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Reading the KAF-8300 CCD datasheet, I came across what looks like an emitter-follower to buffer the analog video output, see below:

enter image description here


Also from the datasheet:

Charge presented to the floating diffusion (FD) is converted into a voltage and is current amplified in order to drive off-chip loads. The resulting voltage change seen at the output is linearly related to the amount of charge placed on the FD. Once the signal has been sampled by the system electronics, the reset gate (RG) is clocked to remove the signal and FD is reset to the potential applied by reset drain (RD). Increased signal at the floating diffusion reduces the voltage seen at the output pin. To activate the output structure, an off-chip load must be added to the VOUT pin of the device. See Figure 5.


Analog isn't really my area of expertise, I do know some bits and pieces though.

So, if this actually is an emitter follower, it will have high input impedance, low output impedance, voltage gain of (approximately) unity, high current gain.

Follow-up questions:

  1. How were the resistors' values chosen?
  2. Does the Iout of 5.4mA indicate it is acting as a constant current sink?
  3. Why isn't the on-chip charge-to-voltage converter sufficient to drive off-chip loads?
  4. If necessary anyway, why do they recommend a rather crude transistor buffer rather than some sort of precision op-amp, acting as a voltage follower?


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    \$\begingroup\$ I don't have time to go in depth I'm afraid, but the suggested circuit is just an output buffer. You can drive whatever load you want from it. The 3 stage on chip amp will have transistors getting progressively larger at each stage to drive the next larger stage and eventually push signal off chip (those wire bonds and package pins, and pcb tracks represent some large resistances and capacitances compared to the stuff on the chip) so that last stage tranny will be 100s of times larger than the first stage. \$\endgroup\$ – IC_Eng Jan 3 at 16:16

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