IDMT Grading of Dyn Transformers

© Copyright Rod Hughes Consulting Pty Ltd
Rod Hughes Consulting
General Web Site
Innovations and
Solutions Home

A bit about
Rod Hughes
 Link to this page...

The URL in the browser address bar is volatile and may be broken at any time.

To obtain a link to this page, click the <<Share>> button top-right of the screen.


Note - if the navigation pane on the left of this window is not visible, click the 2-pane icon on the top bar

Fault levels on LV vs as seen on HV are complicated by vector groupings, but we still must ensure correct grading of Overcurrent relays on the HV to be slower to operate than relays on the LV for an LV fault.

Supposing we have a 1:1 transformer as Dyn - i.e. the ph-ph voltages on both sides are the same.

  1. When you consider a ph-n fault on the LV (0-0-1 current distribution), it is seen as a ph-ph fault on the HV (1-0-1 current distribution)
  2. On the other hand a ph-ph fault on the LV (0-1-1) will be seen as an odd looking fault on the HV as 1-2-1

But in each case the question is what is the value of "1" ?

Due to Volts per turn,  VHV/THV = VLV/TLV for each winding, we get
Vph-phHV/THV = Vph-nLV/TLV
Vph-phHV/THV = (Vph-phLV/sqrt3)/TLV

Since 1:1 Vph-phHV = Vph-phLV, we are left with
THV = Sqrt3 x TLV

Case1: LV Ph-n fault

A ph-n fault as above will produce
Iph-nLV = Vph-nLV / Zwdg
Iph-nLV = (Vph-phLV / sqrt3) / Zwdg
Iph-nLV = (1/sqrt3) x (Vph-phLV / Zwdg)

on the HV that a 1-0-1 distribution on the HV means the winding current IwHV = the phase current IphHV

Applying Amp xTurn balance
IwHV x THV = Iph-nLV x TLV
IphHV x THV = Iph-nLV x TLV

i.e. due to V/T as above
IphHV x Sqrt3 TLV = Iph-nLV x TLV
IphHV x Sqrt3 = Iph-nLV
IphHV x Sqrt3 =(1/sqrt3) x (Vph-phLV / Zwdg)
IphHV = (1/3) x (Vph-phLV / Zwdg)

So we have the two operating currents fo rteh two relays on HV and LV as
Iph-nLV = (1/sqrt3) x (Vph-phLV / Zwdg)
IphHV = (1/3) x (Vph-phLV / Zwdg)

Rearranging we get
(Vph-phLV / Zwdg) = sqrt3 x Iph-nLV
(Vph-phLV / Zwdg) = 3 x IphHV

3 x IphHV = sqrt3 x Iph-nLV
IphHV = (1/sqrt3) x Iph-nLV
IphHV = (0.577) x Iph-nLV

Therefore the OC relays you need to look at the operating time of the HV relay at .577 pu compared with the operating time of the LV relay at 1 pu.
As an IDMT – no problem!  The HV relay will be much slower due to the much lower current.

Case2: LV Ph-Ph fault

An LV ph-ph fault will produce
Iph-phLV = Vph-phLV / (2 x Zwdg)
Iph-phLV = (1 / 2) x (Vph-phLV / Zwdg)

Applying A-T balance
IwHV x THV = Iph-phLV x TLV
IwHV x Sqrt3 x TLV = [(1 / 2) x (Vph-phLV / Zwdg)] x TLV
IwHV x Sqrt3 = (1 / 2) x (Vph-phLV / Zwdg)
IwHV = (1 / 2.Sqrt3 ) x (Vph-phLV / Zwdg)

On two of the phases,
I1phHV = IwHV

But on the other
I2phHV = 2 x IwHV
I2phHV = 2 x (1 / 2.Sqrt3 ) x (Vph-phLV / Zwdg)
I2phHV = (1 / Sqrt3 ) x (Vph-phLV / Zwdg)

So we have
Iph-phLV = (1 / 2) x (Vph-phLV / Zwdg)
I2phHV = (1 / Sqrt3 ) x (Vph-phLV / Zwdg)

Vph-phLV / Zwdg = 2 x Iph-phLV
Vph-phLV / Zwdg = sqrt3 x I2phHV

sqrt3 x I2phHV = 2 x Iph-phLV
I2phHV = (2/sqrt3) x Iph-phLV
I2phHV = 1.155 x Iph-phLV

So now we must make sure the operating of the HV relay is properly graded.

Therefore, when the LV is seeing 1pu current, the HV is seeing 1.155 pu, i.e the HV relay is operating further to the right on its curve and is therefore much faster than would at first be thought. 

If not correctly graded, the HV would trip first, if not first then quite possibly less than the normal 0.4 second grading time after the LV trip, thus leading to the false conclusion that the fault is inside the transformer (remove transformer, de-tank the winding, inspect …) when it was really outside of the transformer on the LV system/busbars.

Contact Me

Skype: (ping even if showing offline)

Email Me

A phone call is nearly always welcome depending on the time of night wherever I am in the world.
Based in Adelaide UTC +9:30 hours e.g.

April-SeptemberNoon UK = 2030 Adelaide
October-March:Noon UK = 2230 Adelaide

  Office + 61 8 7127 6357
  Mobile + 61 419 845 253

Extra Notes:

No Liability:
Rod Hughes Consulting Pty Ltd accepts no direct nor consequential liability in any manner whatsoever to any party whosoever who may rely on or reference the information contained in these pages.  Information contained in these pages is provided as general reference only without any specific relevance to any particular intended or actual reference to or use of this information. Any person or organisation making reference to or use of this information is at their sole responsibility under their own skill and judgement.

No Waiver, No Licence:
This page is protected by Copyright ©
Beyond referring to the web link of the material and w
hilst the information herein is accessible "via the web", Rod Hughes Consulting Pty Ltd grants no waiver of Copyright nor grants any licence to any extent  to any party in relation to this information for use, copy, storing or redistribution of this material in any form in whole or in part without written consent of Rod Hughes Consulting Pty Ltd.