Test Equipment Anonymous (TEA) group therapy thread

[bitseeker]

Well, now that's good timing.

[bitseeker]

That 199 appears to be in great condition. Does it happen to have a scanner card in it? That'd be a bonus.

I like these old Keithley DMMs with the big LED displays. I have a 196.

Will check, thanks.... what does that do?

It enables you to connect the meter to multiple measurement points and sample them sequentially. It's a trade off between the cost and space of using multiple meters to see everything vs. just one meter and not being able to see multiple measurements simultaneously.

There can be setbacks though. For example, on the Prema 6001 the presence of the scanner card limits you to 100V and you can't remove it without recalibration.

The recalibration requirement might be because of the contact resistance of the relays vs without the card.

Here's the manual for the 199. See the table, "Model 1992 Scanner Option" on page 6. Max voltage on the card is 200V peak vs 425V directly on the DMM.

https://www.manualslib.com/manual/1214565/Keithley-199.html?page=6

[bitseeker]

As a heads up to any who may be interested, apparently someone on the bay of evil has stumbled onto a cache of those HP coffee mugs:


https://www.ebay.com/itm/114327183149?ul_noapp=true&shqty=2&isGTR=1#shId

There are allegedly more than ten available, and the price is not too insane compared to what I've seen them go for in the past.

Oh, my!

[TERRA Operative]

I cleaned the fan and it seems that the layers of dust were keeping it quiet. It now sounds the same as a brand-new clothes dryer running with two kilograms of pea gravel in it. I've add a fan replacement to my to-do list. The fan has 4 wires and there is a discrete circuit on the main digital board that looks like may it could be a brushless controller? I literally have no idea. I'll have to hit the schematics. Another user claims that it looks like the fan in the 3437A which they replaced with a regular 12V unit and some caps. Maybe I could steal some DC from upstream since the transformer and rectifiers are next door.

Yep, that'll be a brushless fan, same type as the 4276A LCZ meter, although the driver is a bit different.

You might get lucky and find a constant V+ and Gnd in those 4 wires to run a standard fan from (I think the 4276A was a bit less than 12V, so I just used a high-flow 12V fan that when undervolted was quiet enough but still with good airflow).
Otherwise, you'll just need to find an appropriate voltage rail to tap into.

[tonyalbus]

That 199 appears to be in great condition. Does it happen to have a scanner card in it? That'd be a bonus.

I like these old Keithley DMMs with the big LED displays. I have a 196.

Will check, thanks.... what does that do?

It enables you to connect the meter to multiple measurement points and sample them sequentially. It's a trade off between the cost and space of using multiple meters to see everything vs. just one meter and not being able to see multiple measurements simultaneously.

There can be setbacks though. For example, on the Prema 6001 the presence of the scanner card limits you to 100V and you can't remove it without recalibration.

The recalibration requirement might be because of the contact resistance of the relays vs without the card.

Here's the manual for the 199. See the table, "Model 1992 Scanner Option" on page 6. Max voltage on the card is 200V peak vs 425V directly on the DMM.

https://www.manualslib.com/manual/1214565/Keithley-199.html?page=6

Thanks for the info guys... on the 199 the card works a bit different i think,
Not sure the 199 is effected that much, with or without the card.
Think this is because of the fact that on the 199, the scanner card seems only a bunch of relays... (i had a 30 sec. look in the manual i could be wrong, and have not opened it)
you need to feed-back yourself, from the relays to the terminals in the BACK.. that means to me, the signal by default on this 199 does not go trough the relays..
so with or without the scannercard the FRONT terminals should not be affected in any way by the card, maybe the terminals on the BACK becouse there will be a slight difference if you messeure direct on the back terminal or go trough the scanner-card...
[edit] the 200 volt limit would be the relays..




That prema 6001 looks very cool btw.. like the dotted display!

[tonyalbus]

@terra

Jackpot!
scanner and IEEE

Seems only TTL and power to the scanner card, i think on the 199, the with withoutcard is not sush a problem.

i am missing a fuse cap in the from, i can steal it from the back and replace the holder on the back if i not find it..

Need to find the connectors for the scanner card

[med6753]

The Type 3A3 Vertical plug-in is a 2 channel amplifier with differential inputs for each channel. If we were to take a triangle wave and apply it to the plus input of one channel and a sine wave to minus input of that same channel what would be the result? Let's find out. Both waveforms are 0.5V p-p @ 5kHz. The vertical is set to 0.2V/div.



Very interesting. They almost completely null each other out. But not quite. There's something there. Let's crank open the gain and see what it is.



Vertical set to 20mV/div. Isn't that a strange and cool looking waveform?    I'm getting a headache just thinking about the math involved here.  I'll let the engineers ponder and debate that. 


' 

[bd139]

Looks like phase modulation.

[BU508A]

The Type 3A3 Vertical plug-in is a 2 channel amplifier with differential inputs for each channel. If we were to take a triangle wave and apply it to the plus input of one channel and a sine wave to minus input of that same channel what would be the result? Let's find out. Both waveforms are 0.5V p-p @ 5kHz. The vertical is set to 0.2V/div.



Very interesting. They almost completely null each other out. But not quite. There's something there. Let's crank open the gain and see what it is.



Vertical set to 20mV/div. Isn't that a strange and cool looking waveform?    I'm getting a headache just thinking about the math involved here.  I'll let the engineers ponder and debate that. 


'

 

These are the first steps for a little smurf into the world of Fourier Analysis.

If the sine and the triangle wave are correlated with a phase of 0° and both are having exactly the same frequency, then you are simply removing the basic sine wave of the triangle wave.

Triangle wave:
https://mathworld.wolfram.com/TriangleWave.html

Explanation of the triangle wave with Fourier:
https://mathworld.wolfram.com/FourierSeriesTriangleWave.html

If the sine wave is not 100% correlated with the triangle wave (phase, frequency etc.) then one can get something you got on your scope.

[med6753]

Both waveforms originate from the same 8038 chip. So aren't they correlated?  In fact, if I recall the sine is derived from the triangle within the chip. 


Fourier analysis? Now I'm getting a migraine. 

[bd139]

They are correlated. The 8038 generates a ramp then shapes it. Phase difference is 0 degrees.

[BU508A]

Both waveforms originate from the same 8038 chip. So aren't they correlated?   


Fourier analysis? Now I'm getting a migraine. 

Not necessarily. Perhaps in a ideal world, but if you are dealing with the real world, you have to take into account the cable length, the phase shifting of the RLC networks, phase shifting of the amplifiers, jitter, limited bandwidth (important for the triangle wave), imperfect sine wave etc. ...

For the headache I recommend Aspirin.

[med6753]

So....in an ideal situation if the waveforms were truly 100% correlated then the trace would be flat line, correct?


Either that or I'm gonna go flat line because this conversation could rapidly go over my head. 

[bd139]

That actually looks like a sine and square wave not triangle

https://www.wolframalpha.com/input/?i=sin%28x%29-squarewave%28x%2Fpi%2F4%29

Edit: Phase relation is wrong there. Can't be arsed to work it out. Haven't even had morning coffee 

Edit: if my bloody function generator had bothered to show up I'd run at test here.

[mansaxel]

Both waveforms originate from the same 8038 chip. So aren't they correlated?   


Fourier analysis? Now I'm getting a migraine. 

Not necessarily. Perhaps in a ideal world, but if you are dealing with the real world, you have to take into account the cable length, the phase shifting of the RLC networks, phase shifting of the amplifiers, jitter, limited bandwidth (important for the triangle wave), imperfect sine wave etc. ...

For the headache I recommend Aspirin.

Am I right in assuming that wavelength has a large influence on the amount a given discrepancy (like cable length or bandwidth) would have? Then, with 5KHz, the requirements posed on the precision would not be as large as if we're talking about say 5MHz?

/just adjusting my perception here.

[BU508A]

So....in an ideal situation if the waveforms were truly 100% correlated then the trace would be flat line, correct?


Either that or I'm gonna go flat line because this conversation could rapidly go over my head. 

No, because a triangle wave is obviously something different than the sine wave. So, if you are going to remove the basic sine wave from the triangle wave, there must be something left over and this is what you are seeing on your screen.

[BU508A]

Both waveforms originate from the same 8038 chip. So aren't they correlated?   


Fourier analysis? Now I'm getting a migraine. 

Not necessarily. Perhaps in a ideal world, but if you are dealing with the real world, you have to take into account the cable length, the phase shifting of the RLC networks, phase shifting of the amplifiers, jitter, limited bandwidth (important for the triangle wave), imperfect sine wave etc. ...

For the headache I recommend Aspirin.

Am I right in assuming that wavelength has a large influence on the amount a given discrepancy (like cable length or bandwidth) would have? Then, with 5KHz, the requirements posed on the precision would not be as large as if we're talking about say 5MHz?

/just adjusting my perception here.

In short: yes.

If you have a triangle wave of 5kHz and a bandwidth of 20MHz then it is obvious, that one could add much more overtones to the resulting triangle wave than compared to a triangle wave at 5MHz. Adding more overtones the better the result is. Adding infinite overtones, then one will get a perfect result.

[med6753]

Both waveforms originate from the same 8038 chip. So aren't they correlated?   


Fourier analysis? Now I'm getting a migraine. 

Not necessarily. Perhaps in a ideal world, but if you are dealing with the real world, you have to take into account the cable length, the phase shifting of the RLC networks, phase shifting of the amplifiers, jitter, limited bandwidth (important for the triangle wave), imperfect sine wave etc. ...

For the headache I recommend Aspirin.

Am I right in assuming that wavelength has a large influence on the amount a given discrepancy (like cable length or bandwidth) would have? Then, with 5KHz, the requirements posed on the precision would not be as large as if we're talking about say 5MHz?

/just adjusting my perception here.

In short: yes.

If you have a triangle wave of 5kHz and a bandwidth of 20MHz then it is obvious, that one could add much more overtones to the resulting triangle wave than compared to a triangle wave at 5MHz. Adding more overtones the better the result is. Adding infinite overtones, then one will get a perfect result.

Just for a point of reference this Type 3A3 plug-in has a B/W of only 500kHz. I'm sure that could affect the results too.

[Cubdriver]

So....in an ideal situation if the waveforms were truly 100% correlated then the trace would be flat line, correct?


Either that or I'm gonna go flat line because this conversation could rapidly go over my head. 

No, what you wound up with makes perfect sense.  You're inputting two equal, correlated waves - one a sine and one a triangle.  When you put one into the positive input and the other into the negative, you effectively put them 180* out of phase.  The resultant wave is the difference between the two, with the same polarity as the sine wave.

What you're inputting:


The difference between the two (shaded area):


(And yes, I know I can't draw a sine wave for crap anymore.  It's been a long time since school!)

-Pat

[ch_scr]

Both waveforms originate from the same 8038 chip. So aren't they correlated?   


Fourier analysis? Now I'm getting a migraine. 

Not necessarily. Perhaps in a ideal world, but if you are dealing with the real world, you have to take into account the cable length, the phase shifting of the RLC networks, phase shifting of the amplifiers, jitter, limited bandwidth (important for the triangle wave), imperfect sine wave etc. ...

For the headache I recommend Aspirin.

Am I right in assuming that wavelength has a large influence on the amount a given discrepancy (like cable length or bandwidth) would have? Then, with 5KHz, the requirements posed on the precision would not be as large as if we're talking about say 5MHz?

/just adjusting my perception here.

In short: yes.

If you have a triangle wave of 5kHz and a bandwidth of 20MHz then it is obvious, that one could add much more overtones to the resulting triangle wave than compared to a triangle wave at 5MHz. Adding more overtones the better the result is. Adding infinite overtones, then one will get a perfect result.

Just for a point of reference this Type 3A3 plug-in has a B/W of only 500kHz. I'm sure that could affect the results too.

Looks pretty close to ideal to me. Good job & nice experiment!
EDIT: Also note that the input waves are scaled to 1.0 peak and the result is 0.2 peak.

[bd139]

So....in an ideal situation if the waveforms were truly 100% correlated then the trace would be flat line, correct?


Either that or I'm gonna go flat line because this conversation could rapidly go over my head. 

No, what you wound up with makes perfect sense.  You're inputting two equal, correlated waves - one a sine and one a triangle.  When you put one into the positive input and the other into the negative, you effectively put them 180* out of phase.  The resultant wave is the difference between the two, with the same polarity as the sine wave.

What you're inputting:


The difference between the two (shaded area):


(And yes, I know I can't draw a sine wave for crap anymore.  It's been a long time since school!)

-Pat

Yep which is why I think this is phase mod:



Excuse mspaint...

f(x) = sin(x) - square(x-someshift)

[BU508A]

I love the smell of Fourier analysis in the morning ...

 

[bd139]

I don't. Still haven't finished my first coffee yet 

[Cubdriver]

As a heads up to any who may be interested, apparently someone on the bay of evil has stumbled onto a cache of those HP coffee mugs:


https://www.ebay.com/itm/114327183149?ul_noapp=true&shqty=2&isGTR=1#shId

There are allegedly more than ten available, and the price is not too insane compared to what I've seen them go for in the past.

Oh, my!

Nine gone, six left.  I wonder if she's surprised that they're selling so quickly?

And on that note, time to hit the sack.  I really need to try to quell my nocturnal tendencies.  (Of course I say that every day, then wind up staying awake until after four AM anyway.    )

-Pat

[Zucca]

Today arrivals:

- a new keyboard from Logitech

Please tell me they are red cherry key.