Using a CCD camera to solve physics problems

  • #1
(Sorry for Length)
1. Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable. You can presume that all of the light of your quasar falls on one pixel. You know that a star with a magnitude of 0 would deliver 1 × 10^9 photons/second to one pixel.
The quantum efficiency of the CCD is 90%. The dark current is 2 electrons/second/pixel, and the maximum signal that a pixel can hold (the full well) is 120,000 electrons. The bias level is 1000 ADUs, and the maximum signal that the A/D converter can read is 65,535 ADUs. If the gain of the CCD is 2 electrons per ADU, what is the maximum length of time (in minutes) that you should hold the shutter of the camera open?

a) the rate at which electrons are generated by the quasar flux
b) the maximum number of electrons recordable by the A/D converter
c) the maximum number of electrons that can be accumulated in a well before it responds non-linearly to flux and the corresponding number of electrons that must be read out taking into account the bias
d) which of the two maxima limits the exposure time
e) in symbolic terms, the total signal read out from a pixel after an exposure of length t taking into account the quasar flux F (per unit time), the dark current D (per unit time), and the bias B.
f) the limiting exposure time, after setting the total signal read out to be the identified maximum

Homework Equations


m2-m1=-2.5log(F2/F1)
Electrons/second=F2*90%
A/D max=ADU*gain-bias


The Attempt at a Solution


  1. a) the rate at which electrons are generated by the quasar flux

m2-m1=-2.5log(F2/F1)
16-0=-2.5log(F2/(1*109photons/second)
F2=3.98*102 photons/second

Electrons/second=F2*90%
e/s=3.98*102 photons/second
e/s=358.2
:.The total number of electrons per second is 358.2

First I took the magnitude of the comparison star, with the magnitude of my quasar (16th) to solve for the flux of my quasar. I then multiplied that by the quantum efficiency (90%) to get my total electrons per second hitting the pixel in question.



b) the maximum number of electrons recordable by the A/D converter


A/D max=ADU*gain-bias
A/D max = 65535ADU*(1000ADU*2electrons)-2000electrons

A/D max= 129070 electrons

To find the A/D converter’s max number of electrons I took the ADU amount, multiplied it by the gain (2 electrons per ADU) and subtracted the bias of 2000 electrons.


c) the maximum number of electrons that can be accumulated in a well before it responds non-linearly to flux (refer to Chapter 3), and the corresponding number of electrons that must be read out taking into account the bias

MAX well capacity for full well
MAX well capacity=120,000 electrons-bias
MAX well capacity=118,000 electrons

*At the recomended 70%
MAX well capacity=120,000 electrons*70%-bias
MAX well capacity=82,000 electrons

Here i did two options: to have a full well, or the recommended 70% to prevent non-linear. I took the max well capacity (120,000 electrons) multiplied that by the 70% to not “overfill” the well, and then subtracted the bias (2000 electrons.)

*Point to note, i was not sure if you subtracted the bias before or after multiplying by 70%.


d) which of the two maxima limits the exposure time

The two maxima that limits the exposure time is the Max well limit and the max amount of ADU the A/D converter can read.
The max well limit because once a well fills up too much (past 70%) you can start to get too much noise/ non-linear responding which would corrupt proper data.
The gain because it sets the pace as to how much

The gain because it limits the amount of electrons per ADU. It is also an adjustable parameter, set on the basis of the nature of the research being undertaken.


e) in symbolic terms, the total signal read out from a pixel after an exposure of length t taking into account the quasar flux F (per unit time), the dark current D (per unit time), and the bias B.

TotalSignalReadout=F-D-B
TSR=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

For the total signal readout you need the electrons coming in per second (F), or “flux” subtracting the noise, or Dark images/frames and the bias.








f)the limiting exposure time, after setting the total signal read out to be the identified maximum.


TotalSignalReadout=F-D-B
TSR=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons
*Solving for s (second)*
TotalSignalReadout=F-D-B
Max=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

84000=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

s=10465.11 OR 10465

s/60=minutes
minutes=10465s/60s
minutes=174.25

Here i used the total signal read out equation, subbed in the maximum amount of electrons obsorbed to solve for s (seconds), and then convert into minutes.


CONCLUSION:


The maximum amount of time, in minutes, that you should keep the shutter open is 174.25 minutes to get an accurate CCD picture of a quasar flux that is the 16th magnitude.


I just need to know if what i came up with makes sort of sense?? I know its long, and im sorry for that. This was the first question given to me after ONE intro class so im completely lost. I dont know if i'm just bad at physics or if my prof has overestimated what we knew in review.

Thank you so much for any help!
 

Answers and Replies

  • #2
Tom.G
Science Advisor
3,704
2,387
:.The total number of electrons per second is 358.2
I agree.

b) the maximum number of electrons recordable by the A/D converter


A/D max=ADU*gain-bias
A/D max = 65535ADU*(1000ADU*2electrons)-2000electrons

A/D max= 129070 electrons

To find the A/D converter’s max number of electrons I took the ADU amount, multiplied it by the gain (2 electrons per ADU) and subtracted the bias of 2000 electrons.
The highlighted lines state different operations. Is there a typo in there somewhere? I agree with your number of electrons though.

Here i did two options: to have a full well, or the recommended 70% to prevent non-linear. I took the max well capacity (120,000 electrons) multiplied that by the 70% to not “overfill” the well, and then subtracted the bias (2000 electrons.)

*Point to note, i was not sure if you subtracted the bias before or after multiplying by 70%.
Can you explain why the bias is subtracted here? (hint: where is the bias stated to be in the problem statement?)

TotalSignalReadout=F-D-B
TSR=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons
Perhaps we are considering the TSR differently, but I'm totally confused here.
I thought the bias was taken into account in an earlier calculation.
Please verify the photons/second used here.
The dark current needs to be handled differently.

Cheers,
Tom

p.s. Nice, complete step-by-step explanation of your steps.
 
  • #3
I agree.
Good good, step one out of the way :P


The highlighted lines state different operations. Is there a typo in there somewhere? I agree with your number of electrons though.
Yeah this was a typo, it was meant to say the ADU multiplied by the gain (2 electrons per ADU) minus the bias (1000ADU*2 electrons) so:
A/D max = 65535ADU*(2electrons per ADU)-2000electrons


Can you explain why the bias is subtracted here? (hint: where is the bias stated to be in the problem statement?)
This is the part i wasnt 100% sure on, but the way i saw it: if a well gets full past 70% the data starts to not be as clear (aka it responds non linearly to the flux)
so what I did was found the full well capacity (120,000 electrons) multiplied that by 70% (to get optimal range) then still subtracted out the bias because its "filler data" that needs to be subtracted at the end, so i figured it still needed to count into the 70% because it is still technically there until we take it away anyways. (though im not sure if that makes sense, i could be completely wrong. it just intuitively made sense to me.)


QUOTE="Tom.G, post: 6055914, member: 581973"] Perhaps we are considering the TSR differently, but I'm totally confused here.
I thought the bias was taken into account in an earlier calculation.
Please verify the photons/second used here.
The dark current needs to be handled differently. [/QUOTE]

The piecewise set up kind of confused me, I took each question as a seperate thing, not a string of questions, so i included the bias in twice I suppose? now that you're mentioning it, it doesnt make sense for me to use bias twice. I think I should take it out of one of the equations but im not sure which one. (probably not this one since the word BIAS is in the question haha.)
But to fix it i would change the total MAX to consider the bias not being taken out yet ( 120,000*70%).
Photons/second (because i needed to put it in terms of FLUX (F) its pretty much just equation a) in its extended form (((1*109photons/second)*90%)), multiplied by the Dark frame rate (2 electrons/second/pixel) minus the bias as well, and then solve for s to get the max exposure time.

Again, im literally just going off of what I *think* is the way to do this, because he didnt give us any formal "this is how u go about a problem like this* and just sauced us the assignment without a single lecture :P . This is day one so im worried haha.
 
  • #4
Tom.G
Science Advisor
3,704
2,387
OK, moving right along...

Yeah this was a typo, it was meant to say the ADU multiplied by the gain (2 electrons per ADU) minus the bias (1000ADU*2 electrons) so:
A/D max = 65535ADU*(2electrons per ADU)-2000electrons
I agree.

so i included the bias in twice I suppose? now that you're mentioning it, it doesnt make sense for me to use bias twice. I think I should take it out of one of the equations but im not sure which one. (probably not this one since the word BIAS is in the question haha.)
Yes, your calculations included BIAS twice when it should occur only once. The problem statement is somewhat ambiguous as to where the bias is applied. I read the textual description as having the BIAS applied to the A/D. Your point about the BIAS being mentioned in question 'e)' and being a characteristic of the sensor is equally valid. In this particular case I don't think it affects the final result due to the various signal amplitudes and simplified circuit description. So include BIAS in whichever (one) place it better fits your thought processes.

Please verify the photons/second used here.
Photons/second (because i needed to put it in terms of FLUX (F) its pretty much just equation a) in its extended form (( (1*109photons/second)*90%))
What stellar object are you looking at? How many electrons did you calculate for it in 'a)'?

The problem text states:
The dark current is 2 electrons/second/pixel
Be aware that Dark Current is a current in addition to any signal that may be present, and just like a signal, it accumulates over the exposure time.
The Dark Current needs to be re-addressed here:
(((1*109photons/second)*90%)), multiplied by the Dark frame rate (2 electrons/second/pixel)
Cheers,
Tom
 

Related Threads on Using a CCD camera to solve physics problems

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
6
Views
2K
Top