Electronic – Charge transfer in CCD


I have a basic doubt about charge transfer between pixels of a CCD. Let's consider this animation (picture taken from wikipedia):

enter image description here

A precise time sequence of applied voltages between gates and bulk allows to move the charge between a pixel (represented by a MOS element) to the next one.

But this picture shows the semplified situation in which only the pixel at left has some charge inside (i.e. only that pixel has been excited from light). What would happen if all three MOS are excited with light? How can we distinguish (in the external circuitry) which pixel has been excited (and how much it has been excited) if their charges are mixed (since the charge of one pixel goes to the next one etc) during the transfer mechanism?

Best Answer

The gates shown in your diagram are not pixels. They are only demonstrating the charge transfer mechanism.

In an image sensor, only every third gate is connected to a pixel (photodiode), and the gates are driven by multi-phase clocks, such that there is always isolation (at least one gate that is "off") between one pixel's charge and the next.

This image, taken from How a Charge Coupled Device (CCD) Image Sensor Works, shows the relationship between gates and pixels:

CCD image sensor physical layout

The charges are moved top-to-bottom in the main "Image Section" part of the array, using clocks \$I\phi 1\$, \$I\phi 2\$ and \$I\phi 3\$. Then, at the bottom ("readout" section), they are moved right-to-left toward the output buffer, using clocks \$R\phi 1\$, \$R\phi 2\$ and \$R\phi 3\$.

Only one or two out of every three MOSFETs contain charge from a particular pixel at any given time.