How does carnot efficiency limit manifest itself in solar cells?

Recommended Posts

Assume you have a black body object at 1000K as a light source and an ideal* PV cell at 300K to capture that light. From a thermodynamic point of view, a maximum efficiency of 70% could be achieved (carnot limit), assuming both elements as black boxes. I am wondering how this figure manifests itself in the PV cell, and why the following reasoning is wrong:

Think of an ideal theoretical PV cell, that has following characteristics: infinite amount of junctions stacked upon to capture 100% of the spectrum, infinite diffusion lengths of charge carriers, fill factor of 1 and no resistance in any electrical element. Given this cell, every incident photon would excite an electron from the VB to the CB in the semiconductor segment that has exactly the band gap energy of the photon. Consequently, every photon (and hence all incoming energy) could be utilized to power an external circuit -> 100% efficiency.

While I am aware that it is impossible to build such a device, I'd expect to obtain the same theoretical efficiency limit via both a "black box" and a "white box" approach. Intuitively, I am not convinced by the second consideration. If this theoretical device would be put in front of the light source, it would cool down as itself emits Q~T_PV^4. I thus thought that the assumption of no recombination at all is even theoretically impossible and that some recombination must occur to release the heat required to sustain temperature. Can anyone give an intuitive explanation of what I am overlooking / getting wrong?

Share on other sites

Posted (edited)
17 minutes ago, Matthew99 said:

Assume you have a black body object at 1000K as a light source and an ideal* PV cell at 300K to capture that light. From a thermodynamic point of view, a maximum efficiency of 70% could be achieved (carnot limit), assuming both elements as black boxes. I am wondering how this figure manifests itself in the PV cell, and why the following reasoning is wrong:

I should like to see a proper thermodynamic workup of what you mean by this.

What is your working 'fluid' ?

What is your working cycle and how is the working fluid recycled ?

Why does the photocell temperature not increase ?

Edited by studiot
Share on other sites

Posted (edited)

To be honest, I am uncertain if this setup could be described in terms of a working cycle, but I found literature (https://www.intechopen.com/chapters/47490) suggesting it. Generally, the notion that no device could beat carnot efficiency seems reasonable to me, so I accepted the depiction given in above citation:

However, the "cycle" I came up with is a bit different:

In my ideal PV cell, I don't see why the working fluid, which I assume to be the electrons, would emit heat after state 3 (Q2 in first picture). The heat emitted comes from the lattice I'd think, but I have to say that I am not a physicist so this is just a layman's guess. Nevertheless, the right hand side of the second image describes of how I imagined the "cycle" of the working fluid to be.

Concerning temperature, why should it increase? Every incident photon "transfers" its energy to the outside electric circuit, so there is no heating going on inside the solar cell.

Edited by Matthew99
Share on other sites

If it’s not a thermodynamic process, you’re not going to be able to apply thermodynamic principles.  A solar cell isn’t converting heat into work - the cell will produce electricity even with no temperature difference. The article basically says this. (“we can conclude that solar cell is a quantum converter and cannot be treated as a simple solar radiation converter”)

The comparison is to a heat engine’s carnot efficiency using the same input.

Share on other sites

First of all, thank you for your replies! However, the underlying question if this ideal PV cell which I described could theoretically beat carnot efficiency is still not clear. If you're saying that efficiency reasoning for thermodynamic cycles do not apply to PV cells, that would mean that the carnot factor does not determine the theoretical upper limit of the ideal PV cell I described. If this is really true, wouldn't this mean that the device converts more energy into work than the exergy content of the incident irradiation is?

Share on other sites

11 minutes ago, Matthew99 said:

First of all, thank you for your replies! However, the underlying question if this ideal PV cell which I described could theoretically beat carnot efficiency is still not clear. If you're saying that efficiency reasoning for thermodynamic cycles do not apply to PV cells, that would mean that the carnot factor does not determine the theoretical upper limit of the ideal PV cell I described. If this is really true, wouldn't this mean that the device converts more energy into work than the exergy content of the incident irradiation is?

AFAIK the maximum PV cell efficiency is around 34%, so you aren’t describing a PV cell.

There is no way to have the light spectrum and bandgap behave as you describe.

Share on other sites

I absolutely understand that the cell I am describing is nonsense from an engineering point of view and impossible to make, but what I am interested in is if I could theoretically devise a cell that beats carnot efficiency - just from a physical perspective.

Share on other sites

42 minutes ago, Matthew99 said:

I absolutely understand that the cell I am describing is nonsense from an engineering point of view and impossible to make, but what I am interested in is if I could theoretically devise a cell that beats carnot efficiency - just from a physical perspective.

I don’t think the issue of *beating* Carnot efficiency arises, because this system is not a heat engine, I.e. it does not rely on converting heat to work. So it isn’t limited by Carnot efficiency any more than, say, a hydroelectric turbine, is it?

Share on other sites

20 hours ago, exchemist said:

I don’t think the issue of *beating* Carnot efficiency arises, because this system is not a heat engine, I.e. it does not rely on converting heat to work. So it isn’t limited by Carnot efficiency any more than, say, a hydroelectric turbine, is it?

I'd tend to disagree with that argument. Gravitational energy is pure exergy, and thus is 100% convertible into work. On the contrary, radiation is just a type of heat transfer between the energy source (i.e., the sun or the 1000K object in my example). Consequently, photons have entropy and therefore their energy should not be fully convertible into work. I am not able to describe why I think that this must be the case in a more sophisticated way than the following but this argument seems convincing to me: If you're saying that the incoming photons have an exergy content of 100% (i.e., do not carry entropy), you are also saying that an imaginary system comprised  of just an object at some temperature T and empty space will decrease in entropy over time as the object emits radiation and thereby cools down. While I am no expert on thermodynamics, I'd not subscribe to the notion that this is physically possible.

Share on other sites

12 minutes ago, Matthew99 said:

I'd tend to disagree with that argument. Gravitational energy is pure exergy, and thus is 100% convertible into work. On the contrary, radiation is just a type of heat transfer between the energy source (i.e., the sun or the 1000K object in my example).

If it's not a thermodynamic process, you can't apply thermodynamic principles to it. So unless the process of your object falling under the influence of gravity depends on the temperature, I don't see how you bring Carnot efficiency into it. It's not a heat engine.

Hydro power is ~90% efficient. There's no Carnot efficiency, since it's not a process driven by heat flow and temperature differences.

12 minutes ago, Matthew99 said:

Consequently, photons have entropy and therefore their energy should not be fully convertible into work. I am not able to describe why I think that this must be the case in a more sophisticated way than the following but this argument seems convincing to me: If you're saying that the incoming photons have an exergy content of 100% (i.e., do not carry entropy), you are also saying that an imaginary system comprised  of just an object at some temperature T and empty space will decrease in entropy over time as the object emits radiation and thereby cools down. While I am no expert on thermodynamics, I'd not subscribe to the notion that this is physically possible.

Not all photons come from thermal sources, or are involved in thermal processes. Photons in a laser, for example, will not have the same properties as the photons coming from a blackbody.

Share on other sites

Posted (edited)
3 hours ago, Matthew99 said:

I'd tend to disagree with that argument. Gravitational energy is pure exergy, and thus is 100% convertible into work. On the contrary, radiation is just a type of heat transfer between the energy source (i.e., the sun or the 1000K object in my example). Consequently, photons have entropy and therefore their energy should not be fully convertible into work. I am not able to describe why I think that this must be the case in a more sophisticated way than the following but this argument seems convincing to me: If you're saying that the incoming photons have an exergy content of 100% (i.e., do not carry entropy), you are also saying that an imaginary system comprised  of just an object at some temperature T and empty space will decrease in entropy over time as the object emits radiation and thereby cools down. While I am no expert on thermodynamics, I'd not subscribe to the notion that this is physically possible.

I’d agree that there is entropy in black body radiation, but I’d take some persuading that a solar cell is a heat engine. If it’s not a heat engine then I don’t think Carnot efficiency applies. For instance we don’t apply Carnot efficiency to the electricity production from an electrochemical cell, do we?

But you pose an interesting question. I found this about analysing the entropy in solar radiation, which seems to address what you have in mind: https://www.nature.com/articles/s41598-017-01622-6

This does not seem to make any reference to Carnot efficiency, I notice.

Edited by exchemist
Share on other sites

On 8/29/2021 at 7:14 PM, exchemist said:

I don’t think the issue of *beating* Carnot efficiency arises, because this system is not a heat engine, I.e. it does not rely on converting heat to work. So it isn’t limited by Carnot efficiency any more than, say, a hydroelectric turbine, is it?

Yes 'beating carnot efficiency' is not appropriate or meaningful in this case +1

@Matthew99FYI

Efficiency is defined as as output divided by input expressed either as a fraction or percentage.

Nothing in the basic definition refers to energy, although the ratio quantity most often used is indeed energy.

It is therefore important to always qualify a statement of efficiency by an explicit statement of the quantity or property measured.

For the three processes you describe, the appropriate terms are emissivity, absorbtivity, and conversion efficiency.

Google will help if you do not know what these mean.

Carnot efficiency refers specifically to a cyclic process in which a working fluid is taken round a (thermodynamic) cycle from one state to another ..to another.. and finally returned to its original state. In the process energy is taken form a heat reservoir at one temperature and a different quantity of energy is added to a different reservoir.
Heat reservoirs do not change temperature during this process, which is why I asked about the photocell heating up. Work is carried out as a result of this process.
Entropy is not required to increase at every stage, only as a net result of one transit around the cycle.

Your diagram shows reservoirs but no working fluid.

Share on other sites

Maybe it might not be useful to use language derived from thermodynamic working cycles, as there is no traditional working fluid which is "recycled" after the extraction of exergy involved. However: As the energy input stream is not "pure exergy" but degraded due to carrying entropy, it should not be physically possible to produce 1kW of electrical power (pure exergy) from 1kW radiation. I just found a paper supporting this argument:

Herein, the carnot limit is derived without using the analogy working fluid "recycling" (Page 6). However, while the results they obtain seem rational, I find I find it difficult to follow their reasoning which is based on relations which are unfamiliar to me as an engineer. What I find particularly bewildering, the paper suggests that the difference in Fermi energy of the solar cell is dependent on the temperature of the radiation source (i.e., the sun in their example). As far as my understanding of semiconductor physics goes, the Fermi energy should be predominately determined by doping and in an theoretically ideal (and practically impossible) solar cell identical to the lower band gap edge for the p type and identical to the upper band gap edge for the n type - thus difference in Fermi energy is equal to band gap as long as current is infinitesimally small. If I read the paper correctly, it indicates that this is not the case. Does anyone have an explanation why this might be the case and if the reasoning done in the paper mentioned is sound?

13 hours ago, exchemist said:

For instance we don’t apply Carnot efficiency to the electricity production from an electrochemical cell, do we?

No, because the exergy output is always limited to the exergy input and in electrochemical cells, the input is not heat or heat related energy but chemical energy, which is mainly comprised of exergy

16 hours ago, swansont said:

Hydro power is ~90% efficient. There's no Carnot efficiency, since it's not a process driven by heat flow and temperature differences.

Same reasoning as above, physical efficiency limit is 100% as energy input does not contain entropy and therefore output is not "obliged" to carry entropy

Share on other sites

On 8/29/2021 at 1:41 PM, Matthew99 said:

However, the "cycle" I came up with is a bit different:

In my ideal PV cell, I don't see why the working fluid, which I assume to be the electrons, would emit heat after state 3 (Q2 in first picture). The heat emitted comes from the lattice I'd think, but I have to say that I am not a physicist so this is just a layman's guess. Nevertheless, the right hand side of the second image describes of how I imagined the "cycle" of the working fluid to be.

Concerning temperature, why should it increase? Every incident photon "transfers" its energy to the outside electric circuit, so there is no heating going on inside the solar cell.

This is not a thermodynamic model of this

On 8/29/2021 at 11:31 AM, Matthew99 said:

Assume you have a black body object at 1000K as a light source and an ideal* PV cell at 300K to capture that light. From a thermodynamic point of view, a maximum efficiency of 70% could be achieved (carnot limit), assuming both elements as black boxes. I am wondering how this figure manifests itself in the PV cell, and why the following reasoning is wrong:

It is simply a discussion of one of the three participants in your proposed process.

On 8/29/2021 at 1:41 PM, Matthew99 said:

Concerning temperature, why should it increase? Every incident photon "transfers" its energy to the outside electric circuit, so there is no heating going on inside the solar cell.

You have a temperature difference of 700oK.

Why would it not increase ?

It is just plain wrong to suggest that then photcell is neither an emitter nor an absorber of radiation.

There is no electric circuit described in your opening post or your thermal model.

Forget exergy, entropy and go back to basics.

If you start from there you might be able to come up with something sensible.

Share on other sites

Posted (edited)
45 minutes ago, studiot said:

This is not a thermodynamic model of this

No it is not. As I mentioned in my last post, I don't see a working fluid being recycled in this setup, therefore the traditional approach treating it as a thermodynamic cycle might be faulty. I grant you that.

45 minutes ago, studiot said:

You have a temperature difference of 700oK.

Why would it not increase ?

As I said in my first post I assume that every incident photon excites an electron from the VB to the CB. This energy is then utilized in an external circuit. As there is no recombination of electron hole pairs and every photon "transmits" its energy to the outer circuit, what exactly is heating the cell?

45 minutes ago, studiot said:

It is just plain wrong to suggest that then photcell is neither an emitter nor an absorber of radiation.

I never claimed the opposite. Every absorbed photon is utilized as electric energy. Due to T_Cell = 300K > 0K it will emit radiation, which is why it would cool down if there is no surrounding, but lets assume the cell is placed on earth and T_Cell is constant at T_Cell=T_surrounding = 300K.

45 minutes ago, studiot said:

There is no electric circuit described in your opening post or your thermal model.

In my opening post I said:

On 8/29/2021 at 12:31 PM, Matthew99 said:

Consequently, every photon (and hence all incoming energy) could be utilized to power an external circuit

In my improvised model, I mentioned "work in outer circuit" (right hand side)

45 minutes ago, studiot said:

It is simply a discussion of one of the three participants in your proposed process.

Could you further elaborate on what you mean by that?

45 minutes ago, studiot said:

come up with something sensible

I do not see why the argument I am advocating for, which is that it is physically impossible to convert radiation to 100% work, is irrational.

Edited by Matthew99
Share on other sites

1 hour ago, studiot said:

go back to basics.

If you start from there you might be able to come up with something sensible.

If you start from basics you would not make statements such as this

22 minutes ago, Matthew99 said:

As there is no recombination of electron hole pairs and every photon "transmits" its energy to the outer circuit, what exactly is heating the cell?

Basics.

1) Heat flows from a hotter body to a colder one.

This is non negotiable.

2) General possible modes of heat transfer are:- Convection, conduction, radiation.

Radiation is the appropriate mode in this case, although since you have not properly described your system, you have not excluded the other two.

Share on other sites

Posted (edited)
23 minutes ago, studiot said:

1) Heat flows from a hotter body to a colder one.

This is non negotiable.

I am not negotiating this in any way. However, just because there is a net heat flow from the sun / 1000K object to the cell does not necessitate latter one increasing in temperature if I take out all incoming energy as work and power for example a battery behind the cell.

23 minutes ago, studiot said:

General possible modes of heat transfer are:- Convection, conduction, radiation.

Even though I never mentioned that the cell is in direct contact with the sun / 1000K object, I assumed that this would be self-evident from a) my description of the setup and b) the fact that we are talking about solar cells.

Edited by Matthew99
Share on other sites

3 hours ago, Matthew99 said:

the energy input stream is not "pure exergy" but degraded due to carrying entropy, it should not be physically possible to produce 1kW of electrical power (pure exergy) from 1kW radiation.

Energy does not "carry" entropy. Energy is a property of other things, not a substance unto itself.

...

And, to the point of some of the other discussion, not all radiation is heat. The linked paper, for example, talks of a "monochromatic sun" which is not physical. A monochromatic source is not thermal; there is no temperature you can assign it, so calculating a Carnot efficiency with that is nonsensical.

The thermodynamic treatment of the energy of a monochromatic source would be work, not heat

Share on other sites

35 minutes ago, Matthew99 said:

I am not negotiating this in any way. However, just because there is a net heat flow from the sun / 1000K object to the cell does not necessitate latter one increasing in temperature if I take out all incoming energy as work and power for example a battery behind the cell.

I have never heard such rubbish.

When I was in primary school we did an experiment where we laid out different colour objects in the sunshine and measured their respective temperatures after a few hours.

Even at the age of 9 I knew your statement to be wrong.

Share on other sites

46 minutes ago, Matthew99 said:

just because there is a net heat flow from the sun / 1000K object to the cell does not necessitate latter one increasing in temperature if I take out all incoming energy as work and power for example a battery behind the cell.

Thermodynamics precludes this possibility. You can't convert heat to work with 100% efficiency. You can't swap out work and heat; they are not identical

Share on other sites

2 minutes ago, swansont said:

Thermodynamics precludes this possibility

So it is not then a possibility.

Share on other sites

1 hour ago, Matthew99 said:

However, just because there is a net heat flow from the sun / 1000K object

Sorry, I used a wrong word - I meant net energy flow in terms of radiation. This was an unwise statement as I do not believe that the proposed cell is possible, I just wanted to point out that if it would be possible that this statement would be true.

17 minutes ago, swansont said:

You can't convert heat to work with 100% efficiency

This is exactly what I wanted to claim in the first place to support the thought experiment I proposed. Sorry if I was unclear. Radiation from the sun cannot be turned into work with 100% efficiency and thermodynamic limits do apply.

16 hours ago, studiot said:

'beating carnot efficiency' is not appropriate or meaningful in this case

Judging from that statement, I assumed you wanted to suggest that a 100% efficient solar cell does not violate the laws of thermodynamics.

32 minutes ago, studiot said:

When I was in primary school we did an experiment where we laid out different colour objects in the sunshine and measured their respective temperatures after a few hours.

Of course, real objects placed in the sun will heat up. However, this imaginary solar cell I was proposing would not do that if this device were possible, wouldn't it?

1 hour ago, swansont said:

The thermodynamic treatment of the energy of a monochromatic source would be work, not heat

Are you absolutely sure on that? I've heard that merging of low energy photons to high energy photons is possible, AFAIK people have been working on that in the past for solar cells  - so there is no law of physics preventing me from merging the whole solar spectrum into one wavelength, is there? If not, then where does the photon entropy go?

Share on other sites

Posted (edited)
27 minutes ago, Matthew99 said:

Of course, real objects placed in the sun will heat up. However, this imaginary solar cell I was proposing would not do that if this device were possible, wouldn't it?

You can't pick and choose which Physics Laws you will allow to operate and which you will suspend.

(That could be a subject for speculation, but this is Modern and Theoretical Physics)

The best you can do is to show that the effect of the Law you wish to suspend is insignificant.

Then you can ignore it.

But you have not done this.

All objects will heat up if placed near to an object at a higher temperature.

They will also radiate heat in their own right, something you originally ignored.

27 minutes ago, Matthew99 said:

Judging from that statement, I assumed you wanted to suggest that a 100% efficient solar cell does not violate the laws of thermodynamics.

I have already said that there are different definitions of 'efficiency', stemming from the basic one.

And you are still ignoring that.

I also said that two such measures are absorbtivity and emissivity, that you are ignoring.

100% absorbtivity is a theoretical possibility.

But 100% absorbtivity does not make for 100% conversion.

Swansont has already told you that any conversion cell can only convert certain frequencies by virtue of the only available conversion mechanism.

Energy at all other frequencies emitted by the emitter and absorbed by the cell will be converted to - heat.

But if you choose to define efficiency as the conversion of just radiation the correct frequency, then yes 100% conversion efficiency is possible.

But as Swansont has said several times and I have now pointed out,

Thermodynamics requires there to be a frequency spread of the radiation from a body at 1000oK.

Since this effect is highly significant at this temperature, it cannot be ignored.

Swansont also said

1 hour ago, swansont said:

The thermodynamic treatment of the energy of a monochromatic source would be work, not heat

That is why there is a Physics Quantity known as 'the work function'.

This is also linked to what is known as the threshold frequency, below which any radiation from the emitter could not generate electricity.

And we have already noted that Thermodynamics requires some of these lower frequencies to be present.

Edited by studiot
Share on other sites

Thank you first of all for the new inputs. Before addressing them, I'd like to be certain that we are agreeing on these points:

• The imaginary solar cell I described in my opening statement is physically impossible
• If it would be possible, it would not heat up from irradiation as it converts the incoming radiative energy into external work

Secondly, did l understand @swansont correctly that your suggest

• turning 1kW of solar radiation into 1kW of electricity is physically impossible, because solar radiation must be referred to as heat, leading to a thermodynamic limit
• turning 1kW of monochromatic radiation into 1kW of electricity is physically possible

?

24 minutes ago, studiot said:

They will also radiate heat in their own right, something you originally ignored.

Originally, I assumed that the cell would be placed on earths surface, in perfect contact with its surrounding which holds it at temperature.

26 minutes ago, studiot said:

I have already said that there are different definitions of 'efficiency', stemming from the basic one.

Sorry for using ambiguous language - what I meant by efficiency is energy conversion efficiency - electrical energy out divided by radiative energy in

27 minutes ago, studiot said:

But 100% absorbtivity does not make for 100% conversion.

Wouldn't 100% absorbtivity lead to every photon producing an electron hole pair in the mentioned ideal cell and thus lead to a 100% energy conversion?

31 minutes ago, studiot said:

Energy at all other frequencies emitted by the emitter and absorbed by the cell will be converted to - heat

This is of course true for real solar cells / objects. But what if you have a theoretical cell which has a band gap for every frequency?

34 minutes ago, studiot said:

But if you choose to define efficiency as the conversion of just radiation the correct frequency, then yes 100% conversion efficiency is possible.

But as Swansont has said several times and I have now pointed out,

Thermodynamics requires there to be a frequency spread of the radiation from a body at 1000oK.

Since this effect is highly significant at this temperature, it cannot be ignored.

Swansont also said

2 hours ago, swansont said:

The thermodynamic treatment of the energy of a monochromatic source would be work, not heat

That is why there is a Physics Quantity known as 'the work function'.

This is also linked to what is known as the threshold frequency, below which any radiation from the emitter could not generate electricity.

And we have already noted that Thermodynamics requires some of these lower frequencies to be present.

I'll have a closer look at this later

Share on other sites

1 hour ago, Matthew99 said:

Are you absolutely sure on that? I've heard that merging of low energy photons to high energy photons is possible, AFAIK people have been working on that in the past for solar cells  - so there is no law of physics preventing me from merging the whole solar spectrum into one wavelength, is there? If not, then where does the photon entropy go?

Merging photons is possible in some materials but it is an inefficient process, and does not work over the entire wavelength range for a given material.

16 minutes ago, Matthew99 said:

Secondly, did l understand @swansont correctly that your suggest

• turning 1kW of solar radiation into 1kW of electricity is physically impossible, because solar radiation must be referred to as heat, leading to a thermodynamic limit

I have repeatedly stated that these are not processes for which thermodynamics is applied, seeing as they are not driven by temperature differences. As such, the thermodynamic limit is not particularly relevant, which is supported by the quantum efficiency being around 34% and the Carnot efficiency being 95%. A solar cell is not a heat engine, the mental gymnastics of a few articles notwithstanding. In general you want to apply the best model to a problem, and thermodynamics is not the best model if it's not a heat engine.  At best, you can apply thermo to parts of a problem

16 minutes ago, Matthew99 said:
• turning 1kW of monochromatic radiation into 1kW of electricity is physically possible

I never said that the process would be 100% efficient

26 minutes ago, Matthew99 said:

Wouldn't 100% absorbtivity lead to every photon producing an electron hole pair in the mentioned ideal cell and thus lead to a 100% energy conversion?

No. Some absorptions would create phonons.

Create an account

Register a new account