Hi everyone! As part of my VFD upgrade project, I have purchased a neat "control" transformer. This xformer accepts my 240 volt line and has a 115V secondary and a 24V tap as well.

The transformer is rated 100VA. Stupid question time (I should know this), how do I determine the acceptable load for the secondary circuits? For example, let's say I have a contactor coil that must close, 115V and 100 mA. Also, a cooling fan, 115V 500 mA. What is the formula for determining the VA loading of this combo?

Any hints appreciated. Thanks.

Swede

2. I would just add them together:

100mA = .1 A
500mA = .5 A

24V * .1A = 2.4 VA for the coil

115V * .5 A =57.5 VA for the cooling fan

3. VA for 100ma @115v & 500ma @ 115v = 600ma (.6 amps * 115 ) = 69VA.
Al
Although the total VA rating is 100VA, you also need to know what the individual winding are rated at , current wise.
Al

4. I'll vote for Al's version.
This means you have 31VA to go.
That leaves you with 1.3 A @ 24V.
And don't count on the 115V to be isolated from the grid! It may be just a midpoint tap on the primary (autotransformer). Check that with a VOM.

• The only problem with a multi-secondary transformer is knowing what the VA rating is for each winding, for a single secondary transformer it can be assumed that the total VA value can be used to calculate the maximum current, but with a multi secondary type, it depends on what percentage of the VA value the windings are wound for, although the voltage is decided by the number of turns, the winding must be of a sufficient diameter for the intended current.
It's not always a choice of just divi-ing up the VA between the windings.
If you have just a 24v tap instead of a separate winding, I would guess that the secondary is wound in just one gauge, which probabally means that will restrict the 24v to fairly low current, In ESjaavik's example, I would guess it would handle the 1.3amp max.
Al

• Hmmm. This transformer has a primary with connections for 208V, 240V, and 277V. Since I have 240V in my shop, I'll connect the 240 across the approprite leads. On the secondary side, the full winding is 120V, and it does have a 24V tap.

To be conservative, given that the wire guage is probably constant on the secondary side, I'll multiply the total secondary load current by 120 for the VA load. This is what I suspected but I did want to double check.

Einar, the 120 is truly isolated from the primary. I think this transformer was made specifically for this type of project... turning 240V into low-current, lower AC voltages for relay actuation, etc.

I really appreciate everyone's help, many thanks.

• Swede, If the 24V is a tap off the 120V output then you can be sure that it VA rating is 24/120 of the full rating. This gives 20VA available for the 24V tap with 80VA remaining on the 120V tap.

20VA divided by 24V = 833mA; 80VA divided by 120V = 666mA

If you used just the 120V output you would have the full 100VA available or 833mA.

From your description it looks like you need 600mA at 120V so a full 833mA will still be available for your 24V supply.

Remember that the VA rating is for all practical purposes the same as it power rating in watts... this transformer has enough to drive a household 100W light bulb from a 240V source.

HTMTC Gary

• Whoops! I mistook that for a 24 volt coil. I agree with Al

• Update on the transformer issue - I made a nice eBay purchase which was delivered yesterday. It was an assembly consisting of 2 ea. 4-pole contactors rated at 25 amps, another control transformer, and a 115V mechanical solenoid.

After cleaning the contacts, the contactors worked great. I think it was scrapped because the internal contacts were dirty and the continuity intermittent. Some light polishing of the contacts (they looked like brazed silver pads) restored the contactors to full use.

This was some sort of 3-phase reversing setup, meaning it has an interlock between the contactors which mechanically prevents both from closing at the same time. The whole thing comes apart as it is DIN rail mounted.

The 30VA transformer doesn't even get slightly warm, with a 500mA load (24V contactor coil). Can the temperature of the transformer be a crude indicator of load? I would assume it would heat excessively as the load approaches the limit.

• You got it Pontiac.

Your 30VA transformer at 24V gives you 1.25A to play with, since you only use 0.5 A you have plenty of headroom. The heating of the transformer is produced by both the current passing through the coils and the heating of the soft iron core induced by the magnetic field which is proportional to the current.
You have 12 W of power going through your transformer which would not produce much of a temperaturre rise in the fairly large heatsink mass of the transformer.

electrically yours, Gary

• Swede, The thing that happens to a transformer when the VA rating is exceeded, the magnetic core reaches saturation and it can no longer transfer the energy to the secondary and excessive heating occurs.
When operating magnetic/inductive devices such as contactors etc on AC, it is important to use some kind of fuse protection due to the fact that if devices such as contactors and solenoids etc ever jam or stick preventing the armature in the device able to operate, the current goes through the roof due to the device losing impedance.
This does not occcur when using DC operated devices as the resistance remains the same whether the device picks up or not.
Al

• Thank you both Al and Gary, and everyone else. Al, I like your avatar. That guy is either getting shocked, or he's seen the price tag on the CNC machine he desperately wants!

I'll stop theorizing, and get into the shop now. The first project will be to install a fuse and contactor set in my rotary coverter safety switch box as a warm-up; then I'll tackle the VFD and the mill's electrical system.