = is=20 needed (based on the calculated 60 Ohms), a large and expensive = component=20 indeed.

In reality, we need no such thing, since the resistor will be in = circuit for=20 a brief period - typically around 100ms, and the amp will (hopefully) = not be=20 expected to supply significant output power until stabilised. The = absolute=20 maximum current will only flow for 1 half-cycle, and diminishes rapidly = after=20 that.=20

The only thing we need to be careful about is to ensure that the = ballast=20 resistor is capable of handling the inrush current. During testing, I = managed to=20 split a ceramic resistor in half because it could not take the current - = this=20 effect is sometimes referred to as "Chenobyling", after the nuclear = disaster in=20 the USSR some years ago, and is best avoided.

It is common for large professional power amps to use a 50W resistor, = usually=20 the chassis mounted aluminium bodied types, but these are expensive and = not easy=20 for most constructors to get. For the above example, 3 x 5W ceramic = resistors in=20 parallel (each resistor being between 150 and 180 Ohms) will give us = what we=20 want, and is comparatively cheap.

For US (and readers in other 120V countries), the resistance works = out to be=20 12 Ohms, so 3 x 33 Ohm 5W resistors should work fine (this gives 11 Ohms = - close=20 enough for this type of circuit).

It has been claimed that the resistance should normally be between 10 = and 50=20 ohms, and that higher values should not be used. I shall leave this to = the=20 reader to decide, since there are (IMO) good arguments for both ideas. = As=20 always, this is a compromise situation, and different situations call = for=20 different approaches.

A 10 ohm resistor is the absolute minimum I would use, and the = resistor needs=20 to be selected with care. The surge current is likely to demolish lesser = resistors, especially with a 230V supply. While it is true that as = resistance is=20 reduced, the resistance wire is thicker and more tolerant of overload, = worst=20 case instantaneous current with 10 ohms is 23A at 230V. This is an = instantaneous=20 dissipation of 5,290W (ignoring other resistances in the circuit), and = it will=20 require an extremely robust resistor to withstand this even for short = periods.=20 For 120V operation, the peak current will 'only' be 12A, reducing the = peak=20 dissipation to 1,440W.

In reality, the worst case peak current will never be reached, since = there is=20 the transformer winding resistance and mains impedance to be taken into = account.=20 On this basis, a reasonable compromise limiting resistor (and the values = that I=20 use) will be in the order of 50 Ohms for 230V (3 x 150 ohm/ 5W), or 11 = Ohms (3=20 x 33 ohm/ 5W) for 120V operation. Resistors are wired in parallel. You = may=20 decide to use these values rather than calculate the value from the = equations=20 above, and it will be found that this will work very well in nearly all = cases,=20 and will still allow the fuse to blow in case of a fault. These values = are=20 suitable for transformers up to 500VA.

This is in contrast to the use of higher values, where the fuse will = (in all=20 probability) not blow until the relay closes. Although the time period = is short,=20 the resistors will get very hot, very quickly. Thermistors may be = helpful,=20 because as they get hot their resistance falls, and if suitably rated = they will=20 simply fall to a low enough resistance to cause the fuse to blow.

Another good reason to use a lower value is that some amplifiers have = a=20 turn-on behaviour that may cause a relatively heavy current to be drawn = for a=20 brief period. These amplifiers may not reach a stable operating point = with a=20 high value resistance in series, and may therefore cause a heavy speaker = current=20 to flow until full voltage is applied. This is a potentially disastrous=20 situation, and must be avoided at all costs. If your amplifier exhibits = this=20 behaviour, then the lower value limiting resistors must be = used.

If flaky mains are a 'feature' where you live, then I would suggest = that you=20 may need to set up a system where the amplifier is switched off if the = mains=20 fails for more than a few cycles at a time. The AC supply to a toroidal=20 transformer only has to 'go missing' for a few cycles to cause a = substantial=20 inrush current, so care is needed.=20

If a thermistor is used, I suggest a robust version, rated for a=20 comparatively high maximum current. 20mm diameter devices are generally = rated=20 for much higher currents than you are likely to need, so will suffer = minimal=20 thermal cycling. A nice round value is 10 ohms at 25=B0C - this does = mean higher=20 peak currents than I suggest above, but you can always use two in series = -=20 especially for 230V operation.

Bypass Circuit=20

Many of the large professional amps use a TRIAC (bi-lateral silicon=20 controlled rectifier), but I intend to use a relay for a number of very = good=20 reasons ...

Relays are virtually indestructible
They are easy to obtain almost anywhere
Useful isolation is provided so control circuitry is not at mains=20 potential
No RF noise or harmonics of the mains frequency are generated. = These are=20 low level, but can be very troublesome to eliminate from TRIAC = circuits
No heatsink is needed, eliminating a potential safety hazard = should there=20 be an insulation breakdown between triac and heatsink

They will also cause their share of problems, but these are addressed = in this=20 project. The worst is providing a suitable coil voltage, allowing = commonly=20 available devices to be used in power amps of all sizes and supply = voltages.

3D"Figure
Figure 1 - Soft-Start Resistors and Relay=20 Contacts

Figure 1 shows how the resistors are connected in series with the = supply to=20 the transformer, with the relay contacts short circuiting the resistors = when the=20 relay is activated. This circuitry is all at the mains voltage, and must = be=20 treated with great respect.

A represents the Active (Live or Hot) lead from the mains switch, and = SA is=20 the 'soft' Active, and connects to the main power transformer. Do not = disconnect=20 or bypass any existing wiring, simply place the resistor pack in series = with the=20 existing transformer.

Do not attempt any wiring unless the mains lead is disconnected, and = all=20 connections must be made so that accidental contact to finger or chassis = is not=20 possible under any circumstance. The resistors may be mounted using an = aluminium=20 bracket that shrouds the connections preventing contact. All leads = should be=20 kept a safe distance from the chassis and shroud - where this seems = impossible,=20 use insulation to prevent any possibility of contact. Construction notes = are=20 shown later in this project. The safety aspect of this project cannot be = stressed highly enough !

The relay contacts must be rated for the full mains voltage, and at = least the=20 full power current of the amplifier. The use of a relay with 10A contact = rating=20 is strongly recommended.

HINT: You can also add a second relay to mute the input = until=20 full power is applied. I shall leave it to you to make the necessary=20 adjustments. You will have to add the current for the two relays = together, or=20 use separate supply feeds if utilising the existing internal power = supply=20 voltage.

Control Circuits=20

If a 12V supply were to be available in all power amps, this would be = very=20 simple, but unfortunately this is rarely the case. Most amps will have = DC=20 supplies ranging from +/-25V to about +/-70V, and any attempt to obtain = relays=20 for these voltages will be met with failure in the majority of cases.=20

An auxiliary supply can be added, but this means the addition of a = second=20 transformer, which will be quite impossible due to space limitations in = some=20 cases. It is still a viable option (and is the safest way to go), and a = control=20 circuit using this approach is shown in Figure 2. This is the simplest = to=20 implement, but the added cost of the second transformer may be hard to = justify.=20 It is pretty much mandatory for Class-A amps though (See Class-A=20 Amplifiers).=20

3D"Figure
Figure 2 - Auxiliary Transformer Control=20 Circuit

This uses simple bridge rectifier, and a small but adequate = capacitor. The=20 control circuit uses readily available and low cost components, and can = easily=20 be built on Veroboard or similar. All diodes can be 1N4004 or = equivalent. Use a=20 transformer with a 9V AC secondary, which will supply close enough to 12 = Volts=20 for this circuit. No regulation is needed, and the controller is a = simple timer,=20 activating the relay after about 100ms. I have chosen a MOSFET for the = switch,=20 since it has a defined turn-on voltage, and requires virtually no gate = current.=20 With the component values shown, the relay will activate in about 100=20 milli-seconds. This can be increased (or decreased) by increasing = (decreasing)=20 the value of R1 (27k). The transformer need only be a small one, since = current=20 is less than 100mA.

Q1 is used to ensure that power is applied to the relay quickly. When = a=20 voltage of 0.65V is sensed across the relay, Q1 turns on, and instantly=20 completes the charging of C2. Without the "snap action", the circuit = will be=20 sluggish, and is not suited to some of the other variations below. Feel = free to=20 use a 2N7000 or similar low power MOSFET if you can get them easily. = These use=20 the TO92 package so are the same size as the small signal = transistor.

NOTE: C1 should be rated at a minimum of = 50V to=20 ensure that the ripple current rating is sufficient to prevent capacitor = heating. Be warned that if the cap gets warm (or hot), then its = reliability and=20 longevity will be compromised.

It is possible to make the relay release much faster, but at the = expense of=20 circuit complexity. A simple logic system could ensure that the circuit = was=20 reset with a single AC cycle dropout, but this would be too fast for = normal use,=20 and quite unnecessary. C1 (marked with a *) will have to be selected = based on=20 the relay. If the value is too small, the relay will chatter or at least = buzz,=20 and will probably overheat as well, due to eddy currents in the solid = core used=20 in DC relays. The capacitor should be selected based on the value that = makes the=20 relay quiet, but still releases quickly enough to prevent high inrush = current if=20 there is a momentary interruption to the mains supply. The value shown = (470uF)=20 will generally be suitable for most applications.

You might want to consider using a mains switch with an additional = set of=20 contacts, so that the second set will short circuit the 12V supply when = power is=20 turned off. Make sure that the switch has appropriate ratings, and be = sure to=20 mark and insulate all connections. This is not really necessary though, = and for=20 a DIY project I'd have to say that it's not recommended because of the = risk.

Where it is not possible to use the transformer for any reason, then the = circuit=20 in Figure 3 can be used. This uses a resistor to drop the supply voltage = for the=20 relay, and has a simple zener diode regulator to supply the control = circuit. The=20 method of determining the resistor values and power for Rx and Ry is = shown=20 below.

3D"Figure
Figure 3 - Control Circuit Using Existing=20 Supply

WARNING: In the event of an amplifier fault at=20 power-on, the fuse may not blow immediately with this circuit installed, = since=20 there may be no power to operate the relay. The current is limited to = 200% of=20 that at normal full power, so the fuse may be safe for long enough for = it to=20 destroy the resistor(s)! The ballast resistors will overheat very = quickly, and=20 if you are lucky they will fail. If you don't like this idea - = USE THE=20 AUXILIARY TRANSFORMER.=20

I very strongly suggest the auxiliary transformer = - it is=20 MUCH safer!

The first calculation is based on the supply voltage, and determines = the=20 current available to the zener. This should be about 20mA (it is not too = critical). Since the zener is 12V, use the following formula to obtain = the value=20 for Rx ...

R =3D (Vcc - 12) / I (4) Where Vcc is the = voltage=20 of the main positive supply rail, I is current

Example. The Vcc (the +ve supply rail) is 50V, so ...

R =3D (50 - 12) / 0.02 =3D 1900 Ohms (1.8k is quite=20 acceptable)

Power may now be determined as follows ...

P =3D (Vcc - 12)=B2 / R (5)

Again, from the example above ...

P =3D (50 - 12)=B2 / 1800 =3D 38=B2 / 1800 =3D 1444 / 1800 = =3D 0.8W

A 2W resistor (or two 3k6 1W resistors in parallel) is indicated to = allow a=20 safety margin. Where possible, I always recommend that a resistor be at = least=20 double the expected power dissipation, to ensure long life and cooler = operation.=20 It may be necessary to select different resistor values to obtain = standard=20 values - not all calculations will work out as neatly as this. Remember = that the=20 20mA is only approximate, and anything from 15 to 25mA is quite = acceptable.=20

The relay coil limiting resistor (Ry) is worked out in a similar = manner, but=20 first you have to know the resistance of the relay coil. This may be = obtained=20 from specifications, or measured with a multimeter. I have details of a = suitable=20 relay that has a 12V DC coil, and has a claimed resistance of 285 Ohms. = Coil=20 current is therefore ...

I =3D Vc / Rc (6) Where Vc is coil = voltage=20 and Rc is coil resistance
I =3D 12 / 285 =3D 0.042A = (42mA)

Using the same supply as before, formula 4 is used to determine the=20 'build-out' resistance ...

R =3D (50 - 12) / 0.042 =3D 904 Ohms. 1k Ohms will be = fine here=20 (less than 10% variation)

Power is determined using equation 5 as before ...

P =3D (50 - 12)=B2 / 1000 =3D 38=B2 / 1000 =3D 1444 / 1000 = =3D 1.4W

If the coil current is calculated with the resistor in place, it is = found=20 that it is 39mA - this is a variation of about 7%, and is well within = the=20 tolerance of a relay. A 5W resistor is indicated, as this has a more = than=20 generous safety margin. These resistors will be very much cheaper than a = transformer, and require less space. Wasted power is not great, and is = probably=20 less than that lost in a transformer due to internal losses (small = transformers=20 are not very efficient).=20

With relays, it is often beneficial to use a power saver circuit, = where an=20 initial high current pulse is used to pull the relay in, and a lower = holding=20 current is then used to keep it energised. This is very common in relay=20 circuits, and can provide a saving of about 50%. The basic scheme is = shown in=20 Figure 4 with some typical values for the relay as mentioned in the = text. I have=20 based my assumptions on the relay I have - I tested this part = thoroughly, since=20 it is very difficult to make calculations based on an electro-mechanical = device=20 such as a relay - there are too many variables. If you want to use this = method,=20 then I suggest that some experimentation is in order. Typically, the = relay=20 holding current will be between 20% and 50% of the pull-in current - = generally=20 at the lower end of the scale.=20

3D"Figure
Figure 4 - Power Saving Relay = Circuit

The values shown are those estimated for the 12V 285 Ohm relay - = yours will=20 be different! Do not mess about with this method if you are unsure = of what=20 you are doing. Failure of the relay to operate will cause the ballast = resistors=20 to overheat, with possibly catastrophic results (See below). This method = can=20 also be used with Class-A amps, as it is possible to make sure that the = relay=20 activates even on the lower voltage present while the ballast resistors = are in=20 circuit. (Although I strongly suggest the separate power supply circuit = for=20 Class-A, see Class-A=20 Amplifiers, below.)=20

Notice that the power savings are across the board. The relay feed = resistor=20 now will dissipate 0.8W instead of 1.4W, and the auxiliary limiting = resistor can=20 be a 0.5W type - instantaneous dissipation is only 0.7W, and that is for = a very=20 short time. The feed resistor is now 2k2 instead of 1k, but an extra = capacitor=20 and resistor are the price you pay. The capacitor can be used in the = circuit of=20 Figure 3 too, and will force a large current at turn on. This will not = save any=20 power, but will most certainly ensure that the relay pulls in reliably.=20

A Few Test Results
The relay I used for testing is a 24V = type -=20 this in itself is of little consequence, since it can easily be = re-calculated or=20 re-measured for a 12V unit. A coil resistance of 750 Ohms means that at = nominal=20 supply voltage the relay draws 32mA. I measured the pull-in current at = 23.5mA=20 (typically about 65% of the nominal rating), and drop-out current was = 7.5mA, or=20 about 25% of rated current.

Using the 12V relay mentioned above, this would translate to = (approximately -=20 these are educated guesses)=20

Nominal current - 42mA
Pull-in Current - 28mA
Drop-out Current - 10mA

Most (all?) relays will hold in perfectly well at 1/2 rated current, = and I=20 would suggest that this is as low as you should go for reliability. If = you don't=20 feel like including it, the resistor in series with the electro can be = omitted.=20 Sure this will pulse a 12V relay with 50V, but it won't care. Personally = I=20 suggest that a series limiter be used, calculated to provide an = instantaneous=20 current of 150% of the relay's nominal rating - this will protect the = cap from=20 excessive current. For a 12V unit (as above), this would mean a maximum = current=20 of 60mA and a holding current of 20mA.

Because of the vast number of variables, I shall leave this to your=20 experimentation - Please do not ask me to calculate the values for you, = because=20 I won't. It is entirely the reader's responsibility to determine the = suitability=20 of this (or any other) project to their individual needs. If in any = doubt, use=20 the auxiliary transformer method.

Construction Notes=20

As described above, electrical safety is paramount with a circuit = such as=20 this. Figure 5 shows a suggested method of mounting the input ballast = resistors=20 that ensures that the minimum of 5mm creepage and clearance is = maintained when=20 the resistors are mounted, and still provides good thermal contact with = the case=20 and protection from fingers or other objects coming into contact with = the=20 mains.

3D"Figure
Figure 5 - Suggested Resistor = Mounting

This arrangement may be a little over the top, but feel free to use = it if you=20 want to. The aluminium bracket clamps the resistors firmly in position, = and the=20 plate above and below (which needs to be 5mm shorter than the resistor = bodies)=20 maintain clearance distances. It is imperative that the resistors cannot = move in=20 the bracket, and a good smear of heatsink compound will ensure thermal=20 conductivity.

The alternative is to obtain one of the bolt-down aluminium bodied = resistors.=20 This is obviously much simpler than making up a bracket. In case you are = wondering why all this trouble for resistors that will be in circuit for = 100=20 milliseconds, the reason is safety. The cover will keep fingers away, = and stops=20 the resistors moving about. It also provides a measure of safety if the = relay=20 does not operate, since dissipation will be very high. Since the = resistors will=20 get extremely hot, simply wrapping them in heatshrink tubing will do no = good at=20 all because it will melt. The idea is to prevent excessive external = temperatures=20 until the resistors (hopefully) fail and go open circuit. The method = used with=20 the P39 PCB is simpler again - 3 x 5W resistors are mounted on an = auxiliary=20 circuit board. I have yet to see or hear of a resistor failure.

The relay wiring is not critical, but make sure that there is a = minimum of=20 5mm between the mains contacts and any other part of the circuitry. = Mains rated=20 cable must be used for all power wiring, and any exposed connection must = be=20 shrouded using heatshrink tubing or similar. Keep as much separation as = possible=20 between any mains wiring and low voltage or signal wiring.

The connections to the ballast resistors are especially important. = Since=20 these may get very hot if the relay fails to operate, care must be taken = that=20 the lead will not become disconnected if the solder melts, and that = there is=20 sufficient solder to hold everything together and no more. A solder = droop could=20 cause a short to chassis, placing you or other users at great risk of=20 electrocution. An alternative is to use a screw-down connector, which = must be=20 capable of withstanding high temperature without the body melting.

Do not use heatshrink tubing as insulation for the incoming power = leads to=20 the ballast resistors. Fibreglass or silicone rubber tube is available = from=20 electrical suppliers, and is intended for high temperature = operation.

Class-A Amplifiers

NOTE: I strongly suggest that the = auxiliary=20 transformer method is used with a Class-A amp, as this will = eliminate any=20 possibility of relay malfunction due to supply voltages not being = high=20 enough with the ballast resistors in = circuit.

Because of the fact that a Class-A amp runs at full power all the = time, if=20 using the existing supply you must not go below the 200% suggested = inrush=20 current limit. In some cases, it will be found that even then there is = not=20 enough voltage to operate the relay with the input ballast resistors in=20 circuit.

If this is found to be the case, you cannot use this method, or will = have to=20 settle for an inrush of perhaps 3-5 times the normal full power rating. = This is=20 still considerably less than that otherwise experienced, and will help = prolong=20 the life of the supply components, but is less satisfactory. The = calculations=20 are made in the same way as above, but some testing is needed to ensure = that the=20 relay operates reliably every time. See note, above.

Special Warning=20

In case you missed this the first time: In the event of an amplifier = fault at=20 power-on, the fuse may not blow (or at least, may not blow quickly = enough to=20 prevent damage) with this circuit installed, since there may be no power = to=20 operate the relay. If you don't like this idea - USE THE = AUXILIARY=20 TRANSFORMER. The fuse might only blow after the relay closes, = but at=20 least it will blow. 100ms is not too long to wait. =20

This circuit by its very nature is designed to limit the maximum = current at=20 power on. If there is no power to operate the relay, the ballast = resistors will=20 absorb the full mains voltage, so for my example above will dissipate = over 900W!=20 The resistors will fail, but how long will they last? The answer to this = is a=20 complete unknown (but "not long" is a good guess). Thermistors may or = may not=20 survive.

The reliability of the relay circuit is paramount. If it fails, the = ballast=20 resistor dissipation will be very high indeed, and will lead to it = overheating=20 and possibly causing damage. The worst thing that can happen is that the = solder=20 joints to the resistors will melt, allowing the mains lead to become=20 disconnected and short to the chassis. Alternatively, the solder may = droop, and=20 cause a short circuit. If you are lucky, the ballast resistors will fail = before=20 a full scale meltdown occurs.

Make sure that the mains connections to the resistors are made as = described=20 above (Constructio= n=20 Notes) to avoid any of the very dangerous possibilities. You may = need to=20 consult the local regulations in your country for wiring safety to = ensure that=20 all legalities are accounted for. If you build a circuit that fails and = kills=20 someone, guess who is liable? You!

It is possible to use a thermal switch mounted to the resistor = cover=20 to disconnect power if the temperature exceeds a set limit. These = devices=20 are available as spare parts for various household appliances, or = you may=20 be able to get them from your normal supplier. Although this may = appear to=20 be a desirable option, it is probable that the resistors will fail = before=20 the thermal switch can operate.=20

WARNING: The small metal bullet shaped=20 non-resetting thermal fuses have a live case (it is connected to = one of=20 the input leads). Use this type with great caution !! Also, be = aware that=20 you cannot solder these devices. If you do, the heat from = soldering will=20 melt the wax inside the thermal fuse and it will be open circuit.=20 Connections should use crimped or screw = terminals.

PCB Version=20

The circuit diagram for the PCB version of this project is shown = below. It=20 uses a small transformer, and mains switching is only required for the = small=20 transformer, and the circuit takes care of the rest. The relays have a = standard=20 footprint, and should be available (almost) everywhere.

3D"Figure
Figure 6 - PCB Version of Soft Start/ Mains=20 Switch

A 9V transformer is needed, having a rating of around 5VA. The DC = output is=20 close to 12V, and will activate the relays reliably. The circuit has a=20 reasonably fast drop-out and stable and very predictable timing (approx = 100ms).=20 The PCB has space for 3 x 5W resistors (or a suitable thermistor), and = the=20 circuit has been used on 500VA transformers with great success. The = other=20 comments above still apply (of course), but this circuit simplified the=20 construction process considerably.=20

Feel free to use a thermistor instead of the resistors, but = only if=20 the thermistor is rated for high enough peak current. If you use a 25 = ohm=20 thermistor with 230V mains, assume worst case instantaneous peak current = of 13A.=20 With 120V mains, a 10 ohm thermistor will allow a maximum peak of just = under=20 17A. The thermistor (or resistor) used must be able to handle the peak = current=20 without failure.

Full details, bill of materials, etc. for the PCB version of P39 are=20 available on the secure server, along with detailed construction guide = and mains=20 wiring guidelines.

=20

Projects=20 Index

Main Index

Copyright Notice. = This=20 article, including but not limited to all text and diagrams, is = the=20 intellectual property of Rod Elliott, and is Copyright (c) 1999.=20 Reproduction or re-publication by any means whatsoever, whether=20 electronic, mechanical or electro- mechanical, is strictly = prohibited=20 under International Copyright laws. The author (Rod Elliott) = grants the=20 reader the right to use this information for personal use only, = and=20 further allows that one (1) copy may be made for reference while=20 constructing the project. Commercial use is prohibited without = express=20 written authorisation from Rod Elliott.

Page Created and Copyright (c) 06 Dec 1999./ = Updates: 03 Apr=20 - Modified suggested startup current./ 30 Jan 2000 - added warning about = non-resetting thermal fuse./ 18 Apr 06 - corrected errors and = inconsistencies./=20 12 Nov 2010 - added extra info about thermistors.

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