Trying to understand a broken preamp stage on a Symetrix 528E

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CurtZHP

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Joined
Mar 21, 2005
Messages
634
Location
Allentown, PA
I've got this Symetrix 528E sitting on the bench with a busted mic input stage.  The line input works fine, as well as all the other processing stages.  The mic input will not pass a signal, but the "clip" light stays on.  The radio station I'm fixing this for says the DJ grabbed the mic boom for the mic this was connected to, got a static shock, and that was the last they heard from it.

Popped the cover.  Of course, it's an SMT nightmare!  :'( :'(

Looking at the schematic....
http://www.symetrix.co/wp-content/uploads/2016/02/528E_B01-1998.pdf

At first, I suspected the op amp (NJM2068M) took a hit.  I'm measuring 15VDC on the output pin.  (That would explain the clip light...)  I'm measuring about -10VDC on pins 2 and 3, which sort of makes sense, given that the only things between those pins and the -15V supply is a couple resistors (R8 and R9).

What I'm thoroughly confused about is what happens just upstream of the chip.  Q8 and Q9, a pair of 2N4403's, appear to provide the balanced input to the chip.

I somehow managed to replace the NJM2068 with a plain old 5532 (same pinout, etc.), but the symptoms remain; so now I'm wondering about those transistors.  But I'm sorry to admit I have no clue what is going on around them, especially where Q10 (TIP30A) is concerned.  That's obviously the phantom supply, but what else is it doing?  And I've never seen that sort of gain control before.

I suppose I could blindly replace components, but I rather think I should learn what I'm looking at here first. 

Help?

:-[




 
It´s a very mundane opamp with transistor input configuration (check page 27 of paper by that corp
https://tinyurl.com/y5ps32hh)

the tip is a current source, I can´t explain why they have done it this strange way.
There are no protection diodes so a static discharge will easily kill the input transistors (e.g. exceeding base emitter voltage of 2n4403) replacing them would be the first thing to try, don´t substitute 2n4403, they are there because they are lower noise (lower base resistance) than most.

Hopefully someone more knowledagable than me will chime in ;D
 
CurtZHP said:
I suppose I could blindly replace components, but I rather think I should learn what I'm looking at here first. 

Help?

:-[
I applaud that approach... (learning how).

This is a good opportunity to hone your troubleshooting skills.

#1 with power off you can check the transistor diode junctions with the diode function in most modern VOMs. To check PNPs transistors as used here, the minus (black) lead touches the base, you should measure one diode drop (300-600mV) at both collector lead, and emitter leads with the + (red) VOM lead. In some cases this measurement can be impacted by parallel conduction paths but the low voltage of this test should work OK in this case.

#2 If all the transistors check OK, next step it to test the op amps. This requires measurement  with power applied. Op amps are pretty simple creatures with massive DC gain, i.e. a tiny DC voltage difference between the op amp + and - input will result in huge voltage at the op amp output. This is a simple way to find faulty op amps. Look for ones where their output voltage state conflicts with the input voltage commands. Say the op amp output is pegged to one rail or the other...  To be a valid op amp state for pegged to positive rail, the + input only needs to several mV positive wrt the - input. This measurement can be complicated by small typical input offset voltage errors, but a faulty op amp will often exhibit gross directional output errors (like output positive with - input more positive than + input or vice versa). 

Like Duke suggested we can probably guess from looking at the circuit node voltages, and eventually you can too. In operation the bases of Q8 and Q9 should be close to 0V. Therefore their emitters should be roughly one diode drop higher or around +0.6V. Their collectors will be at some similar negative voltage, single digit volts above the - rail (I am too lazy to calculate exactly how many volts).

Voltages that are not close to what I predict should be inspected further for faults causing the deviation.

Try to figure it out yourself first but if nothing obvious jumps out you can markup a schematic with DC voltages of the sundry nodes, and post it here.

Welcome to the party.... Troubleshooting is the first lesson in learning how to design.

JR
 
L´Andratté said:
There are no protection diodes so a static discharge will easily kill the input transistors (e.g. exceeding base emitter voltage of 2n4403) replacing them would be the first thing to try, don´t substitute 2n4403, they are there because they are lower noise (lower base resistance) than most.

Funny you should mention that.  A later version of the schematic (and the design) actually has protection diodes across those transistors.  I had inadvertently downloaded the later version and wasted a lot of time trying to find those diodes. 

Apparently, the guys at Symetrix got tired of replacing transistors.
 
Martin Griffith said:
I'm a bit confused by their phantom power, it doesn't use the "standard" 6k81 resistors, what are the advantage of doing it this way? Also the lack of input protection (zener)diodes
They are lacking a couple protection schemes but probably get away with it until they don't.

That phantom is dropped down by a series 15V zener from 60V so roughly 45V. Fed common mode the microphone probably doesn't know the difference.

One difference that might be considered a benefit is they don't use the typical 22uF DC blocking capacitors in series with each input leg. That way with no mic plugged in the input transistors will float up to +45V but they are fed by a current source from +60V.

I do not consider needing a 60V rail a "benefit", but it is yet another way to skin that Cohen cat.

JR

PS: The voltages I gave before for troubleshooting are with phantom power off.  :-[ 
 
CurtZHP said:
Funny you should mention that.  A later version of the schematic (and the design) actually has protection diodes across those transistors.  I had inadvertently downloaded the later version and wasted a lot of time trying to find those diodes. 

Apparently, the guys at Symetrix got tired of replacing transistors.
Clamp diodes reverse across base-emitter is generally to prevent them from zenering (at roughly 7V reverse Vbe bias). When (low noise transistors (cough***) are allowed to zener they can become noisy.

Mic preamp protection clamps are usually to ground  and from op amp input to the rails. Static hits to an unprotected circuit like that can take out the op amp. 

The 4403 is considered more of a general purpose transistor but was better than average and popular for low noise designs back in the 70s (Motchenbacher and Fitchen wrote them up in the classic "Low Noise design" text). .

JR
 
Martin Griffith said:
Haha,  any more modern recommendations, rather than the expensive MAT03 and THAT type offerings?
My favorite low noise devices (from ROHM 2BS737, 2SD786) went obsolete around the turn of the century.. :'(

These days most manufacturers just use off the shelf preamp ICs.

Some people are looking at ZTX851 which are not even touted as low noise (nor was the 4403). The 851 is a 5A transistor so low Rbb for the high current carrying capability. Since the Rbb is a noise source they are low noise. Dedicated low noise devices are higher beta, but I heard good gossip about 851s. 

There are other alternatives out there,,, but I have not used any, or wouldn't. Modern manufacturing charges by the touches or "pops" so  one IC that replaces a pile of separate components and does it the same or better is a stone cold winner.

JR
 
OK, had a chance to measure some voltages.  Here's what we have....

With the unit powered on:
Q8 Base:  0.032VDC
Q8 Emitter:  -4.3VDC
Q8 Collector:  -10.08VDC

Q9 Base:  0.061VDC
Q9 Emitter:  0.673VDC
Q9 Collector:  -10.68VDC

Opamp Pins:
1:  14.15VDC
2:  -10.74VDC
3:  -10.06VDC
4:  -15.31VDC
5:  0.004VDC
6:  0.613VDC
7:  -13.98VDC
8:  14.89VDC


Did the diode test on Q8 and Q9 using the diode setting on my Fluke meter.
Q8 base-to-emitter:  0.617V
Q8 base-to-collector:  0.629V

About the same on Q9.  Within 1/1000 of a volt.

I had to test the transistors in-circuit.

The first thing that jumps out at me is the mismatch between the emitter voltages on Q8 and Q9.  They are both fed from the collector of Q10 (TIP30A) via a pair of 2K resistors.  Shouldn't they read the same?  Fiddling with the gain control had no effect on this, for what that's worth.

As far as the voltages on the opamp....
There's only a .7V differential between pins 2 and 3, but as John said, that's enough to swing the output all the way positive, if I understood correctly.


 
CurtZHP said:
OK, had a chance to measure some voltages.  Here's what we have....

With the unit powered on:
Q8 Base:  0.032VDC
Q8 Emitter:  -4.3VDC
Danger will robinson, something is wrong... emitter should be slightly positive.

Is voltage at R11 also at -4V?  if not may be bad solder
Q8 Collector:  -10.08VDC

Q9 Base:  0.061VDC
Q9 Emitter:  0.673VDC
Q9 Collector:  -10.68VDC

Opamp Pins:
1:  14.15VDC
2:  -10.74VDC
3:  -10.06VDC
4:  -15.31VDC
valid state for that input V
5:  0.004VDC
6:  0.613VDC
7:  -13.98VDC
valid... + input is 0.6V lower than - telling op amp to swing hard negative.
8:  14.89VDC


Did the diode test on Q8 and Q9 using the diode setting on my Fluke meter.
Q8 base-to-emitter:  0.617V
Q8 base-to-collector:  0.629V

About the same on Q9.  Within 1/1000 of a volt.

I had to test the transistors in-circuit.

The first thing that jumps out at me is the mismatch between the emitter voltages on Q8 and Q9.  They are both fed from the collector of Q10 (TIP30A) via a pair of 2K resistors.  Shouldn't they read the same?  Fiddling with the gain control had no effect on this, for what that's worth.

As far as the voltages on the opamp....
There's only a .7V differential between pins 2 and 3, but as John said, that's enough to swing the output all the way positive, if I understood correctly.
yup opamp is probably OK... but emitter voltage of Q8 is AFU.

For a slightly more advanced troubleshooting trick, you can use ohms law (I=E/R) to calculate how much current is flowing down R11 and supposedly into the Q8 emitter.  The voltage drop across R11/2k will equal the current flowing into that emitter?

More evidence Q8 collector is at -10V  with R15 connected to -14V , alternately the collector of Q9 has R16 connected to +15V so 25V/30k for some 83uA...

5V across both R8 and R9 looks like roughly 5ma of collector current from each.

The emitter voltage for Q8 is definitely wrong, but the collector voltage looks like it is almost working.

Could be a partially damaged 4403, bad solder connection, or bad measurement.

You can figure exactly how much current is flowing into the two emitters by measuring voltage drops across R10 and R11, and applying Ohms law.

Or just replace Q8...  8)  It looks like a weird failure mode but static voltage shocks can cause partial breakdowns in semiconductors.

JR

 
Took a closer look at that.

The collector voltage of Q10 is 7.65VDC.  That's what shows up on that end of R11.  The other side of R11 measures -4.13.  (That's an 11.78V difference...)

I came up with 0.006A through R11.  I probably did that wrong...  :-[

But, like you suggested, it's looking more and more like Q8 took the hit.  Fortunately, I happened to find an unused 2N4403 in the parts bin!  It's a TO-92 case, so it'll be a little brain surgery to put it where the existing SMT part is.

 
CurtZHP said:
Took a closer look at that.

The collector voltage of Q10 is 7.65VDC.  That's what shows up on that end of R11.  The other side of R11 measures -4.13.  (That's an 11.78V difference...)

I came up with 0.006A through R11.  I probably did that wrong...  :-[
math seems correct  12v/2k=6mA

but transistor should be turned off with -4V e-b, so 4403 is likely toast


But, like you suggested, it's looking more and more like Q8 took the hit.  Fortunately, I happened to find an unused 2N4403 in the parts bin!  It's a TO-92 case, so it'll be a little brain surgery to put it where the existing SMT part is.
not brain surgery, and electrons don't care about the package.

Afterwards check that they are close enough for the servo range. Voltage at TP8 not saturated to one rail or the other.

Reworking smd is not trivial without hot air tool but possible. I used to add a bunch of solder and get all three solder pads melted then slide off the part with tip of the iron. Clean up the excess solder before replacing the part.

JR
 
JohnRoberts said:
not brain surgery, and electrons don't care about the package.

Afterwards check that they are close enough for the servo range. Voltage at TP8 not saturated to one rail or the other.

Reworking smd is not trivial without hot air tool but possible. I used to add a bunch of solder and get all three solder pads melted then slide off the part with tip of the iron. Clean up the excess solder before replacing the part.

JR

Did a similar repair once before with an opamp.  The trick there was to get at least two pins soldered so the chip wouldn't move around while doing the rest.  At least here, there's only three pins.
 
CurtZHP said:
Did a similar repair once before with an opamp.  The trick there was to get at least two pins soldered so the chip wouldn't move around while doing the rest.  At least here, there's only three pins.
For me removing the old smd component was the hard part.

JR
 
Surgery was a success!

Swapped Q8, and she's back in business.  Sounds great!  Thanks for all your help!
 

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