Measurement of vacuum diode electron flow

In summary: S1 is turned on, the voltage goes up and stays there.I already did that, I noticed that the power supply works just fine under normal conditions, it's only...when S1 is turned on, the voltage goes up and stays there.In summary, the power supply increases voltage when S1 is turned on.
  • #1
Abimbola1987
83
13
Dear Sirs,

After my theoretical post on vacuum diode electron flow, I decided to try and measure it physically. I didn't have a vacuum tube, so I improvised with some tungsten wire, aluminum sheets and some screws and the result was usable. Obviously the tungsten wire fails after some time due to the lack of a vacuum envelope.

Screenshot-from-2019-02-22-17-59-33.png


Question: What causes the increase in Iz current when S1 is on?
 

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  • #2
Abimbola1987 said:
Question: What causes the increase in Iz current when S1 is on?
Because the cathode is being heated by the current from the Vf source? Or am I reading your schematic incorrectly?
 
  • #3
berkeman said:
Because the cathode is being heated by the current from the Vf source? Or am I reading your schematic incorrectly?

Dear berkeman,

You are reading the schematic correctly, The cathode is being heated by Vf.

But it is the Iz current associated with Vp that increases, which I don't really understand why it does so.
 
  • #4
Abimbola1987 said:
But it is the Iz current associated with Vp that increases, which I don't really understand why it does so.

I will try venture a guess, could it be that; since the Ip/Ic is constant, it must mean all diodes are forward biased all of the time?
 
  • #5
It looks like the Iz power source is what is pumping the high voltage through the transformer. With no heating of the cathode, there is no measurable diode current. Heating the cathode facilitates diode current to flow, which requires that the power source supply that extra power to push the electrons across the diode gap. The input current from Iz with no diode current is just the overhead to run the HV pumping mechanism with no diode current added...
 
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  • #6
Iz indicates that the power supply draws power on no-load, which may be due to transformer characteristics or inefficiencies in the ZVS driver (not sure what that is). Anyway, the effect you see is just a PSU artifact.
 
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  • #7
Surely the right thing to do would be to measure the Anode - Cathode voltage. What goes on on the right hand side of the circuit is not certain but the V/I characteristic of the 'non-vacuum diode' can be found that way.
 
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  • #8
berkeman said:
It looks like the Iz power source is what is pumping the high voltage through the transformer. With no heating of the cathode, there is no measurable diode current.

Yes.

berkeman said:
The input current from Iz with no diode current is just the overhead to run the HV pumping mechanism with no diode current added...

And yes.

berkeman said:
Heating the cathode facilitates diode current to flow, which requires that the power source supply that extra power to push the electrons across the diode gap.

I don't really understand what you mean by that, could you please elaborate?
 
  • #9
tech99 said:
Iz indicates that the power supply draws power on no-load, which may be due to transformer characteristics or inefficiencies in the ZVS driver (not sure what that is). Anyway, the effect you see is just a PSU artifact.

I would say like berkeman that the power on no-load is just overhead for maintaining the voltage. So yes, transformer characteristics.

ZVS is an acronym for Zero Voltage Switching, or flyback driver or whatever, please see typical schematic below.

zvs.png
 

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  • #10
sophiecentaur said:
Surely the right thing to do would be to measure the Anode - Cathode voltage. What goes on on the right hand side of the circuit is not certain but the V/I characteristic of the 'non-vacuum diode' can be found that way.

Would the V/I characteristic help determine why Iz increases when S1 is on?

sophiecentaur said:
What goes on on the right hand side of the circuit is not certain

You have both ends of the right hand circuit, something apparently happens in the diode bridge, and the effect of that can be measured on it's DC supply Vp.
 
  • #11
Abimbola1987 said:
Would the V/I characteristic help determine why Iz increases when S1 is on?
No but it splits the problem into two. If you are 'testing' the power supply then it's hardly the best thing to use an undefined load. Fault (or anomaly) finding involves isolating as many different parts of a circuit as possible.
I'd be inclined to measure what happens with a simple resistive load right across the power supply, then a resistive load downstream of the transformer.
What are you measuring Iz with? Have you the facility to look at the waveforms involved?
 
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  • #12
Dear sophiecentaur

sophiecentaur said:
No but it splits the problem into two. If you are 'testing' the power supply then it's hardly the best thing to use an undefined load. Fault (or anomaly) finding involves isolating as many different parts of a circuit as possible.

I absolutely agree in that approach.

sophiecentaur said:
I'd be inclined to measure what happens with a simple resistive load right across the power supply, then a resistive load downstream of the transformer.

I already did that, I noticed that the power supply works just fine under normal conditions, it's only when an external current source like the 'tube' is present that it draws more current than it should.

If you look at the HV circuit i.e. diodes+tranformer+ZVS+Vp in that order, it is logical that when something happens around the diodes it will translate into current draw from Vp, so for starters you could ignore what is in between diodes and Vp, and concentrate on the 'interface' between the HV and LV parts of the entire circuit, and that is the diodes?

sophiecentaur said:
What are you measuring Iz with? Have you the facility to look at the waveforms involved?

Vf is DC, so I measured Iz with a simple ammeter. However I do have the facility to look at waveforms should that be necessary.

I blew up my storage scope the other day, so here is a old school screen dump from my old scope.

anode-voltage.jpg


anode-voltage-load.jpg

.
 

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  • #13
Now I managed to stabilize the supply, and it does not change waveform regardless of S1 being on or off.

anode-voltage-stabilized.jpg


And this is the schematic

Screenshot-from-2019-02-23-18-07-36.png


Now it draws 0.7A due to the load of the 10M resistor which gets hot, but the increase in Iz current when S1 is on, is more or less similar to the one in post #1
 

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  • #14
Abimbola1987 said:
it is logical that when something happens around the diodes
Would it be daft to suggest replacing the diodes one at a time? This the last chance saloon so anything is worth trying.
 
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  • #15
sophiecentaur said:
Would it be daft to suggest replacing the diodes one at a time? This the last chance saloon so anything is worth trying.

Ah... "Last Chance Saloon" - Where everybody knows your name Here are no daft suggestions, however I have trouble finding some alternatives other than the same type of diodes CL01-12, which I have tried without any improvement resulting from that.
 
  • #16
sophiecentaur said:
Would it be daft to suggest replacing the diodes one at a time? This the last chance saloon so anything is worth trying.

I found some http://www.vn-experimenty.eu/datasheety/polovodice/2CL77.pdfin my stock it's more noisy than with the CL01-12 diodes in post #13, however Iz still increases as before when S1 is on.

Anode-Cathode Voltage *0.001
IMG-20190224-165202.jpg
 

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  • #17
Abimbola1987 said:
I found some http://www.vn-experimenty.eu/datasheety/polovodice/2CL77.pdfin my stock it's more noisy than with the CL01-12 diodes in post #13, however Iz still increases as before when S1 is on.
Another thing; looking at the heater circuit. I would have placed Sf in the other leg so that the cathode was always in circuit, whether or not the 'battery' Vf is heating the filament. The Ic current has to pass through the battery when Sf is open circuit for your implementation. Depending on what the 'battery' actually consists of, you could imagine a possible effect. (Hardly, though, if it really is a battery). Another last chance idea :smile:. There could be another unintentional current path to Earth that's not on the schematic.

PS What are the voltages around the circuit? Problems measure high volts perhaps.
 
  • #18
sophiecentaur said:
Another thing; looking at the heater circuit. I would have placed Sf in the other leg so that the cathode was always in circuit, whether or not the 'battery' Vf is heating the filament. The Ic current has to pass through the battery when Sf is open circuit for your implementation.

There is only Ic current when S1 is on. But I have tried move the S1 and I have tried reverse the polarity of Vf, and I have tried using a variac as Vf: But the result is the same, when S1 is on Iz current increases.

sophiecentaur said:
Depending on what the 'battery' actually consists of, you could imagine a possible effect. (Hardly, though, if it really is a battery)

I used a real 12V SLA battery.

sophiecentaur said:
Another last chance idea :smile:. There could be another unintentional current path to Earth that's not on the schematic.

I doubt that very much, I'm very careful with isolation and grounding when it comes to HV.

sophiecentaur said:
PS What are the voltages around the circuit? Problems measure high volts perhaps.

I will measure that tomorrow, as it is getting too late in the evening here for working with HV. And I don't believe there are any measuring problems regarding voltage, it's all DC(ish) below 10KV with very low currents. It would have been a different story if it were arbitrary waveforms in the KA range with reactive components.

I will also check my qualified guess in post #4 i.e. when S1 is on, the Ip/Ic current is constant, it must mean all diodes are forward biased all of the time, and that in turn means there must be a current path (not necessary a full shortcurcuit) between the legs of the transformer. How's that for a last chance?
 
  • #19
Abimbola1987 said:
I used a real 12V SLA battery.

Abimbola1987 said:
that in turn means there must be a current path (not necessary a full shortcurcuit) between the legs of the transformer.
Sounds like a DC path through the transformer?
 
  • #20
What is the turns ratio of the transformer? Is the current in the primary a surprise?
 
  • #21
Abimbola1987 said:
I used a real 12V SLA battery.

Abimbola1987 said:
that in turn means there must be a current path (not necessary a full shortcurcuit) between the legs of the transformer.
Sounds like a DC path through the transformer?
 
  • #22
Abimbola1987 said:
I doubt that very much, I'm very careful with isolation and grounding when it comes to HV.
It the whole thing 'floating' or is one side tied to earth?
(I'm, only going through the same thought processes that you almost certainly have.)
 
  • #23
sophiecentaur said:
It the whole thing 'floating' or is one side tied to earth?

The whole thing is earthed, see schematic. I even measured the current (Ie) going to earth, and it is zero.

sophiecentaur said:
What is the turns ratio of the transformer? Is the current in the primary a surprise?

I don't know, it's a standard CRT flyback transformer. No the current on the primary is as expected.

In the meantime I switched the scope to digital mode so I can freeze frame, and I removed R1 from the circuit. Then I measured around, and it seems I was sort of right, D3 and D4 are forward biased ALL of the time, but it has nothing to do with the state of S1/Ip/Ic.

The question is now, why are D3 and D4 forward biased ALL of the time??

EDIT: ALL measurements are relative to Earth/Ground
(Typo: the last trace is obviously CH3)
all-voltages.jpg
 

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  • #24
sophiecentaur said:
Sounds like a DC path through the transformer?

You were right too, I made a mistake, aka assumption, that it was a plain HV transformer, however it turned out to be a DST Diode-Split Transformer with plenty of paths to earth. I need to find another transformer, so the case is temporarily closed while I regroup.

DST-Zeilentrafo.png
 

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  • #25
Abimbola1987 said:
so the case is temporarily closed while I regroup.
We got there in the end! God luck finding a suitable Tx. :smile:
 
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  • #26
Abimbola1987 said:
Question: What causes the increase in Iz current when S1 is on?
The full-wave rectifier power supply is a source of emf. Mentally replace the entire supply with a high-voltage battery connected across plate (+) to cathode (-). It's obvious that if (conventional) current is to flow (plate to cathode) then the battery will carry current which has to be supplied by the battery to obey Kirchhoff. So an ammeter in series with the battery will register a finite current if the filament is excited. The 200 ma without the heated cathode is because your power supply is inefficient. A real battery would supply zero current in the absence of a heated cathode (filament). The net 100 mA supplied through Vp satisfies E Ip = Iz Vp (power conservation) where E = plate to cathode voltage. So E Ip ~ Iz Vp with E ~ (100mA x Vp)/200uA.
 
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  • #27
Studying the 'scope waveforms coupled with the description of your 'open air tube', have you considered effects of air temperature and humidity on your data readings? I seem to remember ambient humidity effecting readouts while 'pushing' RF output from a magnetron to open air. Old time vacuum tube device readouts seemed a bit crazy until the circuit warmed up to a stable operating temperature. Navy literature from the era when experiments such as yours were cutting edge are replete with designs that worked well indoors but became uncertain at sea and aboard early aircraft presumably due to changing atmospheric conditions.

Historical note: during the 1970's government vacuum tube HV power supplies -- particularly rectifiers -- were replaced with solid state devices even when retaining tubes in the primary circuits, not unlike your designs.
 
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  • #28
rude man said:
The full-wave rectifier power supply is a source of emf. Mentally replace the entire supply with a high-voltage battery connected across plate (+) to cathode (-). It's obvious that if (conventional) current is to flow (plate to cathode) then the battery will carry current which has to be supplied by the battery to obey Kirchhoff. So an ammeter in series with the battery will register a finite current if the filament is excited.

Thank you very much, this is an explanation that I can subscribe to! You just saved me hours of work on my circuit, case closed permanently.

I think you should change you tag to "clever man" ;-)
 
  • #29
Klystron said:
Studying the 'scope waveforms coupled with the description of your 'open air tube', have you considered effects of air temperature and humidity on your data readings? I seem to remember ambient humidity effecting readouts while 'pushing' RF output from a magnetron to open air. Old time vacuum tube device readouts seemed a bit crazy until the circuit warmed up to a stable operating temperature. Navy literature from the era when experiments such as yours were cutting edge are replete with designs that worked well indoors but became uncertain at sea and aboard early aircraft presumably due to changing atmospheric conditions.

Historical note: during the 1970's government vacuum tube HV power supplies -- particularly rectifiers -- were replaced with solid state devices even when retaining tubes in the primary circuits, not unlike your designs.

Yes I did, but I didn't know the magnitude of the effect until I built the thing. It obviously is not suitable for high-fidelity applications, if you e.g. breathe on it you can register significant fluctuations in the current, so for my experiment, I tried not to breathe on it and covered it with a glass jar, which was sufficient for my application.
 
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  • #30
Abimbola1987 said:
Thank you very much, this is an explanation that I can subscribe to! You just saved me hours of work on my circuit, case closed permanently.

I think you should change you tag to "clever man" ;-)
Thank you and keep experimenting! All of physics rests on experimentation.
 

1. What is a vacuum diode?

A vacuum diode is an electronic device that allows the flow of electrons in only one direction. It consists of two electrodes, an anode and a cathode, placed in a vacuum-sealed glass tube.

2. How is the electron flow measured in a vacuum diode?

The electron flow in a vacuum diode is measured using a device called an ammeter. The ammeter is connected in series with the cathode and anode, and it measures the current flowing through the diode.

3. What is the significance of measuring electron flow in a vacuum diode?

Measuring the electron flow in a vacuum diode is important for understanding the performance and efficiency of the diode. It also helps in analyzing and troubleshooting any issues that may arise in the diode.

4. What factors can affect the measurement of electron flow in a vacuum diode?

The measurement of electron flow in a vacuum diode can be affected by factors such as temperature, voltage, and the quality of the vacuum in the tube. These factors can alter the behavior of the electrons and affect the accuracy of the measurement.

5. How can the measurement of electron flow in a vacuum diode be improved?

To improve the measurement of electron flow in a vacuum diode, it is important to maintain a stable temperature and voltage, as well as ensure a high-quality vacuum in the tube. Using precision instruments and following proper measurement techniques can also help improve the accuracy of the measurement.

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