Howto find input amps, from output amps on transformer?

[sendkeys]

[ynneb]

This is a question for Al_the_man.
In the mean time I wil have a crack at it.

If your output is 69volts 30 amps then I guess the tranny is 69 x 30= 2070 VA
So you divide 2070 by 120 = 17.25 Amps is drawn on the primary side.

Because of transformer losses I guess you should add 10% to that figure. Making your primary about
19 Amps

I am sure there is a better formula, but this should give you a round about clue, until Al makes a visit.

[Al_The_Man]

Benny, You are close, but the VA rating will be 1500 as the 50v secondary will be supplying the 30 amps . The method you used had factored in the energy stored in the capacitors.
12.5 amps primary
Al

[Bloy2004]

I was wondering how the 30amp output was determined...and if the transformer can withstand that amperage either short-time or as continuous? Did you hook someting up that drew the 30 amps?

Was the rating of 30amps a given?
I get the feeling I'm really missing the boat here.....

[arvidb]

If the transformer is rated 30 A @ 50 VAC, then you are seriously overloading it if you draw 30 A after rectification.

30 A * 50 VAC = 1500 VA.

30 A * 69 VDC = 2070 W. Which is a 38% overload.

A safe value would be 1500/69 = ca 21,7 A.

This is under the assumtion that the transformer is rated at 30 A @ 50 VAC. If it is rated at > 41,4 A @ 50 VAC then it's ok to draw 30 A after rectification.

Arvid

[jfirey]

The method of rectificatoin needs to be taken into account as well. When using the basic bridge rectifier and capacitor (VAC * 1.414 - diode losses= VDC) , current draw on the transfomer will be discontinuous. The current will be drawn in surges based on the conduction cycle of the diodes, casing the charge and discharge of the capacitor at the frequency of the line voltage.

For this method of rectification, actual current load on the transformer will be 1.8 x DC current. AVERAGE current will be the same to match the watts to watts balance, but the transformer MUST be able to supply the surge currents without failure. This is an issue if you intend to run the transformer at or near it's peak!

If the transformer hits saturation, or runs too high current for the wire size of the secondaries, you'll may have problems.

Other methods, have other form factors, such as the half wave rectifier with a factor of 1.5, but there are voltage and power conversion sacrifices. See this link for a great in depth tutorial;

http://www.plitron.com/pages/technote.htm

[sendkeys]

Thanks everyone. the peak of this transformers is nowhere close to 30amps its a 5kva rating . 30 amps dc is basicly what my max controllers will be taking at the max.

What im really trying to find out is if im going to go over my 20amp "wall socket" limit also need to get some ac relays ect. and they max at 20-25amps

Guess really without knowing the loss of the transformer and diodes i will never know with out testing. as my 20amp limit if very close to what it could be.

Time to go down buy a amp tester with over 10amps hehe (mine is limit to 8 why i was trying to find out without testing)

Thanks everyone for the help

[jfirey]

If it is rated for 5000VA, as you state, then a bridge rectification/cap setup will yield ~68 VDC after the diode losses. A little reverse engineering gives the following;

Given the 5Kva rating, the above pulse current flow is NOT an issue for the transformer itself. At a 30 amp DC draw, you will see ~24.5 amps on the primary, allowing for NO core losses. Your current peaks will likely be above the trip point, and the lows will be under the trip point. You are in the gray area of the breaker, and trip will be a function of heat and time. I would bet on a tripped breaker.

A secondary load of ~23A puts the primary current under 20 amps on the primary and allows for ~10% core loss. This *should* prevent a breaker trip, as long as the panel does not build up heat at this breaker. NOTE: All breakers will not trip at the same point.

Of course, I could be wrong...