VL4000 looks like no joke

[Jay Ashworth]

The issue was "diameter in feet of the collision detection domain" and it only applied to hubs.

Switches don't have a collision domain, cause each link is point to point and full duplex.

 

[derekleffew]

Reminder: twice the voltage requires half the amperage, and hence a quarter the copper. ...

So #12 AWG will carry 80A (4x20) at 240V? Which NEC table is that?


Sent from my iPad using Tapatalk

 

[Jay Ashworth]

Sure, Derek; that's what I meant.

"a device of a given wattage which can run at 240V will draw half as much current at that voltage, and can therefore use wondering which is much lighter in gauge."

I believe, but did not look up, that it is approximately a quarter as much copper by weight.

 

[bishopthomas]

Am I the only one who finds it peculiar that the unit is offered with 2P&G, L5-20, and 5-15 plugs when it requires "Standard AC power distribution from 200 - 240VAC, 50/60 Hz." ? Hopefully this is simply a misteak on the prelim datasheet.

HAHAHA!! Misteak! Good one, Derek!

 

[Jay Ashworth]

Naw, Tom. The IEC C14 inlet, common on such gear, is rated for 240V; you can just replace the power cord if you have 240V to plug it into.

 

[Wood4321]

The issue was "diameter in feet of the collision detection domain" and it only applied to hubs.

Switches don't have a collision domain, cause each link is point to point and full duplex.

Fair enough, but Switches cost quite a bit more than a hub, and there is increased latency in a switch vs a hub.

Once again, this latency does matter! If you have hundreds of RGB Led fixtures on a network, this latency can be visible.

I don't see it as likely for manufacturers to include a 2 port switch in each fixture at any point in the near future. Too much cost, and too much latency per jump.
Now, a single port node, that I could see, especially if you could output DMX from the fixture, while inputting Artnet or Sacn.

 

[Jay Ashworth]

Not anymore: you can't *buy* a hub anymore; everything is a switch. And 2-port switch chips are, essentially (in terms of manufacturing a device) free; nearly all VoIP phone sets have them, for example.

Hubs are single-speed; any port that is 10/100(/1000) is necessarily a switched port.

The latencies are microseconds, as far as I know, especially with cut-through; I believe that's fractional-DMX-frame, yes?

That said: yes, you want the actual end-fixtures on something that's multi-drop, cause even microseconds do add up. That's probably largely why DMX is based on RS-485 -- that, and the inherent EMI immunity good 485 driver/receivers provide; 485 was designed for industrial use.

 

[Wood4321]

Not anymore: you can't *buy* a hub anymore; everything is a switch. And 2-port switch chips are, essentially (in terms of manufacturing a device) free; nearly all VoIP phone sets have them, for example.

Hubs are single-speed; any port that is 10/100(/1000) is necessarily a switched port.

The latencies are microseconds, as far as I know, especially with cut-through; I believe that's fractional-DMX-frame, yes?

That said: yes, you want the actual end-fixtures on something that's multi-drop, cause even microseconds do add up. That's probably largely why DMX is based on RS-485 -- that, and the inherent EMI immunity good 485 driver/receivers provide; 485 was designed for industrial use.

Right, the components are cheap, the software to implement them are not.
Anyhow, I think overall we are saying the same thing, that it doesn't really make much sense to do this in the fixture..

 

[TJCornish]

Sure, Derek; that's what I meant.

"a device of a given wattage which can run at 240V will draw half as much current at that voltage, and can therefore use wondering which is much lighter in gauge."

I believe, but did not look up, that it is approximately a quarter as much copper by weight.

That's not how it works, but your confusion is understandable.

Power = current * voltage. A 1000w load at 120v requires 8.33 amps. A 1000w load at 240v requires 4.166 amps. Electrical conductors are rated at a particular ampacity - a 12ga wire can handle 20A, no matter the voltage of the circuit (subject to situational detratings, yada yada). A 120v circuit can supply a maximum of 2400 watts of power before the current of of the circuit exceeds 20A. A 240v circuit can supply a maximum of 4800 watts of power before the current of the circuit exceeds 20A. This is a linear relationship, and doubling the voltage at a given current doubles the power.

The confusion comes when you combine Ohm's law and Joule's law and do derivatives like P = V^2/R, and now we see a square term. For a given resistance, power indeed increases at the square of the voltage. What gives? The confusion comes from the fact that the resistance doesn't stay constant as you increase the voltage.

For example,
If you are designing a heater and wish that heater to dissipate 1000 watts, the resistance of your heater needs to be 14.4 ohms if it will be supplied by 120 volts (120v^2/1000 = 14.4). If your supply voltage is 240v, the resistance of your heater will be 57.6 ohms (240^2/1000 = 57.6).

Understanding the reason electrical power distribution runs at high voltage requires a more accurate circuit diagram. You have a supply voltage, a load resistance, and you have a wire resistance.

Using our heater example of 1000w, let's assume the circuit wiring resistance is 1 ohm:

- At 120v, our circuit resistance is 15.4 ohms - the 14.4 ohm resistance of the load, plus the 1 ohm resistance of the wire. In this case, 6.5% of the power is wasted in the wire - 60 watts, and our 1000w heater will only put out 876 watts.

- At 240v, our circuit resistance is 58.6 ohms - the 57.6 ohm resistance of the load, plus the 1 ohm resistance of the wire. In this case, the wire is a much lower percentage of the load impedance, meaning less loss, and more power delivery to the load. Now the circuit loss is less than 2%. Only 17 watts are now lost in the wire, and we get 968 watts to our load.

 

[Jay Ashworth]

Which part, exactly, were you alleging I was wrong about?

Cause it appears we agree on all the parts I actually mentioned...

Except perhaps how much copper is necessary not to be overloaded by a given single load, and how much less that weighs at the higher voltage, on which I noted I was winging it.

"will put out only 876 watts" seems wrong to me as well: the heater is going to draw whatever attachment cap wattage it's rated for, plus or minus a little; if the resistance of the supply wiring is notably high, that means that breaker-point load would be *higher*, no?

If the attachment-cap voltage is dropped by that resistance, then the current draw at that spot goes *up* to maintain Ohm's Law...

Or is there still something I'm missing?

 

[TJCornish]

Which part, exactly, were you alleging I was wrong about?

Cause it appears we agree on all the parts I actually mentioned...

Except perhaps how much copper is necessary not to be overloaded by a given single load, and how much less that weighs at the higher voltage, on which I noted I was winging it.

The "1/4 of the copper" part.

"will put out only 876 watts" seems wrong to me as well: the heater is going to draw whatever attachment cap wattage it's rated for, plus or minus a little; if the resistance of the supply wiring is notably high, that means that breaker-point load would be *higher*, no?

If the attachment-cap voltage is dropped by that resistance, then the current draw at that spot goes *up* to maintain Ohm's Law...

Or is there still something I'm missing?

In my example of a 1000 watt heater, adding resistance in the supply wiring will decrease not only the fraction of power going to the load compared to what's lost in the wire, but will reduce the total wattage of the system as well. 120 volts into 14.4 ohms is 8.33 amps and 1000 watts. 120 volts into 15.4 ohms - heating element plus supply wiring, is 7.8 amps, for a total power of 936 watts for the system; 876 of those go to the heater, the other 60 are lost in the supply wiring. For a "dumb" load, there is no automatic adjustment factor (other than whatever fudge factor the manufacturer put in to the product to assume some supply losses, though this is a fixed quantity) to compensate for the total system resistance - it's directly Ohm's law.

Loads that are regulated in some fashion may behave like you describe - a mover with a universal power supply will increase current consumption to maintain a fixed power. Induction motors do this as well - as voltage drops, current increases, within certain parameters, which is why low voltage can damage them, but that wasn't the example I gave.

 

[Jay Ashworth]

I did note that it was an estimate. Regardless how much less copper it takes for a given wattage load at a higher voltage, I think we'll have to agree it *does* take less copper...