Derive the expression for the condition for negative differential mobility of transferred effect on devices.

The generation of negative resistance in a Gunn diode is explained by the valley current. The passive electronic device absorbed or dissipated power and active devices generated the power. Power across the passive devices is I²R and across the active devices is -I²R. This negative resistance of the active device generated the power or generated the frequency. In single semiconductor device like Si (silicon) or Ge (germanium) cannot be generated the power or frequency (no negative resistance) because there are no valleys in the conduction band. The Gun diode is made by combination of trivalent element Ga (Gallium) and pentavalent element As (Arsenic). Some others compound semiconductor materials such as InP (Indium phosphate), CdTe (Cadmium Telluride), GaAsP (Gallium Arsenide Phosphate) and ZnSe (Zenc) can be generated negative resistance.

Fig 17. Energy band diagram of GaAs

In Gallium Arsenide (GaAs) compound two valleys is existed on the conduction band. The band gap energy between valance and conduction band is E_g=1.4 eV. The energy difference between upper valley and lower valley is E_V=0.36 eV. At room temperature and zero biasing voltage the thermal energy at the lower valley electron is 0.026 eV. The thermal energy at the lower valley is less compare to the E_V causes no electron transfer from lower valley to the upper valley. At room temperature the mobility and effective mass of the electron is high and low respectively. The mobility is μ_L=0.5 m²/V.s and effective mass is m^*_L=0.072. With the increase of biasing voltage, the electron gets accelerated under electric field as a result the kinetic energy of the electron is increased. In this condition the electrons from the lower valley will be transferred to the upper valley. As a result the mobility and effective mass of the electron at the lower valley is decrease and increase respectively. On the other hand due to transfer of electron from lower valley to the upper valley, redistribution of electros between two valleys of the conduction ban is occurred. The mobility and effective mass of electron at the upper valley is decrease and increased respectively and it becomes μ_H=0.01 m²/V.s and m^*_H=1.2. This negative differential mobility characteristic of GaAs is used for amplification or generation of microwave.

If m_L, μ_L, N_L be the effective mass, mobility and carrier concentration respectively for lower valley, while m_H, μ_H, N_H are the respective quantities for the upper valley, then the steady state conductivity

\(\sigma=q(\ \mu_L\ N_L+\mu_HN_H)\) … (1)

Where q is the charge of the carrier

Now N=N_L+N_H=total carrier concentration

When electrons are accelerated under high electric field E, then their effective temperature rises above the lattice temperature. Thus electron density N and mobility μ are both function of electric field E.

Differentiating equation (1) with respect to E, we have

\(\frac{d\sigma}{dE}=q\left(\mu_L\frac{dN_L}{dE}+\mu_H\frac{dN_H}{dE}\right)+q\left(N_L\frac{d\mu_L}{dE}+N_H\frac{d\mu_H}{dE}\right)\) … (2)

Again N=N_L+N_H … (3)

Differentiating equation (3) with respect to E, we have

\(\frac{d}{dE}\left(N_L+N_H\right)=\frac{dN}{dE}=0\)

Because total numbers of carriers cannot change with electric field

\(\frac{dN_L}{dE}=-\frac{dN_H}{dE}\) … (4)

Again mobility μ α E^p, where p and K are constant

mobility μ=K E^p

\(\frac{d\mu}{dE}=K\ \frac{d}{dE}\left(E^p\right)=KpE^{p-1}\)

\(\frac{d\mu}{dE}=Kp\frac{E^p}{E}=\frac{p\mu}{E}\) … (5)

Substituting the values of equations (4) and (5) in equation (2)

\(\frac{d\sigma}{dE}=q\left(\mu_L-\mu_H\right)\frac{dN_L}{dE}+q\left(N_L\mu_L+N_H\mu_H\right)\frac{p}{E}\) … (6)

Again current density J=σ E

Differentiating with respect to E

\(\frac{dJ}{dE}=\sigma+E\frac{d\sigma}{dE}\)

\(\frac{1}{\sigma}\frac{dJ}{dE}=1+\frac{\frac{d\sigma}{dE}}{\frac{\sigma}{E}}\) … (7)

For the negative resistance , the current density J should decrease with the increase of electric field E.

Therefore \(\frac{dJ}{dE}\) is negative and is possible only when the right hand side of equation (7) is less than zero.

\(1+\frac{\frac{d\sigma}{dE}}{\frac{\sigma}{E}}<0\)

\(-\frac{\frac{d\sigma}{dE}}{\frac{\sigma}{E}}>1\) … (8)

Dividing equation (6) by equation (1), we have

\(\frac{\frac{d\sigma}{dE}}{\sigma}=\frac{q\left(\mu_L-\mu_H\right)\frac{dN_L}{dE}+q\left(N_L\mu_L+N_H\mu_H\right)\frac{p}{E}}{q(\ \mu_L\ N_L+\mu_HN_H)\ }\)

Writing \(\frac{N_H}{N_L}=M\) and multiplying both sides by E, we have

\(-\frac{\frac{d\sigma}{dE}}{\frac{\sigma}{E}}=-\left[\frac{\left(\mu_L-\mu_H\right)}{{(\mu}_L+{M\mu}_H)}\left(\frac{E}{{(N}_L}.\ \frac{dN_L}{dE}\right)+p\right]\)

From equation (8), we have

\(\left[\frac{\left(\mu_L-\mu_H\right)}{{(\mu}_L+{M\mu}_H)}\left(-\frac{E}{{(N}_L}.\ \frac{dN_L}{dE}\right)-p\right]>1\)

To maintain the inequality of equation (9) μ_L>μ_H and \(\frac{dN_L}{dE}\) should be negative. This proves that the electron at the lower valley has higher mobility and density of the lower valley electron decreases with increase of electric field that means the electron transfers to upper valley.