VCE Physics Question Thread!

[Kel9901]

Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?

Thanks!

EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.

there is a lot of stuff in the electronics+photonics section of heinemann that is irrelevant. this is one of those things; phototransistors aren't on the course.

to answer your question, the formula for the energy of a single photon is E=hf, where h is Planck's constant and f is the frequency (which is also equal to speed of light/wavelength). you'll learn about it in unit 4

[Floatzel98]

quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.

The graphs for these are most interesting in the negative voltage region (whichever of photovoltaic/photoconductive this corresponds to, i cant remember) because of this light-dependent current property, and the graphs usually have multiple lines telling us what current leaks through for a certain intensity. since these lines are flat, higher reverse-bias voltages dont affect the leakage current - it is only dependent on light intensity (for an ideal photodiode anyway)

the other mode isnt as interesting and i cant remember but i think it just functions like a regular diode when in forward bias, at least that's what the graphs suggest for voltages > switch on voltage.

So, the exact same thing happens as with a normal diode in reverse bias. Except that when light hits it it kind of decreases the resistance so that a small amount of current can go through? Or does it kind of create a current from the light that hits it (Is that possible?)? I'm still kind of confused though. If we placed it in a circuit with just another resistor, with no light hitting it, it would just stop the circuit as a normal diode does since it is in reverse bias. But what happens when you add light to it then in this example? Because there is light, it lets a current through and the circuit works as normal?

Going from the IV characteristic graph of a photodiode, the positive voltage part would only ever really happen if we put the photodiode in forward bias, right? And even if you did it would just act like a normal diode. And if thats right, i'm still confused about the negative voltage region of the graph. Why are there so many different lines exactly? And they still have 'infinite' resistance like a normal reverse bias diode, but they still let current through?

there is a lot of stuff in the electronics+photonics section of heinemann that is irrelevant. this is one of those things; phototransistors aren't on the course.

to answer your question, the formula for the energy of a single photon is E=hf, where h is Planck's constant and f is the frequency (which is also equal to speed of light/wavelength). you'll learn about it in unit 4

I think i understand the energy of a photon part, but how do they introduce the Number of photons part into P = E/t? I still don't understand it too much. I know you said phototransistors aren't in the course (thanks for that), but i would still like to know how they did this if you can explain it.

Thanks guys

[silverpixeli]

So, the exact same thing happens as with a normal diode in reverse bias. Except that when light hits it it kind of decreases the resistance so that a small amount of current can go through? Or does it kind of create a current from the light that hits it (Is that possible?)? I'm still kind of confused though. If we placed it in a circuit with just another resistor, with no light hitting it, it would just stop the circuit as a normal diode does since it is in reverse bias. But what happens when you add light to it then in this example? Because there is light, it lets a current through and the circuit works as normal?

Going from the IV characteristic graph of a photodiode, the positive voltage part would only ever really happen if we put the photodiode in forward bias, right? And even if you did it would just act like a normal diode. And if thats right, i'm still confused about the negative voltage region of the graph. Why are there so many different lines exactly? And they still have 'infinite' resistance like a normal reverse bias diode, but they still let current through?

I'm not sure on the current-being-created vs current-being-allowed-through thing, my guess would be the latter though. either way it behaves the same, the current going in that part of the circuit is the photocurrent, travelling out the back of the reverse bias photodiode (so same way as voltage is applied)

the different lines each correspond to a different level of light hitting the photodiode, that line represents the I-V behaviour of the diode under that level of light (each line is usually labelled in watts/metre^2, or 'lux' which is a unit of illumination)

[Kel9901]

I think i understand the energy of a photon part, but how do they introduce the Number of photons part into P = E/t? I still don't understand it too much. I know you said phototransistors aren't in the course (thanks for that), but i would still like to know how they did this if you can explain it.

Thanks guys

Well, each photon has a certain amount of energy (E_p=hf). The total energy of the photons is the number of photons (N for the sake of this) times the individual energy of a photon- ie E=NE_p. This energy is equal to the power of the light times the time- ie E=NE_p=Pt. In this case, time is 1 second (since you want to find out the number of photons per second), so it becomes NE_p=P, or N=P/E_p.

[Floatzel98]

Well, each photon has a certain amount of energy (E_p=hf). The total energy of the photons is the number of photons (N for the sake of this) times the individual energy of a photon- ie E=NE_p. This energy is equal to the power of the light times the time- ie E=NE_p=Pt. In this case, time is 1 second (since you want to find out the number of photons per second), so it becomes NE_p=P, or N=P/E_p.

That makes a lot more sense now. Thanks a lot

I'm not sure on the current-being-created vs current-being-allowed-through thing, my guess would be the latter though. either way it behaves the same, the current going in that part of the circuit is the photocurrent, travelling out the back of the reverse bias photodiode (so same way as voltage is applied)

the different lines each correspond to a different level of light hitting the photodiode, that line represents the I-V behaviour of the diode under that level of light (each line is usually labelled in watts/metre^2, or 'lux' which is a unit of illumination)

Thanks again, but just a few more questions to clarify some stuff. So, if no light hits the photodiode, the circuit will not conduct. But the more intense the light is, more current will pass through and will work like normal diode in a circuit, except with smaller amounts of current than what comes from the power supply. It can't pull current through unless light is hitting it, and then the current level varies depending on the intensity of the light? And lastly, the obligatory, what are the used for, question. The only thing they can do is let small amounts of current through based on the intensity of the light around it. Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question. But another question kind of arises, how sensitive are photodiodes, or even LDR? Do you actually have to shine a light on them to get them to work? When you use photodiodes etc.., how is the intensity changed? Also, how long does it take. For example if i went from shining a light on a photodiode to putting my finger over the sensor (effectively not letting it conduct at all if i understand properly), how long would it take to turn off, or at least go down to a lower current? For some reason i don't think it would be very instant.

Thanks again. I should pay you since you are basically tutoring me through physics

[silverpixeli]

Thanks again. I should pay you since you are basically tutoring me through physics

I do tutor physics but I'm full at the moment so I can't take you on as a student and you'll have to keep getting it here for free, sorry,

And lastly, the obligatory, what are the used for, question. The only thing they can do is let small amounts of current through based on the intensity of the light around it. Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question. But another question kind of arises, how sensitive are photodiodes, or even LDR? Do you actually have to shine a light on them to get them to work? When you use photodiodes etc.., how is the intensity changed? Also, how long does it take. For example if i went from shining a light on a photodiode to putting my finger over the sensor (effectively not letting it conduct at all if i understand properly), how long would it take to turn off, or at least go down to a lower current? For some reason i don't think it would be very instant.

First of all, you're right that it's not instant, but photodiodes are pretty fast. I think they have microsecond response times (compare with LDRs which are only millisecond-fast, so 1000 times slower.)

In many applications all you need is a tiny signal, such as in signal transmission, you can have a small current flashing off or on to transmit a code through a circuit. That small current could later be amplified by the way, by feeding it into the base of a transistor or something.

The fast response time is super useful in receiving signals that are sent through optic fibres. These light signals flash on and off really fast (created by a flashing LED or laser at the other end) and a photodiode can grab these changes and respond accordingly, at least as fast as microseconds.

Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question.

So that's the 'dark current' you're hearing about. Evidently, the chemistry behind a photodiode means that it lets a few micro-amps through even when there's no light, so long as there's a reverse voltage across it.

As for the intensity thing, the unit of intensity is lux=W/m^2 (watts/metre^2) and remember that a watt is just a joule every second. So it's a rate of illumination of a certain area. So if you increase the area you absorb more energy every second (power) than before but it's still the same intensity. If you wait longer, you dont have more intensity because it's energy PER second.

Im not sure if this answers all your questions so keep asking for clarifications if you need.

[Floatzel98]

I do tutor physics but I'm full at the moment so I can't take you on as a student and you'll have to keep getting it here for free, sorry,

First of all, you're right that it's not instant, but photodiodes are pretty fast. I think they have microsecond response times (compare with LDRs which are only millisecond-fast, so 1000 times slower.)

In many applications all you need is a tiny signal, such as in signal transmission, you can have a small current flashing off or on to transmit a code through a circuit. That small current could later be amplified by the way, by feeding it into the base of a transistor or something.

The fast response time is super useful in receiving signals that are sent through optic fibres. These light signals flash on and off really fast (created by a flashing LED or laser at the other end) and a photodiode can grab these changes and respond accordingly, at least as fast as microseconds.

So that's the 'dark current' you're hearing about. Evidently, the chemistry behind a photodiode means that it lets a few micro-amps through even when there's no light, so long as there's a reverse voltage across it.

As for the intensity thing, the unit of intensity is lux=W/m^2 (watts/metre^2) and remember that a watt is just a joule every second. So it's a rate of illumination of a certain area. So if you increase the area you absorb more energy every second (power) than before but it's still the same intensity. If you wait longer, you dont have more intensity because it's energy PER second.

Im not sure if this answers all your questions so keep asking for clarifications if you need.

Thanks for everything i really appreciate it. I didn't mean the dark current part of a diode that lets current through with no light. I mean that (if you look at the graph in the spoiler) there is still current flowing through with 0 voltage across it. Like when there is 2mW there is always just under 1mA even in the transition from postive to negative and even at 0 voltage as well. That's why i asked about the created vs being allowed through thing before. I mean now when i think about it, even if it did make the current somehow, there would still be no voltage pulling the current. Obviously the diode isnt a battery and the light doesn't charge it up. It doesn't provide its own voltage and there is no voltage being applied but there is still current going through.

Spoiler

[Adequace]

I just got my results back for my 1/2 electricity SAC and managed to get 99%. I just wanted to say thanks for your help.

[odeaa]

I just got my results back for my 1/2 electricity SAC and managed to get 99%. I just wanted to say thanks for your help.

nice man! i got like 60 or something on that ahaha, luckily managed to get my head around it this year

[Kel9901]

I just got my results back for my 1/2 electricity SAC and managed to get 99%. I just wanted to say thanks for your help.

grats!

[paper-back]

A central loop of wire lies inside a larger loop, which is connected to a battery. Current flows around this outer loop. The resistance of the outer loop is increasing. Determine the direction of the conventional current inuded in the loop

Why and how does the resistance of the loop change the direction of an induced current?

[silverpixeli]

A central loop of wire lies inside a larger loop, which is connected to a battery. Current flows around this outer loop. The resistance of the outer loop is increasing. Determine the direction of the conventional current induced in the loop

Why and how does the resistance of the loop change the direction of an induced current?

assuming that the battery provides a fixed voltage, increasing resistance in the outer loop means that current ill be decreasing.

this does not change the direction of any induced current in the inner loop, but it does determine the direction of induced current by Lenz' law.

[paper-back]

assuming that the battery provides a fixed voltage, increasing resistance in the outer loop means that current ill be decreasing.

this does not change the direction of any induced current in the inner loop, but it does determine the direction of induced current by Lenz' law.

Thanks for responding silverpixeli

How does it determine the direction of the induced current by Lenz' law? Will the direction of the new induced current always oppose the direction of the original current prior to increasing the resistance?

[silverpixeli]

Thanks for responding silverpixeli

How does it determine the direction of the induced current by Lenz' law? Will the direction of the new induced current always oppose the direction of the original current prior to increasing the resistance?

Actually Lenz' law says that it will oppose the change of field brought about by the increased R/reduced I.

Say I in the outer loop is clockwise, then by the right hand rule the field due to I on the inside of the outer loop will be into the page.

When I decreases (R increases) this field still points into the page, but it now has less magnitude. Keep in mind that magnetic field is a vector. So which was way this field changing? It's becoming less into the page so the direction of the change is out of the page (think about it. consider to see the maths behind it)

So is the change in field is out of the page, the change in flux in the inner loop is out of the page. Flux is in the same direction as field, so this step of the argument is trivial and just comes about because I was talking about field before, now I'm talking about flux, which is just field scaled by area.

Lenz' law says that the induced current in the loop will oppose this change in flux. i.e. it will be directed so that it induces its own field to counter the change. Since the change is out of the page, we want a current that produces a field into the page. By the right hand rule again, that's a clockwise current.




This argument has a few steps but it's the kind of argument you go through for every single induction question, so it's worth taking the time to understand it.

[odeaa]

Hey guys, what is the best scientific calculator for physics? I have been using my green ti from year 7,but I find it sucks for gravity questions because you can't see the whole screen