Re: Still trying to solve this K6100 HV

Your schematic with marked voltages is helpful. One red flag is the values shown for Q901 base/emitter being 23.1/14.5. The base/emitter junction forward drop is that of a diode so it would never exceed 0.5V to 0.7V and 0.6V is typical. The excessly large difference of 8.6V indicates a blown transistor.

The job of Q900 is to reduce the power voltage delivered to the HVT primary winding to the level required to achieve the proper output voltage. It is obviously conducting too hard at present. The negative feedback loop for the regulator begins with a string of components D901, R912, ZD902, R910, R918, R911. Since you measured 36V at the junction of R910/R918, it would appear that the entire voltage divider string is conducting okay. When the output voltage goes up, the base voltage of Q903 should also increase. This increases conduction of Q903. As Q903 emitter outputs more current, the voltage drop of R909 increases and the Q903 base/emitter voltage maintains its typical forward biased diode voltage drop of approximately 0.5V to 0.7V. This results in an increase in base current on Q902 (if resistor R915 is okay). As the collector of Q902 conducts more, it pulls down the voltage at the base of Q901 by conducting more current through resistors R905 and R904. As the base voltage of Q901 falls, its emitter voltage also falls by the same amount, again because the base/emitter junction, when forward biased, will always be that of a diode, about 0.5V to 0.7V. The same rule holds for the base/emitter junction of Q900 and so the output voltage of Q900 also will fall. This delivers a lower supply voltage to the primary winding and so the output voltage will fall.

Of course when the output voltage is too low, the opposite happens. The base voltage of Q903 falls, it conducts less collector/emitter current, the base voltage of Q902 falls and it conducts less collector/emitter current. The base voltage of Q901 rises and its emitter follows. The base voltage of Q900 rises and its emitter follows and the supply voltage increases.

I suggest based on your measured values that Q901 is fried and is leaking. It also appears that the feedback loop is not working at Q902. Note that the collector of Q902 has a very high voltage close to the supply level, indicating that it has turned completely off. It is trying to allow Q901 and Q900 to conduct more in order to increase the voltage output. Since the output voltage is already much too high, it is obvious that there is a problem with one of the following components failing to conduct current:

    - wiper of R918
    - Q903
    - R909
    - R908
    - Q902 (already replaced so maybe okay)

Q901 may be failing soon after powerup due to prolonged excessive base and collector current caused by failure of of Q902 to turn on. Replace (or resolder and confirm condition of) the above listed parts as well as Q901.

Good luck,

William Boucher
http://www.biltronix.com
  ----- Original Message -----
  From: David Shoemaker
  To: vectorlist@vectorlist.org
  Sent: Saturday, October 15, 2011 2:24 AM
  Subject: VECTOR: Still trying to solve this K6100 HV

  I am still digging into this HV unit which is giving me some troubles. Here is what I have observed.

  The collector of q901 is way off of spec and I don't understand the circuit well enough to understand what is causing it. And the base / emitter of Q900 are about double what they should be.

  I have checked the values of all the resistors and they are fine.

  I have replaced q900 with a bu406d (verified good in another HV unit).

  I have replaced q901 & q902.

  Any suggestions? Thanks.

  I have attached a copy of the schematic with my noted values in red.

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Received on Sat Oct 15 12:27:41 2011