Increase in conductivity of intrisic germanium with temperature

[PhysicsTest]

Homework Statement

Consider intrinsic germanium at room temperature 300 Degree K. By what percent does the conductivity increase per degree rise in temperature?

Relevant Equations

$n_i^2 = A_0T^3\exp\frac{-E_{G0}} {kT}$

The intrinsic concentration $n_i$ varies with T as
$n_i^2 = A_0T^3e^{\frac{-E_{G0}} {kT}}$ ---> eq1
The mobility $\mu$ varies as $T^{-m}$ over a temperature range of 100 to 400K. For Germanium, m = 1.66 (2.33) for electrons (holes) as per book.
The conductivity is given by $\sigma = (n\mu_n + p\mu_p)q$ ---> eq2
Step1: Calculate $n_i$ at 300K, 400K
Step2: Calculate the $\mu_n, \mu_p$ at 300K, 400K
Step3: Calculate the desired parameter.
When i perform Step1 i get a different value for $n_i$ than the standard value mentioned in the book for Ge $2.5*10^{13}$
Calculating $n_i$ at 300K by substituting in eq1
$n_i^2 = A_0T^3e^{\frac{-E_{G0}} {kT}}$
$A_0 = 6.022 * 10^{23}$
$T = 300$
$E_{G0} = 0.785$
$k = 1.38 * 10^{-23}$
$n_i^2 = 6.022*10^{23} * 300^3 * e^\frac{-0.785} {1.38*10^{-23}*300}$
$n_i = 6.6*10^{15}$ What is the mistake, it differs from the standard value?
Step2:
$\mu$ varies as $T^{-m}$
$m = 1.66 = m_{e300} \text{ electrons} ; m=2.33\text{ holes}=m_{n300}$
As per the standard in the book $\mu_n = 3800=\mu_{n300}; \mu_p=1800=\mu_{p300}$ at 300K
$\mu_{e300} = K_e*T_{300}^{-m_{e300}}$ ->eq3
$\mu_{n300} = K_n*T_{300}^{-m_{n300}}$ -> eq4
$K_e = \frac{\mu_{e300}} {T_{300}^{-m_{e300}}}$ --> eq5
$K_n = \frac{\mu_{n300}} {T_{300}^{-m_{n300}}}$ --> eq6
At 400K
$\mu_{e400} = K_e*T_{400}^{-m_{e400}}$ -> eq7
$\mu_{n400} = K_n*T_{400}^{-m_{n400}}$ -> eq8
Are the eq(7), eq(8) correct?

 

[PhysicsTest]

Sorry it is big mistake in equations 7 and 8. I should have taken 301Deg Kelvin instead of 400 Deg K.
At 301K
$\mu_{e301} = K_e*T_{301}^{-m_{e301}}$ -> eq7
$\mu_{n301} = K_n*T_{301}^{-m_{n301}}$ -> eq8

 

[Delta2]

I am not familiar with the equations relating to this problem but anyway here is my input:

Something looks wrong either with the value of $k=1.38\cdot 10^{-23}$ or with this exponential $e^{\frac{-0.785}{kT}}$ because after I do some algebra I get for the exponential that it is equal to $e^{-0.189\cdot 10^{21}}$. I mean the exponent of this exponential is unusually large: it is not of the magnitude of 21 but of the magnitude of $10^{21}$.

Check that you have the value of $k$ in the correct units. And also check the exact form of the exponential at the equation of $n_i^2$

 

[phyzguy]

This is why it is good to write out the units. You need to make sure that your units match. Your value of EG is in electron Volts (eV), but you value of k is in J/K. You need to use k in eV/K, or multiply k by the conversion factor between eV and J. Also, are you sure A0 is Avogadro's number in that equation?

 

[Delta2]

Using $k=8.617\cdot 10^{-5}\frac{eV}{K}$ I get $n_i=1.027\cdot 10^9$... still different from the book value..

 

[phyzguy]

Using $k=8.617\cdot 10^{-5}\frac{eV}{K}$ I get $n_i=1.027\cdot 10^9$... still different from the book value..

As I said earlier, A0 in this equation is not Avogadro's number. Your A0*T^3 is basically the available density of states per unit volume. Again, if you carried the units, you would see that it can't be Avogadro's number, since Avogadro's number is dimensionless, and ni has units of cm^(-3). I suggest you do some more reading on intrinsic carrier concentration. Here's a possible link to get started:
http://galileo.phys.virginia.edu/classes/312/notes/carriers.pdf

 

[PhysicsTest]

As per the book $A_0$ is a constant independent of T. For me to understand the derivation it will take some time. To solve the problem i have done the following method
$\sigma_{300} = (n_{300} \mu_n + p_{300} \mu_p)q$ ->eq11
$\sigma_{300} = n_{i300} (\mu_{n300} + \mu_{p300})q$ ->eq12
$\sigma_{301} = n_{i301} (\mu_{n301} + \mu_{p301})q$ ->eq13
$\mu_{n300} = T_{300}^{-m_n}$ ->eq14
$\mu_{p300} = T_{300}^{-m_p}$ ->eq15
$\mu_{n301} = T_{301}^{-m_n}$ ->eq16
$\mu_{p301} = T_{301}^{-m_p}$ ->eq17
% increase in conductivity is
$\frac{\sigma_{301} - \sigma_{300}} {\sigma_{300}}$ ->eq18
$\frac{T_{301}^{1.5} e^{-\frac{E_{G0}} {kT_{301}}}*(T_{301}^{-m_n} + T_{301}^{-m_p}) - T_{300}^{1.5} e^{-\frac{E_{G0}} {kT_{300}}}*(T_{300}^{-m_n} + T_{300}^{-m_p})} {T_{300}^{1.5} e^{-\frac{E_{G0}} {kT_{300}}}*(T_{300}^{-m_n} + T_{300}^{-m_p})}$ ->eq19
$T_{301} = 301K; T_{300} = 300 K$
$E_{G0} = 0.785*1.6*10^{-19} J = 1.25 * 10^{-19}$
$k = 1.38 * 10^{-23} m^2 kg s^{-2} K^{-1}$
$m_n = 1.66, m_p = 2.33$
Substitute in eq19.
Numerator: $2.6*10^{-15}$ Denominator $3.23*10^{-14}$
Hence the increase in conductivity is $\frac{2.6*10^{-15}} {3.23*10^{-14}}*100 = 8$

 

[phyzguy]

(1) Why don't you calculate A0 given the known intrinsic carrier concentration of 2.5E13 @ 300K?
(2) When you calculated $n_i$ from $n_i^2$, you took the square root of the $T^3$ term to make it $T^{1.5}$ , but you didn't take the square root of the exponential term. What is the square root of $\exp^{\frac{-E_G}{kT}}$?

 

[Delta2]

I am afraid @phyzguy is right in his comment (2) above. It is $$\sqrt{A^nB^m}=\sqrt{A^n}\sqrt{B^m}=A^{\frac{n}{2}}B^{\frac{m}{2}}$$
assuming $A,B$ are positive real numbers, and $n,m$ real numbers.

Other than that your work looks correct, but as I said before I don't know a lot about semiconductor physics, so I literally got no clue what equations 11-17 are about and if they are correct or not. Taking for granted that they are correct, what you did after that appears to be correct with the mistake noted by @phyzguy of course. I see also that you fixed the units of $E_{G0}$, great.

 

[PhysicsTest]

$\frac{T_{301}^{1.5} e^{-\frac{E_{G0}} {2kT_{301}}}*(T_{301}^{-m_n} + T_{301}^{-m_p}) - T_{300}^{1.5} e^{-\frac{E_{G0}} {2kT_{300}}}*(T_{300}^{-m_n} + T_{300}^{-m_p})} {T_{300}^{1.5} e^{-\frac{E_{G0}} {2kT_{300}}}*(T_{300}^{-m_n} + T_{300}^{-m_p})}$ ->eq19
$T_{301} = 301K; T_{300} = 300 K$
$E_{G0} = 0.785*1.6*10^{-19} J = 1.25 * 10^{-19}$
$k = 1.38 * 10^{-23} m^2 kg s^{-2} K^{-1}$
$m_n = 1.66, m_p = 2.33$
Substitute in eq19. $\frac{11.932 *10^{-8} - 11.046 * 10^{-8}} {11.046 * 10^{-8}} = \frac{88.6} {11.046} = 8.02$
To find $A_0$
$n_i^2 = A_0T^3e^{\frac{-E_{G0}} {kT} }$
$n_i^2 = 2.5*10^{13}, T=300K, E_{G0}=1.25*10^{-19}J, k=1.38*10^{-23}m^2kgs^{-2}K^{-1}$
$A_0 = \frac{n_i^2} {T^3e^{\frac{-E_{G0}} {kT} }} = 1.2*10^{19}$

 

[phyzguy]

I haven't checked all of your numbers, but I think this is correct. One thing that makes these calculations easier is to remember that at 300K, kT = .026 eV.

 

[Delta2]

About your calculation of the constant $A_0$, you replace $n_i^2$ with $2.5\cdot 10^{13}$.

The book says that $n_i^2=2.5\cdot 10^{13}$ or that $n_i=2.5\cdot10^{13}$?

 

[PhysicsTest]

I am really sorry and thankyou for correction it says about $n_i$. I replaced and corrected the answer.
To find $A_0$
$n_i^2 = A_0T^3e^{\frac{-E_{G0}} {kT} }$
$n_i = 2.5*10^{13}, T=300K, E_{G0}=1.25*10^{-19}J, k=1.38*10^{-23}m^2kgs^{-2}K^{-1}$
$A_0 = \frac{n_i^2} {T^3e^{\frac{-E_{G0}} {kT} }} = 2.98*10^{32}$

 

[Delta2]

Up to now i can't see your correction in post #10. Perhaps it doesn't let you edit the post?

 

[sysprog]

Up to now i can't see your correction in post #10. Perhaps it doesn't let you edit the post?

Perhaps if you hit Ctrl-F5 you might get a refreshed (i.e. non-cached) version of the page on which you read the post.