Where to the electrons go from the CRT face?
 

Been wondering this for a long time. I choose a particular example to eliminate spurious reasons that I've found in my searches - the 7JP4 CRT in an Hallicrafters that uses a heavy rubber ring to hold the front of the CRT in place.

After the electrons have hit the phospor, imparted the energy from the accelleration, what is the path to complete the beam current loop? How do you not end up building up a charge on the surface that ends up deflecting and repelling the electrons?

No outer coating on the CRT. internal coating has a seperation from the phosphor.

Dang, that is a good question at first thought. But apparently, since there's no DC return path to the HV supply, the electons' total energy is being absorbed and re-emitted as light.

Its not the energy one has to think much about .... most of the
energy ends up as heat.

What probably happens to the charge (on non-aluminized tubes)
is that secondary emission initially sends more electrons out
backwards from the screen than come in forwards in the electron beam.
They are at low energy and get sucked up by the graphite coated
HV potential in the bell. This would build up a positive charge on
the phosphor, eventually resulting fewer secondary electrons,
resulting in a feedback-enforced regime of one electron in at high
energy, one back out at low energy.

Doug McDonald

There IS a DC return path, or the phosphor would eventually charge negative to the point where no electrons would land. The question here is to identify the return path, which I think is via secondary emission. I also think that in the case of a non-aluminized tube, the secondary electrons go to the internal dag, and in an aluminized tube they just go to the aluminum coating.

How about these tubes, with no internal coating:

http://www.earlytelevision.org/prewar_crts.html

http://www.fas.harvard.edu/~scidemos...eCrossCRT.html

With those examples, I guess the question is - what does the electron do once it has struck the phosphor and imparted its kinetic energy to the excitation of the phosphor atoms. Does it bounce back at a slower velocity and drift back to the nearest anode? Is there a weak surface current along the glass back to an anode surface?

'Bouce' also including secondary emission.

The current eventually returns to the most positive electrode. The unknown is what charges can accumulate on the phosphor and glass envelope.

But how does it return? does it float back to an anode? by a charge spreading on the glass? my question is - how does all that current complete the circuit from the impact on the screen to the nearest anode surface?

In electromagnetically deflected CRTs like the 10BP4 the secondary emission electrons are drawn to the internal aqua-dag coating which is connected to the HV second anode button that connects to the HV rectifier which is hooked to the flyback.

On electrostatic deflection CRTs like the 7JP4 the deflection plates have several hundred to a few thousand volts of positive potential applied to them. The negative electron beam is attracted to them and which ever plate is more positive the scanning beam will move toward(and touch if the plate is ran too positive). The secondary emission electrons are drawn to the relatively positive deflection plates and complete the beam current loop through the HV supply connected to the plates.

EDIT: (missed the last couple of posts when posting the above) The secondary emission does not need the glass to conduct it. Electrons can travel freely in a vacuum(otherwise the scanning beam could not form) so they can just take the shortest unobstructed path between the point on the screen, where they are formed by secondary emission, to the most positive element in the vicinity(the deflection plates in an electrostatic tube or the second anode in electromagnetic tube).