600 Watt Transformer / How Many Amps?

Discussion in 'Landscape Lighting' started by MEXANDME, Sep 1, 2011.

1. MEXANDMELawnSite Senior Memberfrom MYRTLE BEACH,SCPosts: 396

Hypothetically, a 600W TF fully loaded (at 85%) would pull 42.5 amps (600W x 85% = 510W / 12V) on the low voltage side of the TF.

Does this mean that the TF pulls 4.25 amps on the line voltage side (510W / 120V = 4.25 A)?

Sorry for such a basic question.

Thanks / Regards,

Mex

2. Lite4LawnSite Gold Memberfrom Indianapolis, INPosts: 3,113

Yes, that is correct. It could vary slightly based on your actual line voltage, but you have the basic idea there.

3. starry nightLawnSite Silver MemberMale, from Ohio's North CoastPosts: 2,765

Tim, I sent you a PM.

4. David GretzmierLawnSite Gold Memberfrom Fayetteville,ARPosts: 3,645

This answer is not so simple- A couple of other phantom amp things that are hard to nail down. You can add up all the amps on a 400 buck fluke true rms meter in the LV box, and check the imput line voltage on the trans, and then do the math, and then check the loop if you have one on the trans. and it never is what it is supposed to be. a few reasons- not only are all trans not 100% efficient, they may be 95% efficient at a 100 watt load and 80% efficient at a 250 watt load and so forth. I have seen different efficiencies on different commons in the same trans. further, to do the true amp math on the LV side you need to look at each tap voltage on each wire, not just divide by 12.

and trans are notorious for being very efficient at higher voltages- 120-125, and then progressively worse as line voltage drops to 115 to 105.

I've seen trans amp loads amp up a bit when running it through the photo cell and timer.

ambient temperature, and age of bulbs and fixtures in the system tend to change ( usually up ) trans load over time.

thus the 80% load rule. it tends to cover you for multi tap, line voltage loss, efficiency loss in the trans. But I am tending to stay closer to 50-60%, for the reason I have seen good 1200 watt trans show a 10 amp line load with a bulb wattage of 700 watts. long runs, high amp load on wire, reliance on the 17-22volt tap, can really stress a trans.

5. steveparrottSponsorfrom Tampa, FLPosts: 1,199

Excellent points, David. One other - it's not correct to use the terms 'watts' in these equations. Volt-amps from primary to secondary is what you want to compare. Watts is only a reference to the actual power used at the lamps. Volt-amps include additional energy consumed by wire loss, voltage tap used, and power factor (for LED's).

6. MEXANDMELawnSite Senior Memberfrom MYRTLE BEACH,SCPosts: 396

Thanks to all of you for the very timely and informative replies.

Guys like you make this forum great!!!

Regards,

Mex

7. David GretzmierLawnSite Gold Memberfrom Fayetteville,ARPosts: 3,645

point taken steve. watts is a term that most home owners can understand, and many times they ask me how many watts a trans is consuming, or how many kilowatts so they can figure thier electric bill cost. I take the amps on the loop and multiply times volts at the socket to give them the number and write it down on the paper inside. I am always surprised how that number changes over the years even with no changes to the bulbs, sytem or upgrades.

8. steveparrottSponsorfrom Tampa, FLPosts: 1,199

I think you meant to say amps at the loop (if you mean the primary loop) times 120V to equal total wattage consumed by the transformer.

To expand the topic, we all need to shift our thinking for calculations of LED systems. We can't use V x A = W to determine watts on an LED system (because of LED power factor - pf). We need to use V x A = W/pf.

For example: Take a system with (20) 6W LED's that have a pf of .85. Disregarding losses due to wire, the homeowner pays for 120W but the amp reading at the transformer secondary would be (W/pf)/12V = Amps. In this case, (120/.85)/12 = 11.8 A.

In case there's confusion about the units - keep in mind that watts are equivalent to volt-amps, so the equation is actually (VA/pf)/V=A. Volts cancel out to give amps. Pf is a unitless number between 0 and 1.

Practically speaking, to determine homeowner watts for an LED system, take an amp reading at the primary loop, multiple times 120V, then multiply by power factor.

Example: 5.0 amps at the primary loop with an LED system of .85 pf.

5.0A x 120V x .85 = 510W.​

Last edited: Sep 3, 2011