Thanks guys.
I just ordered (three) PP11010.0AOBV (the one Steve linked).
John
@gunrunnerjohn posted:Here's the Sensata-Airpax Breaker Datasheet, so tantalizing close, yet seemingly unavailable. You used to be able to get the instant trip models from Digikey, but now they only stock the slow ones.
I actually just picked a set of the 8A version of these up from Mouser to put them through their paces. They only had two and are listed as a non-stock item.
I also picked up a set of these:
https://www.mouser.com/Product...2t12iIyE3HB32g%3D%3D
They are 10A but have very similar specifications.
Its surprising there isn’t at least one variety of the instantaneous trip Sensata/AIRPAX that are normally stocked. They have about a thousand different versions with different actuators, mounting style, and termination. I’d like illuminated rocker switches but had to settle for the push to reset style. Nothing seems to be available in quantity.
Would those 10 amps be good to but on each handle of a z4000?
Thank you gentlemen. Your discussion is fascinating, educational, and sometimes over my head.
Here’s what I’ve decided for the 1032: 5 or 6 amp breaker of some kind for each terminal, U, B, & C
11 is the magic number
C-B is a constant 11 volts (10.9 on my multimeter). U-B is variable: 0-11 I’ll draw a smaller amount than 11, closer to 5 or 6, but it makes sense to me to use the number 11, already in play and the high end of the U-B variation.
Returning to RJR’s formula, 75 x 0.75 = 56.25 divided by 11 = 5.1 amps
Don, you are misapplying my formula. It is for use only to determine the amps which the secondary windings of the transformer, or any portion of them, can carry continuously. An amp is an amp, and has the same heating effect upon a given length of wire, regardless of the voltage being applied. The portion of the secondary windings which is carrying your load will be overheating. Bottom line is that 5 amps is too much for that transformer.
A 5-amp breaker on each terminal may not limit current everywhere in the secondary to 5 amps
@Don Baird posted:Here’s what I’ve decided for the 1032: 5 or 6 amp breaker of some kind for each terminal, U, B, & C
Those breakers will never trip in this application with a 1032.
Shoot! I thot I had it. I’m sorry I misunderstand and mis-applied your formula, RJR, but glad to know.
I’ve spent the last hour watching uTube videos trying to understand this stuff. However, I’m a theologian, not an electrician or even a handyman (as I’ve said, still in electrical kindergarten).
I’m tempted to guess something lower than 5 amps and pray it works! But we theologians know that temptation can be dangerous, and I suspect you all think guessing is dangerous as well. I’m stuck.
It makes me wonder, Bob, do I even need to worry about it “in this application?”
As a theologian, you should know that the quickest way to end temptation is to give right in. 
I'd try for 4 amp breakers. Does your test meter have an AC amps range? If so, you could check your lighting load and have greater certainty.
If you don't often have derailments with the trolley, chepaest way out would be 4-amp fuses.
Thanks😀
iim in Cincinnati for my grand kids soccer games and track meets. When I get home to Canton I’ll get the meter out. In the meantime I’m “stepping out in faith” and ordering the fastest 4 amp breakers I can find.
My layout is under construction but I can run the appropriate tests.
I am a little Surprised by your recommendation as 4 amps doesn’t seem much lower than five, but thankful nonetheless.
Are you guys saying now that those 10A circuit breakers Steve linked (PP11010.0AOBV) that I ordered yesterday are inadequate? Or am I misreading the comments.
John
@Craftech posted:Are you guys saying now that those 10A circuit breakers Steve linked (PP11010.0AOBV) that I ordered yesterday are inadequate? Or am I misreading the comments.
John
John, whether these 10A breakers are appropriate for your set-up depends on what you plan to connect them to. What transformer to you intend to connect them to and how many trains, accessories, etc will be fed by each circuit at the same time?
@SteveH posted:John, whether these 10A breakers are appropriate for your set-up depends on what you plan to connect them to. What transformer to you intend to connect them to and how many trains, accessories, etc will be fed by each circuit at the same time?
I have a KW to run one conventional Postwar Lionel, or Williams steamer, or Williams GP38on one side.
Other side to run one Bump n Go Trolley or an MTH subway set (which didn't arrive yet).
I use the accessory terminals on the KW, but I have a separate doorbell transformer which I will put into service for accessories.
John
For sizing the breakers for a KW, apply the formula I gave Don (above).. I cannot picture a doorbell transformer having the power to run any accessory.
John, as I understand it the KW is a 190W transformer. Assuming that 190W is input and applying RJR's 75% ouput efficiency calculation, that means it would put out about 143 Watts. If you're running conventionally, lets say that you maybe you don't always run them at full throttle, so at a moderate speed that might be about 12-14 volts. So in a derailment, additional current is drawn from the transformer until it reaches 10 Amps and kicks the breaker. So 14 volts x 10 Amps = 140 Watts, that works. Let's say your Loco is flying along at 18 volts and there's a derailment. This would require 180 theoretical watts being drawn from the transformer for a tiny fraction of a second until the external 10A breaker trips. Can the KW do this? I'll let someone else answer that question. My guess is that a 7 to 8 amp breaker would be a better choice for the KW.
@SteveH posted:John, as I understand it the KW is a 190W transformer. Assuming that 190W is input and applying RJR's 75% ouput efficiency calculation, that means it would put out about 143 Watts. If you're running conventionally, lets say that you maybe you don't always run them at full throttle, so at a moderate speed that might be about 12-14 volts. So in a derailment, additional current is drawn from the transformer until it reaches 10 Amps and kicks the breaker. So 14 volts x 10 Amps = 140 Watts, that works. Let's say your Loco is flying along at 18 volts and there's a derailment. This would require 180 theoretical watts being drawn from the transformer for a tiny fraction of a second until the external 10A breaker trips. Can the KW do this? I'll let someone else answer that question. My guess is that a 7 to 8 amp breaker would be a better choice for the KW.
Instead of putting more thought into it I jumped on the relative bargain and quickly ordered three of the 10A. Live and learn. But I never run any trains at high speed. I don't like the way they look or operate (especially on turnouts). Half way is high for me.
Thanks,
John
SteveH said it. 10 amp is suitable for a postwar ZW (essential) or an MTH Z4000 or Lionel PH-180 (redundant to internal breakers). ABout 7 is right for a KW.
Steve, I do not agree with one part of your analysis. The formula I gave established a maximum continuous amperage through any portion of the secondary windings. If you are running at a lower than full voltage, the wattage being produced by the transformer will be less, but the amperage through the portion of the secondary windings that you have tapped off is determinative. In other words, if you set the voltage at 6 volts and are pulling 15 amps, while that is only 60 watts, a portion of the secondary winding is severely overloaded.
Don, I see Digikey has a 4-amp breaker listed,
@SteveH posted:...180 theoretical watts being drawn from the transformer for a tiny fraction of a second until the external 10A breaker trips. Can the KW do this? I'll let someone else answer that question. My guess is that a 7 to 8 amp breaker would be a better choice for the KW.
The KW breaker is sized to trip in 15-45 seconds into .2 ohms on the A-U posts set to draw 20 amps.
@RJR posted:SteveH said it. 10 amp is suitable for a postwar ZW (essential) or an MTH Z4000 or Lionel PH-180 (redundant to internal breakers). ABout 7 is right for a KW.
Steve, I do not agree with one part of your analysis. The formula I gave established a maximum continuous amperage through any portion of the secondary windings. If you are running at a lower than full voltage, the wattage being produced by the transformer will be less, but the amperage through the portion of the secondary windings that you have tapped off is determinative. In other words, if you set the voltage at 6 volts and are pulling 15 amps, while that is only 60 watts, a portion of the secondary winding is severely overloaded.
Don, I see Digikey has a 4-amp breaker listed,
RJR, I'm trying to better understand what you're telling me about the current draw at a lower voltage. Just to clear up a minor point, I think you meant that 6V x 15A = 90W? Mainly what I'm not understanding is the concept of how 15 amps (regardless of voltage) could be drawn through the PP11010.0AOBV "instant" trip 10 External Amp breaker we're discussing now?
@ADCX Rob posted:The KW breaker is sized to trip in 15-45 seconds into .2 ohms on the A-U posts set to draw 20 amps.
Rob, thank you for that additional information about the KW. It's unclear to me if you realize the External Preaker currently being discussed is the Sensata / AIRPAX PP11-0-10.0A-OB-V If you are following that, then will you please elaborate a little more on your previous reply?
https://www.onlinecomponents.c...0aobv-10090622.html#
Steve, I used 15 amps just as an example. Plug in 10 and my example still works. Bottom line is notthe voltge whichyou have set or the wattage being drawn, but the amperage through any portion of the secondary coil windings.
On the plus side, Online Components hadn't shipped yet so I simply replied to the order confirmation expressing my concerns and got an immediate response. They changed my order immediately.
Hello John,
Thank you for reaching out!
I definitely understand the concerns.
PN: PP11-0-7.50A-OC-V we have available stock and is between 7-8 amps.
Would this part be able to work for your project?
If there is anything else I can help you with, let me know. Have a great rest of your day!
Best,Kianna Minley | Customer Service Representative
Phone: 888-906-8217
Visit: onlinecomponents.com
@SteveH posted:@ADCX Rob posted:@SteveH posted:...180 theoretical watts being drawn from the transformer for a tiny fraction of a second until the external 10A breaker trips. Can the KW do this? I'll let someone else answer that question. My guess is that a 7 to 8 amp breaker would be a better choice for the KW.
The KW breaker is sized to trip in 15-45 seconds into .2 ohms on the A-U posts set to draw 20 amps.
Rob, thank you for that additional information about the KW. It's unclear to me if you realize the External Preaker currently being discussed is the Sensata / AIRPAX PP11-0-10.0A-OB-V If you are following that, then will you please elaborate a little more on your previous reply?
Yes, the KW can put out a tremendous amount of current for a short duration.
@RJR posted:Steve, I used 15 amps just as an example. Plug in 10 and my example still works. Bottom line is notthe voltge whichyou have set or the wattage being drawn, but the amperage through any portion of the secondary coil windings.
Basically the breaker needs to be sized to protect the windings, but still allow “full use” of the output potential, based on efficiency. Right?
For everyone else, breaker size does not matter for a short circuit.** In that case, the trip curve matters, and for the sake of the electronics in our modern DCS and TMCC locomotives, faster is better.
Edit: ** within reason of course. For the low voltage wiring we’re talking about with our trains it doesn’t. However, if you look closely at the specifications you’ll see a maximum current rating as well.
@ADCX Rob posted:Yes, the KW can put out a tremendous amount of current for a short duration.
Rob, thanks for clarifying.
@Craftech posted:On the plus side, Online Components hadn't shipped yet so I simply replied to the order confirmation expressing my concerns and got an immediate response. They changed my order immediately.
PN: PP11-0-7.50A-OC-V we have available stock and is between 7-8 amps.
John, I'm glad you caught that in time and they were able to change your order.
@rplst8 posted:Basically the breaker needs to be sized to protect the windings, but still allow “full use” of the output potential, based on efficiency. Right?
For everyone else, breaker size does not matter for a short circuit. In that case, the trip curve matters, and for the sake of the electronics in our modern DCS and TMCC locomotives, faster is better.
So, if there are a sufficient # of TVS diodes installed, these would protect the sensitive modern electronics from voltage spikes....
@RJR posted:Steve, I used 15 amps just as an example. Plug in 10 and my example still works. Bottom line is notthe voltge whichyou have set or the wattage being drawn, but the amperage through any portion of the secondary coil windings.
And, I thought that having a correctly rated (based on the KW transformer specs), inline "instant" trip (lacking sufficient time details in its product spec sheet, which leads me to think probably ~10 mS or less) breaker would essentially "instantly" limit current flowing to the tracks and do a better job of protecting the transformer (primary and secondary) windings from overheating.
I would also tend to think that (with the correctly rated instant breaker) any excessively high current flow in the transformer secondary would be so brief, that heat damage would be very unlikely due to the "instant" trip function of the external breaker when it's limit is exceeded.
Am I missing something?
RJR I'm still not getting what it was you initially disagreed with my Reply to John about the 10A instant breaker being too high a value and instead recommending the same basic alternative as what you said. Maybe it's possible my meaning was misunderstood.
To be very clear, I'm not arguing anything. Just restating what I have come to understand so far. I'm here both try to help others and to learn. So, please consider taking the time to carefully explain what you mean if my understanding of any of this is incorrect.
@SteveH posted:So, if there are a sufficient # of TVS diodes installed, these would protect the sensitive modern electronics from voltage spikes....
You only need one, as long as it's in the right place. Optimum placement would be onboard the equipment you wish to protect.
@ADCX Rob posted:You only need one, as long as it's in the right place. Optimum placement would be onboard the equipment you wish to protect.
EDITED Reply: Rob, I understand see the validity of this assertion, even though I suspected that in addition to stating the truth you were also being humorous (see below). Based on your and @rplst8 replies I'm in the process of trying to better understand the physics of how the transients caused by the rapid switching of the motor coil current discharge (and in some cases capacitors) during a derailment cause high frequency waves to propagate in all directions through the affected conductors (track, Loco & other electronic circuits). The prospect of adding TVS to every electronic circuit presently seems daunting, but may well be the best solution.
Original Reply: Rob that's funny. Unless you have 1000 Locomotives like some here seem to have. 
In which case I've read that having one TVS diode adjacent to every power feed going to the track [EDIT assuming a feed every 10 track sections or 10 feet] is a good second best solution compared to installing them in every Locomotive, electronically equipped piece of rolling stock, track connected electronic accessory, and remote switch.
Would you agree?
@SteveH posted:Rob that's funny. Unless you have 1000 Locomotives like some here seem to have.
In which case I've read that having one TVS diode adjacent to every power feed going to the track is a good second best solution to installing them in every Locomotive, electronically equipped piece of rolling stock, track connected electronic accessory, and remote switch.
Would you agree?
I’ve read that too, but others with much better post grad degrees than me have said that unless the TVS is mere inches... maybe millimeters from the sensitive devices, they offer little to no protection in the derailment scenario.
Based on their reasoning and then my own reading elsewhere, I tend to agree with them. TVS at the power feed to the track will protect against normal transient spikes from the power source, but not the chaos that happens on a derailment where the load is repeatedly connected and disconnected in rapid succession.
@rplst8 posted:... breaker size does not matter for a short circuit.** In that case, the trip curve matters, and for the sake of the electronics in our modern DCS and TMCC locomotives, faster is better.
@SteveH posted:So, if there are a sufficient # of TVS diodes installed, these would protect the sensitive modern electronics from voltage spikes....
Yes, but see my other reply, and remember that during a short circuit, nearly all breakers, even an oversized slow one, will comfortably protect the transformer. This is because short circuit current is so high.
You don’t need instantaneous breakers to protect the transformer.
@RJR posted:If you don't often have derailments with the trolley, chepaest way out would be 4-amp fuses.
RJR, I’ve discovered a bunch of in-line auto fuse holders and a collection of different blade type fuses including 4-amp fuses. Am I correct that these are of no use to me because they are intended for a 12 volt DC auto setting?
@Don Baird posted:Am I correct that these are of no use to me because they are intended for a 12 volt DC auto setting?
It depends... Many automotive fuses are rated up to 32 volts, some are higher around 48 volts. Sometimes there is a rating stamped on them.
Here’s an example spec sheet, you can see at the top that the interrupt rating is 1000A @80V.
https://www.mouser.com/datashe...ade_fuses-523215.pdf
Here’s another mini auto style rated for 32V.
https://www.mouser.com/datashe...atasheet-1077557.pdf
SteveH, sorry if I'm not clear. What I am saying is that, in sizing a breaker, do not include any factor for what voltage you have the variable voltage control set at. As I read your post, you were calculating wattage based on the voltage you were outputting. I am focusing on determining the maximum current through any point in the secondary wiring.
Let me try to illustrate,(with recognition of the fact that we have many readers who aren't familiar with the basics) starting with basic transformer theory. The input and output voltages of a transformer depends on the ration of the number of turns of wire in the primary and secondary windings. Suppose the primary of a transformer has 1200 turns and is plugged into a 120-volt receptacle. Suppose the secondary has 240 turns. The output voltage will be 24 volts. Let's call one end "U" and connect it to a terminal. Let's call the other end "A". and connect it to a terminal . Put a voltmeter across terminal A-U and you'll get 24 volts.
Suppose, as in Lionel postwar transformers, there is a finger which, by turning a knob, slides across the secondary winding, starting at or near the U end. Connect this finger to a terminal labeled "B". Turn the knob until it is contacting the 80th turn (measured from U). A voltmeter across B-U will register 8 volts. A voltmeter across A-B would register 12 volts.
Let's say this transformer is rated 275 watts. Applying the formula I had way above: 275 x 0.75= 203 watts. The maximum voltage output is 24 volts. Divide 203 by 24 yields 8.6 amps. [amps x volts = watts] This is the maximum current that the wire in the secondary circuit can carry without undue heating.
Now suppose you are using only terminals U & B. You cannot say well I am only using 8 volts, and 203/8=26, so I can pull 26 amps. You still can only pull 8.6 amps, to avoid overheating the stretch of the secondary that is between U and B.
Going beyond, in the postwar ZW, there are 4 such fingers. If the finger for terminal A (right handle) is set at 8 volts and the terminal for terminal D (left handle) is set to 18 volts, you can also get 10 volts from terminals A-D. There is no internal protection. So you have A feeding one loop and D another. If you have those handles set per this example, when the loco crosses from one loop to another, with one roller on one circuit and the other roller on the second chircuit, you do have a dead short---you never want to stop a train that way. If perchance you stop with a lighted passenger car crossing the fibre pin, the short will flow through the car wiring and may well overheat the internal wiring, which is usually very fine. I also have experienced this overheating the roller springs, essentially destroying their tension. Solution is to have both handles at same voltage when crossing over.
On the subject of breaker/fuse trip speed, a thermal device, such as a thermal breaker or fuse, will not open immediately upon its rating being exceeded. Trip time can be fairly long on slight overloads, which can cause heating on the layout at the point of a derailment. Until recently, the cost of magnetic breakers was out-of-sight, and on my layout thermal breakers have worked well; BUT, I use smaller rated breakers so they heat up and trip faster. On my PH-180 circuits I have never had a external thermal breaker open before the PH-180's internal breaker. On my PW ZWs circuits, I've never had the ZW internal breaker open first. On my Z4000 circuits, either the internal 10-amp or the external 7.5-amp may trip first, depending on the load.
"I would also tend to think that (with the correctly rated instant breaker) any excessively high current flow in the transformer secondary would be so brief, that heat damage would be very unlikely due to the "instant" trip function of the external breaker when it's limit is exceeded" Correct.
Hopefully, I'm clear. RJR
RPLST8/Don: Don't sweat the voltage ratings. They indicate the voltage which the device can safely break. Put a 32-volt fuse to break a 240-volt current, and you may get a small explosion as the power jumps the gap in the fuse or breaker. Put a 120-volt fuse on a 16 volt circuit, and it will work just fine. I would say a 12-volt auto fuse is fine on a layout, or in a TIU..
Do note that there are short circuits and there are short circuits. A derailment causes a short, but if the wheel causing the short is a pinpoint contact with the 3rd rail, there may be enough resistance the the wheel heats and it'll be awhile before a thermal breaker/fuse opens. There may not even be enough current to open a 10-amp breaker or fuse.
@rplst8 posted:I’ve read that too, but others with much better post grad degrees than me have said that unless the TVS is mere inches... maybe millimeters from the sensitive devices, they offer little to no protection in the derailment scenario.
Based on their reasoning and then my own reading elsewhere, I tend to agree with them. TVS at the power feed to the track will protect against normal transient spikes from the power source, but not the chaos that happens on a derailment where the load is repeatedly connected and disconnected in rapid succession.
Ryan, Thank you for following up on this. I can sort of envision the possibility of what you're saying and would like to read more about this subject. Are there any online resources you can recommend?
At the same time, I also wonder that if with the combination of the "Instant" load breaker removing the source of the EMF in ~10mS or less and the TVS probably clamping the voltage faster than the breaker, how much high voltage current actually flows in the other direction to the comparatively high impedance inputs of the sensitive electronics in question.
But to reiterate, I'd like to learn more from others who have researched and detailed documentation (not just anecdotal tales) of their findings in this area. Your recommendations would be better than my haphazard internet searching.
@SteveH posted:But to reiterate, I'd like to learn more from others who have researched and detailed documentation (not just anecdotal tales) of their findings in this area. Your recommendations would be better than my haphazard internet searching.
For starters I would read the thread by @Adrian! on the TIU failures he saw at his club out in California. Let me find it...
@SteveH Here it is:
https://ogrforum.com/...between-aghr-and-mth
Here’s the specific reply where he talks about why closer to the thing you want to protect is important...
https://ogrforum.com/...74#79142442698148174
Mind you, this was a scenario at a club with lots of youngsters running trains and people being less than careful. Their layout has 5 TIUs IIRC, and like a mile of track. If this is your basement layout and you are of the careful sort, you may not have to worry that much.
@RJR posted:SteveH, sorry if I'm not clear. What I am saying is that, in sizing a breaker, do not include any factor for what voltage you have the variable voltage control set at. As I read your post, you were calculating wattage based on the voltage you were outputting. I am focusing on determining the maximum current through any point in the secondary wiring.
Let me try to illustrate,(with recognition of the fact that we have many readers who aren't familiar with the basics) starting with basic transformer theory. The input and output voltages of a transformer depends on the ration of the number of turns of wire in the primary and secondary windings. Suppose the primary of a transformer has 1200 turns and is plugged into a 120-volt receptacle. Suppose the secondary has 240 turns. The output voltage will be 24 volts. Let's call one end "U" and connect it to a terminal. Let's call the other end "A". and connect it to a terminal . Put a voltmeter across terminal A-U and you'll get 24 volts.
Suppose, as in Lionel postwar transformers, there is a finger which, by turning a knob, slides across the secondary winding, starting at or near the U end. Connect this finger to a terminal labeled "B". Turn the knob until it is contacting the 80th turn (measured from U). A voltmeter across B-U will register 8 volts. A voltmeter across A-B would register 12 volts.
Let's say this transformer is rated 275 watts. Applying the formula I had way above: 275 x 0.75= 203 watts. The maximum voltage output is 24 volts. Divide 203 by 24 yields 8.6 amps. [amps x volts = watts] This is the maximum current that the wire in the secondary circuit can carry without undue heating.
Now suppose you are using only terminals U & B. You cannot say well I am only using 8 volts, and 203/8=26, so I can pull 26 amps. You still can only pull 8.6 amps, to avoid overheating the stretch of the secondary that is between U and B.
Going beyond, in the postwar ZW, there are 4 such fingers. If the finger for terminal A (right handle) is set at 8 volts and the terminal for terminal D (left handle) is set to 18 volts, you can also get 10 volts from terminals A-D. There is no internal protection. So you have A feeding one loop and D another. If you have those handles set per this example, when the loco crosses from one loop to another, with one roller on one circuit and the other roller on the second chircuit, you do have a dead short---you never want to stop a train that way. If perchance you stop with a lighted passenger car crossing the fibre pin, the short will flow through the car wiring and may well overheat the internal wiring, which is usually very fine. I also have experienced this overheating the roller springs, essentially destroying their tension. Solution is to have both handles at same voltage when crossing over.
On the subject of breaker/fuse trip speed, a thermal device, such as a thermal breaker or fuse, will not open immediately upon its rating being exceeded. Trip time can be fairly long on slight overloads, which can cause heating on the layout at the point of a derailment. Until recently, the cost of magnetic breakers was out-of-sight, and on my layout thermal breakers have worked well; BUT, I use smaller rated breakers so they heat up and trip faster. On my PH-180 circuits I have never had a external thermal breaker open before the PH-180's internal breaker. On my PW ZWs circuits, I've never had the ZW internal breaker open first. On my Z4000 circuits, either the internal 10-amp or the external 7.5-amp may trip first, depending on the load.
"I would also tend to think that (with the correctly rated instant breaker) any excessively high current flow in the transformer secondary would be so brief, that heat damage would be very unlikely due to the "instant" trip function of the external breaker when it's limit is exceeded" Correct.
Hopefully, I'm clear. RJR
RJR, Very clear and an excellent explanation! I really appreciate the thought, time, and knowledge you put into it. Thank you.
My take away from what you are saying, is that a PW style transformer's Wattage rating is given at it's maximum possible output voltage. Using a PW ZW's actual specs as a real word example: (275W*0.75)/20V=10.3125A. I think you're saying this is the maximum current output it can safely deliver, regardless of voltage setting, and that this is because the secondary winding (wire) will begin overheating if this maximum rated current is exceeded in any part of that winding, regardless of the wiper position on the secondary winding (which determines the voltage output). Correct?
Notes to @ADCX Rob and @RJR. I recently edited my last replies to each of you, based on rereading them again this morning and with a fresh perspective. I look forward to your replies.
Also thank you to @rplst8 for posting the information for additional reading about TVS location. I'm still trying to get my head around this.
@ADCX Rob posted:You only need one, as long as it's in the right place. Optimum placement would be onboard the equipment you wish to protect.
@SteveH posted:Rob, I understand see the validity of this assertion, even though I suspected that in addition to stating the truth you were also being humorous...
Well, not really, that is just the optimal positioning. A good compromise for our purposes is one TVS per power district minimum.
