Calculating Reverse Bias Voltage for Alpha Particle Detection

[neural_jam]

Hi all, me again!

I'm trying to calculate the current created when an alpha particle strikes a silicon p-n detector, but I'm getting some strange numbers, so I was wondering if I was doing something fundamentally wrong...

My problems start when I try to calculate the reverse bias voltage required to create a depletion region equal to the range of alpha particles in Si, which I understand to be 7.29x10^-5m.

I use the following equation:

d= $(\frac{2 \epsilon V}{e N})^{\frac{1}{2}}$

where d is the depletion region (m) = 7.29x10^-5
ε is the ionisation energy of Si @ 300K (eV) = 3.62
e is the charge on an electron (C) = 1.602x10^-19
N is the charge carrier number density (m^-3) = 10¹⁹
and V is the bias voltage.

rearranging I get

V = $\frac{d^{2}eN}{2\epsilon}$

So far, so simple.

However, for some reason, after plugging the numbers in I get ~1.18x10^-9V for the bias voltage, which seems to be very small

So can anyone tell me what I'm doing wrong? Or is it right and my expectation of a voltage in the 10s of volts wrong?

Thanks!

 

[AndyW]

Hi there!

It looks like you have the right equation, but there are a few things you might want to double check. First, make sure you are using the correct units for each variable. For example, the ionisation energy of Si is typically given in eV, not just in volts. You may need to convert it to joules before plugging it into the equation. Also, the charge carrier number density should be in units of cm-3, not m-3. So you may need to convert that as well.

Additionally, make sure you are using the correct value for the permittivity of silicon (ε), which is 11.7 ε0, where ε0 is the permittivity of vacuum (8.85x10-12 F/m). This will affect your calculation of the bias voltage.

Lastly, keep in mind that the equation you are using assumes ideal conditions and does not take into account any other factors that may affect the depletion region or the voltage required to create it. So your calculated value may be an approximation and may not match the actual voltage required in a real experiment.

I hope this helps! Let me know if you have any other questions.