# Boltzmann constant in formulas

1. Apr 16, 2012

### aarnes

Hi, I see that Boltzmann constant comes in different forms like: k=8.62*10-5 eV/K and also k=1.38*10-23J/K.
Which one should I use in , say formula for intrinsic carrier concentration ni = sqrt(Nc*Nv)*e-Eg(T)/2kT ?

2. Apr 16, 2012

### Bob S

kT has units of energy; Joules or electron volts. Always remember that an exponent has to be unitless. If the numerator has volts, then kT is in eV.

Sometimes you will see kTB which is noise power times bandwidth.

The noise power is kTB where k= 1.38 x 10-20 millijoules per deg kelvin, T=293 kelvin, and B(bandwidth in Hz)= 1 MHz

So noise power is 1.38 x 10-20 x 293 x 106 Hz= 4 x 10-12milliwatts per MHz = -114 dBm per MHz.

Add 3 dB noise figure to get -111 dBm per MHz

Last edited: Apr 16, 2012
3. Apr 16, 2012

### aarnes

Thank you, Bob! I used the eV form before but I saw some different results on the web and just wasn't sure why is there always a slightly different value for ni, depending on which website you look.

4. Apr 16, 2012

### Dickfore

Instead of $k_ B = 8.62 \times 10^{-5} \, \frac{\mathrm{eV}}{\mathrm{K}}$, people usually find it convenient to remember the following number:
$$k_B = \frac{1 \, \mathrm{eV}}{11600 \, \mathrm{K}}$$

(notice that $(8.62 \times 10^{-5})^{-1} = 1.16 \times 10^4$, so the above are equivalent)

5. Apr 16, 2012

### Dickfore

At room temperature $T \approx 293 \, \mathrm{K}$, the value $k_B \, T \approx 25 \, \mathrm{meV}$.

6. Apr 16, 2012

### aarnes

On the web you usually see 300K as room temperature. I guess it depends on one's preference? :D

7. Apr 16, 2012

### Dickfore

Right, that is why i used only 2 significant figures in the final result and the approximate sign.

8. Apr 16, 2012

### rbj

boy, you sure are good at using HTML markup. i never knew you could get a subscript in the superscript. anyway, it might look better with LaTeX

$$n_i \ = \ \sqrt{N_c N_v} e^{-\frac{E_g(T)}{2 k T}}$$

now, to answer your question, you want you $kT$ quantity to be in the same units as the $E_g$ quantity. if $k=8.62 \times 10^5$ eV/K then $T$ better be in Kelvin and $E_g$ better be in eV.