Brush up on the theory before designing a high power Class‐E amplifier: Page 5 of 15

3.2. Device’s electrical stress

The output capacitance C₁ reaches the max voltage when its current becomes zero (Figure 3):

i_C(θ_m) = 0 → Equation 11

From Equation 3 follows:

I_DC - I₁sin(θ_m + φ) = 0 → Equation 12

Using the boundary conditions for class-E (Equation 6 and Equation 7) in Equation 5 (voltage across C1) we get:

I_DC = (2/π)cosφI₁ → Equation 13

I_DC = 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(θ_m + φ) + sin(φ) = 0 → Equation 16

 Using the Prosthaphaeresis identities, we get:

θ_m = -2φ → Equation 17

Finally, we obtained the popular equations for peak voltage and current for the active device:

V_pk = V_c(θ_m)= 2πφV_DC = 3.562V_DC → Equation 18

I_pk = I_DC + I₁ = I_DC - I_DC/sinφ = 2.862I_DC → 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.