3.2. Device’s electrical stress
The output capacitance C1 reaches the max voltage when its current becomes zero (Figure 3):
iC(θm) = 0 → Equation 11
From Equation 3 follows:
IDC - I1sin(θm + φ) = 0 → Equation 12
Using the boundary conditions for class-E (Equation 6 and Equation 7) in Equation 5 (voltage across C1) we get:
IDC = (2/π)cosφI1 → Equation 13
IDC = sinφI1 → Equation 14
Comparing the two equations, for a duty cycle of 50%, we obtain:
φ = arctan(-2/π) = 0.563 rad → Equation 15
In Equation 12 replacing IDC with Equation 14, we get:
sin(θm + φ) + sin(φ) = 0 → Equation 16
Using the Prosthaphaeresis identities, we get:
θm = -2φ → Equation 17
Finally, we obtained the popular equations for peak voltage and current for the active device:
Vpk = Vc(θm)= 2πφVDC = 3.562VDC → Equation 18
Ipk = IDC + I1 = IDC - IDC/sinφ = 2.862IDC → Equation 19
For a duty-cycle of 50%, the peak voltage at the drain terminal exceeds the supply voltage of the circuit by more than three times.
Dr. Raab in (2) has presented all the equations that govern an idealized Class-E RF power amplifier. In particular, he has proved that a duty-cycle of 50% represents an optimum in terms of output power capability. Moreover, the device’s breakdown voltage determines the maximum allowable supply voltage and it is in direct relation to the output power.