Can someone please explain to me what a volt amp (VA) is. I have an extensive background in electronics and must admit I’ve never heard of a VA. I can only assume they really mean watts. I say this because of the formula P=I*E. Where P=watts or power, I=current or amps, and V=volts. So if I have a power pack that provides 7VA it’s really providing 7watts or 63milliamps of current (7=110*.063). 7=watts, 110=input or line voltage and .063 is the current. It doesn’t take a brain surgeon to figure out that the higher the VA rating of a power pack the more trains you can run. However, why invent a new term like VA when there was already a perfectly good term that would have worked. Another question-has anyone ever tried connecting two power packs in a parallel configuration to get a higher power rating? Again referring to the laws of electronics one should be able to tie two 7watt power packs in parallel and get 14watts of power as opposed to tying them in series and just increasing the voltage. This higher power would allow you to run more trains. Of course you would need to be careful as to not provide too much wattage or you might end up melting your track. Mark
From the Computer User's High Tech Dictionary: Definition for: volt-amps Volts times amperes; an electrical measurement. For direct current, one volt-amp equals the same power as one watt. Or you could say watt is a volt-amp. (No question mark.)
Peirce is right. However, in AC power relationships the watt is a vector product and I forgot exactly what the factor is, (square root of two, I think), such that a watt and a volt-amp are NOT equal in AC relationships, only DC. The term is not new, however it is used more widely to describe power systems than computer systems. Most aircraft AC generators are rated in KVA, or Kilo (1000) Volt-Amps. Each AC generator on the Lockheed L-1011 was rated at 90 KVA. You sure don't need that for your layout!
Yes, getting deep into mucky stuff here. In DC it's Watts, in AC it's VA. If the power factor of the AC is = 1.0, then W=VA. If p.f. is less than 1 then W is less than VA. If p.f. is more than 1, you are in the wrong dimension . Wot's p.f. you say? You don't really want to know because you may end up with imaginary values involving the square root of -1 . Seriously, in an AC system the voltage and current can be out of sync (phase) which means that although you may measure (e.g) 200V and 10A (2000VA) the real power available to do something useful, like light a bulb, will be less (maybe only 1800W). For utilities this is a big issue because in the above example the W result is only 200V x 9A. The 'lost' 1A is called wattless power because it doesn't light your house, BUT it does exist in the system and heats cables and transformers. We micro users can generally ignore it and say W=VA=W, but when you are dealing in mega-watts it is a major concern. You might have heard the phrase "power factor correction" - it means getting the p.f. near to 1 and thus eliminating the wattless curent . [ 29 September 2001: Message edited by: Mike Sheridan ]</p>
Therte's a factor you need to remember in the VA (volt-amp) rating of a power supply. You need to remember that the biggest factor in the rating of a power supply is the heat being dissapated by the power supply. For instance, I've got an 9000 power supply, rated at 19 VA. It has both a voltmeter and an ammeter. After a short while it'll trip out on overload with only three locomotives, but three heavily loaded locomotives pulling 12 coal cars and 8 ore cars carrying the real load on 4% grades. The load on the power supply is under 2 amps at around 9 volts. Technically within the limits of the power supply, but too much heat for it to handle. I'm in the process of beefing up the heat sink to handle the extra heat. I'm also toying with the idea of building an external "booster" to give me a lot more current to run stuff. Mark
Mark, I assume that this PS unit is capable of about 16 to 20 volts (let's say 19V) in which case it would be rated for 1 Amp at 19 Volts. Only using 9 Volts doesn't necessarily mean you can pull 2 Amps. If the output is pulse width modulated (PWM) then you should (?) be OK (for reasons I won't try and explain here), but if not then the 1 Amp limit still applies. The reason is that although you are only 'using' 9V at the rails the unit is probably still producing 19V and losing the other 10V worth as heat (as you say). This means that taking 2A at the track may actually be running the thing at 38VA .
