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I am a bit confused on a few things regarding photodiodes. Here are two circuits that I am referring to:

Circuit1

or  

Circuit 2

For the second circuit the capacitor is not of importance.

Here are my fundamental questions:

  1. Why do we need an external voltage?
  2. Why do we need a resistor?
  3. What is exactly happening to the resistor when light hits the photodiode?
  4. Why is the voltage measured between the photodiode and the resistor?
  5. Why doesn't the external voltage influence the voltage from the photodiode (assuming the voltage from the photodiode is relatively low compared to the external voltage)?
  6. For the second photo: what would be the need for a capacitor here? What would it change?

I don't expect all the questions from one user to be answered but I would definitely appreciate some insight or perhaps some recommendations what I should read to fill in some of the gaps. I know the questions seem rudimentary but I need those "Aha!" moments right about now.

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  • \$\begingroup\$ This is a very crude circuit (essentially unusable) and is not the circuit you use for photodiodes. You need a transimpedance amplifier circuit. \$\endgroup\$
    – DKNguyen
    Mar 22, 2019 at 19:51
  • \$\begingroup\$ But to answer your questions: (1) You don't. But you can operate photodiodes in a reverse-bias which decreases their capacitance at the expense of noise for higher bandwidth (higher speed) operation. It's pretty much unusable in this circuit though. (2) The resistor is supposed to convert the photodiode current to a voltage that you can read. (3) Because you are measuring the voltage across the resistor. (5) See 2. (6) To reduce noise since AC signals see the cap as a short so bypass the resistor to ground (and thus do play into the voltage reading when you measure voltage across the resistor) \$\endgroup\$
    – DKNguyen
    Mar 22, 2019 at 19:55

2 Answers 2

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Why do we need an external voltage?

Because it makes the photodiode more sensitive, it lowers its capacitance, and it makes its capacitance more predictable.

Why do we need a resistor?

Because a photodiode acts like a current source in parallel with a diode. So when it is reverse-biased, it just acts like a current source.

schematic

simulate this circuit – Schematic created using CircuitLab

What is exactly happening to the resistor when light hits the photodiode? Why is the voltage measured between the photodiode and the resistor?

Current flows through the photodiode when light hits it. It then flows through the resistor and generates voltage.

Why doesn't the external voltage influence the voltage from the photodiode (assuming the voltage from the photodiode is relatively low compared to the external voltage)?

There is no "voltage from the photodiode". There is current from the photodiode, that develops voltage across the resistor.

For the second photo: what would be the need for a capacitor here? What would it change?

It shorts out high-frequency current from the photodiode, so the RC combination acts like a low-pass filter.

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  • \$\begingroup\$ "Because a photodiode acts like a current source in parallel with a diode. So when it is reverse-biased, it just acts like a current source." What do you mean in parallel with a diode? Since there is a current without the photodiode - there must also be a current flowing there (assuming no operational amplifier). Why doesn't that current have an effect on the current from the photodiode? \$\endgroup\$
    – mikanim
    Mar 22, 2019 at 20:24
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    \$\begingroup\$ How is there a current without the photodiode? Without the photodiode there's an open circuit. A photodiode acts like --- I did not say "a photodiode is". The photocurrent does have an effect on the current from the photodiode. The current from the photodiode is a tiny bit of dark current, plus the photocurrent. \$\endgroup\$
    – TimWescott
    Mar 22, 2019 at 21:55
  • \$\begingroup\$ But since the photodiode blocks the technical current, there is not current flowing through the circuit right? And since current flows in photodiodes in the blocking direction, the only current we should see is the current from the photodiode, right? \$\endgroup\$
    – mikanim
    Mar 23, 2019 at 9:20
  • \$\begingroup\$ I think you need to get a photodiode, a resistor, a battery, and a voltmeter, and do some experimentation. \$\endgroup\$
    – TimWescott
    Mar 23, 2019 at 16:03
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OK !

I have a single photodiode and it is working excellent . I find that operating it without any voltage seems to be the best. Now what I am doing is connecting it to a special circuit that is terminated @ 50 ohms. Whenever I tried using the standard test configuration circuit it did not help. So now that I am seeing excellent results running the diode in passive mode , I am wondering that in so doing if I were to parallel say around 100 of these would I see greater gain ? I do not believe that the bandwidth would decline as there is no voltage being used. Now understand this ! The diodes will be connected to a 150 db amplifier circuit. I am using just one photodiode at the moment . So this will be using the same 150 db amplifier The focus will be going from one photodiode to perhaps 50 photodiodes in parallel.

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  • \$\begingroup\$ Post this as a new question. This site is question/answer format and this is not an answer to the posted question. \$\endgroup\$
    – DKNguyen
    Jan 22, 2020 at 18:11
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    \$\begingroup\$ Dan - Welcome to the site, but you can't ask questions here. Only answers to the original question are allowed here, so this "non-answer" will be deleted. Ask a new question here - you can link to this question for context, if you want. (Also, don't include your signature, email address etc., this breaks a site rule.) Thanks \$\endgroup\$
    – SamGibson
    Jan 22, 2020 at 18:22
  • \$\begingroup\$ @Dan you can add your email to your nameplate, but you can't put it in an answer. Your post also doesn't answer the question. \$\endgroup\$
    – Voltage Spike
    Jan 22, 2020 at 23:05

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