I've bought a broken scope haven't I ?

[aneng]

No... I have signal ground and  -VCC as different connections.  Signal ground is shared between input and output only.  -VCC goes nowhere but pin 4.

You say it won't work unless the input is at least 3V? (0V+3).  The input is from the earphone socket of my phone... That won't reach 3V will it ?

[Andy Watson]

You say it won't work unless the input is at least 3V? (0V+3).  The input is from the earphone socket of my phone... That won't reach 3V will it ?

+3V relative to pin 4.

[Ian.M]

For the OPAMP, the positive and negative supplies must be referenced to ground, and *SHOULD*. be symmetrical about it   You either need a dual output supply,  or a fully floating single output one + a circuit to split it and get a mid-rail 'ground'.   You can do that with the spare half of the OPAMP (and yes, the Maplin catalog has been FUBARed since before the millennium, so use the manufacturer's pinout!), wiring -in to out of that half and connecting two 10K resistors to +in, one to each supply rail, then use out as ground for the rest of the circuit.  This only works properly with a floating supply because if the -ve supply is grounded, connecting any signal ground (e.g. scope ground clip) to the rail-splitter output 'ground'  will short out the OPAMP.

[aneng]

Right, I see.  OK, two 9V batteries it is then!  I'll tackle that this evening, presuming I'm not still underneath the wife's car, swearing profusely!

[aneng]

Just another quick post to reiterate my thanks and appreciation for the help here.  I certainly wouldn't have learned as much as I have if it wasn't for you guys (esp Ian.M).  This sounds a bit like an Oscar speech doesn't it!  I'll save that till the end when I'm holding a working scope at the podium :-)

[MarkL]

I've found something that may or may not be of help..  Where the connectors for each channel leave the Vertical Attenuator Unit board and join the Vertical Preamp Unit board (P3 and P4 on both schematics (pages 95 & 99 of the service manual for the CS-2070 (which so far is identical to my CS-2100!); https://www.dropbox.com/s/ikw5aa81hvlm5ut/trio_kenwood_cs-2070_70mhz_oscilloscope.pdf?dl=0 ), according to the schematic, the pins on each connector P3 and P4 should be (1-6 in order) Out-, Out+, +10V, -10V, NC, GND.

On CH2 (P4), with the scope setup as per the manual (most knobs centred etc), and set identically for both channels, I'm getting -1.3V approx on both the Out- and Out+ pins.  (The +10/-10 V pins are ok on both channels), however the same Out- and Out+ pins on the CH1 connector (P3) are reading -9.7V.

So - either the fault is somewhere around IC12 on the Vertical Attenuator Unit or the erroneous voltages are coming in from the Vertical Preamp Unit.  I was going to try to separate them and read the connectors on each board to see, but I'm not sure whether that could be dangerous to the scope.

One thing you could do is try swapping the cables plugged into P3 and P4 to see if the problem moves to the other channel.  That will at least get us to the right board.

[aneng]

Alas, that's not possible.  They aren't cables - they're push-together headers on the boards.  One board mounts on top of the other, piggy-back style and P3 and P4 are not moveable.

[MarkL]

Alas, that's not possible.  They aren't cables - they're push-together headers on the boards.  One board mounts on top of the other, piggy-back style and P3 and P4 are not moveable.

Doh!  So much for the easy way.

Can you get access to the pins on IC12?  Pins 8 and 9 are OUT+ and OUT- and it would be worth verifying they are also at -9.7V.  (I'm rather suspect of solder connections in this scope now.)  Also check pin 14 +10V supply, and pin 3 -10V supply.

What is the voltage on pin IC12 pin 15?  How does that compare to IC13 pin 15?  That's the input.

Are there any markings on IC12?  The parts listing says it's an ATM-4010, but I'm having trouble finding a datasheet for it.  It's some kind of high speed amplifier.

[aneng]

Here's what we're talking about...  IC12 is the black square IC in the top of the picture.



The only way I'm getting access to the pins of that IC is by powering it up with both piggy-backed boards unscrewed from the chassis.  Given the significant potential for a quick game of 'shorty-shorty-zap-bang' I don't think it would be wise for me to attempt it !

I'm curious why you want me to test on the pins of that IC.  Surely if the Out+ and - on P3 are reading -9.7V then the connected pins on P3 must be the same ?

Edit: I'm still not getting anywhere with this bloody Opamp circuit !  I've got two 9V batteries connected up as +9V - 0V - -9V.  +9V to VCC, -9V to -VCC.  Is the 0V supposed to be connected to signal ground or not ?  I haven't connected it and I'm not seeing any amplification at all - it's like it's just passing the same signal through it at 1:1.

[MarkL]

Ok, that's going to be a bit of a challenge.

I was interested in the voltages around IC12 since I'm a little distrustful of the connectors and solder joints in this scope.  I wanted to verify directly on the chip that's using them.  And on the IC12/IC13 input, it would be useful to know if there's a problem with the JFET input stage that's preceding the amp.

Let's try from the other board.  Can you measure the Q1 base (should be -6.25V), and then the Q1 base to emitter (should be 0.7V)?  Q1 is biasing the input stage consisting of IC1a and IC1b.  It's probably ok, which is why I wanted to start on the attenuator board.

Also, an exact reading on the +10V and -10V supply will be useful.  Everything in the input stages is using them.

[MarkL]

Looking a little more closely at the picture, it looks like you have a shot at getting to the pins on Q1 or R28 on the attenuator board.

The node that's between Q1b and R28 is connected to the amp input (IC12, pin 15).

If you're concerned about shorting something out with your probe, you could try slipping a piece of insulation over the exposed tip and just leave a tiny bit sticking out.  If it's really tight, a solid piece of wire can also be a probe, again with the slightest amount of copper sticking out.  In some cases a piece of wire is better since you can bend it to get around leads and other obstacles.

If you can get to it, try to also measure the input on the other amp (IC13) in the same way.

[Ian.M]

To get access to boards like that, you often have to make up extender cables. If the boards have a lot of connectors, that becomes a total P.I.T.A.

Otherwise, its possible to tack-solder thin insulated wire (e.g. Kynar wire wrapping wire) to the points you want to test and bring it out to something like a 0.1" pitch pin header (to give you something to safely anchor the free wire ends to so they don't short out) so you can probe them with minimal risk of letting the magic smoke out.

Re the OPAMP:  Yes 0V is also signal ground.   I had to simulate it to convince myself that without that connection it would pass the signal unamplified and non-inverted, but it did, so probably that would be enough to get it working.

You should probably take a look at the rest of Forrest Mims' Engineer's Mini-notebook: Op-amps https://archive.org/details/Forrest_Mims-Engineers_Mini-Notebook_Op_Amp_Ic_Circuits_Radio_Shack_Electronics

[aneng]

Will do.  However, it'll have to wait until Friday now as I'm working away for a couple of days.  I'll get back with the readings then.  Thanks !

[aneng]

I just popped back to say I'd had an idea of removing the board and soldering wires to the inaccessible points and reassembling and found you'd already suggested it !  I do get there in the end, just not as quickly as you guys ;-)

Thanks for the link Ian, I'll definitely check it out.

[Ian.M]

The fact that the gain as determined by measuring amplitudes with the scope is not x10, means either you've still got a problem with the OPAMP circuit (e.g. incorrect resistor values) or there is a serious problem with the Y2 input attenuator.  If you used the correct component values, it certainly shouldn't have a gain of over x127 (13.8V/108mV)

Post a good closeup photo of the OPAMP circuit please.

[aneng]

Er...  You may have noticed I quickly deleted my post !  I had the input going in on the wrong side of R1 ! - hence the insane gain.

I've just corrected it and it's all working nicely with a perfect x10 gain.

Presuming the next step was going to be measuring the output from the opamp with a DMM, here are the figures...

Input (measured on the scope, which we know is reading low)= 1.15V p-p (1kHz Sine)
Output (measured on the scope) = 11.5V p-p
Output (measured on DMM) = 4.75V AC.

I'll solder up those extensions to the IC and get some more voltages now...

[Ian.M]

4.75V RMS *2*sqrt(2) is 13.44V pk-pk   ==> your scope is reading 15% low. (ASS-U-MEing the frequency was within the specified input range of your multimeter's AC volts range and that your multimeter is accurate to better than three sig. figures)

[MarkL]

4.75V RMS *2*sqrt(2) is 13.44V pk-pk   ==> your scope is reading 15% low. (ASS-U-MEing the frequency was within the specified input range of your multimeter's AC volts range and that your multimeter is accurate to better than three sig. figures)

It's consistent with the DC readings which were reading about 18% low.

[aneng]

I've managed to get to the board with IC12 and 13 on (Vertical Attenuator Unit).  That wasn't easy!

I'm too tired this evening to attempt soldering on extension leads to the chips, but I have spotted that there are some minor variations between the schematic and the board I have.  If it's required, I'll try to draw a diagram illustrating what, and where they are.  However, I have spotted a thing that looks like a long resistor that is not on the schematic, but goes in the line between the output of pin 8 of IC12 ( the 'OUT+' connection) and the header for P3.  I have absolutely no idea what it is ! - any suggestions ? :



...It's the long brown thing with the single black band and long leads, running down the side if IC12.


There also appears to be evidence of some previous work around IC12 and 13 in the past, judging by the presence of flux...

[MarkL]

That's a zero-ohm jumper.  One black band = 0R.

  https://en.wikipedia.org/wiki/Zero-ohm_link

[aneng]

Well well, I never knew about those !  Cheers Mark.  One other thing (until I get the voltages later) - what's that extra bit connecting Pin 1 to ground ?  It's not on the schematic, but it appears to consist of two diodes in series with the cathodes facing each other, followed by a resistor.  What would that do ?

[aneng]

Can you get access to the pins on IC12?  Pins 8 and 9 are OUT+ and OUT- and it would be worth verifying they are also at -9.7V.  (I'm rather suspect of solder connections in this scope now.)  Also check pin 14 +10V supply, and pin 3 -10V supply.

What is the voltage on pin IC12 pin 15?  How does that compare to IC13 pin 15?  That's the input.

Are there any markings on IC12?  The parts listing says it's an ATM-4010, but I'm having trouble finding a datasheet for it.  It's some kind of high speed amplifier.

Right then...  readings time !  I made up an extension by chopping the last foot off an ethernet cable.  That gave me 8 wires terminating in an RJ45 plug which worked nicely.

With the scope set as per the settings listed on p90 of the manual;

IC12 (i.e. Channel 1)
Pin 8 (OUT+) = -9.73V
Pin 9 (OUT-) = -9.65V
Pin 14 (+10V) = +9.77V
Pin 3 (-10V) = -9.94V
Pin 15 (INPUT) = -1.375V

IC13 (i.e. Channel 2)
Pin 15 (INPUT) = +0.449V

Both channels controls are set identically.  There is obviously something going on with the input as Ch1 is minus 1.375V and Ch2 is plus 0.449V.

[MarkL]

Well well, I never knew about those !  Cheers Mark.  One other thing (until I get the voltages later) - what's that extra bit connecting Pin 1 to ground ?  It's not on the schematic, but it appears to consist of two diodes in series with the cathodes facing each other, followed by a resistor.  What would that do ?

One diode is probably a switching diode, and the other is probably a zener.  If that's right, the pair would be trying to limit the voltage excursions to zener_voltage+0.7V.  And the resistor is added to limit current once the voltage limit is reached.

I think pin 1 is likely still ground, so this extra cricuit is limiting the input pin 15, perhaps to prevent overload and/or to improve recovery time for overly large signals.  Sure looks like an after-thought to the design.

And yes, I also see some variations from your layout and the schematic.  Going off-road.

More in a bit on the measurements...

[MarkL]

IC12 (i.e. Channel 1)
Pin 8 (OUT+) = -9.73V
Pin 9 (OUT-) = -9.65V
Pin 14 (+10V) = +9.77V
Pin 3 (-10V) = -9.94V
Pin 15 (INPUT) = -1.375V

IC13 (i.e. Channel 2)
Pin 15 (INPUT) = +0.449V

Both channels controls are set identically.  There is obviously something going on with the input as Ch1 is minus 1.375V and Ch2 is plus 0.449V.

The discrepancy in voltage is certainly interesting.  It doesn't exonerate IC12, however, since something wrong inside IC12 could be pulling down the input pin 15.

Let's assume IC12 is ok and let's see what the dual JFET Q1 is supposed to be doing.

To start, we have the diode network of D13 + D2 and D12 + D1.  D13 and D12 are both 12V zeners.

D12 cathode is connected to +10, with a pulldown to -10V via R15.  That would make D12 anode -2V (approx).  D13 anode is connected to -10V with a pullup to +10 via R16.  D13 anode is therefore +2V.  The various caps provide filtering.  So, this circuit generates symmetric -2V and +2V reference points.  (Hint: Those are some voltages to check.)

Q1a gate is the input to the first stage, and is connected to the reference points by diodes D1 and D2.  Anything more positive than D13 anode + 0.7V (= +2.7V) will cause current to flow through D1, thereby clamping the input voltage.  Conversely anything less D13 - 0.7V (= -2.7V) will be clamped through D2.

So, the input voltage to Q1a gate is limited to +/-2.7V (approx).  R18 47R doesn't matter because the input current to Q1a is on the order of a few 10's of picoamps.  Your reading of -1.375V is within range, but check that the -2V and +2V reference points are ok.

JFET Q1a is configured as a voltage follower.  The source (the bottom) will be a couple of volts *higher* than the gate (JFETs operate reverse biased).  Q1b is providing the right amount of current so that the source will be pulled back down to 0V when Q1a gate is also at 0V.

How does Q1b know how much current to provide?  Well, that's why Q1a and Q1b are a matched pair.  Q1b's drop from gate to source and then through R29 100R is equal to the same drop on Q1a and R28 100R.

So, that should tell you some other voltages to check.  See if Q1a gate to source (Vgs) is the same as Q1b Vgs.  And if you want to keep me honest, you can check Q2a and Q2b Vgs also.

I know soldering test wires to probe voltages is a pain.  From your photos, it looks like you can get to Q2's leads from the top.  Maybe Q1 if the potentiometer isn't too much in the way.

Here's another thought.  A dynamic test on IC12's input is also possible.  While watching IC13 pin 15 with your DMM, try the 9V battery DC input test on Ch2.  Keep the trace on the screen.  Note the two DMM readings between 0V and 9V.  Then move the DMM to IC12 pin 15 and input into Ch1 and compare results.  The absolute voltages may be different, but the difference between 0V and 9V input should be the same on both channels.

Perhaps do this test first before the above JFET tests since you already have those wires soldered and accessible.

[aneng]

I found a site selling the service manual for the CS-2100A and bought that in the hope that it would be more accurate than the CS-2070 manual I have.  Unfortunately....  it's not !  Certainly on the section of this board anyway, it's identical to the CS-2070, schematic-wise.  The PCB layout seems slightly different though - but again, it doesn't match my board !  I've since found another site that claims to have the service manual for the CS-2100 (i.e. mine) - and although I've paid, the download link hasn't appeared yet.  They reckon it can take up to 24h... so undoubtedly this time tomorrow I'll be emailing them saying "Where's my *&%ing manual " and shortly after asking Paypal for a refund (cynical old sod that I am !).

Meantime, here is a couple of pics of the relevant board...  IC12 (Ch1) is at the bottom right - it's the two rows of 8 pins.



...and the flip side.  Notice the big blobs of solder joining the ground lands/traces either side midway down.  That doesn't look original.



I've already removed my extension wires, so it's no problem to do any test.  I must confess, your last post was a bit advanced for me at this stage in my learning (the explanation of the circuit) - my knowledge isn't developed enough yet to really understand.  However... what test do you think is the best next ? - the 9V battery one ?

Incidentally, here is the service manual for the CS-2100A : https://www.dropbox.com/s/a9dgag80gldz56s/Trio%20CS-2100A%20Service%20Manual.pdf?dl=0