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Bob, just wondering what has been done to that scope? Is it still giving problem in one channel?
Asking because I have 3 2440's, 2 2432's, 1 2432A and a few 2430A,s all waiting on parts to repair and calibrate,
so in a few months I will find out myself what these series scopes can do (or what else is wrong with them?)
Jeff

--
Jeffeelcr

Mark, Dave, Albert,

I’ve taken a closer look at the 2.2K resistors at pin 9. 3 of them show quite a bit of discoloration due to extensive heat…. I shall change them and see…. They may have drastically drifted and some are shorted or opened…

Albert, R59 is fine.

Thank you all for your help fixing this.

On Tue, Mar 22, 2022 at 07:48 PM, Albert Otten wrote:

Hi Stephen,

So you have -161 V at pins 3 of the output tubes, but hardly an output signal.
When there is no nearly short circuit somewhere in the output wiring this
means that the output tubes draw just a very small current during the most
positive voltage level at the grids, pins 2.
Ideally you measure the waveform of Vg1 versus Vk differentially with e.g. a
7A13 or a 7A22. You could also first obtain the waveform of Vg1 in AC. The
average is (should be) zero. Now measure the average of Vg1 with a DMM in DC
mode. Substract the DC Vk from this DC level and then add the observed AC
waveform of Vg1. What are the two levels of Vg1k?
With Vk = -161 V I expect some Ia = 50 mA per tube. In triode configuration of
the EL84 that nominally means Vg1k = -2.5 V (Philips graphs). In our
configuration even a more positive Vg1k.
R59 is really needed as pull-up resistor. Not open circuit?

Albert

Hi Albert,

Unfortunately I do not have a differential amplifier.

With 8 or 9 hours of time difference a lot has happened since I went to sleep! Last night I had got as far as realising that U1724A was sinking current at its output so was not going to be responsible for excessive current drain. Overnight I remembered that it was part of a dual op-amp so I got up early to check what U1724B might be doing and found, not surprisingly, that Harvey had got there before me.

I hope you are now on the home run.

Roger

Ozen,

I just tried that as well, for Ch 1 and Ch 2, in both 1 MΩ and 50 Ω modes, all scales read as 1 MΩ (998 kΩ - 1.002 MΩ) in the 1 MΩ mode, and as 50 Ω (50.3 Ω - 50.6 Ω) in the 50 Ω mode. I only tried the DC coupled modes.

I think I was expecting to see less than 1 MΩ for Ch 1 (in 1 MΩ mode, or less than 50 Ω in 50 Ω mode), but looking at the attenuator schematic I'm not sure how that could actually happen. More importantly, I can imagine a way that stuck closed switches might decrease the 1 MΩ impedance (by switching in the 50 Ω path), but not how anything similar could happen for the 50 Ω mode (that would require that the attenuator basically be shorted to ground, right?).

I need to go back and verify that each scale setting in Ch 1 produces the expected scale change on screen. If some of the switches are stuck I would expect that different scales would register the same voltage level.

-- Jeff Dutky

I noticed a typo below, diode checker should show ~ +0.7V in forward direction, not -0.7:

On Tue, Mar 22, 2022 at 09:21 PM, Jeff Dutky wrote:

Can I just use a multimeter to check then impedance of the scope input?

Hi Jeff,
Multimeter should be fine. I just did the test on my 2467B showing 1M at all ranges.

Ozan

Ozan,

Sorry, I’m running on a bit of a sleep deficit, and I got the sender names confused in the thread.

Can I just use a multimeter to check then impedance of the scope input?

— Jeff Dutky

My measurements put it at about 290 MHz.? I find the -3 dB display frequency, repetitive signal.
Bob

On Tue, Mar 22, 2022 at 07:48 PM, James55 wrote:

This electronics is addictive stuff!

Even though I am falling asleep, couldn't resist one last 'hit' of voltage
readings before passing out.

You'll like this...

With CR1732 removed, R1757 replaced and all pins of U1724 connected, we
finally have over +60v on the +65v test point!

It was 61.8 volts, without any R1736 adjusttment

Great progress. At this stage please do not remove the opamp. TP+65V will fly to 80+ volts.

CR1732 is nothing special and given you have replaced (or will replace) many components with not-exact replacements you can go with 1N4148. Its job is to protect BE junction, most of the time it is off, reverse biased by 0.7V and having an easy life. Same with CR1733, used exactly the same way.

Just to make sure: the diode checker should show ~ -0.7V (for silicon diodes) in forward direction and show open circuit (or high impedance) in the reverse direction. When you get your new diodes you may want to confirm your diode checker works properly.

Ozan

This electronics is addictive stuff!

Even though I am falling asleep, couldn't resist one last 'hit' of voltage readings before passing out.

You'll like this...

With CR1732 removed, R1757 replaced and all pins of U1724 connected, we finally have over +60v on the +65v test point!

It was 61.8 volts, without any R1736 adjusttment

I'll get some diodes tomorrow and let's see what happens next?


Night night



James


zzzzzzzzzzzzzzz

Interleaved:

On 3/22/2022 10:01 PM, James55 wrote:

Hi Ozan and Harvey.
I do believe we are closing in on the issue.

One certainly hopes <grin>

Following the results of lifting the #5, 6, and 7 pins, and realising that the dual op-amps were in fact not as 'isolated' as I had imagined, I thought I would go and have a look at the +15v rail.
This rail has had a short for a while but I hadn't bothered to investigate as the +65v was the priority.

Now that could do nasty things.........

Anyways......

After having a quick probe on the voltages, it was clear that there was nothing on the Q1756E, so I went to have a closer look at the schematic and saw that R1757 (0.6Ω) was the next component after the emitter. To cut a long story short, R1757 was open. Not only that but as I lifted a leg, it turned out to be in two pieces. That would certainly explain the lack of voltage on the 15v TP.

Resistors are not meant to be two piece assemblies.

Look at U1724B. Pin 7 (output) can both source and sink current. In U1724A, it can only sink current. (more on that later)

This has been a revelation that certain op amp outputs can flow both ways. Definitely helps to understand circuitry

Like the output of a hi-fi amplifier, there's a transistor from the op amp output to the + rail, and one from the - rail to the output.

If there's no load, and the output ought to be at, say, 10 volts, then no current flows into the output.? If you're driving a resistive load, then current flows from the output to the load. If, however, you connect to some voltage source (through a resistor), the op amp still wants to keep the output at 10 volts, even though the output pin wants to go to 11.? The op amp will have to sink current to keep the output at the desired level.

Normally, all the current to the +15 volts comes from the +15 volt supply, the 22.5 volts at the + end of C1751. No problem.

Yes, we have that 22.5v

I assume that CR1757 and C1757 are good and there's no load on the +15 at all.

C1757 is a new tantalum and all the jumper connectors as well as the A5 and A7 boards are disconnected.
As for CR1757, after looking into CR1732 (which I'll explain in a minute), I am now thinking that this could be a source of one of the problems. Is a forward voltage of 0.585v too low?

Not necessarily so.? generally about .6 to .7, but it varies with current as well.? Go find a forward voltage curve vs current on a diode.? The knee is not necessarily all that sharp.

I'd still like to see what happens if you disconnect U1724B pin 7, which would keep the op-amp from sourcing current to the 15 volt supply.

At the moment pins 5, 6, and 7 are disconnected

Then what I'd expect is that the voltage at pin 8 on the op amp would be close to 22 volts, especially if the junction of CR1724 and VR1724 is grounded.

Another alternative would be to connect the 15 volt bulk supply o CR1751, which would convert the two transistors to transistors rather than BE diodes.

Not quite sure that I follow, however I'm happy to give it a whirl if todays developments don't make a significant headway.

In a transistor, the BE junction acts like a forward biased diode, and the BC junction acts like a reverse diode.? Putting in current to the base lead effectively makes the composite CE junction leaky (one way to think of it) allowing collector current to flow.? If the collector of the transistor is open, you get a diode formed by the BE junction.? Note that the current in the transistor flows *through* the base region in normal operation. So with a transistor of a beta of 100, for each 1 ma flowing in the collector, you need 10 ua flowing into the base.? The emitter current therefore becomes (Ib + Ic) or 1.01 ma.? WIth no collector current source, you just have a base-emitter junction, which is? a forward biased diode.

@Ozan

It leaves CR1732 as suspect.

Funnily enough, I had measured and re-measured the voltage on that diode several times as It didn't seem right for some reason. I just wasn't really sure what I was looking out for.

When you measure the voltage drop on a diode, the forward voltage drop on silicon ought to be about .6 to .7.? There should be no reverse voltage drop, it should effectively be open (unless a zener, zeners are reverse biased).? Looking at the Q1732 and CR1732, though, they are effectively a diode (CR1732) and a BE junction on a transistor.? They're connected back to back with opposite polarity.? Measured one way, you get the forward drop of CR1732 and the BE junction of the transistor is reverse biased, measure the other way, CR1732 is reverse biased and you see the forward voltage drop of the BE junction.

On to this evening... CR1732 and CR1733 are identical 30v 175mA diodes, so when in diode mode on the DMM, CR1732 measured 0,315v one way and slightly higher in reverse, whilst CR1733 measured 0.679-ish both ways, I pulled CR1732 and sure enough it still has the low readings. Even though I am fully aware that necessary voltage across a silicon junction is around 0.7v,
I realise now that I have occasionally overlooked lower diode readings as long as the device only give a reading in one direction, and that is wrong.

Same situation as above.? Vbe one direction, Vfwd the other. Always measure diodes both ways, then understand why you get what you get.

Last one is CR1732. Not only it should register a diode in forward direction but it should work well in reverse direction. When you do diode check it is important to check both directions.

My case in point.
I do always check diodes in both directions however, it is clear to me that I need to review the different diode types.


So, tomorrow I shall be out again searching for a 1N4152 equivalent as they don't exist here. Any suitable suggestions would be appreciated.
The R1757 has already been replaced.

Work on seeing why you get the readings you do.

Harvey

Once again it is late here but progress is being made.


Thanks to all for the support and input.


James

Looking online for a 40v 150mA 1N4152 replacement, the 1N4148 keeps popping up along with Schottky options.

Hi Ozan and Harvey.

I do believe we are closing in on the issue.

Following the results of lifting the #5, 6, and 7 pins, and realising that the dual op-amps were in fact not as 'isolated' as I had imagined, I thought I would go and have a look at the +15v rail.
This rail has had a short for a while but I hadn't bothered to investigate as the +65v was the priority.

Anyways......

After having a quick probe on the voltages, it was clear that there was nothing on the Q1756E, so I went to have a closer look at the schematic and saw that R1757 (0.6Ω) was the next component after the emitter. To cut a long story short, R1757 was open. Not only that but as I lifted a leg, it turned out to be in two pieces. That would certainly explain the lack of voltage on the 15v TP.

Look at U1724B. Pin 7 (output) can both source and sink current. In U1724A, it can only sink current. (more on that later)

This has been a revelation that certain op amp outputs can flow both ways. Definitely helps to understand circuitry

Normally, all the current to the +15 volts comes from the +15 volt supply, the 22.5 volts at the + end of C1751. No problem.

Yes, we have that 22.5v

I assume that CR1757 and C1757 are good and there's no load on the +15 at all.

C1757 is a new tantalum and all the jumper connectors as well as the A5 and A7 boards are disconnected.
As for CR1757, after looking into CR1732 (which I'll explain in a minute), I am now thinking that this could be a source of one of the problems. Is a forward voltage of 0.585v too low?

I'd still like to see what happens if you disconnect U1724B pin 7, which would keep the op-amp from sourcing current to the 15 volt supply.

At the moment pins 5, 6, and 7 are disconnected

Another alternative would be to connect the 15 volt bulk supply o CR1751, which would convert the two transistors to transistors rather than BE diodes.

Not quite sure that I follow, however I'm happy to give it a whirl if todays developments don't make a significant headway.


@Ozan

It leaves CR1732 as suspect.

Funnily enough, I had measured and re-measured the voltage on that diode several times as It didn't seem right for some reason. I just wasn't really sure what I was looking out for.

On to this evening... CR1732 and CR1733 are identical 30v 175mA diodes, so when in diode mode on the DMM, CR1732 measured 0,315v one way and slightly higher in reverse, whilst CR1733 measured 0.679-ish both ways, I pulled CR1732 and sure enough it still has the low readings. Even though I am fully aware that necessary voltage across a silicon junction is around 0.7v,
I realise now that I have occasionally overlooked lower diode readings as long as the device only give a reading in one direction, and that is wrong.

Last one is CR1732. Not only it should register a diode in forward direction but it should work well in reverse direction. When you do diode check it is important to check both directions.

My case in point.
I do always check diodes in both directions however, it is clear to me that I need to review the different diode types.


So, tomorrow I shall be out again searching for a 1N4152 equivalent as they don't exist here. Any suitable suggestions would be appreciated.
The R1757 has already been replaced.


Once again it is late here but progress is being made.


Thanks to all for the support and input.


James

The 2440 X-Y bandwidth was 300MHz it says at probe tip, also for the 2439 and 2432/A however the maximum single-event useful storage bandwidth is
200 MHz(2440) 100MHz (2432A) 40MHz (2430A) The 2440 service manual says adjust transient responce for greater than or equal to 310 MHz each channel.
Jeff
--
Jeffeelcr

On Tue, Mar 22, 2022 at 05:42 PM, Jeff Dutky wrote:

I do seem to see correct DC attenuation at 50 mV, 500 mV, and 1 V settings
(I've now checked that multiple times). I was expecting to see the wrong
attenuation at DC, was surprised when I did not see that, and am now wondering
if the stuck switches are tarnished or carbonized so that they act like a tiny
capacitor (maybe with high EPR leakage?)

Perhaps one of the 10x paths stays connected to ground (perhaps with a tiny cap as you wrote) even when it shouldn't. Any additional cap will kill the BW as impedance is pretty high in that network. Does the input resistance stay ~ 1M in all modes?

I see what you mean about there being a path through the attenuators even with
all switches open: it would act like ÷100 has been selected, which could
explain what I'm seeing, if the attenuation of a DC signal were as attenuated
as the AC signals.

I had been following that discussion ("Tek 2465 stuck in horizontal sweep")
and had seen your posts and pictures. Your mention of having to repair or
replace the tabs between the attenuator and the damping network has me
worried.

I believe Siggi posted the comments and pictures. I don't have any direct experience on these attenuators. Only attenuators I fixed were on a Tek 485, and replacing all 16 relays on a TDS7104 attenuator. It was interesting to see BW beyond 1GHz was possible with regular looking relays.

Ozan

The 2200 ohm resistors go to the screen, same as the 470pF capacitor. The suppressor is internally connected to cathode.

Dave Wise

From: [email protected] <[email protected]> On Behalf Of Mark Vincent via groups.io
Sent: Tuesday, March 22, 2022 6:09 PM
To: [email protected]
Subject: Re: [TekScopes] Type 106 Saga (Again)

Notice the screens are floating with a 470mmfd to ground. The suppressor has the 2200 ohm resistors to ground. With a floating screen, the plate will barely get a signal. I am going by the first schematic link on the W140 site. I have disconnected the screen on pentodes, signal and output, and the signal was greatly attenuated and distorted. The screens need to be at some negative voltage potential reference to ground and be less positive than the plate. If I am wrong, what did I miss?

Mark

Notice the screens are floating with a 470mmfd to ground. The suppressor has the 2200 ohm resistors to ground. With a floating screen, the plate will barely get a signal. I am going by the first schematic link on the W140 site. I have disconnected the screen on pentodes, signal and output, and the signal was greatly attenuated and distorted. The screens need to be at some negative voltage potential reference to ground and be less positive than the plate. If I am wrong, what did I miss?

Mark

Ozan,

I do seem to see correct DC attenuation at 50 mV, 500 mV, and 1 V settings (I've now checked that multiple times). I was expecting to see the wrong attenuation at DC, was surprised when I did not see that, and am now wondering if the stuck switches are tarnished or carbonized so that they act like a tiny capacitor (maybe with high EPR leakage?)

I see what you mean about there being a path through the attenuators even with all switches open: it would act like ÷100 has been selected, which could explain what I'm seeing, if the attenuation of a DC signal were as attenuated as the AC signals.

I had been following that discussion ("Tek 2465 stuck in horizontal sweep") and had seen your posts and pictures. Your mention of having to repair or replace the tabs between the attenuator and the damping network has me worried.

-- Jeff Dutky

Interleaved, and I think you're found something.? (Watch this, nothing up my sleeve...........)


On 3/22/2022 6:12 PM, James55 wrote:

Just a quick one as out the door but will give more details in a bit.

Harvey,
I didn't yet connect the junction to ground, however you asked to disconnect pins #5, 6 and 7 form U1724B.

The following voltages were then taken;
Pin # 1 - 21.2v
Pin # 2 - 4.12v
Pin # 3 - 9.23v
Pin # 8 - 21.84v

Look at U1724B.? Pin 7 (output) can both source and sink current.? In U1724A, it can only sink current.? (more on that later)

Let's assume that for some reason there's a fault in the +15 volt supply, such as perhaps a short to ground, or a high current drain.

Current from pin 7 of the op amp can go to the base of Q1752, which emitter drives the base of Q1756, which drives the +15 volts.

Normally, all the current to the +15 volts comes from the +15 volt supply, the 22.5 volts at the + end of C1751.? No problem.

Suppose it got disconnected, and that supply is now floating, the +15 volts unregulated.? No current can come from that.? However, we have pin 7 on the op amp conneced through the BE junction of Q1752, which then goes through the BE junction of Q1756 (remember that without a collector voltage they're just diodes), and then to the +15 volt output.? It looks as if the op amp is trying to source all the current needed to make the +15 volts run

I assume that CR1757 and C1757 are good and there's no load on the +15 at all.

(it would make a difference)

So maybe the problem isn't in the +65 volt circuit, but in the +15.

Grounding the anode of CR1724 removes the ability of U1724A to control the voltage of the +65 supply.? IF U1724A were bad, then the +65 would go to the lowest voltage capable of being selected.

I'd still like to see what happens if you disconnect U1724B pin 7, which would keep the op-amp from sourcing current to the 15 volt supply.

Another alternative would be to connect the 15 volt bulk supply o CR1751, which would convert the two transistors to transistors rather than BE diodes.

Interesting thought to consider the *other* half of the chip....

Harvey

So U1724B and circuitry is definitely not helping our cause.

R1735 measured 57.8K and C1735 didn't register as a capacitor on the DMM but is not shorted.

Ozan,
you asked for measurements too. The following are with the above op-amp pins disconnected...

VR1724 - 4.462v
C1724 - 4.468v
R1724 - 0.00v

Removing Q1736 had no effect on the relevant voltages and R1723 measured 15.6K with a leg lifted.

Must dash. Back soon.


James