Surge Protector

Suggest you attend to your website links.

Reply to
Sally
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Reply to
Sally

One can't expect any more from a cheap imported transister sales person

Reply to
a t e c 7 7

**Feel free to list all the known examples of where a standard, off the shelf surge protector has protected equipment. If you want snake oil products (over-priced surge protectors) then you'll need to shop elsewhere.

Learn about 'top-posting' and try to refrain from doing so. It is poor netiquette. Let Google be your friend.

--
Trevor Wilson
www.rageaudio.com.au
Reply to
Trevor Wilson

**I suggest you read this site:

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Pay particular attention to the part on 'top-posting'. I realise that you are (probably) a newbie. NOW is the time to break your bad habits.

I am still awaiting your answer about surge protectors.

--
Trevor Wilson
www.rageaudio.com.au
Reply to
Trevor Wilson

nice sidestep and tap dance you spamming prick , top posting is and has been acceptable for some time and the weakness of your rebutal does nothing to repair your reputation or website.

Reply to
a t e c 7 7

Hi Trevor

It is a silly question because few people would record and document such an event (even if they were aware of it). But over 20 years dealing with events of this kind I would guess I have seen a few hundred successes (and some failures, with very close or direct strikes). I did not record them because I didn't then have the pleasure of foreseeing your question. I have the individual files going back 6 years (Statute of Limitations -- annual cull -- space problems!) but I am not going to hack through every report and each file and make a schedule just to prove a point.

BTW, I agree with your point about "cheap off the shelf"; I have never implied otherwise. The product has to be properly engineered.

I am not a newbie. The very opposite (active on Usenet since about 1984). I choose to top post because I believe that regulars prefer it to ploughing through a whole ream of stuff first. The majority are aware of the context. Things have changed since that netiquette point was established.

Trevor, your website link does not work -- try it!

Reply to
Sally

Yep, looks like

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is missing an index.html (and all the other files too).

Reply to
swanny

Not quite off topic, but it's in the spirit of the contents of this thread

Interesting news item on the Teev last night. Some kid wearing earphones and an ipod in his pocket got hit by lightning.

Now his family are claiming ipods are dangerous, and attract lightning Duh!!

It probably saved the kid's life by creating a better path down the side of his body, rather than through him. Like having a lightning rod - bypass the house, not hit the house

(be interesting if he claimed insurance for his ipod!!)

Trevor - your site link needs some TLC

Argusy

Reply to
Argusy

On Wed, 11 Jul 2007 10:49:41 -0700, w_tom put finger to keyboard and composed:

I would have thought that the converse was true. For the first 2 or 3 mains cycles after switch-on, the lamp draws 10 times its normal operating current, which means that it dissipates 10 times its rated power. This energy impulse warms up the filament, after which time the lamp begins to emit light. Presumably, when the lamp is emitting light it would be absorbing less energy because it would be converting the energy that would otherwise be absorbed as heat.

I accept that the surge *duration* is only microseconds, but the surge

*current* can be 100s or 1000s of amps. Hence the surge *energy* (VI.dt) could be far in excess of what the lamp experiences during switch-on, which is itself a stressful transient. And to make the situation worse, the surge voltage is not clamped (?) as it is in the case of a MOV, so AFAICS the effective impulse is Isquared.R.dt.

Let's assume the surge lasts for 1us and that the current is 1000A. A

100W 240V lamp would have a resistance of around 600 ohms when hot and about 60 ohms when cold.

The surge energy would then be at least 1000 x 1000 x 60 x 1E-6 = 60J.

As this is more than what the lamp experiences during a normal cold start (40J ?), one would expect that the strike would very quickly warm it up. In any case, my Internet research suggests that the duration of a typical strike is at least 20us, so the energy figure would be more like 1200J.

Sorry, but I can't make any sense of this.

Conductivity is something else altogether. You are confusing me by redefining your own terms.

I can't see it.

Sure, it shunts the surge *current* through itself in an attempt to protect the attached equipment, but in so doing it must dissipate an amount of energy equivalent to Vclamp x Isurge x dt.

To put this energy rating in perspective, 714J = 171 calories, which means that the MOV can absorb that amount of energy required to increase the temp of 171 grams of water by 1 degC, or 1 gram of water by 171 degC. As for how much this would raise the temperature of a typical MOV disc, I don't know, but it shouldn't be too hard to estimate.

- Franc Zabkar

--
Please remove one 'i' from my address when replying by email.
Reply to
Franc Zabkar

Time to study datasheets. Learn what the curves are reporting. How you we make an MOV more conductive? We increase it joules. How do we make an MOV dissipate less energy? We make it larger - more joules. It is right there in datasheets.

Again, what is the current rating of a 2mm wire? 20 amps? Then why is that same wire also sufficient for conducting a 50,000 amp surge? Same reason why a hundred amp surge inside the building does not blow out filaments. Same reason why an MOV rated for a 2000 amps surge will dissipate less than 2 watts AND why its wire leads are 0.8 mm diameter. 0.8 mm conducts 2000 amps and does not vaporize? Read datasheets.

Do some numbers. What is the resistance of a cold 100 watt bulb?

15 ohms? 300 amps squared times 15 ohms times 30 microseconds is maybe 40 joules. Well MOVs that are rated at 70 joules must not dissipate more than 2 watts after a 2000 amp surge. Why would the 100 watt light bulb be dissipating any more than 2 watts?

But again, read the MOV datasheet. Even a 0.8 mm wire on that MOV conducts 2000 amps and does not vaporize.

How do we make a MOV more conductive? We increase its joules. What makes a protector better? More joules. How much energy does an MOV dissipate during same a size surge when we increase its joules? Less energy is absorbed.

Again, joules that measures MOV performance measures parameters completely different in a light bulb. Do the numbers. To make a protector absorb less energy; to make it more conductive - then we increase its joules.

We don't want these shunt mode protectors to absorb any energy. However we make neither perfect conducting MOVs nor perfect conducting wires. Both MOVs and wires dissipate energy. Better means less energy dissipated by a wire and by an MOV.

MOVs don't work productively by absorbing surges. MOVs work by doing what wires also do.

Finally back to that 70 joule MOV. While conducting its maximum

2000 amp surge, it may absorb 70 joules. Meanwhile maybe 2400 joules from that same 2000 amp surge is being absorbed in earth. 2000 amps times 40,000 volts times 30 microseconds. 40,000 volts may be across 4 kilometers of earth between the earthed MOV and distant earthborne charges. That same surge did not dissipate over 1000 joules destructively inside the building because the 70 joule MOV shunted energy in 2000 amps elsewhere.

How do we increase energy dissipated in earth? We enlarge that 70 joule MOV to 300 joules. Now the MOV may absorb only 40 joules during that 2000 amp surge. That is even better protection. By increasing joules, we absorb less energy in the MOV - the MOV has been made more conductive.

Reply to
w_tom

On Thu, 12 Jul 2007 22:57:23 -0700, w_tom put finger to keyboard and composed:

The reason that it makes no sense is that both the filament and the MOV are resistive devices, the difference being that the MOV has a knee shaped IV characteristic. Below its rated voltage, the MOV's resistance is extremely high, but above this point the resistance falls sharply. Nevertheless, both devices still dissipate power according to the formula P=VxI, where V is the voltage appearing at the terminals of the device, and I is the current passing through the device. The energy dissipated during a surge would be VI.dt.

Resistivity and conductivity refer to an intrinsic property of a material.

OTOH, resistance is measured in ohms, conductance in siemens. It doesn't matter whether you have a MOV or a wire filament.

I still don't see it.

The amount of energy that a MOV dissipates would be given by VI.dt, ie it is determined by the event, not by the device. For example, a 20mm

275V 1000J MOV and a 1m 275V 1MJ MOV (does it exist?) will both dissipate the same amount of energy when hit by the same surge.

That makes no sense. Assuming that the MOV has a 275V rating, and assuming that the duration of the surge is 20us, then ...

Power = I.V = 2000A x 275V = 550kW Energy = P.dt = 550E3 x 20E-6 = 11J

Can you show me one?

Power = V.I = V x 2000A = 2Watts

Therefore V = 1 millivolt.

Clearly this is absurd.

Because it is a 100W light bulb?

No, the same amount of energy is absorbed, it's just that the bigger MOV will experience a lesser increase in temperature.

No, the amount of energy dissipated in the MOV is determined by V.I.dt.

But the MOV is designed to *clamp* the surge voltage, therefore it must absorb an amount of energy given by Vclamp.Isurge.dt.

Energy = V.I.dt.

If Energy = 0, then V=0, or I=0, or dt=0.

It shunted the 2000A through itself, and clamped the surge voltage to

275V.

The MOV still clamps the surge to 275V. The rest of the surge voltage (40kV - 275V) still dissipates in the earth, or the air, or wires, or wherever. Increasing the rating of the MOV doesn't alter the amount of energy that it is called upon to dissipate.

- Franc Zabkar

--
Please remove one 'i' from my address when replying by email.
Reply to
Franc Zabkar

I am troubled that you are posting without reveiwing MOV datasheets. My posts repeatedly referenced datasheets for good reason.

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Review a 2V250. Seventy joules. Conducts 4500 amps. Dissipates how many watts? Less than 1 watt. Left for you is to get another number from those datasheets - the maybe 0.8? mm diameter wire conducts 4500 amps without vaporizing. For reasons defined below, I have intentionally not provided enough numbers for doing arithmetic.

Myths proclaim MOVs protect by absorbing energy. If MOV's purpose is to absorb energy, then the protector must be a series mode device. It would connect between appliance and wall receptacle like a switch or fuse. But MOV protectors are shunt mode devices. MOV connects just like a light bulb - in parallel.

Series mode devices block (high resistance) to stop surge current. Shunt mode devices shunt (short, conduct, clamp, near zero resistance) to divert current. Clamping means that conductor (the MOV) is better when resistance is lower. Better shunt mode protector absorbs less energy.

Example: light bulb and MOV connect same way in a circuit. But light bulb conducts current without affecting voltage fed to adjacent appliance. MOV shunts current so that voltage to the adjacent appliance is reduced.

Again conductivity of a wire is defined in amperes (discussing siemens is only arguing semantics). Conductivity of an MOV is defined in joules. That example accurately defined the underlying concept.

A 20mm 275V 1000J MOV and a 1m 275V 1MJ MOV will not dissipate same energy when conducting the same surge. But again, study charts in those datasheets. And that is the point. You are speculating without first learning the technology. That 1MJ MOV will dissipate less energy. Saying otherwise means you did not study charts in datasheets or manufacturer application notes.

You also do not know how many watts are dissipated because datasheets were not read. Speculating without first learning manufacturer facts which is why I am explaining less with each post.

If confused, then numbers from that datasheet can say why I have posted in error. Instead you want me to read manufacturer datasheets and application notes for you? If you need me to do that, then experience says this thread will 'waste on' forever. Posted are bottom line facts. If you don't do the work from datasheets, well, I have already spent too much time explaining what manufacturers have already written. Your calculations are invalid because you did not first learn information.

BTW, I intentionally shorted some facts. Each reply says you ignored what was required - study of charts in manufacturer datasheets. You did not first consult manufacturer datasheets or their application notes. Therefore arithmetic, performed without information from those datasheets, is in error. Repeat again ... consult those datasheets.

Wire, light bulbs, and MOVs are resistive devices. Using your logic, then all must do exact same thing. Reality - joules determines MOV conductivity. As MOV joules increase, the MOV becomes more conductive, it becomes an even better shunt mode protector, AND it absorbs less energy. What happens when a shunt mode protector is even better? It absorb less energy. You will never understand that without first learning from those manufacturer charts and application notes.

Read datasheets before do> On Thu, 12 Jul 2007 22:57:23 -0700, w_tom

Reply to
w_tom

On Sat, 14 Jul 2007 00:18:37 -0700, w_tom put finger to keyboard and composed:

Left for you to understand is what the "Transient Power Dissipation" figure actually means. In fact it most likely refers to the average (not instantaneous) power that can be dissipated within the MOV (yes,

*within* the MOV) in response to *repetitive* transients. In other words, while the instantaneous surge power (watts = joules per second) can be of the order of hundreds of kilowatts, the "Transient Power Dissipation" rating is a reflection of the MOV's recovery time.

For example, let's assume that a 20mm MOV has just absorbed a 20us 70J surge. If it is rated at 600mW, then this means that it can tolerate

0.6 joules per second. Therefore the next transient must not arrive within 70/0.6 = 117 seconds, ie about 2 minutes.

The reason is that you clearly don't have a grasp of basic electrical theory.

Your NTE datasheet defines the MOV's energy rating as "the maximum electrical energy which can be dissipated *within* the varistor by a single impulse of 10 x 1000µs current waveform with continuous voltage applied".

If the device to be protected can tolerate 300V, say, then all that a shunt protector needs to do is to clamp the surge to 300V. It does this by diverting the surge current through itself. In so doing, it absorbs an amount of energy given by Vclamp.Isurge.dt.

I suggest that you study other shunt mode protectors such as transorbs, zener diodes, spark gaps, MOSORBs, etc.

Here is a Vishay datasheet.

High Surge Suppression Varistors:

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Note 4 on page 5 confirms what I have been saying about energy absorption, namely that ...

"If Vp is the clamping voltage corresponding to Ip [the peak current], the energy *absorbed* in the varistor is determined by the formula:"

E = K x Vp x Ip x t2

The "Peak Current as a Function of Pulse Width" drawing on page 6 shows a non-square current pulse. For t2 = 20us, we have K=1, and the formula essentially reduces to E = Vclamp.Isurge.dt.

For the sake of convenience I have assumed that a varistor has a perfect knee characteristic, and that the clamp voltage is the same for all surge currents. However, the characteristic is shaped more like an ice hockey stick with a finite slope.

Even so, if we take just one example, a particular 20mm MOV that has a voltage of 300V at 400A will have a voltage of ~310V at 800A. This means that if we connect two small MOVs in parallel, thereby making a larger MOV with twice the energy rating, then an 800A surge that was clamped to 310V by the single MOV will now be clamped to 300V by the two MOVs (since each will carry 400A). Therefore your contention about joule ratings and significantly reduced energy absorption is clearly false.

You can also directly compare a 320V 7mm MOV with a 320V 20mm MOV. The first clamps the surge at 780V while the second clamps it at 740V. That's hardly any difference at all.

That's exactly my point, only I would use the term "clamped".

That makes no sense. You cannot redefine existing electrical terms to suit yourself.

See the above examples. I suggest you study those same datasheets yourself.

You have not understood what the datasheets were telling you. Learn the difference between instantaneous power and average power.

Your "information" is invalid because you do not understand basic electrical principles. Datasheets will not teach you the meaning of energy, power, amperes, joules, volts, etc. That knowledge is assumed.

Done and ditto to you.

See the examples above. Learn what a "clamp" does.

Energy = Voltage x Current x time

See the examples above. Learn what a "clamp" does.

Energy = Voltage x Current x time

See the examples above. Learn the difference between average and instantaneous power, and the difference between watts and joules.

- Franc Zabkar

--
Please remove one 'i' from my address when replying by email.
Reply to
Franc Zabkar

On Sat, 14 Jul 2007 00:18:37 -0700, w_tom put finger to keyboard and composed:

Here is the datasheet for Littlefuse's ZA Varistor Series:

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Notice the references to *average* power as opposed to *instantaneous* power.

For example, note 1 on page 4 states ...

"Average power dissipation of transients not to exceed 0.2W, 0.25W,

0.4W, 0.6W or 1W for model sizes 5mm, 7mm, 10mm, 14mm and 20mm, respectively."

Page 5 talks about "Power Dissipation Ratings":

==================================================================== Should transients occur in rapid succession, the average power dissipation required is simply the energy (watt-seconds) per pulse times the number of pulses per second. The power so developed must be within the specifications shown on the Device Ratings and Specifications table for the specific device. Furthermore, the operating values need to be derated at high temperatures as shown in Figure 1. Because varistors can only dissipate a relatively small amount of average power they are, therefore, not suitable for repetitive applications that involve substantial amounts of average power dissipation. ====================================================================

- Franc Zabkar

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Please remove one 'i' from my address when replying by email.
Reply to
Franc Zabkar

On Sun, 15 Jul 2007 10:03:41 +1000, Franc Zabkar put finger to keyboard and composed:

I should have said that the surge current used in this example was

1000A.

- Franc Zabkar

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Please remove one 'i' from my address when replying by email.
Reply to
Franc Zabkar

Exactly. Now you are starting to grasp why lightning surges typically don't explode light bulbs. Why a 70 joule MOV dissipates so few watts. And why a surge so large to stress an MOV to maximum results in 10 or more times energy absorbed elsewhere. Dissipating most of the energy eslewhere is why MOVs are so effective shunt mode protectors. More energy is absorbed elsewhere (in earth) and less energy is absorbed in inside MOVs.

Joules consumed by a light bulb determines things such as light efficiency. Light bulbs are designed to absorb joules. But absorbing more joules is bad for MOVs. How to reduce the joules consumed by a shunt mode protector? Increase MOV joules rating (same reason we increase wire size). An MOV with more joules means that MOV absorbs less energy - which is exactly what we want for surge protection.

Until you grasp that fact from datasheet charts, then you will not understand what MOVs do. MOV does not provide better protection by absorbing more energy. MOV provides better protection by absorbing less energy AND when shunting more energy to earth. What makes a better protector? More MOV joules means it absorbs less energy. More MOV joules means a better conductive connection (shunt) to earth. What makes a better protection 'system'? Better earthing.

You formulas are correct. But assumed is that (for example) voltage is a mathematically independent variable. What happens when an MOV has higher joules? Then its Vp decreases resulting in more energy shunted to earth and less energy absorbed by the MOV.

In simple terms, MOVs perform a job more like a switch. A more conductive switch, means a better protector. That means more MOV joules so that less energy is absorbed by the MOV.

Series mode protectors operate better by absorbing more energy. Shunt mode protectors operate better by absorbing less energy. MOV is a shunt mode protector. Again, study charts in those manufacturer datasheets. The numbers that confirm the above concepts are in those manufacturer datasheet charts.

As an MOV joule rating increases, then MOV absorbs less energy. Less energy absorbed means longer life expectancy. Effective protectors also shunt surges and remain functional - another fact from those charts. A protectors joules rating must be high enough as to absorb less energy, shunt more energy to earth, and therefore remain functional after direct lightning strikes.

Install protectors to make direct lightning strikes irrelevant. A properly sized (sufficient joules) and earthed protector means protection inside appliances is not overwhelmed and homeowner should never even know the surge existed. Too few MOV joules means the protector absorbs more energy, vaporizes, and may even result in these scary pictures:

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Reply to
w_tom

Exactly. Now you are starting to grasp why lightning surges typically don't explode light bulbs. Why a 70 joule MOV dissipates so few watts. And why a surge so large to stress an MOV to maximum results in 10 or more times energy absorbed elsewhere. That is what we want. More energy absorbed elsewhere (in earth) and less energy absorbed in the MOV.

Joules consumed by a light bulb determines things such as light efficiency. Light bulbs are designed to consume joules. But absorbing more joules is bad for MOVs. How to reduce the joules consumed by a shunt mode protector? Increase MOV joules rating. An MOV with more joules means that MOV absorbs less energy - which is exactly what we want for surge protection.

Until you grasp that fact from those datasheet charts, then you will not understand what MOVs do. MOV does not provide better protection by absorbing more energy. MOV provides better protection by absorbing less energy AND when shunting more energy to earth. What makes a better protector? More MOV joules means it absorbs less energy. More MOV joules means a better conductive connection (shunt) to earth. What makes a better protection 'system'? Better earthing.

Your formulas are correct. But assumed is that (for example) voltage is a mathematically independent variable. Not true. Study the charts. What happens when an MOV has higher joules? Then its Vp decreases resulting in more energy shunted to earth and less energy absorbed by the MOV.

In simple terms, MOVs perform a job more like a switch. A more conductive switch, means a better protector. That means more MOV joules so that less energy is absorbed by the MOV.

Series mode protectors operate better by absorbing more energy. Shunt mode protectors operate better by absorbing less energy. MOV is a shunt mode protector. Again, study charts in those manufacturer datasheets. The numbers that confirm the above concepts are in those manufacturer datasheet charts.

As an MOV joule rating increases, then MOV absorbs less energy. Less energy absorbed means longer life expectancy. Effective protectors also shunt surges and remain functional - another fact from those charts. A protectors joules rating must be high enough as to absorb less energy, shunt more energy to earth, and therefore remain functional after direct lightning strikes.

Install protectors to make direct lightning strikes irrelevant. Properly sized (sufficient joules) and earthed protector means protection inside appliances is not overwhelmed and homeowner should never even know the surge existed.

Reply to
w_tom

I agree with Franc's posts. w_ is totally off in this branch of the discussion.

I would only add that actual measurements by the US-NIST guru on surges found that light bulbs would burn out on a surge (maybe 100 microsecond duration) at about 1500V (US - 120V bulbs). Typically only a single light bulb would burn out (it would bypass enough of the surge to protect the others). Another tidbit - in the US there is arc-- over at panels or receptacles at about 6000V limiting the voltage between power wires. Both voltages are likely to be higher on 230V systems.

-- bud--

Reply to
bud--

Voltage numbers provided by Franc for a 320V 7mm and 20mm MOV are for 1 ampere; not 1000 A. Voltage numbers for a 1000 A spike would be

1060 volts and just over 1000 volts; not 780V and 740V.

More reasonable numbers for these Vishay VDRH07K320 (45 joule) and VDRH20X320 (382 joule) MOVs would be 400A and 100A; in next two paragraphs.

A 45 joule 7mm MOV conducting a 400 A current is 1040 volts. A 382 joule 20 mm MOV conducting same 400 A is only just above 1000 volts. As joules increase, then the MOV becomes more conductive; absorbed energy decreases by more than 3%.

For 100A, voltage numbers for that 45 and 382 joule MOVs are 970V and

910V. Energy absorbed by a higher joule MOV decreases by 5%.

As previously stated, energy absorption decreases when MOV joules increase. More MOV joules means that MOV becomes more conductive. Better is a more conductive MOV. Why? Better shunt mode protectors absorb less surge. Shunt mode protector means more conductive (less power absorbed) is better.

Now to complete an analysis using those datasheet charts. What happens due to that 3% and 5% reduction in energy absorption? Fact from the datasheet that Franc did not grasp.

For a typical 30 microsecond 400A transient, the 45 joule MOV has a life expectancy of slightly more than 10 surges. But a more conductive 382 joule MOV has a life expectancy of 900 surges. With 8 times more joules, better conductivity means the MOV lasts 50 times longer. Just another reason why we want MOVs with better conductivity; to absorb less energy. Life expectancy increases massively.

For the 100A surge, life expectancy of a 45 joule MOV is 2,000 surges compared to 80,000 for the 382 joule MOV. Again, 5% less energy absorbed due to 8 times more joules results in a 40 times increase in life expectancy.

Demonstrated:

1) As MOV joules increase, then MOV conductivity increases. 2) More MOV joules means less energy absorbed. 3) Less energy absorbed also means less degrading of an MOV's crystalline structure, less degrading, and therefore a significant increase in MOV life expectancy. 4) MOV joules increased 8 times caused a 40 or 50 times increase in number of surges; increased MOV life expectancy.

More joules, better conductivity, and less energy absorbed means less MOV degradation during each surge. That is what we want from MOVs. More surge energy shunted elsewhere. Less energy absorbed by an MOV.

Returning to the original point. MOV are not for absorbing surges. We install MOVs to shunt (divert, clamp) that surge elsewhere. As demonstrated earlier, for every joule absorbed by an MOV, maybe 10 or

30 joules are shunted elsewhere. Increased MOV joules makes a better (more conductive) protector. Less energy absorbed by MOVs also results in significantly longer MOV life expectancy.

Using Vishay datasheets, MOVs are not installed to absorb surges. That directly contradicts popular myths. Unfortunately, it does absorb some energy because an MOV is not perfect. Increasing MOV joules means an MOV becomes more conductive and degrades less during each surge. Install more joules in a shunt mode protector for better conductivity and longer life expectancy.

Reply to
w_tom

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