Solartron CD 1400 Oscilloscope Refurbishment

So I did some back of a envelope calculations after disassembling one of the rectifier i found it contained 51 selinium plates, my guess is it should be 50 looking at the data sheet for a K8/50 the peak inverse voltage os a selinium plate is 68 volts so 50 would give 3,400 volts.

The current its rated for is 5mA

And each plates forward voltage drop is around 1v so that would be 50volts each rectifier and 100volts total

On the other hand a 1N4007 can block 1000v reverse so 4 would be appropriate in each replacement and each one drops around 1.1v for a total per rectifier of 4.4 and total of 8.8v

The difference therefore would be 91.2v 

At 5mA the required resistance to drop 91.2 volts is 18.24 kilohm (kΩ) and it would dissipate 0.456 watt 

If I was to use 2 10kohm resistors one in each rectifier at 5mA the drop would be 100v thus the voltage would be a little lower and the power dissipated by each would be 0.5watt.

However if the current was only 3mA the drop would instead be 60v and the power 0.18 watt the output would see a higher voltage by approximately 31.2 volts however I’m guessing the inrush current would be smaller at 3mA so I hope that would be fine?

Thought?

@stevendeacon Hi, I had a quick look at the schematic and here are some considerations to take into account:

So basically you have a 1000V AC (RMS) from the transformers winding output that is rectified  (but in  a special way)  and at the rectified output end you’ll get 3200V or thereabout. Just keep in mind that the 3200V figure might differ. The tolerance is quite permissive. Anyways.

It seems (to me) that C306 (0.07uF) the two diodes MR304 and MR305 (K8/50 each) and the other two identical in value capacitors (C312 and C313)  form a topology called Quasi-Voltage Tripler using a capacitor-charging rectifier:

At first the C306 cap charges on one AC half-cycle, then  the same C306 dumps the charge into C312+C313 and the output via D2

The peak output reaches roughly 3× V peak of the transformer winding (~1000 V RMS × √2 × 3 ≈ 4200 V), minus the diode drops and ripple — which matches what the schematic reports at the CRT ~3200 V DC

This kind of (charge pump) circuit is common in vintage CRT HV supplies, including oscilloscopes and TVs, because it boosts HV while using fewer components than a full multiplier ladders would require and in turn saving some manufacturing costs !

Now in regards with a suitable silicone diodes my idea is 3 diodes of 1N4007 (1000V diodes) in series  but mounted on a connecting strip somewhere safe and with sufficient clearance so to avoid accidental arcing across (or even double or triple heat shrink wrapped). You can also use the UF4007 variant if you want a quicker recovery time but not really needed here (in my opinion). 

As for the resistors I would add one wire wound (ideally 3-5 watts ) resistor per each diode (so 2 resistors needed) and I would place them as following:  R1  connects from C306 to the Cathode of the MR304  and R2 (again) connects from C306 to the Anode of MR305.

Please beware: Do not put a resistor at the DC output — it won’t protect the transformer or the capacitors during charge surges !!!

Now, R1 and R2’s function is (obviously) to simulate the internal resistance of the original selenium diodes therefore keeping a similar level of voltage drop across the new silicone diode (replacements)  and i would say 22–47 Ω, 3–5 W wire-wound each is a good starting point.

The reason for such a high wattage resistors is maybe not immediately obvious but here is the answer: 

Because this circuits (topology) produces a High voltage ”charge-pump action” in which C306 dumps current into the second diode and the capacitor bank (formed by the 2 other caps in series) this means a significant pulse current. Even though is true that the average current is low (yes, well below 5 mA rated current of the selenium diodes) But the ”instantaneous pulse current” when the AC waveform is near peak, and C306 dumps its energy into the output caps through MR305, can be in the order of tens of mA, even >50 mA, for periods of time ranging in the order of microsecond and even low milliseconds.

In other words, even though the total energy moved is small, the ”peak current” during this brief periods is the one that matters the most !

The diode sees ”impulsive high current”  so, the power dissipation is not linear but rather ”SPIKY” and needs some additional headroom therefore a more appropriate resistance range would be 22-47 Ohms  rated @ 3-5watts (to be safe.)

This would be more in line with real-world selenium-to-silicon upgrades in ”high-voltage pulse circuits” like CRTs, radar gear, vintage scopes, and X-ray drivers etc.

Now, how did I get to this resistor value you may ask . Well here is the explanation:

As far as I know and read Selenium diodes typically have 20–50 Ω of ”dynamic resistance” per element under load. Your K8/50 stacks are high-voltage types and,  likely with a ”stack resistance” of ~30–40 Ω per diode, so that gives us a ”starting ballpark”.

C306 (according to the schematic) is 0.07 µF and is rapidly charging/discharging during operation. Assume:

 A Peak AC = 1000 V RMS   or the equivalent of 1414 V peak voltage

The Discharge path is through diode + the newly attached resistor

The initial surge current would be: I=V/R

Keep in mind that we are aiming for an extended ”headroom current” of lets say 40 something to 50 something mA so:​

Let’s try different R values:   for a 10 Ohms the calculated peak current will be  141mA  therefore too little dumping                                                                                  for a 22 Ohms the calculated peak current will be   64mA   acceptable for limited surge !                                                                              for a 33 Ohms the calculated peak current will be   42mA   closer to the selenium diodes behaviour                                                                for a 47 Ohms the calculated peak current will be   30mA   a conservative choice !                                                                                      for a 150 Ohms the calculated peak current will be   9.4mA   over-limiting and restrictive to charging of caps                                                  for a 220 Ohms the calculated peak current will be   6.4mA   undercharges the filter caps !

Please Note: these calculations assume ”idealized first-pulse behaviour” but in real-life practice, filter caps draw more than one pulse, and the surge current occurs every half-cycle.

So this concludes my thought process but if you or anyone else have a different way of dealing with this  I’m happy to hear other views too.

Good luck.

🔥 Do not put a resistor at the DC output — it won’t protect the transformer or the capacitors during charge surges.

@ovi4 thank you so much for your response can I confirm your calculations for resistance values?

Using V = IR or I = V/R to gain the current at the peak voltage to gain peak current 

1414v ÷ 10ohm = 141.4 Amps 

1414 ÷ 10Kohm = would be 141.4 mA

Have I made a mistake here?

My initial thought was to empty the casings of the selinium rectifiers and place the components inside the old tubes to try to keep it looking somewhat original, however this is less important to me than a working oscilloscope so a connecting strip might be better? 

If using 3 1N4007 Diodes, where would you place the two in replacement of MR304 or MR305? With the other having only a single diode. 

Thanks once again for your help

Kind regards 

Steve 

@stevendeacon Hi, Im sorry for the somehow derailed (last part of my message). I do not know what has happened but hope you could still follow the lines. :)))   Anyways.  As for the diodes  I would use 3 in series for each (K8/50 diode)   so there will be 2 series groups of 3 1N4007 diodes.   As for the calculation  you are 100%  right  the resulting value is indeed in Amps not mA  but it is late at night over here and I must have been half way asleep when I wrote the message :))  and I have tried to edit my message shortly after I had it published but unfortunately this site is set in such a way that you only have 1.5 or 2 minutes to edit your own message (if you realised of a mistake or want to add some more details to it). Well  I was in the middle of editing my message but I went past the time threshold while doing it and my edit was no longer valid so it will remain as it is unfortunately but but at least you now know of my mistake. Now having said that please note that even though the K8/50 selenium diodes are rated at only 5mA  their resilience to brief surge charge currents (in the order of Amps !) is far superior to that of these 1Amp silicone diodes. Now you could recalculate the resistor value using the same formula but I would still allow for a current in the magnitude of 3-5 times higher than the 5mA per (old selenium) to cover for those brief surge periods. I would say experimentation is the key !  Start with a comfortably higher value and find a way to measure the 3.2KV output via some voltage divider or something. and start reducing the initial value until you get around 3.2KV. To be fair I don’t really know what would be the best and safe method you could measure it (without a proper high voltage probe) but I’m pretty sure the voltage divider should work (when done properly using large spacing and plenty air gaps between the resistors so to avoid arcing). Good luck.

Welcome! Now I won’t be as in depth as Ovi here, but I have a few things to add. Don’t worry about following values or schematics to a T in old stuff like this. Obviously there are exceptions for certain circuits like timing and RF stuff, but power supplies will not know the difference for a slight change in values. 3 diodes in series is probably a safer solution than using a single diode. You could get single diodes to do the job, but using 3 in series helps keep the reverse voltages at each diode at bay for reliability. Not to mention much more common and cheaper than any special HV diode. The dropper resistors likely aren’t required, but doesn’t hurt to have. There is a Patreon video discussing replacement of seleniums if you want to reference that as well. Do all the major stuff you know is bad like electrolytics and any leaky capacitors in the rest that would cause issues. Once you can get it running without burning it up pick through the remaining issues one at a time.

Board1 Recapped

Here is the progress so far