How does Doherty increase efficiency? Starting from the input, after passing through a power divider, it is divided into two channels. Doherty improves the efficiency, which means that it can improve the efficiency when the back-off power is used. The source means that the circuit element that realizes the load traction is an active device, in Doherty it means the power amplifier tube.
Recently, I have been working on overtime positioning, which has exhausted people. This series of articles has been updated slowly. Today I am going to introduce the classic two-way Doherty amplifier principle, trying not to use formulas, just to explain the principle. Last time, the principle of load traction was roughly explained, explaining the mechanism of load traction to improve efficiency. This time, I will continue with the previous article. I will briefly explain how Doherty improves efficiency, and then talk about active load traction, and then explain how active load traction improves the efficiency of Doherty when it rolls back.
How does Doherty increase efficiency?See Figure 2-1 directly. The picture is a typical two-way Doherty. Let me make a brief introduction. Starting from the input (drawing in a hurry, unmarked in the figure, which is the leftmost node), the signal is divided into two channels after a power divider. The first channel is called the Carrier circuit, also known as the main circuit; the second channel is called Peak Road, call the auxiliary road. These two signals eventually converge at a place called the junction (the intersection of the two signal outputs in the figure) (just like the Yangtze River and the Yellow River share the same source (to be verified), and finally sink to the Wangyang Sea), then Hao The massive flow into the load.
Figure 2-1 Typical two-way Doherty architecture
In fact, Doherty's efficiency improvement refers to the efficiency when it can improve the back-off power. As mentioned in the first round, the current communication signals all have a high peak-to-average ratio, and the power amplification works at the average 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 amplifier will reach 400W, so if you use a 400W class AB amplifier to fall back to 100W, then you are inefficient. Are scared. So for the Doherty architecture, first, the total output power is non-isolated and combined by two (or more) power amplifier tubes. Carrier and Peak as shown in the above picture provide output power together. In this way, the output power of each tube does not need to be so large; second, when the average power is output (when retreating), usually only one power amplifier tube is working (as in the above figure, only the Carrier tube, Peak is off), this tube is in The efficiency when outputting this level of power is high, which is nearly 30% higher than that of ordinary AB class retreat. Take the above picture as an example, let's talk about the working process of Doherty. We fall back from the full power state to the average power. In the full power state, both the carrier and peak circuits are saturated and output. When the output power gradually decreases, the peak circuit is gradually turned off, and the load impedance of the carrier circuit becomes larger than in the saturated working state, so that when the power returns to the average power Although the carrier current decreases when the load is unchanged, the voltage swing increases because the load impedance becomes larger, so that the same output power can be obtained, but the efficiency is greatly improved at this time.
The above explains why Doherty can improve the efficiency at the back-off power-the load impedance becomes larger during the back-off power. Let's talk about the active load traction that its "load impedance becomes larger" depends on.
What is active load traction?Let's look at it separately: active + load traction. Load traction has already been said, so where is the source language? In fact, active means that the circuit element that realizes load traction is an active device, in Doherty, it refers to the power amplifier tube. We make an agreement here that the power amplifiers of Carrier and Peak can be equivalent to current sources (so far it is possible). With this agreement, we 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 drop produced by the two parts of 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 seen from the current source Im toward the load is equal to V / Im, which is equal to the load impedance R. Okay, let ’s assume that the current Ip slowly flows out of the no-current state to the load. At this time, what is the impedance Zm from the current source Im to the load? Or divide voltage by 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 side of the current source Im has always been Im, there is no change), then Zm at this time It 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 circuit 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 the open state; when the auxiliary circuit is gradually opened, the current Ip changes from small to large The apparent impedance of the road changes from R to 2R. In this way, the modulation of the apparent impedance of the main circuit is completed by the transformation of the auxiliary circuit current injection. Long talk is so much that I want to make this active load traction process clear without writing formulas. In fact, the above stack is the following formula (still mathematically concise), please move if you are willing to read it.
Some people have read the mess above, and may be wondering: How does this correspond to the improvement of the Doherty amplifier's regression efficiency? Next, let's talk about how to perform load traction in Doherty (accurately speaking on Carrier Road), and then improve the efficiency of back-off. For the sake of aspect, re-attach Figure 2-1 here.
Let's take the most classic two-way symmetric Doherty. At this time, the power divider is 3dB equal power division, and the main and auxiliary power amplifier tubes are the same (the 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 has no current. At this time, the impedance Rp seen from the junction point to the Peak Road is infinite, which is an open circuit state. When the input signal power slowly increases, the peak circuit begins to open and current flows into the load. As previously analyzed, the impedance Rm seen by the main circuit at this time begins to gradually increase. When both channels are saturated, Rm becomes 2R. This process is the active load traction of Peak Road to Carrier Road. Then some people may have questions at this time: Isn't it said that Doherty is to improve the efficiency in the fallback state (outputting less power)? According to your analysis, it seems to be just the opposite. The output power becomes larger, the load impedance of the carrier circuit becomes larger (the efficiency becomes larger), and the load impedance becomes smaller (the efficiency becomes lower) when the power is backed off (the Peak circuit reduces the output). Very good, in fact, careful students will find that in the Carrier road, there is something called an impedance converter after the output of the power amplifier. This thing is actually a passive circuit, usually replaced by a 1/4 wavelength conversion line in theoretical analysis. Anyone who has studied radio frequency should know that after the characteristic impedance of the 1/4 wavelength conversion line is determined, the impedance at both ends is inversely proportional, that is, the impedance at one end changes from small to large, then the other end changes from large to small.
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