AVR Trouble.

Discussion in 'General Electronics Chat' started by R!f@@, Jun 27, 2015.

  1. R!f@@

    Thread Starter AAC Fanatic!

    Apr 2, 2009
    I am attacking a 3 phase genset AVR.
    I was able to remove the potting and take out the PCB.
    Trouble shooting lead me to an open Resistor, labelled as R2.
    I drew a schematic to show the particular area. AVR.png
    X,Y and Z are the 3 phase 380VAC input from the generator.

    The Issue is R2 is open and to make things works the first color code is gone, totally.

    R3 is 5.1K, Measured out of circuit. Color code is , Gr,Br,Bk,Br,Br,Rd.
    R2.---> ??,Rd,Bk,Or,Br,Rd.
    The first band is unknown.
    Size of body 6mm.

    I wanted to figure out the value by studying the diagram
    The trimmer is the Voltage adjust of AVR. So I guess the bridge is the feed back sensor part and the resistor is used to limit the current to a safe value from the phase voltages.

    Question is what will be the value ?
    I am still trying to figure out.
    I cannot put a higher value to test cause the genset is on another island and I cannot go to it so first try needs to be spot on.
    Any one got any idea?
  2. vrainom

    Active Member

    Sep 8, 2011
    Well, if it's a common value it would either be 220K or 820k. It's a voltage divider feeding a 50v capacitor so unless the trimpot r1 is a really high value I guess it's gonna be 220k.
    R!f@@ likes this.
  3. ian field

    AAC Fanatic!

    Oct 27, 2012
    They look like 5 band resistors, so you could have 3 numerical digits, a multiplier and a tolerance. The tolerance is usually 1% or less on those.

    Not sure whether the bottom band is supposed to be "salmon pink" - but that would denote high stability.
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    Agreed. However, I'd plump for 820k, based on the following:-
    Peak voltage = 380 x 1.414 ~ 537V.
    Required micro input from pot assumed ~5V.
    Pot value assumed <= R4, say 5k.
    For resonable pot adjustment range, assume volts across pot + R4 ~7V.
    Hence 10k drops 7V, hence R2 + R3 must drop ~ 530V, hence R2+R3 ~ 10k x 530/7 ~ 757k, hence R2 ~ 752k.

    Edit: Is R2 really rated for all those volts?
    Last edited: Jun 28, 2015
    R!f@@ likes this.
  5. R!f@@

    Thread Starter AAC Fanatic!

    Apr 2, 2009
    Pot is 2K.
    I too figured I should try 820K. So it is already done. I will sent it to test on Thursday.

    Dunno about that but the info and picture I provided is correct to my knowledge.