1. Cathode ray tube description.   

Cathode ray tubes work via two mechanisms.

(1)
Phosphorescence - the inner surface of the face of the CRT is coated with a thin film of a substance which, when bombarded with electrons, emits visible light for a certain period after the bombardment has ceased.

The light emitted can be of different frequencies = different colours. Green and white being the easiest and hence commonest.

The "persistence" also varies i.e the time taken for the intensity to fall from 100% to 10% of its initial value.

Some fast phosphors which decay quickly

     P31 .038ms      P4  .06 ms

     P1 24.5 ms;     P28 600 ms

If first dot on screen has faded out before last dot is drawn then the screen will flicker therefore persistence must be long enough to last for one screen of information to be drawn.

Phosphors must be immune to burning:

     beam power is 0.1 watts in 1/100 inch.

     -> 1000 watts /inch.

     Therefore could burn phosphor away leaving a single

     "black" dot.

       If we need deflection in 2 directions ie X,Y

       we use 2 electric fields at right angles to deflect

       a single beam of electrons.

If we want to reference some point (X,Y)=(100,230) on the screen, we must apply a voltage to the X deflection circuitry corresponding to a value of 100. Similarly, simultaneously we apply a voltage to the Y deflection circuitry corresponding to a value of 230.

We hope that the function F, which gives the deflection factor, is linear. ie (X,Y) =(100,230) is 1/2 as far from some origin (0,0) as point (X2,Y2) =(200,460).

NOTE: Most functions F are not totally linear but are linear

       over desired range.

If this is not true then a circle or a square drawn may look like:

            


    
      Fig. 2.1 : Results of Non-linear deflection function

Cathode ---> heated to produce electrons as a cloud

Control grid ---> reduces number of electrons leaving the Cathode

               = brightness control.

Accelerators ---> impart enough energy to produce light when they

               strike phosphor.

Focussing System ---> Produce a dot rather than a smear.

               Field focussing is just like lens in a camera.

               Focusses 1/100 th -inch.

Deflection system ---> This is normally magnetic through electro

               magnetic field interaction with electron stream.

               Coils used for magnetics; plates for static deflection.

Phosphor: as described above.

                 All CRT's work like this.

      


    
                  Fig. 2.2 : Cathode Ray Tube.

2. Vector Scan Display.   

Example: Lets draw a box on the screen:

      


    
                 Fig. 2.3 : Vector Scan example

Each X,Y pair is used to generate deflection voltages which will sweep the beam across the face of the screen. Where the beam strikes the phosphor visible light is produced for a certain time (persistence time).

A "move" is just a sweep of beam at zero intensity.

For example if we need to refresh every 1/30th of a second due to the phosphor and we assume that the Vector Scan Display will process draws, moves and text in approximately 100 microseconds per command then
1/30th second = 33 milliseconds = 33000 microseconds

Therefore this Vector Display can only process approximately 330 commands before flicker starts. How many commands could be processed if the phosphor were P4?

The Vector scan display can have a very high resolution (64K x 64K) since electron beam is very fine and can be positioned very accurately.

3. Direct View Storage Tube Display.   

This CRT has a slightly different construction (See Figure 2.5.). Flood cathode produces free electrons. The electron beam now writes to (i.e.stores a charge on) the storage grid:- this is semi-permanent. The resultant charge on the grid attracts the electrons produced by the flood cathode and these will pass the grid and hit the phosphor.

The finer the grid the higher the resolution of the DVST (Direct View Storage Tube). Resolutions up to 4096 x 4096 are currently available for monochrome displays. Colour displays are much more difficult but Tektronix, a leader in the field of DVST's, sells these also. The T4010 is a DVST. (This technology is pretty much dead now.)

To erase any part of the screen, we must erase all of the screen since we have to destroy the charge on the storage grid. Therefore no good for interactive work which requires selective erasure. But needs no memory nor computer power.
Example:

      


    
           Fig. 2.4 : Direct View Storage Tube example

                


    
               Fig. 2.5 : Direct View Storage Tube

4. Raster Scan Display.   

A raster scan device normally has some intelligence associated with it - a microprocessor. The hardware of the raster scan system will cause the electron beam to sweep across the face of the tube line by line. Whenever the micro's "memory" says a "dot" should be displayed, it increases the brightness of the beam from 0 to visible. The memory holds the bits of information indicating whether that co-ordinate is visible or not. Each X,Y co-ordinate corresponds to a location, either a bit, a byte (=8 bits) or a word (16 bits), in memory.

The microprocessors job is to accept commands from the host computer, interpret them and to set the bits, bytes or words in its memory in such a way that the hardware will display graphic images made up of lines or text. That is the microprocessor must have an algorithm for setting bits in its memory that will be displayed as a straight line.

        


    
            Fig. 2.6 : Video Memory of Raster Display

This device might have maximum X & Y co-ordinates of 512 X 512

  = 262144 bits of information.  = 32768  (8 bit) bytes

The first 512 bits (=64 bytes) are "mapped" onto first line.

The second 512 bits are mapped onto second line.

Therefore memory locations are an image of the screen.

If more information per pixel is required then more bits per pixel must be used.
If a pixel can have 8 levels of intensity from 0 up to fully on then it would require how many bits?
If the screen is colour and each pixel can have 256 possible colours how many bits are required to specify each pixel?

Most micro's use their own memory address space as the memory that is mapped onto screen. For colour display may need 262144 bytes! This is outside the addressing range of most early micro's but can use bit planes = 32768 bytes of memory on 8 "separate boards" which can be accessed through special hardware one at a time.

Newer 16 or 32 bit micro's could easily access this much memory and are frequently being used to control colour raster screens such as the Vectrix ( 8086 microprocessor)

Draw a box.

Bits are set semi-permanently so host computer only sends commands once.

Dots can be selectively erased.  Therefore good for animation.

      


    
                    Fig. 2.7 : Raster example

Note:  Good resolution for 256 colours is 1024 by 1024

which would be 1 Megabyte of memory!!


To clear memory in micro ie clear screen.


For i:= start display-area to end-display do memory[i]:=0;

                   (* clear screen *)

This is 1 pixel every 32 nanoseconds. Therefore memory must be capable of being read at 32 ns/ pixel. This is far too fast! However note that one read normally gives us 8 or 16 bits (1 byte or word) and hence we need 8 x 32 or 16 x 32 ns per read. This is 250 ns --> 500 ns which is quite acceptable.

Note that colour graphics devices often use a Colour Lookup Table to get the value of the intensities for the red, green and blue guns from the value stored in the microprocessors memory. This means yet another memory access into the lookup table and hence memory speeds must be even faster. How is a colour lookup table used?