Detailed Analysis of Doherty Amplifier Design: How to Improve Efficiency (Part 2)

How does Doherty improve efficiency? Starting from the input, it is divided into two paths after passing through a power divider. Doherty improves efficiency by improving the efficiency when the power is backed off. The source refers to the circuit element that realizes load traction, which is an active device. In Doherty, it refers to the power amplifier tube.

I have been working overtime to locate the problem recently, which has made me exhausted. The update of this series of articles has been slow. Today I am going to introduce the principle of the classic two-way Doherty amplifier, and try to explain the principle clearly without using formulas. Last time, I talked about the principle of load pull in general and explained the mechanism of load pull to improve efficiency. This time, I will continue with the previous article and first briefly explain how Doherty improves efficiency, then talk about active load pull, and then explain how active load pull improves efficiency during Doherty fallback.

How does Doherty improve efficiency?

Look directly at Figure 2-1. This figure is a typical two-way Doherty. Let me briefly introduce it. Starting from the input (the drawing was hasty, so it is not marked in the figure, that is, the node on the far left), the signal is divided into two paths after passing through a power divider. One of them is called the Carrier path, also known as the main path; the other is called the Peak path, also called the auxiliary path. The two signals finally converge at a place called the combining point (that is, the intersection of the two signal outputs in the figure) (just like the Yangtze River and the Yellow River have the same source (to be verified), and finally converge into the vast ocean), and then flow into the load in a mighty way.

Figure 2-1 Typical two-way Doherty architecture

In fact, when we say that Doherty improves efficiency, we mean that it can improve the efficiency when the power is backed off. As mentioned in the first part, current communication signals have a high peak-to-average ratio, and the power amplifiers all work at the mean power. For example, if the signal peak-to-average ratio is 6dB and the average power is 100W, then the maximum output power of the power amplifier must reach 400W. Therefore, if you use a 400W class AB power amplifier to back off to 100W, you will be afraid of the low efficiency. Therefore, for the Doherty architecture, first, the total output power is non-isolated and combined by two (or more) power amplifier tubes. For example, the Carrier and Peak tubes in the figure above provide output power together. In this way, the output power of each tube does not need to be so large; second, when outputting the mean power (backing off), usually only one power amplifier tube is working (for example, only the Carrier tube in the figure above, the Peak is turned off), and this tube has a higher efficiency when outputting this level of power, which is nearly 30% higher than the ordinary class AB back-off. Taking the above figure as an example, let's talk about the working process of Doherty. We retreat from the full power state to the mean power. In the full power state, both the carrier path and the peak path are saturated output. When the output power gradually decreases, the peak path is gradually turned off, and the load impedance of the carrier path becomes larger than that in the saturated working state. In this way, when the power retreats to the mean power, although the carrier current decreases compared to when the load remains unchanged, its voltage swing increases due to the increase in load impedance. In this way, the same output power can be obtained, but the efficiency is greatly improved at this time.

The above explains why Doherty can improve efficiency at the back-off power - the load impedance increases when the power is backed off. The following is a discussion of the active load traction on which the "increase in load impedance" depends.

What is active load-pull?

Let's look at it separately: active + load traction. We have already talked about load traction, so where does the word active come from? In fact, active means that the circuit elements that realize load traction are active devices, which refers to power amplifier tubes in Doherty. We make a convention here that the power amplifiers of the carrier path and the peak path can be equivalent to current sources (so far, it is possible). With this convention, let's analyze the working process of active traction. As shown in Figure 2-2, the main and auxiliary power amplifiers are equivalent to two current sources, named Im and Ip respectively. The common load impedance of the two is R.

Figure 2-2 Schematic diagram of active load traction

In this way, the voltage on the load is the superposition of the voltage drops produced by the two parts of the current on it. Let's do a scenario simulation now. First, assume that the current Ip is 0, then only the current Im flows through the load R at this time, and the voltage V on the load is Im*R. In other words, the impedance Zm from the current source Im to the load is equal to V/Im at this moment, which is equal to the load impedance R. Okay, then let's assume that the current Ip slowly flows out of the current to the load from the no-current state. At this time, what is the impedance Zm from the current source Im to the load? It's still the voltage divided by the current. At this moment, the voltage on the load is (Im+Ip)*R, and the current is Im (this is very important, because the current flowing into the load from the current source Im side is always Im, without change), then Zm at this time is (1+Ip/Im)*R. Smart you will find that the current source Ip modulates (pulls) the apparent impedance of the current source Im. Assuming that the currents of the two current sources are the same, when the auxiliary current is 0, the apparent impedance of the main circuit is the load impedance R, and the apparent impedance of the auxiliary circuit is in an open circuit state; when the auxiliary circuit is gradually turned on, the apparent impedance of the main circuit changes from R to 2R when the current Ip changes from small to large. In this way, the modulation of the apparent impedance of the main circuit is completed through the transformation of the auxiliary current injection. I have said so much because I want to explain the process of active load traction clearly without writing formulas. In fact, the above is just one formula below (the mathematics is still concise). If you are willing to read it, please move on.

Some people may have a question after reading the above mess: How does this correspond to the Doherty amplifier's improved back-off efficiency? Next, let's talk about how to perform load pulling (or more precisely, the carrier path) in Doherty to improve the back-off efficiency. For the sake of convenience, Figure 2-1 is re-posted here.

Let's take the most classic two-way symmetrical Doherty as an example. At this time, the power divider is 3dB equal power division, and the power amplifier tubes used in the main and auxiliary circuits are the same (matching is also the same). When the input signal is relatively small (that is, when the output power is not large), the Peak circuit is turned off, does not work, and there is no current. At this time, the impedance Rp from the junction to the Peak circuit is infinite, which is an open circuit state. When the input signal power increases slowly, the peak circuit begins to open, and current flows into the load. As analyzed above, the impedance Rm seen by the main circuit begins to increase slowly at this time. When both circuits are saturated, Rm becomes 2R. This process is the active load traction of the Peak circuit on the Carrier circuit. Then someone may have a question at this time: Isn't it said that Doherty improves the efficiency in the fallback state (outputting less power)? According to your analysis, it seems to be the opposite. As the output power increases, the load impedance of the Carrier circuit increases (efficiency increases), and when the power is fallback (the Peak circuit reduces the output), the load impedance decreases (efficiency decreases). Very good. In fact, careful students will find that in the carrier circuit, there is something called an impedance converter after the power amplifier output. This thing is actually a passive circuit, usually replaced by a 1/4 wavelength conversion line in theoretical analysis. Students who have studied radio frequency should know that after the characteristic impedance of the 1/4 wavelength conversion line is determined, the impedances at both ends are inversely proportional, that is, if the impedance of one end changes from small to large, then the impedance of the other end changes from large to small.