Re: CCD Charge clearring, is it required?

[Tom Droege <droege@wwa.com>]

Below is typical of what  is being done in the clear cycle
of the Mark IIIs.  One can speed up the clear cycle somewhat.
But not nearly as much as one would like.  Once the wells fill 
up, which they do when the device is off or exposed to bright
light, charge migrates everywhere.  It takes some time to 
get the whole device to a stable condition.

Tom Droege

At 10:02 AM 1/28/98 -0500, you wrote:
>
>Chris wrote:
>
>> I'll raise this issue as a possible concern.  I don't know
>> if it is really a problem.  What do the CCD experts think?
>> 
>> Seems to me 40 seconds is a long time for a clear operation.
>> I've read a few places that you want to do a clear a few times
>> before each exposure.  (Is this true?)  If so This adds 80 to 
>> 120 seconds of unproductive dead time to the time between 
>> exposures.  Maybe we can just take back to back exposures and
>> not clear?
>> 
>> I am thinking the a lot of that 40 seconds is taken up by the ADC
>> delay.  I think we could skip this.  At least the Mk III driver
>> does not bother to perform a conversion while clearing.  Norman's
>> Mk III code drives the CCD at maximum rate for about three frames
>> and never touches the ADC's "start conversion" pin.
>> 
>> I'd like to clear and open the shutter very quickly so as to
>> minimize the dark current.  With a 40 second time to clear,
>> would a short integration period dark frame contain a "dark
>> gradient"?
>> 
>> Tom Droege wrote:
>> 
>> > 
>> > The only way I can see to clear the CCD is to read it out.  You
>> > probably have to do it several times at the start.  This seems to
>> > be a common problem for CCDs.  I hear grumbling from the Sloan
>> > worker bees about this.  Each read out takes 40 seconds.  You can
>> > determine when it is done by polling on the memory done bit.  Or
>> > later an interrupt.  If it works.
>> > 
>> > One does a read out by sending a message to the stamp that says
>> > "read out the CCD".  All the stamp does, is send a single pulse
>> > to the read out circuitry.  It is off and running for the next
>> > 40 seconds.
>>
>
>The time for a CCD readout is dominated by the readout of the serial
>register:  parallel shift one row, shift 1000 columns, parallel shift
>one row, shift 1000 columns, etc.  Can one not clear the imaging area
>by just running the parallel clocks alone (i.e. 1000 parallel shifts)
>followed by multiple readouts of the serial register (i.e. shift 1000
>columns, shift 1000 columns, shift 1000 columns)?  This clearing is 
>done with no data being sent to the data collector (or the data 
>collector not listening to the data sent).  Even if this method
>is not appropriate for the Sloan Drones, it might be good enough for
>the Mark IV, even if the following image(s) have a slight gradient due
>to leftover charge in the imaging area.  
>
>Roland Vanderspek
>
>
>