[AMPS] Two XFMR in parallel

[David Kirkby]

Firson Maryutenli wrote:
> 
> I have two unit transformer with different manufacturer, one of them are
> 1880 volt and 0.6A RMS secunder, and another are 1870 volt and 0.4A RMS
> secunder. Actual voltage when connected to AC line are 1950 volt and 1890
> volt, measuring by HV Voltmeter.
> 
> I want to connect both of them in parallel in order to have 1A RMS Output.
> What will be happen when I Connected this two transformer in parallel with
> little bit different in output voltage ???.
> 
> Thanks for your suggestion,

I have not done this, but would guess that while you might be able to
increase the output beyond 0.6 A, you won't get 1 A - there is nothing
to force the currents to split in the right ratio. (60% from one, 40%
from the other). In fact, in your case, they would not be anywhere near
correct.

To analyse this properly, you need a lot of data, but the following
might be a *rough* approximation of what will happen. As is common in
transformer analysis, you can consider the losses as to be referred to
one winding only - I'll use the secondary. However, we no nothing about
the reactance of the windings, so this will be only approximate. 

You could consider T1 (1880 on load, 1950 off load, 0.6A) as a perfect
voltage source of 1950 V, in series with (1950-1880)/0.6 = 117 Ohms. 

You could consider T2 (1870 on load, 1890 off load,  0.4 A) as a voltage
source of 1890 V in series with (1890-1870)/0.4 = 50 Ohms. 

At no load, T1 will supply a current to T2, given by
(1950-1890)/(50+117)=0.359 A. In other words, T2 will be under nearly
full load - perhaps over full load, as this is only approximate. 

If I was forced to do this (and I personally would just get another
transformer), I would be tempted to:

1) Rectify the outputs, so one can't feed the other.

2) Apply a series resistance in series with the output from one bridge,
*before* putting in parallel. Hence the load is connected to one of the
bridges directly, but the other one through a series resistance. If that
resistor was chosen correctly, you could split the currents in the
correct ratio. This would produce heat and make the regulation worst.
However, I suspect it would make the best of a bad job. 

One could work out the resistor value (and what bridge it needs to be
on) with a CAD package or by hand, using the above approximations.
However, the best way would be to make some measurements. I could work
it out for you if you wanted, but would rather not, as I don't fancy
solving all the equations - I did enough of that at university !!!
However, if you get stuck I'll do it. In the end though, you would have
to measure it. 

-- 
Dr. David Kirkby Ph.D,
email: drkirkby@ntlwold.com (formally davek@medphys.ucl.ac.uk)
web page: http://www.david-kirkby.co.uk       
Amateur radio callsign: G8WRB
-- 
Dr. David Kirkby Ph.D,
email: drkirkby@ntlwold.com (formally davek@medphys.ucl.ac.uk)
web page: http://www.david-kirkby.co.uk       
Amateur radio callsign: G8WRB

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