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From: basiji@milton.u.washington.edu (David Basiji)
Newsgroups: rec.audio.high-end
Subject: Re: Photo-detectors in CD players
Message-ID: <12824@uwm.edu>
Date: 6 Jun 91 12:53:28 GMT
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acgd@ihlpf.att.com (Andrew Charles) writes:


>I have some questions which may not make sense (due to a lack of technical
>background on my part). However, I'll give it a try:

>What factors affect the output voltage of a photocell such as those used
>in CD players? What factors affect the time it takes a photocell to reach
>its "expected" output?

Actually, your questions are very valid...

>Is it possible that circumstances could arise in a CD player which would
>affect the photocell sufficiently in these ways to either cause a misreading
>of a bit (wrong voltage value reached) or cause "jitter" (photocell reaction
>time is too long or too short)? If the answer to either of these is "yes",
>could these circumstances arise under normal operation if the player is
>not broken? If "jitter" is possible here, can it be large enough to affect
>the sound?

>Another way to approach this is to ask: how can a photocell fail? and if it
>does (or is degrading over time) what symptoms would one expect to see at
>the system level?

A photodiode detector like the ones used in CD players is basically a light
EMITTING diode operated in reverse.  Of course it is optimized for its use,
but the above is essentially true.  Like any simple diode, its current/voltage
characteristics are exponential, that is, as one shines light of the appropriate
wavelength on it and increases the intensity of that light the output voltage
rises from a negative value very quickly to a positive value with very little
current output.  As the intensity of the light increases, the output current
rises dramatically with very little change in output voltage above a nominal
0.6 volts or so.  What the CD's laser does then is to turn on and off a small
current source (the photodiode).  The current is passed through a resistor to
ground and the voltage is measured at the top (diode side) of the resistor to
convert the current to a voltage (V=IR).  That voltage is then passed into a 
device called a Schmidt trigger which outputs 0volts if the input is low (<1.5V)and outputs about 5Volts if the input is higher than 1.5volts.  Now the signal
is in a form useable by the digital circuitry downstream.  
	The photodiodes commonly have response times of less than 1uS which 
corresponds to a frequency limit of 1MHz.  Since the data is sampled at about
50KHz and there are 16bits per sample, one needs a minimum response time of
about 1.3uS.  Obviously, things seem to be cut pretty close, but it's not so
simple.  There'sa truism in engineering that the more you pay for a part the 
better the documentation.  Since I only have a sketchy tech sheet (I'm cheap),
I don't know any more than a single number to characterize the diode's response
rate.  In fact, response time is meaningless.  How is it responding?  Does it
have a maximum and minimum current?  Does the response time tell how long it 
takes to swing between them?  What I want is a slew rate.  This is the time it
takes the diode to change its output current by a fixed amount.  Ideally, I'd
get a graph of slew rate, since it may swing faster between two higher or lower
current values even though the difference is the same.  That way, I could 
operate the diode in its fastest region.  Or, given a slow diode, I could run
it very close to the threshold voltage for the Schmidt trigger so that the diode
wouldn't have to change its output very much to swing the Schmidt trigger.  
The problem is that you increase the noise level since it only takes a small
perturbation to turn a 1 to a 0 or vice versa.  All told, in my experience,
the diodes are faster than they need to be and if there was any problem,
it would probably come from the IC's downstream.  The best thing you can do is
to keep the diode clean so the output current is sufficient to swing the 
Schmidt trigger.

Sorry for the long winded response.

David Basiji