# Photoconductive Semiconductor Switches: Materials, Physics, and Applications
## 2 Theory of PCSS Operation
Incorporating a capacitive source of charge rather than a constant one results in the current-voltage behavior shown in Figure 4. This figure shows current and voltage after optical triggering. The new theoretical results in Figure 4 are in qualitative agreement with the measured data of a comparable measured configuration shown in Section 7 below. In both figures, the current rises after optical triggering and then it decreases. In Figure 4, the current ceases when the capacitive current source is not able to maintain the on-state. The figure shows that the threshold voltage is approximately 40 V for this switch and capacitor combination. The switch voltage remains large because the electrons trapped at the acceptors cause a large internal field that maintains the switch voltage at lock-on. This field enhancement has a primary role in the avalanche injection mechanism. Lock-on-like behavior in GaN has only recently been observed, and its mechanisms are accordingly less well-understood than those giving rise to the phenomenon in GaAs. Additional physical experimental and simulation work is required to understand, and therefore to engineer, the enormous potential for high-gain operation in these direct bandgap devices. The two breakdown mechanisms thus differ substantially. The process in GaAs relies on the fact that the increase in carrier-carrier scattering upon triggering generates the high carrier density needed for filamentation. The increasing current reduces the field and this stabilizes the avalanche mechanism. The high-current-density region exists inside a filament. Outside the filament, the carriers achieve a low carrier density steady state. Both the high- and low-density regions are locally stable and co-exist for the lock-on field. Figure 4: Calculated current–voltage of the \(n\)-\(i\)-\(n\) switch during and after optical triggering while the switch is charged by a capacitor. The GaN mechanism relies on a large density of charged defects at the anode that create a high field region under bias. These initially neutral defects capture electrons during the optical triggering process. The two mechanisms result in different characteristic fields. In GaAs, the electric field is almost uniform throughout the device. However, in the GaN mechanism, the electric field is large near the anode and comparatively low elsewhere. The two mechanisms cause very different behaviors in a circuit. In both cases, the device is assumed to be a simple one-dimensional device with length \(L\). The effects can be understood by considering a switch in a simple circuit consisting of a voltage source \(V_{C}\) (the capacitor), a load resistor \(R_{L}\), and the switch under bias voltage \(V_{S}\). During a quasi-static switch process, the relation between the slowly varying voltage contributions can be expressed as \(V_{C}=IR_{L}+V_{S}\). In this equation, the voltage source is a large capacitor that discharges during the operation of the switch. This simple circuit can be used to discuss the differences in the temporal evolution of GaN and GaAs. In the GaAs mechanism, the electric field \(E_{LO}\) during lock-on is approximately constant. Because the field is constant, the lock-on voltage \(V_{LO}=E_{LO}L\) is also constant, i.e., the lock-on voltage is the switch voltage. Initially, the capacitor voltage \(V_{C}\) exceeds the lock-on voltage. However, at the end, when the capacitor discharges to approximately \(V_{LO}\), the current collapses and the switch opens at \(V_{C}=V_{LO}\). In the case of the GaN switch, the circuit is identical. However, in this case, the field varies within the switch. It must have a large value at the right-hand side (of Figure 1) to induce avalanche injection. In that region, of thickness \(\Delta\), the voltage will be \(V_{LO}=E_{LO}\Delta\). The field \(E_{0}\) in the remainder of the device of thickness \(L-\Delta\) will be much smaller, such that \(V_{0}=E_{0}(L-\Delta)\). The switch voltage is thus \(V_{S}=V_{0}+V_{LO}\).