8116A Code 42

[MarkL]

On BNC tees, Amphenol makes good tees for a reasonable price.  I always like Pomona components, but their tees are about double and I'm not sure worth it.  I would buy new from Mouser or Digikey.

Looking at the end points of C528/C530 in series, you can see it is between the base of Q504 and the collector of Q506.  A quick determination of their effect can be done by imagining a small positive voltage change on the base of Q504.  This would move Q504 closer to turn-off, and reduce the emitter current.  The emitter would move positive, being pulled up by the resistor network R530/R532.

This would in-turn cause the base of Q506 to go more positive, which would then pull down the collector of Q506.  The negative change here would be coupled back to the base of Q504 through C528/C530 which we imagined was going positive, therefore limiting extent of the positive change.

In short, what all this is doing is reducing the slew rate of this stage of the amplifier until the feedback can take affect and keep the waveform true.

On the positive driver side, C529 is performing the exact same function, and contributes to the slew rate control along with the C528/C530.  C529 is adjustable in order to tweak the overall slew rate of this stage.

Note that if C529 has insufficient range, C525 and C530 can be changed as per Table 9-1 with the procedure described on page 9-7.  C525, C551, and C552 also limit the high frequency response of this stage by the shunting the bases of Q503 and Q504.

A 200V rating on C528/C530 is not necessary, plus you have them in series which approximately doubles the voltage rating (ignoring imbalances).  You could substitute 50V, but I would be surprised if the existing 1pF caps were bad and needed replacing.

[Mick B]

WOW! Mark, I can see it so clearly now a great explanation of that circuit. I have my doubts also about anything being wrong with c528 & c530, but I wonder about c529 being off, basing this on scope results while making adjustments on c529 and Bias adjustment R535 for adjustment on Pg.9-6. Maybe I’m overthinking this. Awhile back I changed all the Power transistors. In the process Q501 and Q502 are different Q501 is a TO18 pkg. and Q502 is a TO92 pkg. Q501 is a direct replacement Q502 suppose to be, and the specs are remarkably close, but not exact Part numbers Q501 is a 2N2369A and Q502 is a 2N5771 /with a heatsink attached. Perhaps the problem is here.
I thought it might be fun to play around with c529,c528 & c529 to see what difference it makes to the overshoot and slew rate.   
Before I do that I’m going to get on the HP5005A.

[MarkL]

The power amplifier and other analog sections look like they've had many tweaks applied to get it as perfect as possible as HP engineers went through the design process.  And sometimes they still couldn't depend on consistent enough specifications for components (probably the semiconductors), as evident by Table 9-1.

I wouldn't be surprised if some of the semiconductors were standard parts, but "selected".  The HP Parts List and HP Cross-Ref has this info if you want to check on the ones you replaced with non-HP (generic) part numbers.

[Mick B]

Hi Mark, I have a file that was downloaded as HP Master Xref.pdf Pg.1. Transistor-Diode Cross reference - H.P. Part Numbers to JEDEC Numbers
and it list Q502 as the same number as the manual. I don't remember how I came up with 2N5771 as a replacement just in TO92 pkg. I spent weeks tracking down original part numbers for this and found all the transistors except Q502. I don't know if this means anything in today's part market. Do you have a Xref. besides the ones I have that list the
1853-0405 as anything besides a 2N4209? I will upload the ones I have, for those that might not have them. CORRECTION! only one file the other mentioned above is too large.

Then I found this today.

https://www.microchipdirect.com/product/2N4209
availability: Order now, estimated to ship on 06-Jan-2025
1.  $16.91 per unit

[MarkL]

That appears to be correct that for Q502, 1853-0405 is a 2N4209.  And all the xrefs I have show that 1853-0405 is not a selected part; just a run-of-the-mill 2N4209.

I also have a copy of the xref you posted, and it says 1853-0405 is a "TO-17" case.  I don't know what a TO-17 is, but everything else including the datasheet says it's a TO-18 (metal can).  Probably a misprint.

[Mick B]

Yep, TO18 that's the same conclusion I came to. Compare this datasheet 2N5771 to the 2N4209. also, I have the heatsink on it.
can you believe they want $17. plus, about $8. freight for that 28c part. NO lube and you don't even get a kiss.

Working on the 3 coax cables for the HP 5005A. Trying to reuse their old ferrule and hold-down/strain relief that were bent in the removal process is turning out to be quite challenging. I did find a sheath to use from an 02 sensor I changed out last weekend.
Slow going, but going

[MarkL]

Yep, TO18 that's the same conclusion I came to. Compare this datasheet 2N5771 to the 2N4209. also, I have the heatsink on it.
can you believe they want $17. plus, about $8. freight for that 28c part. NO lube and you don't even get a kiss.
...

Hmmm...  I don't see any differences with the electrical specs on the 2N5771 either, apart from the case as you point out.  However (and even though the thermal resistance characteristics are essentially the same), the 2N4209 has a much higher power dissipation of 1.2W vs. 350mW.

You could measure the quiescent Vbe and the voltage across collector resistor R526 to calculate if you're near or above 350mW.  Unfortunately, if R526 is close to 12.1V, then the parallel VR502 will interfere with the measurement, in which case you could more annoyingly measure across all the emitter resistors R519, R521, R523, R529, and R531 to compute the total collector current.

[Mick B]

Quote from: MarkL on: Today at 04:14:00 pm
You could measure the quiescent Vbe and the voltage across collector resistor R526 to calculate if you're near or above 350mW.  Unfortunately, if R526 is close to 12.1V, then the parallel VR502 will interfere with the measurement, in which case you could more annoyingly measure across all the emitter resistors R519, R521, R523, R529, and R531 to compute the total collector current.

Well, fortunately all those resistors are exposed I could do that. but to what end? The 2N4209 is not available until Jan (maybe) I waited over a year for an opamp to fix a power supply once. Here are the specs on the heatsink, It's better than a poke in the eye with a sharp stick. I Unfortunately am an expert at this comparison.

I have some Sunon 12V fans I could mount it up front where it would blow directly on Q502 and Q501. OVERKILL? 
I will post the heatsink & fan specs
 

[MarkL]

Quote from: MarkL on: Today at 04:14:00 pm
You could measure the quiescent Vbe and the voltage across collector resistor R526 to calculate if you're near or above 350mW.  Unfortunately, if R526 is close to 12.1V, then the parallel VR502 will interfere with the measurement, in which case you could more annoyingly measure across all the emitter resistors R519, R521, R523, R529, and R531 to compute the total collector current.

Well, fortunately all those resistors are exposed I could do that. but to what end? The 2N4209 is not available until Jan (maybe) I waited over a year for an opamp to fix a power supply once. Here are the specs on the heatsink, It's better than a poke in the eye with a sharp stick. I Unfortunately am an expert at this comparison.

I have some Sunon 12V fans I could mount it up front where it would blow directly on Q502 and Q501. OVERKILL? 
I will post the heatsink & fan specs

To what end?  To determine if your substitute is above the max power spec.  We already know a lot of components in this circuit run hot and HP may have designed Q502 to normally run above 350mW.

It's odd that the lower power dissipation of the 2N5771 is not reflected in any other parameter compared to the 2N4209, but I'm not a semiconductor designer.  Maybe there's something in the structure of the die that it can't support more power, or something else I'm overlooking.  But in general, when I select a part I make sure I stay within the specs and not think I know more than the manufacturer.

Fan or heatsink, if you're above 350mW, you're out of spec.  It's your unit - check the power dissipation of Q502 or not.

[Mick B]

Mark, the point of "to what end that would be me thinking out loud
If it's out of spec there is nothing, I can do about it, well maybe a fan, at least until Jan. and then it wouldn't matter. and do I want to spend another $25.00 on it? I'm into it for a bunch already. OF CORSE I DO! I have to remind myself occasionally why". 

 The whole purpose of this is a learning experience, I have a working one with OPT. 001. (we fixed) This has been a wonderful learning experience, with your guidance, I will definitely check the power dissipation of Q502, I had no idea how to do this before yesterday. I just put in this part knowing that the TO92 pkg. was only 350mW this was the closest thing I could find, hence the heatsink, I was waiting for it to go up in smoke, but it didn't. now I am questioning if this is overheating and not going up in smoke but causing distortion resulting in overshoot & high slew rate. 
Be easy on me, I'm in my 70's and just started getting into electronics on a component level a short while ago to occupy myself after physical injuries retired me. I'm a fast learner. most times.
I will post the results for Q502 after repairing the HP5005A, perhaps tommorow. Once again Mark, thanks for your help. I'm just sorry you don't live next door.   

[Mick B]

  I did it! my HP 5005A is alive. That was not fun, but in the end the 3 coax cables were successfully replace with RG316 and a sheath from an old O2 sensor out of one of my Saabs.
After putting it together there were error messages E18 & E15 this turned out to be the ribbon cables connecting the 3 boards together, after reseating them a few times no errors. As far as I've tested is the quick test as far as the test leads go there supposed to measure approximately 100K all 3 are 98.7 ohms winner winner chicken dinner!
Now I can get back to testing the HP 8116A's before I was so rudely interrupted by a faulty coax cable.       

[Mick B]

Hello, I posted this conversation of HP5005A unit in a new thread. Apparently, there is more work to do. 

 https://www.eevblog.com/forum/testgear/hp-5005a-a6-pod/msg5593715/#msg5593715] [url]https://www.eevblog.com/forum/testgear/hp-5005a-a6-pod/msg5593715/#msg5593715[/url]

[Mick B]

Quote from: MarkL on:  August 02, 2024, 04:14:00 pm
You could measure the quiescent Vbe and the voltage across collector resistor R526 to calculate if you're near or above 350mW.  Unfortunately, if R526 is close to 12.1V, then the parallel VR502 will interfere with the measurement, in which case you could more annoyingly measure across all the emitter resistors R519, R521, R523, R529, and R531 to compute the total collector current.

Hi Mark, I have the voltage readings across the emitter resistors.
Setup: Amplitude 999mV. Offset 0V. Frequency 1K. No Waveform. Mode Normal. 0utput on, Bk. lead on the emitter side.
R519 = 031.5mV
R521 = 0.144V
R523 = 031.5mV
R529 = 030.5mV
R531 = 022.0V

NOTE: Readings are the same at 16.0V
I do not know what the formula for this measurement is.
I know this about Q205, but this might be relevant. Attached are pictures of the trigger signal with Overshoot and Preshoot. The readings are jumping around from an OS of 53%~60% and PS of 7%~20%

I also acquired an HP 8007B Pulse Generator for calibrating the HP 5005A, to check the ICs on the HP 8116A 

[MarkL]

So, what we have is five resistors connected to the emitter of Q502.  By computing the current through each and adding them, you can get the current through the emitter.  Then, by measuring the voltage drop from collector to emitter (I said Vbe, sorry I meant Vce), you can calculate the power dissipation in Q502.

Like this:
                   ohms
                   ----
  R519 = 031.5mV / 59    =   0.533mA
  R521 = 0.144V  / 1k    =   0.144mA
  R523 = 031.5mV / 1k    =   0.250mA
  R529 = 030.5mV / 866   =   0.035mA
  R531 = 022.0V  / 1.78k =  12.3mA
                            --------
                            13.262mA

  Power Dissipation = Vce * 13.262mA

This assumes you've measured the resistors all in the same direction (e.g., + probe to the emitter or visa-versa), and you've reported the polarity of the reading.

As for the overshoot/undershoot on the trigger out, do you have another 50 ohm pulse output from another piece of equipment that you can look at to verify this is not a measurement setup problem?  Is this a piece of coax with a terminator on it going to the scope?  Does the overshoot change with the length of coax?

[Mick B]

So, what we have is five resistors connected to the emitter of Q502.  By computing the current through each and adding them, you can get the current through the emitter.  Then, by measuring the voltage drop from collector to emitter (I said Vbe, sorry I meant Vce), you can calculate the power dissipation in Q502.

Like this:
                   ohms
                   ----
  R519 = 031.5mV / 59    =   0.533mA
  R521 = 0.144V  / 1k    =   0.144mA
  R523 = 031.5mV / 1k    =   0.250mA
  R529 = 030.5mV / 866   =   0.035mA
  R531 = 022.0V  / 1.78k =  12.3mA
                            --------
                            13.262mA

  Power Dissipation = Vce * 13.262mA

This assumes you've measured the resistors all in the same direction (e.g., + probe to the emitter or visa-versa), and you've reported the polarity of the reading.

As for the overshoot/undershoot on the trigger out, do you have another 50 ohm pulse output from another piece of equipment that you can look at to verify this is not a measurement setup problem?  Is this a piece of coax with a terminator on it going to the scope?  Does the overshoot change with the length of coax?

Hello Mark, Fantastic it's way under spec. Using the numbers in the quote above.
VCE is 10.627V
Power Dissipation = 10.672V * 13.262mA = 140.935mW
yes? Also when taking voltage readings is it OK for instance, to wright 10.5V and -10.5V is it assumed if there is no minus sign its positive? I ask this because all of meters do this.

As for the overshoot, have a multitude of Tektronix 50 Ohm 2W passthrough adaptors looking at them on my spectrum analyzer they are very flat out to 3.2GHz somewhere between -1 & -2.5 dB down. Using these with different size and length coax makes little difference. I also have another function Gen that has a 50 Ohm output and has very little overshoot and preshoot both 2.6%.

But I just realized Q503 & Q504 also in a TO92 package with heat sinks I will see what these are and check them also. After looking at the schematic this seems a little more complicated, looks like the same resistors and their counter parts but what about R532?
Thanks Mick
 

[MarkL]

Yes, that would be the right calculation.  Clearly the current through R531 dominates.  The power dissipation could also be different at different DC voltages, but we're just trying to get a ballpark here.  140mW is ok.

When adding currents, yes, the sign is critical since that indicates direction.  If the currents were a little closer in this case, and one was of the opposite sign, it could make a significance difference in the emitter current.

Ok, so maybe something is wrong with the trigger output driver.  I don't want to get too far tangent of the initial problem with the main output, but maybe this is worth spending a little time on.  You could start investigating with a terminator on the trigger out directly (with nothing else) and probe R265 with a 10x probe.  Make sure you use a spring ground clip to minimize ground lead inductance effects.  See if you get a similar waveform.  I am not sure where ground is on your waveform captures but the signal should be (approx) 0V to 2.5V with minimal overshoot.

For Q503 current, you could calculate it via R531 and R532, and for Q504 via R530 and R532.  Some of the current is being siphoned off by the bases on Q505 and Q506, but it is probably not significant compared to the total current.