Why Do We Need an External Voltage in Photodiode Circuits?

[mk9898]

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

or

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.

 

[phyzguy]

The way the circuit works is as follows. The diode is reverse biased, so if no light is falling on it, the current is basically zero. So the current through the resistor is zero, so the voltage across the resistor (V=IR) is zero. So all of the applied voltage is dropped across the diode. When you shine light on the diode, it causes a small current to flow which is proportional to the light intensity. So the voltage across the resistor (V=IR again) increases, and this voltage is proportional to the light intensity.

Here are my fundamental questions:

1. Why do we need an external voltage?

Without an external voltage, no current will flow.

2. Why do we need a resistor?

See above. The resistor is basically used to convert the current into a voltage for you to measure.

3. What is exactly happening to the resistor when light hits the photodiode?

See above.

4. Why is the voltage measured between the photodiode and the resistor?

See above.

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)?

Your assumption is wrong,. Typically almost all of the external voltage is dropped across the diode, and the voltage across the resistor is small.

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

The capacitor is used to filter out noise so you get a smoother signal (I think).

 

[mk9898]

The way the circuit works is as follows. The diode is reverse biased, so if no light is falling on it, the current is basically zero. So the current through the resistor is zero, so the voltage across the resistor (V=IR) is zero. So all of the applied voltage is dropped across the diode. When you shine light on the diode, it causes a small current to flow which is proportional to the light intensity. So the voltage across the resistor (V=IR again) increases, and this voltage is proportional to the light intensity.

Is this because we place the photodiode so that the passage direction is blocked? I.e. no current will flow based on the nature of diodes? Just to be clear: are we talking about technical direction or electrons? From the first photo I would assume technical current flow since the diode is placed so that no + would go to - but just want to make sure. Also, we place the photodiode so, since we know that the current flows in the blocking direction, right? When we say current in this direction flows, do we then mean the electrons or again the technical current i.e. the holes?

Without an external voltage, no current will flow.

Why not?

See above. The resistor is basically used to convert the current into a voltage for you to measure.

So that means if we didn't have a resistor there we would have V/R = I and it would go to infinity, right? I.e. the wire would melt.

 

[berkeman]

Without an external voltage, no current will flow.

Not true. Hook an LED to a DVM and shine some light on the LED -- you get a voltage from the photocurrent flowing through the DVM's measurement resistance.

1. Why do we need an external voltage?

There's an important reason to use at least a few volts of reverse bias across the photodiode -- it has to do with bandwidth. Do a little Google searching to see if you can find a good explanation of that, and post it here.

LED under a lamp with DVM measuring the voltage (Red positive lead on the LED anode):

 

[berkeman]

And a bonus Quiz Question for @mk9898 -- Why are ICs packaged in black plastic, instead of clear plastic?

https://cdn.thomasnet.com/insights-images/976a48a1-3114-4692-9356-0fd0efbf9e7b/750px.png

 

[phyzguy]

Is this because we place the photodiode so that the passage direction is blocked? I.e. no current will flow based on the nature of diodes? Just to be clear: are we talking about technical direction or electrons? From the first photo I would assume technical current flow since the diode is placed so that no + would go to - but just want to make sure. Also, we place the photodiode so, since we know that the current flows in the blocking direction, right? When we say current in this direction flows, do we then mean the electrons or again the technical current i.e. the holes?

I don't think I understand your question. In your diagram, the current flow through the photodiode is downwards. In the wires, this means electrons are flowing upward. In the diode, whether the current flow is holes flowing downward or electrons flowing upward depends on where in the diode you are talking about and how the diode is constructed.

Why not?

berkeman is right that current will flow whether there is an applied voltage or not. I was wrong on this point. A better answer is that we want a large applied voltage across the diode so there is a large depletion region, because this increases the sensitivity of the diode.

So that means if we didn't have a resistor there we would have V/R = I and it would go to infinity, right? I.e. the wire would melt.

No of course not. If there is no resistor, the current is limited by the diode. It won't be infinite. But if you eliminate the resistor and measure the voltage across the diode, you will get a very non-linear response.

 

[CWatters]

In case its not obvious a photodiode does not work like a solar panel. It doesn't "create" a voltage like a solar panel. Instead, when you shine light on it it effectively changes its resistance.

 

[berkeman]

In case its not obvious a photodiode does not work like a solar panel. It doesn't "create" a voltage like a solar panel. Instead, when you shine light on it it effectively changes its resistance.

Is that true? Both create a photocurrent from the light...

EDIT -- I think you may be thinking of the light-dependent resistor (LDR), instead of photodiodes:

https://en.wikipedia.org/wiki/Photoresistor

 

[dlgoff]

Both create a photocurrent from the light...

Yep. Here's an image from a page from this excellent web book on http://ecee.colorado.edu/~bart/book/book/contents.htm.

 

[CWatters]

Yes you are right, I was thinking of an LDR.

 

[sophiecentaur]

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?

To sum up:
These questions could nearly all apply to virtually any simple circuit. To 'use' the signal from any transducer, you need to present the following circuit with a suitable bias (volts or current, depending) and 'source resistance'. That design assumes that the circuit is connected to a 'voltage amplifier', which will have a high but unspecific input resistance. The 100kR resistor defines the load on the diode. With no light, there is no photo current from the diode so the output volts will go to Zero. Increasing current will bring the output volts positive (1V per 10μA of photocurrent), reference to 0V, which will suit most amplifier designs.
The purpose of the C, as in many circuits, will Low Pass filter the signal volts to eliminate high frequency noise. The photo current could be very small so the front end circuit could be susceptible to RFI (sparks etc.), without the Capacitor.