Author Topic: Oscilloscope. To repair or not? Fixed. (Read 27099 times)

BradC · « **Reply #50 on:** September 20, 2016, 06:09:03 am »

Quote from: tautech on September 20, 2016, 05:42:09 am

Purists use something like 100 $\$\Omega\$$ but most of us just zap them with a driver.

I use a 60W incandescent bulb. I already have the ES socket with a couple of flyleads and croc clips for when I need a dim bulb tester, so it's handy for the odd cap discharge.

Second the comment about putting something across the cap. I use cheap croc-leads. I've been bitten once or twice by HV caps that have regained some bite over time. I now keep them firmly clipped off. Remember to remove the clips before you power up. That's a mistake you'll only make once.

alsetalokin4017 · « **Reply #51 on:** September 20, 2016, 07:58:48 am »

Cheating is all well and good, you might think ... as long as you have a cheat sheet. But life does not usually come with a cheat sheet. It is better by far to learn the principles and facts you need to accomplish your goal.

The voltage divider resistor values shown on the CRT data sheet are a starting point, and are useful certainly. But in order to understand better the reasons for the original power handling capacity that the scope's designers may have used... and to make sure any substitutions will work for more than a few minutes... you really should try to understand the principles involved, rather than relying on a "cheat sheet".

In other words... that first-grade teacher was wrong, and did you and everyone else in the class a disservice by telling you all basically that it's "OK to cheat" to obtain your desired result. Try cheating on your GRE for example... it only works if you don't get caught.

clay1905 · « **Reply #52 on:** September 21, 2016, 12:14:08 am »

Spent a bit of time this morning tracing the divider circuit more thoroughly. I do wish some businesses would buy more different coloured wires.

Anyway, here it is. A rough sketch showing casual disregard for drawing convention. The two 100K pots are ganged.

Oh boy, the arithmetic is a problem. I'm getting weird answers in attempting a current calculation, but I suspect I'm missing something fundamental here. In the calculation, I chose to halve the values of the pots, because that's the 'normal' condition, but that can't be correct. The divider uses both the full value of two of the pots, as well as tapping off variable voltages. (I'm pretty woeful at arithmetic. When I was studying chemistry a long time ago, the class was tasked with calculating how many molecules were in a litre of a sample. Typical answers took the form NX10^26. I got 2.

) For this calculation, I got 0.000008824, which I think would be Amps if I got it right, but the number looks pretty meaningless to me.
I think I need a bit more guidance in this.

Thanks,
Clay.

StillTrying · « **Reply #53 on:** September 21, 2016, 12:24:57 am »

I make it about 1300V/1.7M $\$\Omega\$$ = about 7.6470588235294117647058823529412e-4 = about 0.76 mA, about twice as much as I would have guessed, about.

rf+tech · « **Reply #54 on:** September 22, 2016, 01:14:45 am »

Clay,

The important detail is that although the pots are mechanically ganged and operate in parallel, they are electrically connected in series.

Since everything is in series, one simply adds up all of the resistor values. Depending on which rotation extreme of the left pot, the total resistance varies from 1.73 MOhms to 1.63 MOhms.

Current through the divider varies between a minimum of 1300 Volts divided by 1.73 MOhms, to a maximum of 1300 Volts divided by 1.63 MOhms.

Using the maximum current from above, the maximum power dissipated by each resistor is I squared divided by R.

Try working out the answers yourself and post back the results. For extra credit, solve for the voltages at each point in the divider.

We'll check your math and set you on the right track.

RF+ Tech

clay1905 · « **Reply #55 on:** September 22, 2016, 02:48:45 am »

Thanks for the reply rf+tech,

I was despairing of getting a result. This project has grown into something far more important to me than just repairing the 'scope, and any help would be enormously appreciated.

Somehow I'm not yet seeing the gaudily coloured elephant in the room. I will knuckle down this arvo and see what I can do. I have this Ohm's circle chart which I've been trying to use but so far the figures seem so far out that I'm rather confused. Maybe I'm misunderstanding the results though. I'll run the numbers through the mill again and post what I'm getting. I'll try to show working so you can see where I'm going off the rails, if indeed I am.

Thanks again,

Clay.

alsetalokin4017 · « **Reply #56 on:** September 22, 2016, 04:21:26 am »

Quote from: rf+tech on September 22, 2016, 01:14:45 am

Clay,

The important detail is that although the pots are mechanically ganged and operate in parallel, they are electrically connected in series.

Since everything is in series, one simply adds up all of the resistor values. Depending on which rotation extreme of the left pot, the total resistance varies from 1.73 MOhms to 1.63 MOhms.

Current through the divider varies between a minimum of 1300 Volts divided by 1.73 MOhms, to a maximum of 1300 Volts divided by 1.63 MOhms.

Using the maximum current from above, the maximum power dissipated by each resistor is I squared divided by R.

Try working out the answers yourself and post back the results. For extra credit, solve for the voltages at each point in the divider.

We'll check your math and set you on the right track.

RF+ Tech

Err.... um..... no.

http://info.ee.surrey.ac.uk/Teaching/Courses/ee1.cct/circuit-theory/section1/powerdis.html
and many other references....

and basic Ohm's Law: P = I x V, where V = I x R, so P = I² x R

clay1905 · « **Reply #57 on:** September 22, 2016, 10:10:40 am »

OK, here come the sums.
Can't get the formatting for the divisions quite right.

163Mohm 173Mohm

0.000797546 0.000751445
1300 )1630000 1300 )1730000

0.08mA 0.075mA

Working left to right in the diagram I posted earlier, and using P=I^2*R :

I^2 = I^2 =
0.0008^2* = 0.00075^2 =
0.00000064 0.000000563

& &

100,000* 100,000*
0.00000064 0.000000563
0.064 0.05625

the 100Kohm pot = 0.064Watts. the 100Kohm pots = 0.05625Watts each.

330,000* 330,000*
0.00000064 0.000000563
0.2112 0.18579

the 330Kohm = 0.2112Watts. the 330Kohm = 0.18579Watts.

200,000* 200,000*
0.00000064 0.000000563
0.128 0.1126

the 200Kohm = 0.128 Watts. the 200Kohm = 0.2112Watts.

1,000,000* 1,000,000*
0.00000064 0.000000563
0.64

and the 1Mohm = 0.64 Watts. and the 1Mohm = 0.563 Watts.

So the suggested wattages are, at least for the final resistor in the chain, a little on the low side, and the 1 Watt rated are an acceptable choice.

Voltages. Using V=I*R, I calculated the drops thus:

100K Ohm 80V 100K Ohm 75V
330K Ohm 264V 100K Ohm 75V
200K Ohm 160V 330K Ohm 247V
1M Ohm 800V 200K Ohm 150V
_____ 1M Ohm 750V
SUM 1384V _______________
SUM 1297.5V

Considering I didn't do enough rounding, I think the final figures aren't too far from actual values. They look quite reasonable, at least to me.

How did I really go?

Clay.

alsetalokin4017 · « **Reply #58 on:** September 22, 2016, 11:45:57 am »

Quote from: clay1905 on September 22, 2016, 10:10:40 am

OK, here come the sums.
Can't get the formatting for the divisions quite right.

163Mohm 173Mohm

0.000797546 0.000751445
1300 )1630000 1300 )1730000

0.08mA 0.075mA

Some decimal point trouble here.

First, you have 163Mohm and 173Mohm when you should have 1.63 Mohm and 1.73 Mohm.

Then your long divisions are actually correct.

1300 / 1,630,000 = 0.000797546
1300 / 1,730,000 = 0.000751445

But now when you divide by 1000 to produce the answer in milliAmps you've misplaced the decimal.
To divide by 1000 simply move the decimal point three places to the right:

0.000797546 A = 0.797546 mA
0.000751445 A = 0.751445 mA

so with rounding and discarding false precision we have 0.80 mA and 0.75 mA as the max and min currents.

Quote

Working left to right in the diagram I posted earlier, and using P=I^2*R :

I^2 = I^2 =
0.0008^2* = 0.00075^2 =
0.00000064 0.000000563

Correct. You have used the correct values for the current here.

Quote

& &

100,000* 100,000*
0.00000064 0.000000563
0.064 0.05625

the 100Kohm pot = 0.064Watts. the 100Kohm pots = 0.05625Watts each.

330,000* 330,000*
0.00000064 0.000000563
0.2112 0.18579

the 330Kohm = 0.2112Watts. the 330Kohm = 0.18579Watts.

200,000* 200,000*
0.00000064 0.000000563
0.128 0.1126

the 200Kohm = 0.128 Watts. the 200Kohm = 0.2112Watts.

1,000,000* 1,000,000*
0.00000064 0.000000563
0.64

and the 1Mohm = 0.64 Watts. and the 1Mohm = 0.563 Watts.

So the suggested wattages are, at least for the final resistor in the chain, a little on the low side, and the 1 Watt rated are an acceptable choice.

All good. However, you should consider being conservative here. Myself, I'd use 2 Watt resistors out of an abundance of caution, at least for the 1M, and I might even use 1/2 Watt for the pots.

Quote

Voltages. Using V=I*R, I calculated the drops thus:

100K Ohm 80V 100K Ohm 75V
330K Ohm 264V 100K Ohm 75V
200K Ohm 160V 330K Ohm 247V
1M Ohm 800V 200K Ohm 150V
_____ 1M Ohm 750V
SUM 1384V _______________
SUM 1297.5V

Considering I didn't do enough rounding, I think the final figures aren't too far from actual values. They look quite reasonable, at least to me.

How did I really go?

Clay.

Close enough. Also, if there is room, you can mount the resistors standing up off the board by their leads a bit to allow for better cooling.

clay1905 · « **Reply #59 on:** September 22, 2016, 12:32:32 pm »

Thanks for the comments, alsetalokin4017.

You'll notice the errors were all about moving from actual numbers, like 1,630,000 to a shorthand version, like 1.63M. It's a bit of a blind spot for me. I seem to be more able to get to base units from the abbreviation.

Now I can see where in the divider circuit I can safely put my probe to measure actual voltage. This might help to figure out why there is still no image, even though all the failed resistors I found in the divider are replaced.

The two fixed value resistors are flying, so as to speak, and 1M connects to a Bakelite tab strip, thence to ground. the 330K just loops across behind the front panel. They're both inside a plastic tube like the originals.

(Pretty pleased I got it mostly right.)

Clay.

tggzzz · « **Reply #60 on:** September 22, 2016, 12:57:08 pm »

Quote from: clay1905 on September 22, 2016, 12:32:32 pm

You'll notice the errors were all about moving from actual numbers, like 1,630,000 to a shorthand version, like 1.63M. It's a bit of a blind spot for me. I seem to be more able to get to base units from the abbreviation.

With all SI calculations, it is best to use the base unit plus exponent notation. Thus:

1630000ohms = 1.63*10⁶ = 1.63e6
2.5cm = 2.5*10^-2m = 2.5e-2m
2.5cm = 25mm = 25*10^-3m = 25e-3m
0.0000000015s = 1.5e-9s = 1.5ns

Additionally, if you are working in an environment where a decimal point might be lost, e.g. on a drawing or when making handwritten notes, then don't use a decimal point but use the multiplier. Thus 1.5nF would be shown as 1n5, 0.15nF as 0n15 or 150p, and 15nF as 15n.

And never use "S" as the unit of time, e.g. nS. "S" means Siemens, i.e. conductance. "s" means seconds, i.e. time.

clay1905 · « **Reply #61 on:** September 22, 2016, 11:30:19 pm »

Thanks, but no.

I work in and around various fields, using different units. Mostly in machine shops where millimetres are the rule, or decimals of. In any case, it's deriving that's where I seem to go wrong most, so working with long decimal strings yield the better results.

Thanks though,
if everyone used exponents I'd probably be better off, but that's not the case.

Clay.

clay1905 · « **Reply #62 on:** September 23, 2016, 12:11:15 am »

OK,
Started to look at where in the divider I could put a DMM probe to measure, but soon realised that if the 1M resistor drops the voltage by 753V then anything before it is out of reach, as my meter can only go up to 500V

So, it looks like I need to get another resistor, maybe around 500K off the top of my head, to interpose between the ground end of the 200K resistor and the chassis.
I've added what I mean to the diagram, as attached. The extra resistor and a dial face indicate where I believe the voltage should be taken, and how. Please excuse the hand drawn resistor and dial.

If things are OK, I should find about 375V using a 500K as above.

Before I let some smoke out, I'd like this confirmed please. It's a bit of a commitment based on newly learnt maths.

Added later-

Was sitting on the verandah when it occurred to me, parallel resistors is what I'm proposing. So I did a bit of reading and saw that wasn't the way. The voltage across parallel resistors is equal, and would fry the DMM.

Thanks,
Clay.

clay1905 · « **Reply #63 on:** September 23, 2016, 02:39:51 am »

It's happened again.

Went to visit a friend who asked me what I've been up to lately. I told him of my adventures trying to repair the 'scope, and how I'm now pushing through the mental block concerning maths.
Long story short as they say. I'm now the proud owner of a Philips Type 5650 oscilloscope. A tiny unit with a 70mm screen. Another one on the shelf for later. I did download a schematic, and this 'scope produces a fuzzy trace. There's hope for it, but not yet.
I think it's time I kept quiet about this. I'm running out of space.

Clay.

Brumby · « **Reply #64 on:** September 23, 2016, 03:10:47 am »

Quote from: clay1905 on September 23, 2016, 02:39:51 am

It's happened again.

Went to visit a friend who asked me what I've been up to lately. I told him of my adventures trying to repair the 'scope, and how I'm now pushing through the mental block concerning maths.
Long story short as they say. I'm now the proud owner of a Philips Type 5650 oscilloscope. A tiny unit with a 70mm screen. Another one on the shelf for later. I did download a schematic, and this 'scope produces a fuzzy trace. There's hope for it, but not yet.
I think it's time I kept quiet about this. I'm running out of space.

Clay.

Too late. You've begun a journey that will occupy every available bit of storage space you have - and then some. All filled with a combination of working gear, repair projects and sources of parts.

StillTrying · « **Reply #65 on:** September 23, 2016, 11:54:01 am »

That's nonsense, come up with another measuring plan.

tggzzz · « **Reply #66 on:** September 23, 2016, 01:36:54 pm »

Quote from: clay1905 on September 22, 2016, 11:30:19 pm

I work in and around various fields, using different units. Mostly in machine shops where millimetres are the rule, or decimals of. In any case, it's deriving that's where I seem to go wrong most, so working with long decimal strings yield the better results.

The evidence you provided indicates the results are unsatisfactory! Given that, only a software "engineer" (and I use the word loosely) would repeat doing the same thing in the hope it would be better next time.

But out of curiousity, given that you prefer "long decimal strings":

which of these capacitors can be purchased/made/modelled: 0.00000001F, 0.000000001F, 0.00000000001Fm 0.0000000001F?
if you have a probe with a 15pF tip capacitance and a 150nH (i.e. 6") ground lead, what is the resonant frequency?
what is the impedance of a 10Mohm/15pF scope probe at 100MHz?
if you use that probe to measure a 100MHz 1V signal from a 1kohm source, what voltage will the scope show?

Once you start to deal with those very real-world calculations, you will see the advantages of the base-unit-plus-exponent method.

alsetalokin4017 · « **Reply #67 on:** September 23, 2016, 02:08:03 pm »

No fair !

For someone who is not very fluent with arithmetic, the use of scientific (aka exponential) notation should wait a bit. That's why I showed my calculations using ordinary decimal notation, that can be punched into a calculator as-is and the results read off as-is.

Certainly scientific notation is "easier" when you are dealing with 9 or more decimal places, and once a person is used to using it --- and getting correct answers --- no one will argue that it shouldn't be used. But that isn't the problem here.

alsetalokin4017 · « **Reply #68 on:** September 23, 2016, 02:11:46 pm »

Quote from: clay1905 on September 23, 2016, 12:11:15 am

OK,
Started to look at where in the divider I could put a DMM probe to measure, but soon realised that if the 1M resistor drops the voltage by 753V then anything before it is out of reach, as my meter can only go up to 500V

So, it looks like I need to get another resistor, maybe around 500K off the top of my head, to interpose between the ground end of the 200K resistor and the chassis.
I've added what I mean to the diagram, as attached. The extra resistor and a dial face indicate where I believe the voltage should be taken, and how. Please excuse the hand drawn resistor and dial.

If things are OK, I should find about 375V using a 500K as above.

Before I let some smoke out, I'd like this confirmed please. It's a bit of a commitment based on newly learnt maths.

Added later-

Was sitting on the verandah when it occurred to me, parallel resistors is what I'm proposing. So I did a bit of reading and saw that wasn't the way. The voltage across parallel resistors is equal, and would fry the DMM.

Thanks,
Clay.

Better to play it safe and simple, and get yourself a high voltage probe for your DMM.

rf+tech · « **Reply #69 on:** September 23, 2016, 03:00:31 pm »

Clay,

Although others disagree about your proposed 500 kOhm measurement resistor idea, let's apply some math to the problem to see why.

0.5 M $\$\Omega\$$ divided by 1.73 M $\$\Omega\$$ equals about a 29% increase in total resistance. How will this affect the measurement results?

If the amount of added resistance is limited to 1% of the total, would this be better or worse?

Can you see the advantage of using a 100:1 ratio?

So two 33 kOhm wired in parallel (to approximate 1% of 1.68 MOhms - the average), and the combination added in series between the bottom 1 MOhm and ground will suffice. The objective is to insert a sampling port that provides a safe measurement method, while minimizing disturbance of the bias network. The reading obtained will show if HV is present and represent 1% of the actual value, just as if a HV probe were used.

If he above technique seems too much of a risk to some, to each his own. To those who are open-mined and willing to learn - this technique is sound.

The limitation here is that the other voltages in the bias network cannot be measured. For those measurements, and future HV measurements on your growing collection of CROs, a HV probe would certainly be a worthwhile investment.

@tggzz - thanks for the sanity check in my previous post. I actually looked at an Ohm's Law pie chart to be sure, only to have a neuron misfire and flip a bit in the following few seconds on the keyboard.

RF+ Tech

edit: Clay,

Quote

Was sitting on the verandah when it occurred to me, parallel resistors is what I'm proposing.

I envisioned the 500 kOhm in series and think that is what you had in mind. Those pesky misfiring neurons ...

tggzzz · « **Reply #70 on:** September 23, 2016, 03:53:59 pm »

Quote from: alsetalokin4017 on September 23, 2016, 02:08:03 pm

No fair !

For someone who is not very fluent with arithmetic, the use of scientific (aka exponential) notation should wait a bit. That's why I showed my calculations using ordinary decimal notation, that can be punched into a calculator as-is and the results read off as-is.

Certainly scientific notation is "easier" when you are dealing with 9 or more decimal places, and once a person is used to using it --- and getting correct answers --- no one will argue that it shouldn't be used. But that isn't the problem here.

Very soon he will need (and I hope want) to do calculations similar to those I mentioned. Doing it properly from the beginning means it is easier in the long term, and there's less chance of someone being discouraged by simple arithmetic errors. There's enough subject matter to learn without also having to worry about arithmetic errors.

But I agree; it is probably better for the OP to get an HV probe for his DMM.

Plus, ideally, an HV scope probe, so he can observe a scope's blanking waveform on the grid

Yup, been there, done that, got the pretty pictures of the CRT's internal fault, and haven't go any scars. Yet.

clay1905 · « **Reply #71 on:** September 23, 2016, 10:47:20 pm »

Well, that cruelled it.

I haven't attempted anything like this for over 45 years. For me maths equals black eyes, blood, scars and welts. I had hoped that the scars were more healed than this but it''s all back, just like yesterday. I can't deal with the pain.

I'm going away to think about where to go, but right now I feel like just dumping the scopes and walking away. The veil has dropped, and that moment of clarity is gone.

Very sad today.

Clay.

clay1905 · « **Reply #72 on:** September 24, 2016, 01:52:58 am »

OK, I've had a think. There's a few things to say.

First, and most important, nobody intended any hurt. So there should be no finger pointing, guilt or any other negative stuff.

I'm surprised by how I reacted this morning. I can see more than ever how the legend of Pandora's box came about. I thought all that stuff was dead and buried. Not so. I'll have to find a way to deal with it, but this is not the forum for such things.

I've decided that I want to carry on and finish what I've started. However, perhaps a few days to get equilibrium back? If you gentlemen will be so kind as to bear with me, I would appreciate that more than I can say. Since I joined here I've achieved what feels to me like a miracle. And I'm so grateful for that alone.
Please though, let me do things the way that works for my muddled mind. I know the results look patchy, but for the time being at least, it's better to do badly over doing nothing.

Meanwhile, I'll read and re-read your post rf+tech, and see what I can digest.

Thanks all for your kindness and care. You're making an immeasurable difference.

Clay.

rf+tech · « **Reply #73 on:** September 24, 2016, 04:30:36 am »

Clay,

Quote

Thanks all for your kindness and care. You're making an immeasurable difference.

Sir, you are most welcome. It is a pleasure to share my knowledge and skill with someone who appreciates this. Never stop learning.

RF+ Tech

pricemp · « **Reply #74 on:** October 15, 2016, 05:09:08 am »

Hi Clay
I am in a similar situation. Ie limited experience but lots of interest and have been given a Jayem Model 555 oscilloscope.
Mine is showing a trace but not functioning as designed. I have been watching a youtube video of restoring an old scope by Mr Carlson see link attached. This is useful in terms of approach although a different make/model. https://youtu.be/RF7CRqb54XM
The good news is that I have an original operating manual (A4 size) for the Jayem model 555. It contains 20 pages in 4 sections - 1) general characteristics and specs, 2) initial setting up procedure, 3) measurement procedures and 4) maintenance and alignment procedures.
At the back there are 9 detailed schematics as appendices:
Power supply & CRT circuit
Vertical amplifier
Time-base trigger
Time-base generator
Time-base timing switch
Horizontal amplifier
Ext. Horizontal amplifier
Voltage calibrator
High voltage rectifier

I am happy to share this with you. Maybe we can colloborate to get our scopes up and running.
Let me know if you are still interested in restoring yours.
Cheers Phil


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Author Topic: Oscilloscope. To repair or not? Fixed. (Read 27099 times)

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