# Formula derivation of Doherty power amplifier from circuit diagram

I'm having a hard time understanding how formulas are derived from the diagram of a simple Doherty power amplifier circuit, as pictured below.

While I understand the concept behind load modulation and thus the necessity to introduce a λ/4 transmission line, I'm having trouble deriving the formulas shared by most of the literature I've explored and that I've pasted below. (Diagrams are in German, U stands for the voltage V.)

The main amplifier and the peak amplifier are represented as current sources I_H and respectively -jI_S. There's a load R_L, Z_H and Z_S are the impedances seen by the two current sources.

QUESTION 1) I understand that, looking from the node between resistance R_L and current source I_s into the transmission line, the high-ohmic impedance of current source I_H will look like a short circuit due to the length λ/4 of the transmission line, but I don't see how this implies that:

That is " the voltage at impedance R_L is only dependent on I_H."

How is this possible considering I_S is there, too? Also, is 'minus j' multiplied with I_H because the transmission line creates a phase shift in I_H?

QUESTION 2) How is following formula for the voltage at current source I_H defined? I've read that superposition is used here, so I guess one current source is observed in the circuit while the other is open-circuited and viceversa, then the two formulas are superposed. Yet, how was this done?

QUESTION 3) Lastly, I'm completely clueless as to why this is the impedance seen by current source I_S:

My results don't agree with what you show. The results would match though if $$\R_L=Z_W\$$. But here is what I did:
Let's turn off (open circuit) current source $$\I_S\$$, then, forward current in the T-line: $$I^+ = I_H$$ Reflection coefficient at the load is: $$\Gamma_L = \frac{R_L - Z_W}{R_L+Z_W}$$ Thus the current into the load $$\R_L\$$ is: $$I_L = (1-\Gamma_L)I^+e^{-j\frac{2\pi}{\lambda}.\frac{\lambda}{4}} = -\frac{2Z_W}{R_L+Z_W}jI_H$$ $$V_{L1} = I_LR_L = -\frac{2Z_WR_L}{R_L+Z_W}jI_H$$ Now turn off the current source $$\I_H\$$ and the T-line acts as a short circuit, so $$V_{L2} = 0$$ Thus, using superposition, $$V_{L} = -\frac{2Z_WR_L}{R_L+Z_W}jI_H$$
Turn off the current source $$\I_S\$$, and the impedance looking into the line is: $$Z_L = \frac{Z_W^2}{R_L}$$ Thus, $$V_{H1} = \frac{Z_W^2}{R_L}I_H$$ Now, turn off current source $$\I_H\$$, the forward current in the line is: $$I^+ = -\frac{R_L}{R_L+Z_W}jI_S$$ Thus, $$V^+ = I^+Z_W = -\frac{R_LZ_W}{R_L+Z_W}jI_S$$ The reflection coefficient at the load is now unity, thus total voltage at load is: $$V_{H2} = (1+1)V^+e^{-j\frac{2\pi}{\lambda}.\frac{\lambda}{4}} = -\frac{2R_LZ_W}{R_L+Z_W}I_S$$ Thus, total voltage using superposition: $$V_H = \frac{Z_W^2}{R_L}I_H-\frac{2R_LZ_W}{R_L+Z_W}I_S$$
The impedance $$\Z_S\$$ is, by definition: $$Z_S = \frac{V_L}{I_S}$$ $$Z_S = \frac{2Z_WR_L}{R_L+Z_W}\frac{I_H}{I_S}$$
• @AndreaToffanin I did not assume a short circuit at the left end. Note that $I^+$ is not the total current flowing in the T-Line, it is the current wave flowing from right to left. When the current wave reaches the left end, it will be reflected. Now since this end is open circuit, the reflected current will cancel the forward current making the total current zero as you are expecting. But the current waves (both forward and reflected) always see the characteristic impedance of the T-Line. Commented May 20, 2020 at 22:18