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HP 1741A Oscilloscope

KrisBlueNZ

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After you've done step 5 in post #137 you can tell whether the fault is in Q11 or VR3.

If the voltage across VR3 withi Q11 removed remains around 25V, VR3 is faulty. Replace it with a 1N5258B - 36V, 0.5W zener diode.
http://www.digikey.com/product-detail/en/1N5258B-T/1N5258BDICT-ND/190874
http://www.digikey.com/product-detail/en/1N5258B-TP/1N5258B-TPCT-ND/950569

If the voltage across VR3 with Q11 removed increases to around 36V, Q11 is faulty. Or possibly there is a very heavy load on the 156V rail. But try replacing Q11. You can use a BD139:
http://www.digikey.com/product-detail/en/BD13916S/BD13916S-ND/975650

It has a different package and a different pin order from the original. You'll need to compare the data sheets to work out how to use it. You probably won't find any suitable transistors in the same package as the original; that package is almost obsolete now.

I suggested the BD139 because it's probably available locally. If you have to use Digikey, there's a much better option: KSC2690:
http://www.digikey.com/product-detail/en/KSC2690AYSTU/KSC2690AYSTU-ND/1050610
It's in the same package as the BD139.
 
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tryppyr

Oct 22, 2013
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I plan on checking that out tomorrow. I doubt Q11 is the problem, though, as I just replaced it a couple of days ago using NTE396. Still, I'll pull it again and see what happens. I would have done it today, but for all the holiday-related running about like a headless chicken.

As an aside, Vetco sells the NTE373, which is sold as equivalent to KSC2690.
 
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KrisBlueNZ

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OK, the NTE396 looks like a suitable replacement. If you've replaced it and nothing changed, VR3 may be faulty, but it could also be a heavy load on the +156V rail. If you can measure the voltage across R18 I should be able to tell you which case it is.

I'm going to have to check out NTE's product range. They seem to make a good range of general purpose transistors that are suitable replacements for a lot of standard and older parts.
 

tryppyr

Oct 22, 2013
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The voltage across R18 is currently 9.1V.

After reading some of the service section of the manual today I was coming to the conclusion that I've got a heavy load somewhere. However, one of the tests the manual recommended was to remove the A14 Interconnect and test again. I did that and the results didn't change. That rules out the Vertical Preamplifier (A3) and Horizontal Sweep (A7)

That suggests that if there is a heavy load it must be in one of the two places that still have links down from the LVPS. One of those is A12 (Gate Amplifier Assembly), and the other is A17 (Storage Board).

When I got the numbers closest to the normal parameters I had both of those boards disconnected.

- Greg
 

KrisBlueNZ

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9.1V across a 1.5k resistor works out to about 6 mA. That is not a huge load. And if there's 77V across C7, that means there's 68V between the collector of Q11 and the 120V rail, so Q11 should easily be able to provide the 36V needed for the +156V rail. I think VR3 must be faulty. Try replacing it with a 1N5258B - 36V, 0.5W zener diode. You should be able to get one locally.

VR3 is shown reversed on the schematic. The new zener diode must be installed so its cathode (the end with the stripe) connects to Q11's base. The original part is most likely connected that way round too.
 

tryppyr

Oct 22, 2013
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Yes, I can get one of the VR3 replacements locally. If I get the time I'll pop by and get one today. Otherwise, I'll get it during a break tomorrow.
 

tryppyr

Oct 22, 2013
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I got the new VR3 installed today and decided it was time to take a full load of readings to see if we can chase down where the load is coming from. I ran the same set of measurements under five load conditions:

1) No load (LVPS not connected to anything)
2) A17 only
3) A12 (with attached HVPS) only
4) A17 and A12 (with attached HVPS) only
5) Full load

The results:
 

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KrisBlueNZ

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There are some weird numbers in the middle column - the +120V, +156V and +48V rails are much too high!

I guess this could be due to back-feeding from the HVPS, if the HVPS is running. What is the HVPS doing at the moment? Is it running and producing the ozone smell and the illuminated neons, or not?

The important numbers are those with all boards connected. You've listed two voltages that are wrong:
"-15V" is -0.5V. Is this the -15V rail on TP1 on the LVPS? If not, where did you take this measurement?
"5V" is 1.4V. Is this the +5V rail on TP2 on the LVPS board? If not, where did you take this measurement?
 

tryppyr

Oct 22, 2013
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All of the readings were taken from the LVPS. Most were taken from the primary rail. the +5 and -100 readings were taken from the connector leading down to A17.
 

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KrisBlueNZ

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OK, I guess we need to check the -15V and +5V regulators on the LVPS.

We have to fix the +5V regulator first because the +5V rail voltage is used by the -15V regulator.

The 5V regulator is a standard linear series regulator using Q5 (off-board) as the pass transistor, with U2 as the controller.

Can you measure the following voltages with the scope powered up and everything connected as normal. When measuring U2 pin voltages, be careful that the probe doesn't slip and short two pins together. If possible, measure the U2 pin voltages on the underside of the PCB. Measure relative to 0V except where stated.

1. Across R30. Should be less than 0.6V. R30 is a "current sense" resistor, connected in series with the load, so the load current can be calculated using Ohm's Law: I = V / R where I is current in amps; V is voltage across R30 in volts, and R is the resistance of R30, in ohms, which is 0.82.

So when you've measured the voltage across R30, divide that value by 0.82 and the result will be the load current in amps.

2. The +5V rail test point.

3. Q5 (off-board) collector. Should be between +7V and +10V roughly.

4. U2 pins 7 and 8. Should both be about +15V (they come from the +15V rail).

5. U2 pin 3. Should be between +4.5V and +5.5V. This pin sets the +5V rail voltage.

6. U2 pin 2. This is the feedback from the +5V rail that U2 uses to tell when the +5V rail voltage is right.

7. U2 pin 6. This is the output from U2 that controls Q5. It should be about 0.7V higher than the +5V test point voltage.

With those numbers, I should be able to figure out the problem with the +5V regulator.


It seems reasonably likely that U2 is faulty. This is a µA723H or LM723H regulator. U1 and U3 are the same.

The ones used on that board are in a 10-pin metal can package, indicated by the "H" suffix on the part number, which is still available from Texas Instruments at http://www.digikey.com/product-detail/en/LM723H/NOPB/LM723H/NOPB-ND/148205 for USD 8.45. The DIP version (http://www.digikey.com/product-detail/en/UA723CN/296-11104-5-ND/382232) is under USD 1.00 but would require an adapter made from a socket and some wires.

If U2 turns out to be faulty, you might want to buy one more in case you need to replace U3 in the -15V regulator as well.

What is the status of the HVPS now? Is it running?
 

tryppyr

Oct 22, 2013
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HVPS is not running (or emitting ozone) at present. Its status is unchanged since our last look at it. It occurred to me after the last batch of tests that I should try a test with only A12 connected to the LVPS, after removing HVPS from the other end of A12. That would give us a more accurate read of whether A12 is causing problems.

I'll have time this evening to run some tests, but I am not sure my hands and eyes are steady enough to get the readings on these round ICs. I'd frankly rather replace them and take the measurements elsewhere.

I found NTE923 is available locally and is listed as equivalent to LM723H. I'll pick up a couple on the way home.
 

KrisBlueNZ

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Yep, NTE comes to the rescue again! I'm surprised they second-source the metal can version!

Yes, you might as well try connecting A12 without the HVPS to see whether the A12 board messes up the voltage rails by itself.
 

GonzoEngineer

Dec 2, 2011
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I'm impressed:D

Although, I don't know that I would expend this much effort on a "Boat Anchor."
 

tryppyr

Oct 22, 2013
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I'm impressed:D

Although, I don't know that I would expend this much effort on a "Boat Anchor."

I completely understand. For me it's a learning experience and a labor of love.

I must say, I've been very impressed with the assistance Kris has been willing to provide. He probably doesn't have the love for this I do, but he sure has the determination.

- Greg
 

tryppyr

Oct 22, 2013
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I just noticed what may be a fundamental problem with my plan to replace A16U2. Take a look at the way the leads are mounted. They aren't soldered in. They seem to be glued in to sockets.

Any advice?

As an aside, I replaced VR2, on the off chance that might help U3 function better.
 

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tryppyr

Oct 22, 2013
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I just wiggled out the old U2, confirming it was glued into place. I'm going to take this to mean I should use a heat sink for each lead before soldering it in.
 

tryppyr

Oct 22, 2013
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I very carefully removed the lugs using a pair of bent hemostats and the soldering iron. No damage was done to the traces that I can detect. I then trimmed the leads so I could put them in sequentially and used the same bent hemostats as the heat sink for each lead as I soldered it it. As you can see I was very sparing with the solder so as not to allow any solder bridges to form. Should be okay, since the manual stated that all holes on the PCB are plated through, so soldering on one side should be equivalent to soldering on both.
 

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KrisBlueNZ

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I suspect the pin sockets were "machine screw" sockets that hold the wires in by friction. If there was something that looked like glue, it might have been some kind of lacquer to seal the hole and prevent corrosion. You probably didn't need to solder the leads in, but it won't matter... unless you ever need to remove them!

I'm finding this thread quite interesting. I hate to see an oscilloscope go to waste, even if it is a boat anchor. And this one is pretty old.
 

(*steve*)

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I just wiggled out the old U2, confirming it was glued into place. I'm going to take this to mean I should use a heat sink for each lead before soldering it in.

Maybe it was an expensive chip at the time? Maybe they needed to be selected for performance in the actual circuit?

I would probably remove the "socket" and solder the replacement straight in.

If the chip needed to be glued into the socket, I'd not be too confident of how well the pins were held in place.
 
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