开云体育

Well, looks like I'll need an external display sooner rather than later... my 576 just arrived with a smashed screen.

Run the exerciser 03 routine.? See the manual.

Regards

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On Tue, May 3, 2022 at 05:32 PM, BryanByTheSea wrote:

2465B Trace Possible Z-Axis Issue

I looked at your video and it is not similar. Mine is showing very regular dots all the way across the bottom of the screen. This usually means the scope is not calibrated.

On Tue, May 3, 2022 at 03:59 PM, lpb612 wrote:

On Tue, May 3, 2022 at 03:58 PM, lpb612 wrote:

There is no DC feedback in the error amplifier TL494/U2. Just R1033 in

series

with C1033. The same happens with U4 - no feedback at all. Therefore the
output voltage V2 and V4 should be either LOW or HIGH, but not in the

middle,

as Figure 3-11 seem to suggest. It can work like this in a bang-bang way

(100%

or 0% duty cycle). But is this how the circuit is supposed to work?

Correction of typo: V2 and V4 should be U2 and U4, the error amplifiers.

---
That is what I guessed in my reply. U4 is wired for over voltage protection so it is OK it be be on/off. When over voltage at the primary happens it needs to turn off the switcher.

U2 is the one that controls the loop. As I wrote in the previous message it is wired more like an integrator, it also has a zero coming from R1033 for compensating the loop, so in normal operation output of U2 is a control voltage and not on/off only. There is no local DC feedback between pins 2 and 3 but global feedback (around the full switcher control loop) still controls DC voltage at the input.

Ozan

I got one of those pots from QService some years ago.

EJP

On Tue, May 3, 2022 at 03:58 PM, lpb612 wrote:

There is no DC feedback in the error amplifier TL494/U2. Just R1033 in series
with C1033. The same happens with U4 - no feedback at all. Therefore the
output voltage V2 and V4 should be either LOW or HIGH, but not in the middle,
as Figure 3-11 seem to suggest. It can work like this in a bang-bang way (100%
or 0% duty cycle). But is this how the circuit is supposed to work?

Correction of typo: V2 and V4 should be U2 and U4, the error amplifiers.

@Jeff I can't remember exactly which rate I settled on, but yep, it did need to be bumped down a bit for the clearest image

@zentronics "Disturb the measurement"? To an extent, sure. "Cause the unit to be all over the map"? Depends how you're doing it. Every time you take a current measurement with an ammeter there is burden voltage, do you consider those all over the map or do you choose your range/size your shunt resistor to minimise the impact? There's a world of difference between simply connecting a multimeter in series with a DUT and placing a low value shunt resistor right at the connections for a specific purpose. And this doesn't at all take into account the other options available like changing shunt value, kelvin sensing or using a current clamp instead (like I mentioned earlier). I agree that if you're doing very sensitive measurements you need to be aware of any issues a shunt may cause, of course, but in context of what I said I was doing (measuring tubes and saving the plots digitally) it's unlikely to be a concern

May be way off topic but I did have what could be a similar situation many years ago with what appeared to be dots on the screen trace. I did a quick utube video "2465B Trace Possible Z-Axis Issue", solution was the Grid Bias adjustment. May or may not be related. If I recall this is a adjustment done on the HV board.

Zen,

I did make some progress with the 067-0521-00 Calibration Fixture on my 547 and need only remove a small dip or 'anti-bump,' if I may, in the first 10 nS of the top of the square wave. Otherwise, I seem to have the requisite 7 nS rise time per the instruction manual and proper squareness of the waveform. Once this is all settled, I will try to tame the square waves coming out of the Tek 106 & 107.

What about the Bodner for the 453? It seems like that scope would be a reasonable candidate while conceding what D. Wise had noted about the waveform so generated by the Bodner. Of course, I suspect that it would be better for me to mend the 106 properly so that I could get the proper square wave for the 453, but as I have never touched the HF response of that scope, I am tempted to leave it alone for the moment.

PMF

On Tue, May 3, 2022 at 01:58 PM, lpb612 wrote:

Can you help me with the following two questions? I have had difficulty
figuring out an answer by myself.

1) The PWM control signal is taken from the REF line and 5V REF of TL494
through R1045 and R1046. However the REF line relative to PIN 7 of TL 494
(namely REF2 in later versions) is not a DC voltage. It is a switching voltage
following the switching pattern of Q1050. Does C1035 serve as a filtering
capacitor, together with the Thévenin equivalent resistance R1045||R1046?

Yes, C1035 acts as a filter capacitor as far as I can tell.

2) There is no DC feedback in the error amplifier TL494/U2. Just R1033 in
series with C1033. The same happens with U4 - no feedback at all. Therefore
the output voltage V2 and V4 should be either LOW or HIGH, but not in the
middle, as Figure 3-11 seem to suggest. It can work like this in a bang-bang
way (100% or 0% duty cycle). But is this how the circuit is supposed to work?

U4 is just for over voltage protection, it is inactive in normal operation. U2 controls the switcher operation.

C1033 is a large cap, you can think of U2 as an integrator rather than a comparator. DC level at input is set by switcher control loop (let me know if this needs more explanation). Optocoupler U1040 and feedback signal from sheet <10> regulates +5VD.

Ozan


There are two control loops R

On Tue, May 3, 2022 at 09:03 AM, Martin wrote:

mine is working BUT one of the pots is mechanically broken. I tried to repair,
in vain. Couldn't find a replacement neither.

I did a quick search on ebay (mod pot in business and industrial) and came up with lots of good hits. Depending on the resistance values, you might find exactly what you need.

I've contacted the people now making the mod pot line, hoping to get a replacement for an HP part. I found their pricing to be extreme.

On Tue, May 3, 2022 at 03:05 PM, Ozan wrote:

Good analysis and great idea to simulate the start up section. Your spice
netlist is missing Q1062 and the load at its emitter. For inverter driver to
function Q1062 should be able to support at least the current through R1066
and R1068 (in alternate phases) in addition to other components. R1069 is too
large to supply such current. With a single 2k from Q1062 emitter to (REF)
node there is not enough voltage across C1066 and at the emitter of Q1062 for
proper driver operation without CR1062-CR1065 working.

This is one of those Aha moments for me. Thanks for the correction. It makes perfect sense to me now.
With just the R1066 and R1068 as load, C1066 only charges to a bit less than 3V. More load (U1062-U1064) will lead to even lower voltage. So the initial push using C1072 to get C1066 charged is necessary.

It also removes another doubt for me. When Q1062 is not considered, C1066 would charge to about 40V at maximum. However, this exceeds its 35V rating. With Q1062 considered, that will not happen.

Can you help me with the following two questions? I have had difficulty figuring out an answer by myself.

1) The PWM control signal is taken from the REF line and 5V REF of TL494 through R1045 and R1046. However the REF line relative to PIN 7 of TL 494 (namely REF2 in later versions) is not a DC voltage. It is a switching voltage following the switching pattern of Q1050. Does C1035 serve as a filtering capacitor, together with the Thévenin equivalent resistance R1045||R1046?

2) There is no DC feedback in the error amplifier TL494/U2. Just R1033 in series with C1033. The same happens with U4 - no feedback at all. Therefore the output voltage V2 and V4 should be either LOW or HIGH, but not in the middle, as Figure 3-11 seem to suggest. It can work like this in a bang-bang way (100% or 0% duty cycle). But is this how the circuit is supposed to work?

Thanks so much.

Luca

On Tue, May 3, 2022 at 10:30 AM, lpb612 wrote:

My current understanding of the start up process is as follows:
1. C1025 is charged using the following paths

T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- C1072 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- C1066 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- R1061 -- VR1062 -- T1020
PIN6

Initially the first 2 paths are used. After C1072 reach about +40V and C1066
reaches 13V in about 250ms, the only path that is active is through the Zener
Diode VR1062 because it is 12V limited. It will take about 2 seconds for C1025
to reach +21V.

2. After C1025 reach about +21V relative to REF, Q1021 and Q1022 saturates,
and TL494 (U1030) starts to work. The initial cycles are almost 100% duty
cycle. And Q1050 conducts. The current will flow through CR1072 to further
charge C1072 towards regulated line voltage (325V).

The manual states that the initial current pulses creates enough secondary
voltage to charge C1066 to its working voltage, and INTERTER DRIVER starts to
work. However, it seems that C1066 is already charged in the initial stage to
about +13V before C1025 is charged to +21V. So I have some doubts here as to
whether C1072 is really needed to act as a initial load. I think the circuit
should start even if it is removed. It does provide a path for the current, in
case the INVERTER DRIVER is not working yet.

Good analysis and great idea to simulate the start up section. Your spice netlist is missing Q1062 and the load at its emitter. For inverter driver to function Q1062 should be able to support at least the current through R1066 and R1068 (in alternate phases) in addition to other components. R1069 is too large to supply such current. With a single 2k from Q1062 emitter to (REF) node there is not enough voltage across C1066 and at the emitter of Q1062 for proper driver operation without CR1062-CR1065 working.

Ozan

Martin,
One more possibility...
There is a place in New Jersey, State Electronics, that apparently is making modular potentiometer assemblies similar to the ones used in Tek equipment.
Here is a URL to their "Mod Pot" catalog:
The catalog looks like they can build to order from stock pieces, they do list multi-section potentiometers with dual concentric (.125/.078 inch) shafts.
I don't know how spendy, however it could be an option. The catalog talks about supplying prototype quantities.
gwerl

The following is a simulation program for the charging circuit. It should run using NGSPICE.

* Charging circuit for C1025

Vin 1 0 PULSE(0, 325V, 10u)
R1020 1 2 270K
C1025 2 3 100u IC=0V
D1034 3 4 Diode
R1046 4 0 240K

D1072 3 5 Diode
C1072 5 0 3.3u IC=0V

R1069 5 6 36k
C1066 6 0 4.7u IC=0V

R1061 6 7 2k
D1062 0 7 Dzener

.MODEL Diode D
.MODEL Dzener D BV=12V

.control
tran 1m 5s
plot V(2)-V(3) V(5) V(6)
.endc

.END

My current understanding of the start up process is as follows:
1. C1025 is charged using the following paths

T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- C1072 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- C1066 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- R1061 -- VR1062 -- T1020 PIN6

Initially the first 2 paths are used. After C1072 reach about +40V and C1066 reaches 13V in about 250ms, the only path that is active is through the Zener Diode VR1062 because it is 12V limited. It will take about 2 seconds for C1025 to reach +21V.

2. After C1025 reach about +21V relative to REF, Q1021 and Q1022 saturates, and TL494 (U1030) starts to work. The initial cycles are almost 100% duty cycle. And Q1050 conducts. The current will flow through CR1072 to further charge C1072 towards regulated line voltage (325V).

The manual states that the initial current pulses creates enough secondary voltage to charge C1066 to its working voltage, and INTERTER DRIVER starts to work. However, it seems that C1066 is already charged in the initial stage to about +13V before C1025 is charged to +21V. So I have some doubts here as to whether C1072 is really needed to act as a initial load. I think the circuit should start even if it is removed. It does provide a path for the current, in case the INVERTER DRIVER is not working yet.

3. Once INTERTER DRIVER starts to work, the secondary windings of T1050 will generate voltages to power TL494 the PWM controller and the INTERTER DRIVER itself.

On Mon, May 2, 2022 at 03:17 AM, TV7 wrote:

...
When using the A sweep by itself the trace is perfect. When using the B trace
by itself it is again, perfect. When you use alternate mode the A sweep
contracts as per the photo. The top waveform is the A.

If you have another scope and can look at collectors of Q465 and Q475, do you see 0V to 27.5V pulses with opposite phases? Perhaps one of the components (e.g. R474) drifted around that area and in switching state pulses are not reaching to their target voltage.

Ozan

Martin,
Check here:
They are listing the pot for $12.95.
I don't know anything about the vendor, maybe you will get lucky there...
gwerl

hi David,

VK2seb had two YouTube videos on an LCD upgrade to an HP 141 SA that looked like a promising start for the Tek 576.
I do not see the Part 2 YouTube video on this project ( ? ),
- hope he did not take it down ?

Your idea is good, the Tek 576 seems worthy of such data collection for comparison work, aging issues, & repairs.
I have ordered the VKseb LCD's to try out on the Tek 576, but due to supply chain issues, they have not arrived yet.
I am not certain if there might be other LCD - CRT replacement project blogs that might provide insight for this Tek 576 project ?

See:


Seb Holzapfel is in Oz & his YouTube page has his email address.

Let us know about your progress !
rick

**
Previous Ref's to this topic: possible Tek 576 USB port to CPU ?
/g/TekScopes/topic/69490299#164213
/g/TekScopes/topic/86718765#187594
/g/TekScopes/topic/69626540#163154
/g/TekScopes/topic/90831293#192998

**

Well, I thought I was done but now the dotted line has reappeared across the bottom, meaning, I think, that it is not calibrated. I still get no errors on bootup though. I was checking various voltage to see how accurate it was after calibration, and it is off. Most of the voltages read low, some maybe 10% low, so I guess it does need calibration (again), but how does the scope know that? Any ideas what could be wrong and why calibration doesn't result in calibrated ranges?

Admittedly my test equipment isn't the greatest, but if it is off, wouldn't the scope just be off the same way? As I mentioned in another post, I had to tweak the 1V and 10V calibration voltages to get by the limit errors. I had to make them a little lower. I assumed that the PSG9080 waveform generator output was a little high, causing the limit error.