In the attached image I have calculated the current flowing as 0.11A

Can anyone advise if this is an error in the solutions, my professor has not replied to my emails so any help would be greatly appreciated.

I know that forward biased we ignore the "-1" term.
To get to the answer, I used:

Io = 2x10^-13
q = 1.6x10^-19
V = 0.7
k = 1.38x10^-23
T = 300


Thanks!

 

I also got 109 mA. I used the shortcut of taking kT/q to be 25.9 mV at 300 K.

Note that every one of the values you give has units that you neglected to include. That will haunt you some day (likely soon).

 

In the attached image I have calculated the current flowing as 0.11A

Can anyone advise if this is an error in the solutions, my professor has not replied to my emails so any help would be greatly appreciated.

I know that forward biased we ignore the "-1" term.
To get to the answer, I used:

Io = 2x10^-13
q = 1.6x10^-19
V = 0.7
k = 1.38x10^-23
T = 300


Thanks!


View attachment 176664

HI,

When i do a straightforward calculation with the model they give you (if they gave you that model) i get 0.114 amps which is close to the other calculations you both did.
However, the text hints at a more complete Spice model which includes the intrinsic carrier concentration. This makes me wonder if they did not leave that factor out of the model equation and thus we are all getting results that very "significantly" from the text. I quote the word 'significantly' because although it is a significant difference in current it would not take much of a difference to get that answer with a change in the intrinsic carrier concentration which also varies with temperature. Probably something like 2 or 3 percent.
The only way to be sure unfortunately is to get a hold of the good professor that gave this question out because it may be possible that he either gave you the wrong model or didnt care if your result varied from the text. If you cant get him then we will have to assume some things and see what we can boil out of this using pure deductive reasoning.

 

HI,

When i do a straightforward calculation with the model they give you (if they gave you that model) i get 0.114 amps which is close to the other calculations you both did.
However, the text hints at a more complete Spice model which includes the intrinsic carrier concentration. This makes me wonder if they did not leave that factor out of the model equation and thus we are all getting results that very "significantly" from the text. I quote the word 'significantly' because although it is a significant difference in current it would not take much of a difference to get that answer with a change in the intrinsic carrier concentration which also varies with temperature. Probably something like 2 or 3 percent.
The only way to be sure unfortunately is to get a hold of the good professor that gave this question out because it may be possible that he either gave you the wrong model or didnt care if your result varied from the text. If you cant get him then we will have to assume some things and see what we can boil out of this using pure deductive reasoning.

I'm pretty sure that the comment about the reverse leakage current being proportional to n_i² is intended to serve as a hint as to how to use the reverse leakage current value at 300 K to get the reverse leakage current value at 273 K.

 

Hi everyone, thank you all for your input. When I was messing about with it again I got the answer shown in the solutions, this is just an error right?

I think they have converted the top and bottom to eV but the electron charge value of 1.6x10^-19 cancels on top and bottom of the equation.

I managed to complete the rest of the question with this value, just doesn't seem correct to me!

Many thanks!

 

Hi everyone, thank you all for your input. When I was messing about with it again I got the answer shown in the solutions, this is just an error right?

I think they have converted the top and bottom to eV but the electron charge value of 1.6x10^-19 cancels on top and bottom of the equation.

I managed to complete the rest of the question with this value, just doesn't seem correct to me!

Many thanks!

View attachment 176858

Hi,

What is your reasoning, theory, or source for why they might have wanted to do that?

 

Hi,

What is your reasoning, theory, or source for why they might have wanted to do that?

All I can think is that they want to divide by the same units, I still haven't received a reply from my professor so I can only guess.

Does this seem like a mistake to you?

Thanks

 

All I can think is that they want to divide by the same units, I still haven't received a reply from my professor so I can only guess.

Does this seem like a mistake to you?

Thanks

Hello,

I guess you mean top and bottom of the exponential part?

In the top we have q*V
in the bottom we have k*T

Now what are the units (without adding anything else that you dont see in the formula) in the top, and what are they in the bottom, and what do you get when you divide the top by the bottom?
Show the base units.
q ?
V ?
k ?
T ?
q*V ?
k*T ?

 

Hi everyone, thank you all for your input. When I was messing about with it again I got the answer shown in the solutions, this is just an error right?

I think they have converted the top and bottom to eV but the electron charge value of 1.6x10^-19 cancels on top and bottom of the equation.

I managed to complete the rest of the question with this value, just doesn't seem correct to me!

Many thanks!

The problem is that they are being way too sloppy with their work.

kT/q at 300 K is not 25 mV (what you are showing in the equation that "works out"), but rather much closer to 25.9 mV.

k = 1.38064852E-23 J/K
q = 1.6021766209E-19 C
T = 300 K

kT/q = 25.85199101 mV

At the very least, they should be using 26 mV.

It may sound like a small difference, but it is a small difference in an exponent.

 

Hi,

Yeah i was wondering where they got that 0.025 from too. I guess it's just a round about value used for an exercise.

 

Hi,

Yeah i was wondering where they got that 0.025 from too. I guess it's just a round about value used for an exercise.

It's known as the thermal voltage and is a very useful and important parameter in the function of many devices.

 

It's known as the thermal voltage and is a very useful and important parameter in the function of many devices.

Hi,

Oh ha ha, you serious? If you really think i dont know that then you must have not been paying attention in class

Really it was the value that surprised me just like for you. It's usually quoted as 26mv or 25.85mv or 25.86mv or something like that. So 25mv sounded strange to me too.

 

It's known as the thermal voltage and is a very useful and important parameter in the function of many devices.

Hi,

Hey i hope you know i was kidding around with my last reply i know that these web questions and ideas come at you from out of the blue sometimes and the context is not always immediately apparent. So i still appreciate the comments and it also helps to clarify for other readers here to when it is talked about again.