Talk:Phototube

Electronics Low‑importance

	Electronics portal This article is part of WikiProject Electronics, an attempt to provide a standard approach to writing articles about electronics on Wikipedia. If you would like to participate, you can choose to edit the article attached to this page, or visit the project page, where you can join the project and see a list of open tasks. Leave messages at the project talk pageElectronicsWikipedia:WikiProject ElectronicsTemplate:WikiProject Electronicselectronic articles
Low	This article has been rated as Low-importance on the project's importance scale.

Technology Start‑class

	Technology portal This article is within the scope of WikiProject Technology, a collaborative effort to improve the coverage of technology on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.TechnologyWikipedia:WikiProject TechnologyTemplate:WikiProject TechnologyTechnology articles
Start	This article has been given a rating which conflicts with the project-independent quality rating in the banner shell. Please resolve this conflict if possible.

Move proposal

This page should be combined with and then redirected to photomultiplier tube. —Preceding unsigned comment added by Aardvarkleg (talk • contribs)

Oppose – a photomultiplier tube is a more modern and rather specialized extension of the phototube; let's expand and illustrate this article instead. Dicklyon (talk) 03:17, 16 October 2009 (UTC)[reply]

Experience

It is a deeply weird feeling to hook up a phototube such as the one on the right in the illustration and to see the meter show a voltage produced by *empty space* - you can look into the tube, there's obviously nothing there - and yet there's current flowing! Mind, when I did this it was with a 10 megohm input digital multimeter, but still- it was memorable. If you get the chance, do this; it reassures your faith in E = h*nu. For some reason this freaked me out a lot more than ordinary vacuum tubes...--Wtshymanski (talk) 13:49, 9 June 2011 (UTC)[reply]

'Generation' of current.

The photoelectric effect itself may produce a small current, but this article is not about the photoelectric effect. It is about (what the Americans call) a phototube, or specifically (in English) a 'Photoemmisive cell' - that is a valve (or tube) that has a photoemissive cathode. Consulting the characteristics for all the 12 different photoemissive cells manufactured by Mullard around 1960 (both vacuum and gas filled), they all show one constant characteristic - that the anode current is zero for all levels of illumination in the absence of any voltage bias on the anode. That is to say: that the device itself does not generate any current in the absence of a voltage. The presence of a voltage causes the device to conduct current when a suitable voltage is applied between the anode and cathode dependant on the level of illumination striking the photo emissive cathode.

If one wants to be really pedantic, it can be shown that the device technically does generate a extremely small e.m.f. (and hence a current in a load resistance - or even a short circuit) in the absence of any other biasing voltage, but such a current will be extremely small and several orders of magnitude below a 'few microamps". Certainly not enough to show on a characteristic curve with a scale of 5 microamps (a typical photocell current). In the same way a thermionic diode will produce a very tiny e.m.f. when its cathode is heated (though in this case it may well be lost in leakage of the heater current). If you wish to revert the article to claim that a photoemissive cell actually generates a current of a few microamperes in the absence of any bias, then please provide a verifiable reference that trumps my reference that this is not the case. (Reference: Mullard Technical Handbook Volume 4 Section 4:Photoemissive Cells - 1960 edition).

And "... of the order of ..." is entirely the correct gramatical form. "... of the order on ..." is entirely meaningless. 109.145.22.224 (talk) 15:19, 17 April 2012 (UTC)[reply]

There's only so many ways to get electrons to jump out of a metal plate. You can apply a high voltage and strip 'em off by field emission. That's not how a phototube works. You can make the plate really hot and have thermionic emission. That's not how a phototube works. YOu can make an arc, but that's really just inciting thermionic emission. You can initiate a cascade like a Townsend discharge, but that's not how a vacuum-filled tube works either. You can shine a light on a plate and the photoelectric effect knocks electrons loose - they move, hey presto, it's generating current. Take the light away and they don't generate any current. --Wtshymanski (talk) 17:01, 17 April 2012 (UTC)[reply]

Pedantically, you are right. You can charge a capacitor (eventually) by shining ultra-violet light onto a metal plate connected to one of the terminals (having earthed the other), though whether the net charging current exceeds the inherent leakage is another matter. But as already stated, we are not discussing a single plate. We are discussing a photo emissive cell. The cell generates no current worth mentioning merely from shining a light onto its cathode. 109.145.22.224 (talk) 17:14, 17 April 2012 (UTC)[reply]

Need for a correction?

The article says "Gas filled devices are more sensitive but the frequency response to modulated illumination falls off at lower frequencies".

Shouldn't that be _higher_ frequencies? I seem to recall from about 65 years ago that phototubes for playback of movie sound tracks had frequency responses that were not much above the audio range, perhaps, rather akin to the gas-focused CRTs such as the Western Electric 224-B.

Regards, Nikevich 09:58, 6 March 2015 (UTC)