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A novel model of avalanche current generation in the GaN HEMT equivalent circuit

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We consider the large-signal equivalent circuit of a field-effect transistor (FET) with characteristics and parameters calculated using a two-dimensional (2D) quasi-hydrodynamic model while taking into account the electron velocity overshoot. In accordance with this equivalent circuit when applied to an AlGaN/GaN high-electron-mobility transistor (HEMT), the avalanche current (in the feedback branch) is zero at all operating points. However, this contradicts the significant difference seen between the results of time-domain simulations of a radiofrequency (RF) amplifier in which the transistor is simulated using a 2D model that does versus does not include the avalanche multiplication of the charge carriers. We propose a new model for the avalanche current generation, based on such simulations including the avalanche multiplication and on the theory of impact-ionization avalanche transit-time (IMPATT) diodes. This model allows the synthesis of amplifiers with high power-added efficiency (PAE).

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Correspondence to Gennadiy Z. Garber.

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Appendix 1: The method for calculating the functions \(\varphi_{0} (y)\) and \(n_{0} (y)\)

To formulate the conditions at the left and right boundaries of the transistor active subregion (at \(x = x_{\rm l}\) and \(x = x_{\rm r}\) shown in Fig. 2), consider the model stated by Eqs. (2)–(7) in the one-dimensional approximation under thermodynamic equilibrium: \(\partial /\partial x = \partial /\partial t = 0\), \(T_{\rm e} = T_{0}\), and \(j_{ny} = j_{nuy} = 0\). Equation (3) gives

$$ \frac{{kT_{0} }}{q}\frac{{\text{d}}}{{{\text{d}}y}}\left[ {\mu_{0} (y)n_{0} (y)} \right] = - \mu_{0} (y)F_{ny} (y)n_{0} (y), $$

where \(\mu_{0} (y)\) is the low-field electron mobility and \(n_{0} (y)\) is the electron concentration. The last subscript 0, as in \(\varphi_{0} (y)\) below, stands for thermodynamic equilibrium. Using Eqs. (7) and (8), we have

$$ \frac{{kT_{0} }}{q}\frac{{\text{d}}}{{{\text{d}}y}}\left[ {\mu_{0} (y)n_{0} (y)} \right] = \mu_{0} (y)n_{0} (y)\frac{{\text{d}}}{{{\text{d}}y}}\left[ {\varphi_{0} (y) + X_{\rm e} (y)} \right], $$

where \(\varphi_{0} (y)\) is the electric potential.

Dividing both sides of this equation by \(kT_{0} \mu_{0} n_{0} /q\) yields

$$ \frac{{\text{d}}}{{{\text{d}}y}}\ln \left[ {\mu_{0} (y)n_{0} (y)} \right] = \frac{{\text{d}}}{{{\text{d}}y}}\left[ {\frac{{\varphi_{0} (y) + X_{\rm e} (y)}}{{kT_{0} /q}}} \right]. $$

Integration yields

$$ \ln \left[ {\mu_{0} (y)n_{0} (y)} \right] = \frac{{\varphi_{0} (y) + X_{\rm e} (y)}}{{kT_{0} /q}} + C, $$

where C is a constant of integration. If \(\varphi_{0} (0)\) = \(X_{\rm e} (0)\) = 0 and \(n_{0} (0)\) = \(N_{\rm d} (0)\), we have

$$ n_{0} (y) = \frac{{\mu_{0} (0)N_{\rm d} (0)}}{{\mu_{0} (y)}}\exp \left[ {\frac{{\varphi_{0} (y) + X_{\rm e} (y)}}{{kT_{0} /q}}} \right]. $$

Substitution of this expression into Eq. (6) yields the following equation:

$$ \frac{{\text{d}}}{{{\text{d}}y}}\left( {\varepsilon \frac{{{\text{d}}\varphi_{0} }}{{{\text{d}}y}}} \right) = Q(y), $$

where \(Q(y) = q\left\{ {\frac{{\mu_{0} (0)N_{\rm d} (0)}}{{\mu_{0} (y)}}\exp \left[ {\frac{{\varphi_{0} (y) + X_{\rm e} (y)}}{{kT_{0} /q}}} \right] - N_{\rm d} (y)} \right\}\).

We solve the last equation with the boundary conditions \(\varphi_{0} (0)\) = 0 and \({\text{d}}\varphi_{0} /{\text{d}}y\) = 0 at \(y = y_{\rm m}\) by the quasilinearization method for the Shockley–Poisson equation, according to Chap. 3 of Ref. [9]. The resulting function, \(\varphi_{0} (y)\), and the function \(n_{0} (y)\) calculated by formula (16) are plotted in Fig. 7.

Appendix 2: The method for solving the \(i_{{{\text{am}}}}\) generator equation with the condition of periodicity

Write Eq. (15) in the canonical form

$$ \frac{{{\text{d}}z}}{{{\text{d}}t}} + P(t)z = R, $$

where \(z = i_{{{\text{am}}}} (t)\), \(P(t) = \frac{{1 - \Phi^{\prime}[u_{{{\text{gs}}}} (t),u_{{{\text{ds}}}}^{{}} (t)]}}{{\tau_{\lambda } }}\), and \(R = \frac{{I_{{\text{r}}} }}{{\tau_{\lambda } }}\). The coefficient \(P(t)\) is periodic with period T. To obtain a cyclic solution, suppose that \(z(t)\) is a linear combination of the following two functions:

  • \(u(t)\), solution of Eq. (17) with zero initial condition, \(u(0)=0\), and

  • \(v(t)\), solution of Eq. (17) with zero right-hand side, \(R=0\), and unity initial condition, \(v(0)=1\).

Substituting \(z(t) = \alpha u(t) + \beta v(t)\) into Eq. (17), we obtain \(\alpha = 1\). The condition of periodicity, \(z(0) = z(T)\), gives \(\beta = \frac{u(T)}{{1 - v(T)}}\).

The initial value problems for \(u(t)\) and \(v(t)\),

$$ \begin{aligned} \frac{{{\text{d}}u}}{{{\text{d}}t}} & = - P(t)u + R,\quad u(0) = 0, \\ \frac{{{\text{d}}v}}{{{\text{d}}t}} & = - P(t)v,\quad v(0) = 1, \\ \end{aligned} $$

are considered on a uniform grid constructed in \([0,\,\,T]\) for analyzing the IFET equivalent circuit. (In our case, this grid has 73 nodes, including the boundary nodes.) To solve these initial value problems, we use the method in Sect. 5.2 of Ref. [9], which is based on the quadratic spline construction.

Using the obtained \(u(t)\) and \(v(t)\), we calculate the avalanche current using the formula

$$ i_{{{\text{am}}}} (t) = z(t) = u(t) + \frac{u(T)}{{1 - v(T)}}v(t). $$

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Garber, G.Z. A novel model of avalanche current generation in the GaN HEMT equivalent circuit. J Comput Electron (2020). https://doi.org/10.1007/s10825-020-01452-2

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  • Full 2D model
  • Harmonic balance method
  • Large-signal equivalent circuit
  • Quasi-hydrodynamic model
  • RF amplifier
  • Splines