Iconoscope Design and Function

[Rinehart]

I have sent the following questions to one or two people already as a private message, so please don't be offended if you're reading it again here. It's not that I am impatient or dissatisfied, but in retrospect I realize that I should have posted it here in the first place, and that's what I am doing now.
There are three main questions that I have:
1) How exactly did the Iconoscope produce its signal? My understanding is that in the original concept, (fig. 1) each cesium granule on the mosaic would act as a capacitor, the amount of charge stored being proportional to the intensity of the light shining on it. The purpose of the electron beam would be to impart energy to the granule to overcome the resistance of the insulator and discharge it to the common conductor on the other side. This method however also produced a lot of secondary emission, leading to a lot of signal noise. Later refinements (fig. 2) incorporated the secondaries into the signal by collecting them and running them through the anode ring to complete the circuit, and this somehow was linked to increased sensitivity. But if I've read the diagram right, there is a DC power source connected to the output from the anode ring on one side and through a resistor to the signal from the photoelectric element and then on to the preamplifier, so how does this return the electrons to their source?
2) I understand that Iconoscope tubes had considerable variation is the peak sensitivity of the photoelectric element across the spectrum of visible light, not only from one make and model to the next, but amongst individual tubes of the same make and model. This was caused by the manufacturing process of the scanning target being a considerably difficult one for the time and so it wasn't possible to control the outcome as well as would be desired. Therefore two different TV cameras photographing the same thing would have rendered the image in different shades of grey. Is this true, and if it is, was the difference noticeable enough that the broadcaster had to correct
for it, and how would they go about doing this?
3) Around 1940, RCA put on the market an Iconoscope tube, Model 1847, designed specifically for the amateur market. This tube featured a two-sided target, which considerably simplified things, because you wouldn't have to correct for keystone distortion. Why didn't RCA incorporate this into its professional cameras?
I suspect that I have misunderstood more than a couple of things here, and also that the answers are complicated ones. So if that is the case, could you point out some articles or reference works I could read?

[Sandy G]

It's Majick. And SECRET-We could tell you, but then we'd hafta KILL you....(grin) Seriously, the depth of talent here should be able to explain it, & I'd be interested myself, too !

[old_tv_nut]

Donald Fink explains the operation of the iconoscope in Television Engineering, 2nd Ed, 1952, pp.102-104.

1) [Paraphrising Fink] The simple and wrong explanation is that each element of the mosaic takes charge from the scanning beam depending on the photoelectric charge of the element. If this were true, it would produce a sensitivity gain of about 150,000 times compared to a non-storage device (image dissector). However, experiments show that the electron beam in the iconoscope has practically infinite resistance, that is, it always delivers a constant current to the mosaic. What is varying is the amount of secondary emission depending on the voltage on each mosaic element, which in turn depends on how much photoemission has taken place. Each element is also being discharged by a shower of secondary electrons from all the other elements. The result is that the sensitivity of the iconoscope is only about 100 times that of an image dissector, since the variation between a light and dark "pixel" is only the variation in secondary emission.

I think the improvement from using a collector ring is in having fewer secondary electrons fall back on the target, thus increasing the net charge on the mosaic elements, although Fink simply assumes that a collector is used.

By the way, the beam does not "overcome the resistance" of the insulator. The mosaic elements are miniature capacitors and current flows by charge/discharge with a change in electric field, but no electrons flow through the support plate.

2) All photo-emitters are delicate - if a substance gives up electrons easily, it usually means it is chemically somewhat unstable as well; dependent on the exact formulation and treatment in manufacturing, etc. In fact, when I worked on a flying-spot video player using film and photomultipliers, we found that taking a photomultiplier out of its box under bright fluorescent lighting would damage the photocathode and significantly increase the noisiness of the tube. So, I believe the sensitivity of the tubes could easily be different from unit to unit.

I don't know about differences in spectral peak. I think tubes of similar formulation would be similar, but perhaps differ somewhat. According to Fink, the usual iconoscope mosaic had high sensitivity in the red and near infrared, which would be modified to reduce infrared response by a final step of evaporating a small amount of silver onto the mosaic after the cesium deposition.

3) I don't know what the construction of the two sided iconoscope target consisted of. If it had similar globules on an insulator, that would mean the insulator had to be transparent. Perhaps it was too difficult to make in the large size for the studio iconoscope.

[Rinehart]

Ah very good, thank you for that. As I said, I suspected that I had made a fundamental mistake, and there it was: the electrons don't pass through the insulator. I had always envisaged it as something like what you get with a spark plug or spark-gap transmitter, when the voltage rises high enough the electrons jump from one electrode to the other. Of course this produces a lot of heat, but through the miracle of magic thinking, I assumed that didn't matter.
So let me see if I've understood:
1) in targets with photoemissive materials, light falling on them creates free electrons which escape from the material altogether, the number of which depends on the intensity of the light.
2) these electrons eventually do settle back on the material, almost all of them at some other point.
3) this process is constantly occurring all over the plate, with the result that on average the distribution of electric charge over the plate as a whole corresponds to the image. (Perhaps the expression "electron image" comes from this?)
4) the purpose of the electron beam is to deliver a constant electric current to the mosaic, one that can move across it without needing a physical conduit, like an electric wire.
5) at the moment that the beam strikes the mosaic, emitted electrons from other points on the mosaic can't settle on the point the beam strikes. This causes a discontinuity; the amount of charge that can't settle there corresponds to the number of photoelectrons released at that point.
6) this causes current flow to the signal plate, and because the beam moves over the mosaic in a very precise way, the signal is a representation of the electron image.
7) unfortunately the scanning beam itself can cause electrons to be liberated from the mosaic, and these are called secondary electrons. These also are sent to the signal plate, resulting in a lot of signal noise.
8) to reduce the number of secondaries sent to the signal plate, there is a secondary electron collector which will rout them to the anode to complete the circuit instead of being sent to the amplifier and then to the transmitter.
Is this correct?
In terms of the non-uniform peak sensitivity, I think it was in William Eddy's or Thomas Hutchinson's book. Somewhere in preparing for Christmas I've misplaced a bunch of my notes, so I'm not sure where I read it.
Also, thanks for explaining why the mosaic was given a final treatment with evaporated silver. I had heard they did this, but didn't know why.
Can I ask you a couple more questions? The first concerns an unimportant problem in an oscilloscope and doesn't pertain to a camera tube, but which is a vital problem for a picture tube: how did they get around the problem of varying the intensity of the electron beam without changing the cross-sectional area of it?
And secondly, what kind of oscillator produces the sawtooth wave that drives the deflection circuits?

To answer your question about the RCA amateur iconoscope tube, the insulator was in fact transparent, and I should have mentioned that it was. The article I read didn't give any details as to what it was made from. (It also mentions that the signal electrode was transparent, but doesn't give any details about that, either.) This wasn't the only difference, however: for one, it only produced a 120-line raster without interlacing, and the secondary anode voltage was only about 600V, so perhaps that has something to do with it?
Could you send me the publishing details of the textbook you have quoted? I'll try to see if I can get a copy from Bookfinder.com.
As always, thanks for your help.
PS: Oh hang on, there's one more I have thought of: it seems that in the early days of TV, to create the effect of a dark space like an office in the middle of the night, they used the normal blinding light but simply reduced the gain on the camera signal. I suppose that it's the equivalent of Day For Night shooting in film. Does anyone know how convincing this was? If it's anything like its cinema counterpart it would have been dreadful, but you never can tell.

[old_tv_nut]

Originally Posted by Rinehart

Ah very good, thank you for that. As I said, I suspected that I had made a fundamental mistake, and there it was: the electrons don't pass through the insulator. I had always envisaged it as something like what you get with a spark plug or spark-gap transmitter, when the voltage rises high enough the electrons jump from one electrode to the other. Of course this produces a lot of heat, but through the miracle of magic thinking, I assumed that didn't matter.
So let me see if I've understood:
1) in targets with photoemissive materials, light falling on them creates free electrons which escape from the material altogether, the number of which depends on the intensity of the light.
<YES>
2) these electrons eventually do settle back on the material, almost all of them at some other point.
<Yes, in the iconoscope. In the image orthicon they are propelled to the target and do not resettle on the photocathode.>
3) this process is constantly occurring all over the plate, with the result that on average the distribution of electric charge over the plate as a whole corresponds to the image. (Perhaps the expression "electron image" comes from this?)
4) the purpose of the electron beam is to deliver a constant electric current to the mosaic, one that can move across it without needing a physical conduit, like an electric wire.
<The action is to deliver a constant current. It was probably the original intention to deliver a variable current depending on the charge on each element, but it didn't work that way.>
5) at the moment that the beam strikes the mosaic, emitted electrons from other points on the mosaic can't settle on the point the beam strikes. This causes a discontinuity; the amount of charge that can't settle there corresponds to the number of photoelectrons released at that point.
< At the moment the beam strikes an element of the mosaic, the element emits a number of secondary electrons that is influenced by the charge (voltage) on that element. The variation in this secondary emission is unfortunately much less than the total charge that an element could accumulate if there were no electrons raining back on it the rest of the time.>
6) this causes current flow to the signal plate, and because the beam moves over the mosaic in a very precise way, the signal is a representation of the electron image.
<The instantaneous change in charge causes current to flow to the signal plate...>
7) unfortunately the scanning beam itself can cause electrons to be liberated from the mosaic, and these are called secondary electrons. These also are sent to the signal plate, resulting in a lot of signal noise.
< It's not just that the secondary "rain" causes noise, but it also cancels out most of the signal since it discharges the mosaic elements to a great extent.>
8) to reduce the number of secondaries sent to the signal plate, there is a secondary electron collector which will rout them to the anode to complete the circuit instead of being sent to the amplifier and then to the transmitter.
<yes>

Can I ask you a couple more questions? The first concerns an unimportant problem in an oscilloscope and doesn't pertain to a camera tube, but which is a vital problem for a picture tube: how did they get around the problem of varying the intensity of the electron beam without changing the cross-sectional area of it?

<Careful design. Sorry to be flippant, but gun design is very complex and I would have to study it further to give an answer of HOW the desired spot distribution is obtained. No design is perfect, and spot size does change with current, but the effect should be minimized. The other aspect that should be minimized is any extended halo, which would produce a fog around highlights. Oscilloscope guns typically have a good central spot, but also have a halo that makes them unsuitable for pictorial applications.>

And secondly, what kind of oscillator produces the sawtooth wave that drives the deflection circuits?
< The power stages are different for electrostatic vs. magnetic deflection. Horizontal magnetic deflection in particular depends on using the yoke inductance to turn a voltage pulse into a current sawtooth. Vertical and and horizontal electrostatic deflection typically use a linear amplifier. Vertical magnetic deflection is basically linear, but will include some pulse voltage to overcome the yoke inductance. Several different types of oscillators are possible to drive the power stages.>

To answer your question about the RCA amateur iconoscope tube, the insulator was in fact transparent, and I should have mentioned that it was. The article I read didn't give any details as to what it was made from. (It also mentions that the signal electrode was transparent, but doesn't give any details about that, either.) This wasn't the only difference, however: for one, it only produced a 120-line raster without interlacing, and the secondary anode voltage was only about 600V, so perhaps that has something to do with it?
Could you send me the publishing details of the textbook you have quoted? I'll try to see if I can get a copy from Bookfinder.com.
As always, thanks for your help.
PS: Oh hang on, there's one more I have thought of: it seems that in the early days of TV, to create the effect of a dark space like an office in the middle of the night, they used the normal blinding light but simply reduced the gain on the camera signal. I suppose that it's the equivalent of Day For Night shooting in film. Does anyone know how convincing this was? If it's anything like its cinema counterpart it would have been dreadful, but you never can tell.

answers interpolated between < >