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Interesting fault in 535A
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Hi all,
I recently fixed a 535A (actually a RM35A) with some interesting and deceptive symptoms. The scope powered up OK after 10 years of storage but apart from a green flash when the relay pulled in there was no display. All the LV supplies were in spec and the beam finder neons worked properly. The HV rectifier filaments lit and the oscillator screen was at +91V (normally +85). Checking with a HV probe revealed that the cathode of the CRT was at -1150 V (should be -1350) and the grid at -1450 V thus cutting off the CRT. When I jumpered the grid and cathode the CRT lit up nicely with -1150 on both electrodes. Definitely something fishy in the HV section. The grid supply diode V822 looked like it had been replaced and the cathode supply diode V862 looked pretty grubby and I wondered if it was weak. Bridging it with a suitable Si diode restored function alright, but both supplies were low at -1150 V, not what you would expect if V862 really was weak. The grid supply filter cap C820 had been replaced with a modern HV disk ceramic and it was only after I looked a bit more carefully that I noticed the two cathode supply filter caps, C842 and C831 tucked away under the rectifier diodes. They were still the original Sprague black beauties! Fortunately I had a couple of modern HV ceramics in the parts drawer and after a few minutes with the soldering iron and silver solder, in they went. That was all it took to restore perfect operation. The cathode supply test point was running a little high but it adjusted to spec nicely. What a beautiful scope it is, with a razor sharp display! What was interesting was the behaviour with the Si diode in place. That would have increased the current available to the cathode circuit but didn't have the effect of raising the cathode supply voltage because the caps were just too leaky to sustain a higher voltage than -1150. Instead, the grid supply fell to meet it. Why? I can only explain it by the leaky cathode caps pulling so much current when fed with the Si diode that they loaded down the oscillator and reduced the power available to the grid circuit. I'd be interested to see any alternative explanations. Morris |
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Albert
Hi Morris,
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Here is a different interpretation. A leaky C842 will reduce the cathode voltage, in this case to -1150 V. About 20 uA leakage would be sufficient for that. The extra Si diode makes no different since regulation was still active. But with the extra Si diode the cathode supply will be more efficient. That will reduce, loosely speaking, the duty cycle of the oscillator. That in turn will reduce the grid voltage. If you measure cathode voltage with a multimeter, you will see that the trace gets darker (depending on the load produced by the multimeter). Not because the cathode gets more positive, but because the grid gets more negative. Albert Hi all, |
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No explanation, Morris, but I'd like to remind all that I have
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a bag 6.8nF/4kV and 10nF/4kV polyester film caps that make fine replacements for the HV BB's. I'm only asking $0.50 each; that's a break-even price. I have more than I will ever use. Dave Wise -----Original Message----- |
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Thanks Albert,
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I think that's pretty much what I thought was happening however I don't know that the added load changes the duty cycle of the oscillator. As the load increases on that oscillator the effect would be to drop the output amplitude rather than the duty cycle. That oscillator and its control loop is not a digital switching circuit and has a resonant tank which maintains the output close to a sine wave. As the cathode voltage was always under spec regardless of the presence of the Si diode I think the feedback circuit would be trying to increase the oscillator power under all circumstances. Maybe with the added diode it was a bit less under spec and so the feedback would pushing it a bit less hard. Unfortunately I didn't note the exact voltage, just that it was too low. Morris --- In TekScopes@..., "Albert" <aodiversen@...> wrote:
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Thanks for the offer Dave, but after adding shipping down here to Oz I'm sure the local or ebay prices are not too bad in comparison.
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Regards, Morris --- In TekScopes@..., David Wise <david_wise@...> wrote:
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Albert
Hi Morris,
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I said "loosely speaking" since it's more complicated indeed. But the point is that with heavier cathode load, the oscillator is forced by the control loop to work "harder" in some sense, in order to keep the cathode at the desired voltage level. But that "harder" working has effect on the grid circuit since the grid circuit didn't need that "harder" working. The grid voltage gets more negative. I'm quite sure I measured this some time ago. The negligible positive change in cathode voltage when extra loaded is far less than the increase in negative voltage between grid and cathode. I measured the cathode voltage permanently with a HV probe and then touched a second HV probe to the cathode to increase the load. I would describe "harder" working as a combination of larger osc tube anode current and smaller anode voltage (= higher primary voltage) during the conducting state of the tube, and also a larger duty cycle of the conducting state. I know it's not simply a matter of fully conducting or completely shut-off, but this is the idea. The conducting state roughly corresponds to the distorted bottom part of the sine wave. Albert --- In TekScopes@..., "Morris" <vilgotch@...> wrote:
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Sorry, Morris, I didn't notice where you were.
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Carry on. Dave Wise -----Original Message----- |
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