Negative Voltage Regulation in Transformer

Discussion in 'General Electronics Discussion' started by QwertyXP, Nov 4, 2013.

1. QwertyXP

5
0
Oct 26, 2013
How can a transformer have negative voltage regulation?
A few texts state that a leading power factor can give rise to negative power factor, but I have not been able to figure out how this is possible.

One vague explanation in my mind is that capacitance across the secondary terminal will somehow cancel out the leakage reactance on the primary side, thus resulting in a higher primary voltage and hence higher secondary voltage as well.

Can somebody please describe more accurately.

2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
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Jan 21, 2010
I googled "transformer negative voltage regulation" and found many explanations, from the in-depth mathematical approach to the purely descriptive.

Have a quick look and see if there is one which describes this in the form that suits you.

If not, tell us what comes closest and where you need more information.

3. QwertyXP

5
0
Oct 26, 2013
I think I've somewhat become used to the idea of negative voltage regulation after going through lots of articles and also drawing the phasor diagram myself. The main issue seems to be resolved now.

However, I still have confusions as far as the related calculations are concerned. For example, if you have a look at the following link:
http://yourelectrichome.blogspot.com/2011/07/voltage-regulation-of-transformer.html

According to example 1, full load current = KVA rating/secondary voltage specified in the question statement. Thus in this case, he has assumed 125V to be the full load voltage. However, in another place (part ii), he considers 125V to be the no load voltage while calculating the secondary terminal voltage.

Is this a mistake, or did he make this assumption because the value of current would not differ by too much even if no load voltage were used in the calculation?

4. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
It is likely that 125V was chosen because whilst the actual value is unknown, it would be similar to this. Since we can assume the actual value will be very close to this, any calculation based on that estimate will be quite close.

Although, given that the actual value has been already calculated it seems odd not to have used it...

Perhaps someone else can give a more authoritative answer. Whilst I understand the concepts involved here, I don't pretend that I have ever needed to do these calculations myself.