# Simple Question about Photodiodes [Current and Voltage]

1. Mar 13, 2005

### Crumbles

Would someone be right in saying that a photodiode produces a current as output and that to be able to make use of this current, you ABSOLUTELY need to convert it into a voltage by using a current to voltage op-amp circuit?

I've heard this from someone but to be honest doesn't make much sense to me. I mean if you get a current from the photodiode, surely it means you have a voltage present. In other words, if you have a current, it inherently implies you have a voltage and vice versa. So, I don't understand the need for a 'current to voltage' circuit unless what you want to do is amplify the signal from your photodiode.

So if anyone has any understanding of this 'current to voltage' concept, please enlighten me!

2. Mar 13, 2005

### Davorak

The photo diode can act as a current source and produce a certain amount of current(with in limitations) when it is exposed to a certain intensity of light.

A current to voltage amplifier is used be because:
1. Easier to deal with voltages
2. Simpler to build a current to voltage amplifier then a current to current amplifier.

3. Mar 13, 2005

### Crumbles

Thanks for trying to explain this Davorak. I kinda see what you mean. If I understand correctly, a current source is different to a voltage source. I think this is the bit I am confused about. Let's say you have a current source A and a voltage source V. If you connect A to a circuit, would you not inherently get a voltage across that circuit? And vice versa, if you connect a V across a circuit, would you not inherently get a current through the circuit?

Eventually I think my question boils down to: What is the difference between a voltage source and a current source if when you connect either of them to a circuit, you get both a voltage and a current?

4. Mar 13, 2005

### KingNothing

5. Mar 13, 2005

### Davorak

An Ideal Voltage source can source any amount of current. So if you connect the two leads of an Ideal Voltage source you would get infinity current :surprised . This is a physical impossibility of course as so is an voltage source.

An Ideal current source can produce any amount of voltage. So if you connect the two leads of a voltage source with an infinite resistance you would get an infinite voltage. This is also not physically possible.

At the extremes for real voltage or current source the ideal model does not work.

The real voltage source is molded by an ideal voltage source in series with a resistor R.

The real current source is molded by an ideal current source in parallel with a resistor R.

These real models are the simple ones a more complex model would have capacitors and inductors as well.

6. Mar 13, 2005

### Crumbles

Thanks for making this clear guys! I think it makes sense now.

7. Mar 13, 2005

### chroot

Staff Emeritus
A "current source" is one whose current is either constant, or varies according to some specific relationship. A "voltage source" is one whose voltage is either constant, or varies according to some specific relationship. The other value is assumed to vary based on the impedance connected to the source.

A "current signal" is a signal in which the information is encoded by the amount of current flowing. A "voltage signal" is a signal in which the information is encoded by the voltage present.

You can convert between current and voltage signals with a simple resistor. There is usually no need for an op-amp.

- Warren

Last edited: Mar 13, 2005
8. Mar 11, 2009

### Vincent90

Hi guys ! I need to convert the current received of photodiode to a voltage and I need to build sensor circuit .
What device should i use to convert current to voltage ?
Where can I get the complete circuit of Photodiode ?

9. Mar 11, 2009

### cabraham

Very good summary KingNothing. I've been describing I & V as "mutually inclusive" for the last 7 years on several forums including this one and got flamed for it. I'm glad the mutual inclusion concept is taking hold.

A photodiode, herein called PD, can operate in either mode, photovoltaic PV, or photoconductive, PC. In PV mode, the PD is terminated in an open or very high impedance and exhibits voltage source behavior where the voltage value is proportional to incident light.

In the PC mode, the PD is terminated in a very low impedance. This low Z could actually be the 2 inputs to an op amp (traditional voltage feedback topology) since the servo loop maintains a virtual short. The behavior is that of a current source where the current value is proportional to light intensity. The feedback resistor across the op amp terminals results in a voltage source output. The PD current Ipd, times Rfdbk equals Vout.

Claude

10. Mar 11, 2009

### Bob S

The basic principle of a phototube is to amplify the signals from photoelectrons leaving the photocathode and hitting the first dynode, and this signal being amplified and hitting the anode. Suppose the amplification is G and there are 1 million photoelectrons per sec. Then the output anode current is 1.6 x 10^-10 G milliamps. The is a current source, and all of the information about the number of photoelectrons is in this current. If you terminate this current in a 50 ohm resistor, the average voltage would be 8 x 10^-12 G volts. If you were integrating the current with a megohm termination, then the voltage would be 1.6 x 10^-7 G volts. I have personally used 14 stage phototubes (RCA 6810) with 10^7 gain so this would then be over a volt. In summary two things 1) all the information is in the output current, not voltage; and 2) the anode is a current source, not a voltage source.

11. Mar 11, 2009

### Vincent90

Hi Can anyone give the correct explaination of " N level of thresholding " ?
There is two mode of photodiode ( Photovoltaic mode , Photoconductive mode ) Which mode must I choose for sensing the amount of environmental light ?

12. Mar 12, 2009

### cabraham

I use the PC mode with the PD placed across the 2 input terminals of an op amp. This is a transresistance amp, and the feedback resistor times the PD current equals the output voltage. By placing the PD at virtual zero volts due to the op amp servoing the 2 inputs to near zero, the dark current is minimized, improving accuracy and dynamic range.

Claude

13. Mar 15, 2009

### Vincent90

Can anyone explain me how to measure N-level thresholding in light sensor ( photodiode ) circuit ?