Control transformer in reverse

[gar]

180916-2017 EDT

Back in post #52 I made a mistake in calculating the transformer internal impedance as viewed from the 24 V winding. The change in 24 V winding current with the 25 ohm load should have been 21.1/25 = 0.884 A instead of 0.17 A, and thus the calculated impedance on the 24 V winding side should be 5.5/0.884 = 6.52 ohms. Reflecting this to the 120 winding produces a result of 6.52*4.51^2 = 133 ohms.

So in my last test where the bulb current is 0.1 A on the 120 winding side I get a calculated change in voltage of about 133*0.1 = 13 V. The measured change was 108.4 - 93.5 = 14.9 V. Some of this 14.9 is a result of the drop in my test voltage source. Basically a good correlation.

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[gar]

180916-2055 EDT

LarryFine:

No. But now I have.

With no load on the 120 winding the 24 winding requires 26.8 V to get 120. This ratio is 26.8/24 = 1.12 . From a previous post I had a 24 to 120 measured ratio of 4.487 , using this to calculate the 120 winding voltage from a source of 24.0 V produces 24*4.487 = 107.7 .

120/107.7 = 1.114 . A good correlation with 1.12 .

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[LarryFine]

How about achieving 120 with the bulb load?

 

[gar]

180916-2217 EST

LarryFine:

I have just put that stuff away. But you can do a rough calculation.

My corrected internal impedance is 133 ohms. Thus, with a 15 W load current is about 0.125 A at 120 V. Internal impedance voltage drop is about 133*0.125 = 17 V. Thus, input would need to be about 137/120 = 1.14 times 26.8 = 31 V. Does this get us into corre satutation problems? I don't off hand know.

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[Skokian]

Transformer Ratios.

Transformer Ratios.

Note that the winding ratio of the 120 Vac to 24 Vac transformer will be less than 5:1.

This is done to account for the full load voltage drop in both the primary and secondary. Hence, at no load or light load, with exactly 120.0 Vac input, the output will be higher than 24 Vac. Typically it would be from 25.2 (+5%) to 26.4 Vac (+10%).

Hence, when used in reverse with exactly 24 Vac into the secondary, the primary will be from 114 Vac (-5%) to 108 Vac (-10%) - and this is with no load or light load.

The same will be true for control transformers. However, this "compensation" is always greater for small transformers than larger ones. This is because smaller transformers dissipate their internal heat better than larger ones and so the percent of winding resistance is always higher for commercial smaller transformers than larger ones.

Hope this helps some.

 

[Jraef]

Yes, runs only .05ma.

Reason for request : Trouble light wire was CL2 Lo VO wire run in same conduit as other Power limited cable. The suggestion is to be able to re-use the status/strobe light.

.5ma is the device at 120v

.12@120v

Thank you all for your inputs.


Sorry , My OP was about was it ok to use a control trans in reverse , nothing more nothing less.
I gave the current requirements as was told to me.
I verified in person only yesterday.
What more do you want from me. Blood or something.

Lot's of conflicting info here, but I think I might see the issue.

A LOT of standard indicator lights use a 6S6 bulb that is rated for 6W at 120V, either mini-candelabra base or bayonet base. 6W at 120V is 0.05A. I think someone erroneously reported that to the OP as "0.05mA", which made it off by a factor of 1000...

Bottom line if it is a 6S6 6W indicator lamp and you have a transformer capable of 20W, that is not going to be an issue.

TECHNICALLY however, article 450.11 of the NEC requires that you be able to show that a transformer as capable of being back fed.

(B) Source Marking. A transformer shall permitted to
be supplied at the marked secondary voltage, provided that
the installation is in accordance with the manufacturer's
instructions.

Chances are low that a doorbell transformer will have instructions for that use.