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No voltage on pins 8-19 on SX-1050 awr 103 board????!!!!

Steve 1050 said:
No work has been done on 104 board diodes 3&4 were check and tested good. All mentioned transistors are "cool" to the touch. C10 is also "Cool".Voltage on R1=.487V, R19=.213V . I think transistors were packed loose.
Click to expand...

That means the positive side current is 48.7 mA and the negative side is 21.3 mA which are quite reasonable numbers.

Something is suspect, and only the fuses, the actual wire connections and the diodes are in between the capacitors and the transformer windings on the awr-104 board.

Swap FU1 and FU2 and see if the voltages on pins 2 and 3 stay the same... that will exonerate the fuses...

I would take ac (ground referenced ) voltage readings at the transformer's orange wire connections to the board at pins 13 and 14, and both sides of each fuse's clips FU1 and FU2, and then at the actual diode leads.

If there is a high resistance in the negative supply side the drop in the ac voltage should show.

The other possibility is a diode that reads ok under test, but under load fails. The surge that blew the 10 ohm resistors could have only left a shred of the diode junction functional, a high resistance shred. Hard to spot without a scope. And then there are those 0.01uF caps across the diodes, they could be either VERY leaky or behaving weirdly....

It looks like a simple full wave rectifier bridge and capacitor circuit, but there's a lot to go wrong. In your place I would just replace all 4 diodes and ohm out the current paths between the orange wires, the diodes, the caps, and the pins going off the board. Something has gone high resistance...

The most important thing is that it's hard to hurt a transformer and the orange wires carry the SAME AC voltage(s on EACH side)!! so the (resultant) rectified and filtered dc voltage should be the same(magnitude on each side, but opposite polarity, our problem is that these are NOT).
(Zerojunk PM'd me this:" I'm sure you know this but the sentence might be confusing, DC should be 1.414 times AC", thus the red additions... sorry.... )


Then when you open up r1 and r19 OR pins 2 and 3 the dc load on the caps on the awr-104 board will be removed, and both the voltages should be mirror images of each other. IF not, we HAVE to get that right first, and we blame the rectification, if re-applying the load again drops the voltage, we look for high resistance circuit paths....
 
Mark, I took the requested readingsthey are as follows:
Pin13=59.9V
Pin14=59.6V

D2 A=.073 C=78.8V
D3 A=-55.8V C=.061V
D4 A=-55.8V C=.073V
D5 A=.073V C=78.9V

When R1 & R19 are opened Voltages are Pin 7= -80.1V
Pin 6= 80.2V

readings at each end of fuses same as pins 13,14
 
edit: (bottom line... top actually) I just typed and erased several paragraphs... PLEASE JUST REPLACE D3 and D4. d2 and d5 also to be safe.


Steve 1050 said:
Mark, I took the requested readingsthey are as follows:
Pin13=59.9VAC
Pin14=59.6VAC
Click to expand...

Assuming the readings were taken with R1 and R19 still connected. it means the transformer is OK. These are AC voltages, so Pin13=59.9VAC

Steve 1050 said:
readings at each end of fuses same as pins 13,14
Click to expand...

Assuming the readings were taken with R1 and R19 still connected. it means the circuit to the fuse clip is OK. No voltage drops.

BUT you stopped just short of confirming the answer!!!

Steve 1050 said:
D2 A=.073VAC?? C=78.8VDC This goes to the +80 supply and is OK
D3 A=-55.8VDC C=.061VAC?? This goes to the -80 supply and is LOW!!
D4 A=-55.8VDC C=.073VAC?? This goes to the -80 supply and is LOW!!
D5 A=.073VAC?? C=78.9VDC This goes to the +80 supply and is OK
Click to expand...

When R1 & R19 are in circuit, Voltages are Pin 6= 78.8V Pin 7= -55.8V

These are dc VOLTAGES, showing the voltage drop we are trying to eliminate. We need to read the AC voltages entering into the diode at the leads themselves, as indicated by the red highlighting.
Thus if the AC voltage is correct at the diodes yet the dc voltage is incorrect either the diodes are bad or the solder connections from the diodes to the pc board foil traces are bad.

Steve 1050 said:
When R1 & R19 are opened Voltages are Pin 6= 80.2V Pin 7= -80.1V
Click to expand...

When the load is removed BOTH the capacitors are able to fully charge. There's a "kink" in the garden hose...

Please make AC voltage readings with R1 and R19 CONNECTED, at D2's anode, D5's anode, D3's cathode and D4's cathode.
If they are the same as pins 13 and 14, the wiring up to the diodes is OK.

I just typed and erased several paragraphs... PLEASE JUST REPLACE D3 and D4. d2 and d5 also to be safe.

The PC board pattern is SO SIMPLE, a high enough resistance to cause a twenty three volt voltage drop at 21 milliamps is next to impossible (23v / 0.022a = 1000 ohms!!). Just because the diodes checked ok with the tiny test current from the DMM doesn't mean they ARE OK.

Remember, that r1 and r19 were both DESTROYED, and it takes a LOT of energy to do that. And for a short period of time, the transformer CAN push that much energy through, and it pushed ALL that energy THROUGH D2 D3, D4 and D5.
 
Mark I have to go out, I will replace said Diodes and make voltage checks when I return my bad on original voltage checks I wasn't paying attention,Duh!
 
New measurements on pin13=60.2Vac pin14=60.1vAC
D2 Anode=60.2Vac
D3 Cathode=60.2Vac
D4 Cathode=60.1Vac
D5 Anode=60.2Vac

all readings with R1, R19 in circuit
Diodes 2,3,4,and 5 are new.
 
Steve 1050 said:
New measurements on pin13=60.2Vac pin14=60.1vAC
D2 Anode=60.2Vac
D3 Cathode=60.2Vac
D4 Cathode=60.1Vac
D5 Anode=60.2Vac

all readings with R1, R19 in circuit
Diodes 2,3,4,and 5 are new.
Click to expand...

Good AC voltages, and new diodes, I breathlessly await the voltage readings on pins 2 & 3 on the awr-103 board, as well as the voltages on 7,8,9 10 & 11 if we have better negative voltages on pin 3 ( a mirror image of voltage on pin 2?).

The voltages on pins 2 & 3 will indicate if we at least have that problem fixed.
 
Steve 1050 said:
Pin Voltagesare as follows:
Pin2=78.4
Pin3=-55.8
pin7=-18.6
Pin8=-42.8
Pin9=-18.9
Pin10=-15.38
Pin11=-15.38
Click to expand...

ARRRRRRRRRRRRRRRRRRRRGHHHHHHHHHHHHH

All these requested readings (below) assume that r1 and r19 are connected.

Read the DC voltages on the awr-104 (diode board) at: d3 anode, d4 anode, AC voltages at d3 cathode and d4 cathode.

read the dc voltage at pin 7 on the awr-104 board
switch the meter to read ac voltages and measure the ac voltage present on:
awr-104 pin 7 (looking for stray AC at a DC point here, sort of substituting for an oscilloscope at C9)(to compare)
awr-104 pin 6 (looking for stray AC at a DC point here, sort of substituting for an oscilloscope at C10)
the presence of ac voltages at these DC points would indicate a lot of ripple NOT being filtered/stored by the capacitor C10. (or c9)

Be prepared to swap the two 470uF/80v (c9 & c10) capacitors, and if the positive voltage now is low and the negative voltage is normal, then the capacitor is at fault. Pay CLOSE attention to the polarities of both capacitors when they are re-installed. On one the negative is grounded (c9) and the other the POSITIVE is grounded (C10).

The only other reasonable explanation is that C10 has lost a large portion (90%??) of it's value.

Once the negative voltage is correct, then we'll explore if and why the awr-103 pin 7's voltage is too BIG (-18v vs -13v), which influences directly the -60v on pin 8 of the awr-103 board.

(just in case, part # 140-XRL100V470-RC @ $1.03 each , in stock, at mouser 470uF/100v)
 
Steve 1050 said:
Mark, I feel your pain, I'm as frustrated as you are.
Voltages are:
Pin 7= -57.1VDC 21.7VAC
Pin 6= 80.7VDC .362VAC
Click to expand...


:banana: We just NAILED it !!! :banana: C10 is baaaad !!! high ac voltage, low dc voltage compared to the mirror image circuit.

You can swap C9 and c10 so that we can work on the -60v supply while waiting for the replacement caps ( or accumulate other parts to add to the order).

Once the negative voltage is correct, then we'll explore if and why the awr-103 pin 7's voltage is too BIG (-18v vs -13v), which influences directly the -60v on pin 8 of the awr-103 board.


BTW: c10 could be part of the ORIGINAL problem.

It's 2:20 am here, getting on to bedtime.... (yawn) nite nite....
 
How did it go? A few moments after I last posted, the power went off, and stayed off or several hours.... fun.... the UPS allows an orderly shutdown but no extended running time...

The neighbors across the street had juice.... AGAIN!!!!!!!........ of the last couple of outages THIS YEAR, that has been the pattern.... at least it hasn't been raining, so I didn't have to worry about water in the basement.
 
swapping caps gives us the following voltages;
Pin2= 55.8
pin3= -80.0
pin7= -32.8
pin8= -78.8
pin9= -40.3
pin10= -33.6
pin11= -33.6

Aren't power outages fun...Not! atleast you hav a ups, around here no ups storm starts computer goes off. However unexpected outages do happen, one word - BACKUPS!
 
Steve 1050 said:
swapping caps gives us the following voltages;
Pin2= 55.8
pin3= -80.0
pin7= -32.8
pin8= -78.8
pin9= -40.3
pin10= -33.6
pin11= -33.6
Click to expand...

This post is like the old grade school TRICK test, don't do anything until the end..

Ok, we got the -80, is r1 and r19 connected? I thought it would be a little lower (like -75), but with the +55 we KNOW that cap is BAD.

So, now we concentrate on pin 7, did you get extras of D2, the 13v zener diode? There is way too much voltage across it. It is acting like the transistors Q5, Q6 & Q7 are feeding it current instead of reading it as a reference point. I suspect that it is either very hot or very blown. (swagging edit: it occurs to me it may be in backwards?? gotta think that one through) The current to activate D2 NORMALLY comes from R12, the 7.5k ohm resistor. BUT we can't pull the three transistors without providing some voltage to R12. That is accomplished by jumpering temporarily the collector and emitter of the empty Q5 pads. Since pin 8 is disconnected, nothing is hurt by the higher voltage. Two caps (C14 & C15) are in some jepoardy (already) but there is a 5k 50% voltage divider isolating C14 and Q8's removal protects C15.

BUT: Until we tame the voltage on pin 8 , I think it would be best to remove Q8 and keep the elevated voltages away from the rest of the receiver, it's easier than disconnecting pins 9, 10 & 11 .

The difficulty I originally feared has come to pass, and that is that is that there was NO reference in the manual as to the layout of the leads for the 2sa628a (Q6 & Q7) transistors. In that case I usually verify each and every solder pad's designation visually AND with an ohmmeter by locating the other components connected to that pad in the circuit diagram.

Then I would confirm each pad's location and the lead layout of the transistor and also test that transistor out of circuit.

SO:
1) pull Q5, Q6, Q7 & Q8 , test them and identify the leads in case we get another transistor layout mixed in....
2) temporarily jumper together the collector and emitter pads of Q5 and then see if the voltage on pin 7 comes closer to -13v, if not we need another diode, you can temporarily swap in a 14, or try the original 13 and see if works as a substitute.
3) once -13v is achieved at pin 7, the Q5 temporary jumper can be removed
4) verify that the silk screened lead designations on the printed circuit board are accurate.
5) discuss results to that point
 
do you want voltage with all Transistors removed? All leg designations are correct for all transistors (diode Check),all transistors test Good
 
markthefixer said:
SO:
1) pull Q5, Q6, Q7 & Q8 , test them and identify the leads in case we get another transistor layout mixed in....
2) temporarily jumper together the collector and emitter pads of Q5 and then see if the voltage on pin 7 comes closer to -13v, if not we need another diode, you can temporarily swap in a 14, or try the original 13 and see if works as a substitute.
3) once -13v is achieved at pin 7, the Q5 temporary jumper can be removed
4) verify that the silk screened lead designations on the printed circuit board are accurate.
5) discuss results to that point
Click to expand...

The next step was to get a viable -13 or -14 volts on pin 7 of the awr-103 board. That involved finding a working zener diode among the parts you have, installing it to get a reading, removing bad ones and finding another to try, to get that -13 to-14 volts on pin 7. The Q5 jumper was to get some voltage to R12 to energize D2.

Once that was done, then good Q5, Q6 & Q7 could be installed correctly.

Then if the voltage on pin 8 isn't around -60 (-13v zener) to -65 v (-14v zener) then votages on the e, b & c of Q5, Q6 & Q7 are needed.

Otherwise if the voltage on pin 8 was in range, Q8 could go in.
 
Steve 1050 said:
with the Diodes I have Voltages range from 5.68VDC to 35.2VDC. WTF???!!! Fourth down do we punt?
Click to expand...

If you don't have anything around 13 v ( getting -13 is just done by putting the diode (D2) in with the cathode (band) to ground and the anode connected to pin 7) go with the closest lower voltage. Even if the Q5, Q6 Q7 circuit is regulating at -30v instead of -60, it IS still regulating, and once the correct voltage diode is installed, things will be ok.
 
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