Photoelectric Effect Saturation Current

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  • #1
embphysics
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Hello all,

I have been searching the internet for quite some time now, and have been unsuccessful in finding some article that will explain WHY there is a saturation current. All I have found is the definition of the saturation current. I don't understand why the current becomes constant as you put the emitting electrode at a greater and greater positive electric potential.
 

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  • #2
embphysics
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To add to the above post, I keep finding sources say that all of the electrons get collected on the one plate, and that is why there is a saturation current. This, however, does not make much sense.
 
  • #3
Drakkith
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As light strikes the emitting electrode, electrons are ejected. But not all of the ejected electrons are collected at collecting electrode since they don't all get ejected in the same direction. Increasing the positive potential of the collecting electrode attracts these electrons to it (the collecting electrode has a positive potential, not the emitting electrode), and after a certain amount of potential all the ejected electrons are being collected and a further increase in potential does nothing.
 
  • #4
embphysics
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Why does increasing the potential to more positive values not affect anything? From my understanding, I = V/R; so, an increase in voltage would cause an increase in current.
 
  • #5
Drakkith
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Ah, I see the issue. The current is solely from the ejected electrons being captured by the collecting electrode. The collecting electrode is simply kept at a positive potential. No current is flowing through it unless it picks up electrons ejected from the emitting electrode. So if you keep the intensity of the light striking the emitting electrode steady, then the current is also steady at some max value. Once the positive potential of the collecting electrode is enough to collect all of the ejected electrons, increasing it further does nothing as there are no more electrons to collect.
 
  • #6
ZapperZ
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Why does increasing the potential to more positive values not affect anything? From my understanding, I = V/R; so, an increase in voltage would cause an increase in current.

You have water coming out of a faucet. Initially, you had a very small container such a bottle that had a very small opening. So you could collect some of the water, but not all. Some of the water spilled out of the opening of the bottle and you couldn't collect that.

So you chose a bottle with a bigger opening. You now collect more since you lost a lot less of the water.

And you continue to get a bigger and bigger opening. Eventually, you got a bucket where you've collected everything that came out of the faucet. Now, do you think by getting an even bigger bucket, you are going to collect even more water? You've collected everything that's flowing out of the faucet per unit time as it is. Why would having a bigger and bigger bucket makes any difference?

The only difference now, if you want to get even more water per unit time, is to open the faucet more i.e. increase the intensity of light hitting the cathode, so that the amount of charge per unit time being emitted increases.

Zz.
 
  • #7
embphysics
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I don't understand, why does the electrons stay on the collecting electrode? Why don't they move through the circuit, come back to the emitting electrode, and repeat the process? And from my understanding, by increasing the voltage, you are simply increasing the rate at which this process happens, that is, the current.
 
  • #8
ZapperZ
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I don't understand, why does the electrons stay on the collecting electrode? Why don't they move through the circuit, come back to the emitting electrode, and repeat the process? And from my understanding, by increasing the voltage, you are simply increasing the rate at which this process happens, that is, the current.

What electrode? Why would electrons "stay" on an electrode?

This now appears to be a different question that you started with!

Zz.
 
  • #9
embphysics
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I am terribly sorry. I am simply trying to understand the photoelectric effect, and why a constant current is attained when you put the collecting electrode at a higher positive potential relative to the emitting electrode. Generally, whenever I have read about the photoelectric effect, the experimental set-up consists of a battery connected to two electrodes contained within a vacuum chamber.
 
  • #10
ZapperZ
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I am terribly sorry. I am simply trying to understand the photoelectric effect, and why a constant current is attained when you put the collecting electrode at a higher positive potential relative to the emitting electrode. Generally, whenever I have read about the photoelectric effect, the experimental set-up consists of a battery connected to two electrodes contained within a vacuum chamber.

It is, but I can't figure out what you are asking now. Why are you having trouble understanding even my water faucet analogy?

Zz.
 
  • #11
embphysics
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Well, in the experiment, the ammeter is what measures the current between the electrodes. But how can it even measure a current if none of the electrons ever go through ammeter, because all of the electrons are collected in the "bucket" and never go down the drain, that is, never go through to the rest of the circuit?
 
  • #12
ZapperZ
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You are missing something here.

There is a completely, closed loop of current when there's electron flowing from the anode to the cathode. Without any photoelectrons, there is an open circuit between the anode and the cathode. The batter is simply applying positive potential to the anode. But without any electrons hitting the anode, there's no current flow. Look at the circuit for the typical photoelectric effect carefully.

Zz.
 
  • #13
sophiecentaur
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Why does increasing the potential to more positive values not affect anything? From my understanding, I = V/R; so, an increase in voltage would cause an increase in current.

If you are 'stuck on' I = V/R, then this could be your problem. You cannot expect a photo diode to behave like an Ohmic Resistor (i.e. it's not a piece of metal). It can be described better as a 'current source', which means the current is (more or less) independent of the applied PD. Why? Only one electron can be emitted per photon arriving - that is the limit. The rest of the electrons on the surface of the cathode would need several kV to tear them off but the thought experiment doesn't go that far.
 
  • #14
embphysics
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Hmm, I am still not certain if I understand. What if we did not illuminate the electrode, but only applied a potential difference, would there be a current?
 
  • #15
ZapperZ
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Hmm, I am still not certain if I understand. What if we did not illuminate the electrode, but only applied a potential difference, would there be a current?

No. You have an open circuit(*).
photoel.gif


There's no connection to complete the circuit between the anode (collector plate) and cathode (emitter plate). You have a vacuum there!

Zz.

(*) This assumes that we have not gotten into the field-emission regime, which is typically the case in a standard photoelectric effect experiment.
 

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  • #16
embphysics
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Oh, so there is a gap between the anode and cathode, and the only way electrons can go from one plate to the other is if light strikes a photon? So, the positive potential can't actually generate a current, but can only "help" the electrons move more quickly?
 
  • #17
ZapperZ
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Oh, so there is a gap between the anode and cathode, and the only way electrons can go from one plate to the other is if light strikes a photon? So, the positive potential can't actually generate a current, but can only "help" the electrons move more quickly?

The electrons coming out of the cathode almost go in all different directions. Read what Drakkith said in his post (#3). If you want to collect all of them, you have to apply a potential so that the electrons are attracted to the anode. The higher the potential, the more electrons are attracted to it. You increase the potential NOT because you want the electrons to move to it quickly, but you want to collect those that are going in the 'wrong' direction away from the anode. At some point, the potential is large enough that all those stray electrons are corralled to move towards the anode.

Zz.
 
  • #18
embphysics
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Okay, I understand now. Thank you all for the help, I greatly appreciate it.
 
  • #19
dk_glysys
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if potential can stop electrons then why can't it increase it and make current more
 
  • #20
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The potential considered here was always increasing their speed - but it does not increase their number and therefore does not change their rate (=current) once all emitted electrons contribute to current flow.

This thread is from 2013, if you have more questions please open a new thread.
 
  • #21
manasi bandhaokar
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I have similar problem.if we increase potential and are able to pull the electrons to the collector then Increasing the potential further should increase their speed( i understand that their number remains constant) .if their speed is increase they should cover the circuit faster and hence in by i =q/t as t is decreasing and q remains same hence i should increase.where am i wrong?
 
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  • #22
ZapperZ
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I have similar problem.if we increase potential and r able to pull the electrons to the collector then Increasing the potential further should increase their speed( i understand that their number remains constant) .if their speed is increase they should cover the circuit faster and hence in by i =q/t as t is decreasing and q remains same hence i should increase.where am i wrong?

But at some point you are going to bump into the limit on the number of electrons per second reaching your anode because that is as many electrons that can be created per second.

Zz.
 
  • #23
Drakkith
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I have similar problem.if we increase potential and are able to pull the electrons to the collector then Increasing the potential further should increase their speed( i understand that their number remains constant) .if their speed is increase they should cover the circuit faster and hence in by i =q/t as t is decreasing and q remains same hence i should increase.where am i wrong?

Think about it this way: If the electrons are cars leaving a parking garage and going home, then even if the cars go faster, the rate at which they pass by someone sitting on the side of the road (cars per minute) is still limited by how many come out of the garage per minute. If you have 50 cars leave the garage every minute, then the maximum rate at which they can pass that person is 50 cars/min. Accelerating up to a higher velocity for their trip home puts a larger gap between each car that cancels out any gain by the increased velocity.
 
  • #24
sophiecentaur
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.if their speed is increase they should cover the circuit
They only cross the vacuum space faster. Once they get into the wire they are back at their 1mm/s speed. All that will have happened is that the 'effective length' of the circuit has changed. Nothing to get excited about and no paradoxes. :wink:
 
  • #25
manasi bandhaokar
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Got it!
thanks everyone.
 

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