ccdinvertersensortransistors
I'm trying to use a ccd linear image sensor (µPD3753) and I'm using the application circuit example in its datsheet, however I'm new at using these type of sensors. Can someone please explain why they're using a hex inverter and not just one (this is needed because the sensor has 2 clocks and they must be inverted in order for the device to work) also why is the pnp transistor needed.
In this answer I showed that the optimal series resistor value is
\$ R_S = \sqrt{R_{MIN} \cdot R_{MAX}} \$
where \$R_{MIN}\$ and \$R_{MAX}\$ are the minimum and maximum resistances of your sensor. The Sparkfun page speaks of between infinity and 300 kΩ. If we assume 10 MΩ as maximum value then your series resistor should be 1.8 MΩ (rounded to nearest E12 value).
These are pretty high values, too high for an ADC, which, like Mathieu says, likes a impedance of less than 10 kΩ. So you'll need to buffer the divider with an unity gain buffer opamp:
Note that a common opamp may have an input bias current as high as 1 µA, and this could distort your reading. A CMOS opamp like the MCP600x has a much lower input bias current, 1 pA typical for the MCP600x, which won't deteriorate your reading.
The optical SETI project at Harvard uses multi-pixel PM tubes (64 pixels per tube, 16 tubes total) from Hamamatsu. Would these be suitable for your project?
Best Answer
...because you need 3 (not just one; their basic function is buffering the signal and just inverting by the way) and they come in packages containing six.
...because the CCD output signal is high impedance (cannot provide much current). The transistor circuit is an emitter follower that leaves voltage about the same (just adds about 0.6V but that doesn't matter at all in this application, because the signal will have a more or less arbitray offset anyway because of "dark current") but has low output impedance (can provide much more current).