# Photodiode

Discussion in 'General Electronics Chat' started by crosbis2, Oct 14, 2009.

1. ### crosbis2 Thread Starter New Member

Oct 14, 2009
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A photodiode is often described as a current source in photovoltaic mode. the formula for developing a voltage across a photodiode diode is V = S. P . R
V= v out
S= photodidoe responsivity with wavelength
P= laser/ light power

What i can't understand is placing a 10k resistor as R instead of a 1k resistor will give me a 10 times increase in voltage. so for the same input light intensity I've just increased my output voltage.........great- but doesn't it violate conservation of energy, where is the catch?

2. ### steveb Senior Member

Jul 3, 2008
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Yes, it does violate the conservation of energy, hence there must be underlying assumptions. The formula is derived with the assumption that the current is forced through a transimpedance amplifier/load that will maintain about zero volts across the photodiode. A transimpedance amplifier is a powered device that converts a current source into a voltage source, hence energy conservation is not violated in reality.

If you place a simple resistor across the photodiode, the photodiode is no longer a simple ideal current source, and as the resistance is increased, the formula becomes less accurate, and eventually fails completely.

3. ### Ratch New Member

Mar 20, 2007
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crosbis2,

There is no catch. There would be if voltage were energy, but it is not. You can have a very high voltage involved with a very small amount of energy like a comb through hair. Or a low voltage involved with a large amount of energy like a automobile battery. To understand voltage better, you should read the first post of the link below.

Ratch

4. ### BillB3857 Senior Member

Feb 28, 2009
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If you put a 1 ohm resistor across a AAA dry cell you will read lower voltage from that battery than if you put a 10K resistor across it. You need to take the equivalent internal resistance of the current source.

5. ### steveb Senior Member

Jul 3, 2008
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While this is true, it is not relevant for the OP's question.

What is mentioned here is that current is S*P and voltage is S*P*R. Hence electrical power is S^2*P^2*R which can be arbitrarily large as R is increased and eventially exceeds the optical power input P.

So, the OP is correct to question the formula, but he needs to understand that it is derived under the assumption of the photodiode driving a transimpedance amplifier which has its own power input.

As BillB points out, if you just stick a resistor directly on a photodiode, you load it down, and the quoted formula does not apply in this case.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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Once again we see the effects of applying a formula outside the conditions for it to hold true.

An ideal current source can indeed produce the effect noted as it has infinite internal resistance and provides infinite voltage across the series combination of its internal resistance and the load resistance, supplying the required current in the process.

Of course a photodiode is not an ideal current source, except over a small range, as noted in the manufacturer's literature and by others here.

7. ### BillB3857 Senior Member

Feb 28, 2009
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Do you really mean infinite? I thought a perfect current source would have ZERO resistance. Theoretical generators show a coil in series with a resistance that represents the departure from a perfect generator.

8. ### steveb Senior Member

Jul 3, 2008
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Yes, studiot really meant infinite. An ideal current source has infinite resistance. A good non-nonideal current source would have a finite, but high resistance (or impedance) in parallel with the ideal source.

The generator model with a low resistance and inductance in series is better thought of as a voltage source representation. The coil is assumed to be the effective series inductance, the resistance is the copper (typically) resistance of the wire and an EMF is generated through a Faraday Law effect. Of course, a non-ideal voltage source, like this, can also be represented as an equivalent current source through a transformation. However, an ideal generator with no losses (resistive, hysteresis and eddy currents) is fundamentally a voltage source by nature. The resulting current is dictated by the load, and this current induces a reaction force/torque (i.e. through the Lorentz force law) back at the mechanical system driving the generator.

9. ### BillB3857 Senior Member

Feb 28, 2009
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Now I'm really confused. If the current from the source must return to and flow through the source, how does infinite resistance allow that flow? Does an ideal source act as a single point source with no need for return to source? If so, I can understand the concept.

10. ### steveb Senior Member

Jul 3, 2008
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I agree. It is a little confusing.

From a conceptual point of view, the definition of an ideal current source is something that provides the required current for any load. Hence, the voltage is determined by the load. The charges flow out one terminal, and all of them flow through the load. Then they all flow back in trough the other terminal. Now ask; what resistance can be in parallel with the ideal current source and not interfere with this perfect operation? The answer is an infinite resistance. If you put zero resistance across the current source, then all current flows throught the short and none through the load.

To be specific, consider the photodiode case in this thread. A reverse biased diode (the photodiode) looks like a high resistance. Light hits the active region and generates a hole-electron pairs (i.e. the charges). These charges are swept away by the high electric field from the reverse biasing voltage and they must flow back around through the load to complete the circuit. Here is an example where a high resistance source is providing current.

Another specific example is a bipolar junction transistor. The collector base junction is reverse biased and looks like a high resistance from the collector side. However, the collector current drives the load on the collector side as a current source since the current is controlled by the base emitter driving circuitry and is very insensitive to the voltage drop on the collector load, when in the linear region.

11. ### hobbyist Distinguished Member

Aug 10, 2008
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A voltage source is modeled with its internal resistance in series, with the generator.

A current source is modeled with it's internal resistance in parrallel with the generator.

12. ### steveb Senior Member

Jul 3, 2008
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There is also another way to look at the issue, and it might be more intuitive.

We usually think of current and voltage sources as sources and not loads. However either of these can be a load. Generally, we can plot voltage versus current for any load device and the slope of the curve indicates the resistance. This even works for nonlinear loads, however the slope is considered to by the dynamic (or AC) resistance.

Now, plot a graph of voltage versus current for both the constant voltage load and the constant current load. The constant voltage load has a slope of zero and the constant current load has a slope of infinity. Hence, an ideal voltage source has zero resistance and an ideal current source has infinite resistance.

Now, what about a constant power load? Plot it, and you will see that it has a negative resistance.

Last edited: Oct 16, 2009
13. ### BillB3857 Senior Member

Feb 28, 2009
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I totally agree with the parallel resistance being infinite in an ideal current source, but what about the SERIES resistance? That is the resistance I was referring to in my initial response in this thread. I'm still confused!

Feb 28, 2009
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15. ### steveb Senior Member

Jul 3, 2008
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The series resistance is infinite. Please refer to my previous post number 12 above. Make a plot of voltage versus current for an ideal current source used as a load. No matter what voltage you apply to the constant current load, the current is always the same. The slope of the line is infinite indicating infinite resistance.

The link you provided seems to be implying a voltage source which has low source resistance usually, and is zero in the ideal case. If you plot voltage versus current for a constant voltage load, the slope of the line is zero indicating zero resistance.

16. ### BillB3857 Senior Member

Feb 28, 2009
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I guess my confusion is based upon the difference between a Voltage source and a Current source. I had always considered them to be one in the same. You can have voltage without current, but can't have current without voltage.