Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems.

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  #1   Report Post  
Robert Wolcott
 
Posts: n/a
Default Testing a photodiode

I have a silicon photodiode and I'm unsure if it is working or not. How
would I go about testing it? Things I have done so far:

Measured the resistance - 50 megs in either orientation, regardless of
lighting condition.
Measured resistance with a 5mw diode laser shining on it.

Anything else I should do? How do these typically fail?

Thanks,
Bob


  #2   Report Post  
klasspappa[remove]
 
Posts: n/a
Default

A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current to
voltage converter...

Robert Wolcott wrote:
I have a silicon photodiode and I'm unsure if it is working or not. How
would I go about testing it? Things I have done so far:

Measured the resistance - 50 megs in either orientation, regardless of
lighting condition.
Measured resistance with a 5mw diode laser shining on it.

Anything else I should do? How do these typically fail?

Thanks,
Bob



  #3   Report Post  
Sam Goldwasser
 
Posts: n/a
Default

"klasspappa[remove]" writes:

A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current
to voltage converter...


Even simpler: Just hook it to a multimeter on mA and shine a light on it.

With your laser diode, the output should be somewhere around 0.2 to 0.5 mA/mW.

For more accuracy, reverse bias it with a few V and measure the current.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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Robert Wolcott wrote:
I have a silicon photodiode and I'm unsure if it is working or not.
How would I go about testing it? Things I have done so far:
Measured the resistance - 50 megs in either orientation, regardless
of lighting condition.
Measured resistance with a 5mw diode laser shining on it.
Anything else I should do? How do these typically fail?
Thanks,
Bob

  #4   Report Post  
Robert Wolcott
 
Posts: n/a
Default

I'll give that a try. This diode is part of a relative power meter on my
yag laser. The meter is currently not working and I'm not sure if it is the
diode or the circuit. The circuit can be seen at:

http://oregonstate.edu/~wolcottr/Yag...0schematic.jpg

Tests I have run include:

Measured the voltage supplied to the diode via BNC connector and it is 5.0
volts
placed a 1k resistor across the BNC connector and the power meter was
pegged.
Hooked up the photodiode to the circuit and shined a 5mw diode laser into it
and there was no reading (direct illumination).

Does this sound about right?

Thanks,
Bob




"Sam Goldwasser" wrote in message
...
"klasspappa[remove]" writes:

A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current
to voltage converter...


Even simpler: Just hook it to a multimeter on mA and shine a light on it.

With your laser diode, the output should be somewhere around 0.2 to 0.5
mA/mW.

For more accuracy, reverse bias it with a few V and measure the current.

--- sam | Sci.Electronics.Repair FAQ Mirror:
http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:
http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:
http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the
excessive
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Robert Wolcott wrote:
I have a silicon photodiode and I'm unsure if it is working or not.
How would I go about testing it? Things I have done so far:
Measured the resistance - 50 megs in either orientation, regardless
of lighting condition.
Measured resistance with a 5mw diode laser shining on it.
Anything else I should do? How do these typically fail?
Thanks,
Bob



  #5   Report Post  
Sam Goldwasser
 
Posts: n/a
Default

"Robert Wolcott" writes:

I'll give that a try. This diode is part of a relative power meter on my
yag laser. The meter is currently not working and I'm not sure if it is the
diode or the circuit. The circuit can be seen at:

http://oregonstate.edu/~wolcottr/Yag...0schematic.jpg

Tests I have run include:

Measured the voltage supplied to the diode via BNC connector and it is 5.0
volts


Is the input 5 VDC or 12 VDC? 5 VDC wouldn't peg a 1 mAFS meter since
there is a 10K current limiting resistor.

placed a 1k resistor across the BNC connector and the power meter was
pegged.


Sounds consistent with 12 VDC input and the circuit working properly.

Hooked up the photodiode to the circuit and shined a 5mw diode laser into it
and there was no reading (direct illumination).


Do the tests above but sounds like it's bad.

Does this sound about right?


Yep. You can also test the photodiode on the diode check range of your
DMM. It should test like a silicon diode. I bet it tests open both ways.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
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  #6   Report Post  
Robert Wolcott
 
Posts: n/a
Default

I measured the voltage across the bnc connector (disconnected from diode)
again and it was 5.015 volts. The diode was open in both directions with my
diode tester (Fluke 177). What usually causes these to fail?

Thanks,
Bob


"Sam Goldwasser" wrote in message
...
"Robert Wolcott" writes:

I'll give that a try. This diode is part of a relative power meter on my
yag laser. The meter is currently not working and I'm not sure if it is
the
diode or the circuit. The circuit can be seen at:

http://oregonstate.edu/~wolcottr/Yag...0schematic.jpg

Tests I have run include:

Measured the voltage supplied to the diode via BNC connector and it is
5.0
volts


Is the input 5 VDC or 12 VDC? 5 VDC wouldn't peg a 1 mAFS meter since
there is a 10K current limiting resistor.

placed a 1k resistor across the BNC connector and the power meter was
pegged.


Sounds consistent with 12 VDC input and the circuit working properly.

Hooked up the photodiode to the circuit and shined a 5mw diode laser into
it
and there was no reading (direct illumination).


Do the tests above but sounds like it's bad.

Does this sound about right?


Yep. You can also test the photodiode on the diode check range of your
DMM. It should test like a silicon diode. I bet it tests open both ways.

--- sam | Sci.Electronics.Repair FAQ Mirror:
http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:
http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:
http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the
excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above
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  #7   Report Post  
Sam Goldwasser
 
Posts: n/a
Default

"Robert Wolcott" writes:

I measured the voltage across the bnc connector (disconnected from diode)
again and it was 5.015 volts. The diode was open in both directions with my
diode tester (Fluke 177). What usually causes these to fail?


They can just fail on their own. In that circuit, current is safely
limited by the 10K resistor. Not much other than excessive current
or excessive reverse voltage could damage a low speed photodiode.

Just install any cheap photodiode with a large enough active area and
see if it works. A $2 one from Digikey would likely work just fine.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above is
ignored unless my full name is included in the subject line. Or, you can
contact me via the Feedback Form in the FAQs.

Thanks,
Bob


"Sam Goldwasser" wrote in message
...
"Robert Wolcott" writes:

I'll give that a try. This diode is part of a relative power meter on my
yag laser. The meter is currently not working and I'm not sure if it is
the
diode or the circuit. The circuit can be seen at:

http://oregonstate.edu/~wolcottr/Yag...0schematic.jpg

Tests I have run include:

Measured the voltage supplied to the diode via BNC connector and it is
5.0
volts


Is the input 5 VDC or 12 VDC? 5 VDC wouldn't peg a 1 mAFS meter since
there is a 10K current limiting resistor.

placed a 1k resistor across the BNC connector and the power meter was
pegged.


Sounds consistent with 12 VDC input and the circuit working properly.

Hooked up the photodiode to the circuit and shined a 5mw diode laser into
it
and there was no reading (direct illumination).


Do the tests above but sounds like it's bad.

Does this sound about right?


Yep. You can also test the photodiode on the diode check range of your
DMM. It should test like a silicon diode. I bet it tests open both ways.

  #8   Report Post  
Bradley1234
 
Posts: n/a
Default


"klasspappa[remove]" wrote in message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or decrease it.
When the p type and n type silicon meet, the charges mix to form a neutral
region which is resistive. if the applied electricity is on one side it
makes that neutral region expand out, if its the other way the neutral
region is compressed

But silicon has unique properties where it can convert energy from light or
heat into current. In the photodiode's case if I recall, it can use light
or heat to control that neutral region


  #9   Report Post  
Sam Goldwasser
 
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Default

"Bradley1234" writes:

"klasspappa[remove]" wrote in message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or decrease it.
When the p type and n type silicon meet, the charges mix to form a neutral
region which is resistive. if the applied electricity is on one side it
makes that neutral region expand out, if its the other way the neutral
region is compressed


Not really.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above is
ignored unless my full name is included in the subject line. Or, you can
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  #10   Report Post  
Bradley1234
 
Posts: n/a
Default


"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"klasspappa[remove]" wrote in message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current

to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or decrease

it.
When the p type and n type silicon meet, the charges mix to form a

neutral
region which is resistive. if the applied electricity is on one side it
makes that neutral region expand out, if its the other way the neutral
region is compressed


Not really.


Then please to explain me how silicon diodes work? All this time I thought
it was the depletion region but its something else?

If my simplified explanation of the solid state physics involved in PN
junctions is incorrect Im grateful to learn where Im wrong.



--- sam | Sci.Electronics.Repair FAQ Mirror:

http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:

http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:

http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the

excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above

is
ignored unless my full name is included in the subject line. Or, you can
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  #11   Report Post  
Sam Goldwasser
 
Posts: n/a
Default

"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"klasspappa[remove]" wrote in message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a current

to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or decrease

it.
When the p type and n type silicon meet, the charges mix to form a

neutral
region which is resistive. if the applied electricity is on one side it
makes that neutral region expand out, if its the other way the neutral
region is compressed


Not really.


Then please to explain me how silicon diodes work? All this time I thought
it was the depletion region but its something else?

If my simplified explanation of the solid state physics involved in PN
junctions is incorrect Im grateful to learn where Im wrong.


Please Google "Photodiode principles". There were 33,100 hits but
the first one is adequate. Thanks.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above is
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  #12   Report Post  
Bradley1234
 
Posts: n/a
Default


"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"klasspappa[remove]" wrote in

message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a

current
to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or

decrease
it.
When the p type and n type silicon meet, the charges mix to form a

neutral
region which is resistive. if the applied electricity is on one

side it
makes that neutral region expand out, if its the other way the

neutral
region is compressed

Not really.


Then please to explain me how silicon diodes work? All this time I

thought
it was the depletion region but its something else?

If my simplified explanation of the solid state physics involved in PN
junctions is incorrect Im grateful to learn where Im wrong.


Please Google "Photodiode principles". There were 33,100 hits but
the first one is adequate. Thanks.


I did, and it still explains similarly to what I said. Ive worked on the
design of semiconductors, lithography, processing, doping, circuits design,
and enough circuit testing to be content with, and have trained and shared,
possibly the wrong information with dozens and dozens of people. maybe
more.

I enjoy studying, but to say just go do a google search? Id expect any such
searches will simply verify and validate what I said.



--- sam | Sci.Electronics.Repair FAQ Mirror:

http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:

http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:

http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the

excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above

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  #13   Report Post  
Bradley1234
 
Posts: n/a
Default

Since there isnt an edit to these things, what can happen over the years is
that topics we used to work in daily, if we havent done significant work in
them for a while we can lose track of certain details. While I strive to be
accurate I realize its possible that Im mistaken. having a background
itself does not imply accuracy, only to say that Ive studied something in
it.

If a person tells me Im mistaken? Im grateful but will appreciate knowing
how so we can examine details and I can correct my understanding. Im the
type who will buy lunch/dinner for that favor, or shave my head in public if
Im wrong, etc...gotta make it interesting and funny.

The first thing to go is your eyesight, the second thing is your memory, and
I cant remember what the third thing was




Please Google "Photodiode principles". There were 33,100 hits but
the first one is adequate. Thanks.

--- sam | Sci.Electronics.Repair FAQ Mirror:

http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:

http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:

http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the

excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above

is
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  #14   Report Post  
Sam Goldwasser
 
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Default

"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"klasspappa[remove]" wrote in

message
...
A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a

current
to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or

decrease
it.
When the p type and n type silicon meet, the charges mix to form a
neutral
region which is resistive. if the applied electricity is on one

side it
makes that neutral region expand out, if its the other way the

neutral
region is compressed

Not really.

Then please to explain me how silicon diodes work? All this time I

thought
it was the depletion region but its something else?

If my simplified explanation of the solid state physics involved in PN
junctions is incorrect Im grateful to learn where Im wrong.


Please Google "Photodiode principles". There were 33,100 hits but
the first one is adequate. Thanks.


I did, and it still explains similarly to what I said. Ive worked on the
design of semiconductors, lithography, processing, doping, circuits design,
and enough circuit testing to be content with, and have trained and shared,
possibly the wrong information with dozens and dozens of people. maybe
more.

I enjoy studying, but to say just go do a google search? Id expect any such
searches will simply verify and validate what I said.


Did you do the search? Here is the first sentence of the first hit:

"When these diodes are exposed to photons of energy greater than 1.12 eV
(wavelength less than 1100 nm) electron-hole pairs (carriers) are created.
These photogenerated carriers are separated by the p-n junction electric
field and a current proportional to the number of electron-hole pairs
created flows through an external circuit."

That is the most simple explanation of how a photodiode works. It's not
a variable resistor but a a light to current converter.

A device like a cadmium sulfide photocell is a variable resistor, controlled
by how much light falls on it. That's totally different than a photodiode.

I agree that a lot of Web search returns are useless - in fact this one
goes on to talk about other things that aren't directly relevant, it is
a start for many thing.

I'll let others comment on whether your explanation is similar to this one
but unless I missed something, you didn't even bother to mention anything
about a photodiode, only a normal PN diode used in an electrical circuit.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above is
ignored unless my full name is included in the subject line. Or, you can
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  #15   Report Post  
Bradley1234
 
Posts: n/a
Default

Okay, way back somebody said a diod is a diod, not a resistor, so the theme
became what is a diode, not what is a photodiode.

So thats why I mentioned the electrical properties of a diode being used for
a variable resistance. Of course its not a bidirectional resistance, but by
any other name becoming a variable resistance to current flow. (yes non
linear) I described the simple effect of recombination in the depletion
region at the PN junction that changes based on bias, to which you said this
was incorrect.

In photodiodes? If the photodiode is biased and light is applied the amount
of current flow should change. Net effect its resistance has changed.

The website mentioned shows a non standard photodiode, and yes I did the
search, interesting product line



A diod is a diod, not a resistor.
See it as a current source, and put it into a opamp to form a

current
to
voltage converter...


A diode is a variable resistor.

the circuit reacts around the depletion region to increase or

decrease
it.
When the p type and n type silicon meet, the charges mix to form

a
neutral
region which is resistive. if the applied electricity is on one

side it
makes that neutral region expand out, if its the other way the

neutral
region is compressed

Not really.

Then please to explain me how silicon diodes work? All this time I

thought
it was the depletion region but its something else?

If my simplified explanation of the solid state physics involved in

PN
junctions is incorrect Im grateful to learn where Im wrong.

Please Google "Photodiode principles". There were 33,100 hits but
the first one is adequate. Thanks.


I did, and it still explains similarly to what I said. Ive worked on

the
design of semiconductors, lithography, processing, doping, circuits

design,
and enough circuit testing to be content with, and have trained and

shared,
possibly the wrong information with dozens and dozens of people. maybe
more.

I enjoy studying, but to say just go do a google search? Id expect any

such
searches will simply verify and validate what I said.


Did you do the search? Here is the first sentence of the first hit:

"When these diodes are exposed to photons of energy greater than 1.12 eV
(wavelength less than 1100 nm) electron-hole pairs (carriers) are

created.
These photogenerated carriers are separated by the p-n junction electric
field and a current proportional to the number of electron-hole pairs
created flows through an external circuit."

That is the most simple explanation of how a photodiode works. It's not
a variable resistor but a a light to current converter.

A device like a cadmium sulfide photocell is a variable resistor,

controlled
by how much light falls on it. That's totally different than a

photodiode.

I agree that a lot of Web search returns are useless - in fact this one
goes on to talk about other things that aren't directly relevant, it is
a start for many thing.

I'll let others comment on whether your explanation is similar to this one
but unless I missed something, you didn't even bother to mention anything
about a photodiode, only a normal PN diode used in an electrical circuit.

--- sam | Sci.Electronics.Repair FAQ Mirror:

http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:

http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites:

http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the

excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above

is
ignored unless my full name is included in the subject line. Or, you can
contact me via the Feedback Form in the FAQs.





  #16   Report Post  
Sam Goldwasser
 
Posts: n/a
Default

"Bradley1234" writes:

Okay, way back somebody said a diod is a diod, not a resistor, so the theme
became what is a diode, not what is a photodiode.

So thats why I mentioned the electrical properties of a diode being used for
a variable resistance. Of course its not a bidirectional resistance, but by


In no way, shape, or form, is a diode generally used like a resistor or
behaves like a resistor.

any other name becoming a variable resistance to current flow. (yes non
linear) I described the simple effect of recombination in the depletion
region at the PN junction that changes based on bias, to which you said this
was incorrect.


I said: "Not really" as an explanation of how a photodiode worked. And while
it some reasonable information about what happens when voltage is applied
to a PN junction with respect the electrons and holes and width of the
depletion region, etc., I doubt anyone who didn't already know how a diode
worked would have after reading it. There was no attempt made to relate
the textbook description to circuit behavior.

In photodiodes? If the photodiode is biased and light is applied the amount
of current flow should change. Net effect its resistance has changed.


The current is proportional to optical power and is more or less independent
of applied voltage. That's not what is normally considered to be a resistor.

The website mentioned shows a non standard photodiode, and yes I did the
search, interesting product line


Yes, but the first sentence was a good summary of how a silicon photodiode
works.

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  #17   Report Post  
Bradley1234
 
Posts: n/a
Default


"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

Okay, way back somebody said a diod is a diod, not a resistor, so the

theme
became what is a diode, not what is a photodiode.

So thats why I mentioned the electrical properties of a diode being used

for
a variable resistance. Of course its not a bidirectional resistance,

but by

In no way, shape, or form, is a diode generally used like a resistor or
behaves like a resistor.


AHA, here is a problem, a diode changes its resistance, if you say it
doesnt, then please explain the effect on the current flow while its in the
circuit. If you ignore how the process takes place internal to the PN
junction and observe the diode operates, why would it not simply be a
component that appears to change resistance?

Ive never said a diode can used as a replacement for a resistor, the
resistor is typically linear, a diode is non linear, for a voltage applied
it is not a constant resistance unlike a common resistor, but I say the
diodes reaction is resistive.



any other name becoming a variable resistance to current flow. (yes non
linear) I described the simple effect of recombination in the depletion
region at the PN junction that changes based on bias, to which you said

this
was incorrect.


I said: "Not really" as an explanation of how a photodiode worked. And

while
it some reasonable information about what happens when voltage is applied
to a PN junction with respect the electrons and holes and width of the
depletion region, etc., I doubt anyone who didn't already know how a diode
worked would have after reading it. There was no attempt made to relate
the textbook description to circuit behavior.



Its my observation that many people do not understand WHY a diode works, but
most everyone knows the diode tends to block current in one direction and
permit it in the other. Ask them why? I offered a simple model of the
depletion region which changes based on bias, So maybe your saying not
really was saying a photodiode doesnt work that way, when I was describing a
standard diode?

Ask people why does an LED emit light? its a very interesting and hard to
explain phenomenon, I would estimate that most people do not know exactly
where the light comes from or why. and hard to explain meaning some
foundation of chemistry, minority/majority carriers, etc

What does a transistor do? its simply a controlled resistor also. Ive
known "experts" say a transistor generates power, it creates power where
none existed before, power comes out, nobody knows why, probably the crystal
creates the energy.




In photodiodes? If the photodiode is biased and light is applied the

amount
of current flow should change. Net effect its resistance has changed.


The current is proportional to optical power and is more or less

independent
of applied voltage. That's not what is normally considered to be a

resistor.

The current? so we agree that current changes. Im saying its because the
resistance changes, not because power is created inside a diode that is
adding to the circuit. Are you saying optical power is transferred right
out of the diode? It comes in as light and is converted to current?

Im saying light affects the depletion region that causes a change in
resistance, its proportional to incident optical power but assuredly not
linearly proportional



The website mentioned shows a non standard photodiode, and yes I did the
search, interesting product line


Yes, but the first sentence was a good summary of how a silicon photodiode
works.


But Im still wondering how Im wrong, perhaps we were talking apples and
oranges, I said a pn junctions resistance changes by the variable depletion
region, you said not really. So then am I still wrong? or do we agree on
theory?


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Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
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  #18   Report Post  
Sam Goldwasser
 
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"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...


The current is proportional to optical power and is more or less
independent of applied voltage. That's not what is normally
considered to be a resistor.


The current? so we agree that current changes. Im saying its because the
resistance changes, not because power is created inside a diode that is
adding to the circuit. Are you saying optical power is transferred right
out of the diode? It comes in as light and is converted to current?


So how do you explain that a photodiode can operate in photovoltaic mode
with no bias and generate power?

Im saying light affects the depletion region that causes a change in
resistance, its proportional to incident optical power but assuredly not
linearly proportional


But the current in a photodiode is very linearly related to optical power
until the device saturates.

Each photon generates one or more electron-hole pairs. That is where the
current comes from.

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  #19   Report Post  
Bradley1234
 
Posts: n/a
Default


"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...


The current is proportional to optical power and is more or less
independent of applied voltage. That's not what is normally
considered to be a resistor.


The current? so we agree that current changes. Im saying its because

the
resistance changes, not because power is created inside a diode that is
adding to the circuit. Are you saying optical power is transferred

right
out of the diode? It comes in as light and is converted to current?


So how do you explain that a photodiode can operate in photovoltaic mode
with no bias and generate power?


I dont deny this, its the effect silicon has which is "special" there is
current flow from the conversion, it doesnt need bias, but if there is bias,
the converted energy, as I understand it, is typically used to control the
size of the depletion region, which is the resistance


Im saying light affects the depletion region that causes a change in
resistance, its proportional to incident optical power but assuredly not
linearly proportional


But the current in a photodiode is very linearly related to optical power
until the device saturates.


Okay, here we are both describing a graph in general terms. If there is a
point where the device saturates, its no longer linear after that. I call
this nonlinear, if a typical photodiodes response contains a nearly linear
region? Then its sometimes very linearly related to optical power



Each photon generates one or more electron-hole pairs. That is where the
current comes from.


Well this is the ideal response, but not the actual response. To get to a
point to be able to get 1 to 1 reaction? This is the photon counter
problem. To be able count photons takes a very precise, temperature
controlled, and expensive photodiode, like APD (avalanche photodiode) which
isnt perfect. But in a general photodiode, its not necessarily efficient
where each photon gets converted.

The conversion causes current, but in typical sensor circuits Im guessing
the application is usually to control a bias current in a known linear
region, so it would be in the input to an opamp, with some bias and by the
photodiode varying resistance would provide varying current in response to
incident light, that output would feed an a/d converter and to be accurate,
it would require calibration

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http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ:

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  #20   Report Post  
Sam Goldwasser
 
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Default

"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...
"Bradley1234" writes:

"Sam Goldwasser" wrote in message
...


The current is proportional to optical power and is more or less
independent of applied voltage. That's not what is normally
considered to be a resistor.

The current? so we agree that current changes. Im saying its because

the
resistance changes, not because power is created inside a diode that is
adding to the circuit. Are you saying optical power is transferred

right
out of the diode? It comes in as light and is converted to current?


So how do you explain that a photodiode can operate in photovoltaic mode
with no bias and generate power?


I dont deny this, its the effect silicon has which is "special" there is
current flow from the conversion, it doesnt need bias, but if there is bias,
the converted energy, as I understand it, is typically used to control the
size of the depletion region, which is the resistance


The photodiode response has nothing to do with the size of the depletion
region. Changing the amount of light on a photodiode doesn't affect the
depletion region significantly. Light generates charge carriers which
pass through the depletion region.

Im saying light affects the depletion region that causes a change in
resistance, its proportional to incident optical power but assuredly not
linearly proportional


But the current in a photodiode is very linearly related to optical power
until the device saturates.


Okay, here we are both describing a graph in general terms. If there is a
point where the device saturates, its no longer linear after that. I call
this nonlinear, if a typical photodiodes response contains a nearly linear
region? Then its sometimes very linearly related to optical power


Yes, over several orders of magnitude. Even a resistor is nonlinear once
it starts smoking.

There is a fundamental difference between a device which is non-linear
and one which has a linear operating region with some limits where it
becomes non-linear.

Each photon generates one or more electron-hole pairs. That is where the
current comes from.


Well this is the ideal response, but not the actual response. To get to a
point to be able to get 1 to 1 reaction? This is the photon counter
problem. To be able count photons takes a very precise, temperature
controlled, and expensive photodiode, like APD (avalanche photodiode) which
isnt perfect. But in a general photodiode, its not necessarily efficient
where each photon gets converted.


I can go to Digikey and buy a $2 photodiode which will demonstrate linearity
over several orders of magnitude. Is it perfect? No, but darn close. Is
it a photon counter? No. But on average, it's doing a decent job of
counting large numbers of photons.

The conversion causes current, but in typical sensor circuits Im guessing
the application is usually to control a bias current in a known linear
region, so it would be in the input to an opamp, with some bias and by the
photodiode varying resistance would provide varying current in response to
incident light, that output would feed an a/d converter and to be accurate,
it would require calibration


Go do an experiment. Take a photodiode, almost any one will do, Rip it
out of your PC mouse if you like. Put a reverse bias on it of a few V
and a mA meter in the circuit. Now get your calibrated light source and
try it at various intensities. The current will be very nearly linear
with respect to optical power up a point. That point is where secondary
effects come into play. But for a typical 1x1mm active area, a couple
mW should be fine. You can try a laser pointer or LED on it.

Except for a fixed dark current (may be a uA or so), the current will
be linear with optical power.

Now remove the bias but leave the mA meter in place. Repeat the experiment.

In some applications, the photodiode feeds into the summing junction of
an op-amp, between the + and - inputs, with a feedback resistor to set
the transimpedance gain. There is no voltage across the photodiode.
(OK, there is a uV or nV across it depending on the open loop gain of
the op-amp but an ideal op-amp with infinite gain and exactly 0 V
across the photodiode would work as well.)

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
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