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2021 Electronic Design Competition Question C—Three-Port DCDC Converter

 
Overview
1: Preface
  This work is the result of the efforts of my teammates and has nothing to do with me (dog head). I only sleep, eat, and sleep every day. They were so awesome during the preparations for the competition. I am also very grateful to Lichuang and Jialichuang for buying (bai) and (piao) things. like!
 
2: Personal introduction to
Mango Crisp,
a fresh graduate from an unknown university who
is a game, sleeping, and eating enthusiast who
participated in the
20-year e-sports competition and wants to sleep in
the 21-year e-sports competition...
 
3: Project analysis
  we are getting When answering questions, you generally need to read the entire text first to find all the tasks that need to be done.
Ro4mtHoTiT5twNyF1Ff2FPaj0ywWGWWkBFWa2rOa.pngHere we can learn 5 points about the task of the question:
   1: Simulate photovoltaic cells to supply power to the load and charge the battery pack.
   2: Simulate photovoltaic cells and battery packs to supply power to the load together.
   3: The two modes switch to each other under different circumstances.
   4: The output Uo is constant, 30V.
   5: Overall work frame diagram.
 
From the requirements, we usually aim for full marks, so we must combine all requirements to formulate our plan.
zj28zDGy1E2DObKhO8No4oMF65X3QFMYEqoiUbIc.pngIcYu5yk7BB8G1G2rrYVTNnjOUNp7y2lGjYcAU0nj.png      Judging from the results of our analysis, we need Uo to be constant, the efficiency to be as high as possible, and to achieve full-range maximum power tracking. Then we analyze when it is mode 1 and when it is mode 2. In playing the requirements of 1, we can see the specific requirements for switching:
       1: Switching at 55-25. Let's do the math: divide it in half; the output is always constant, and the power is 30*1.2=36W. At 55V, the input is 55/2/10=2.75A, and the power is 2.75*25=75.625W; at 25V, the input is 25/2/10=1.25A, and the power is 1.25*12.5=15.625W.
       2: Switched at 35V, 1.2-0.6A. Still half; the output power is 30*1.2-0.6=36W-18W. When 1.2A, the input is 35/2/10=1.75A, the power is 61.25; when 0.6A, the input is 35/2/10=1.75A, the power is 61.25W;
after analysis, it can be known that when high power is input, The battery must consume power in the charging state, which is mode 1. When inputting low power, the battery discharges to supply power to the load, which is mode 2.
   The input voltage of the photovoltaic cell is between 25-55V, calculated according to the voltage division of the load and internal resistance, and the maximum power tracking is required in the question, so the range is between 12.5-27.5, and the load end output voltage is required to be stable at 30V. So the input needs to be boosted. Mode 2 requires photovoltaic cells and battery packs to jointly power the load. The maximum voltage of the battery will not exceed 17V. Therefore, if we want to obtain a voltage of 30V, we must also increase the voltage of the battery, which requires a boost circuit. The question requires that the battery pack must be charged and discharged, so the system uses bidirectional DC-DC control.
 
Conclusion: Photovoltaic charges the battery and supplies power to the load at high power input, and photovoltaic and battery power the load at low power input. Photovoltaics are always boosting.
 
4: Circuit design analysis
system diagram:
9nmR41fUznMFPvXYFHbvzPPDBJNuOSOw4IGt8fag.pngAccording to the requirements of the topic, what we need to do is boost MPPT and bidirectional DCDC. It is necessary to collect the values ​​​​of U1, I1, Uo, Io, Ub, and Ib. These are collected for pid. The overall circuit is divided into two large blocks, one for boosting MPPT and the other for bidirectional DCDC.
aTfjAibIowNr2Sp8p9HRDGscJF1HLcYUC3pNCXqE.pngBased on Baidu's results, we chose the most basic boost circuit, as shown in the figure below:
PX5YybEDrGBXykFnSJ48hzFMV9KsGEyX8lZtTD2Q.png 
What we need is very simple, a suitable inductor, a suitable MOS tube, and a suitable diode.
What is appropriate? Let’s first calculate the parameters we need and just plug them in. Because Vi changes in
the inductance
Ez9RpDSb8ADuI9P6wXUYtqmWFiD86B7ymGfhRTXp.png, we take an intermediate value of 20V, which means that the inductance we need is approximately 330UH.
MOS tubes
ScYSvtLhSFiWJM7Mq166BNCJAK3Cb4f3SXRfwF0x.pngcan easily find the MOS tube we want from the various important parameters of Lichuang Mall: STB15810
vGtPNcT9w6L8WBv3oA5pRdO1PTyfM9oShgfbWIFC.pnglooks at the brief parameters, it is a very good tube. Just choose any
 
diode that meets the requirements. MOS tube drive circuit design Bidirectional DCDC design: Because this thing came out 15 years ago, I personally think it is a pretty good blog, and we have learned a lot. Link: https://blog.csdn.net/weixin_42121843/article/details /99288773So students who prepare power supply will basically make a set of bidirectional DCDC. Our simplified diagram: Current detection design: Voltage detection design: Auxiliary power supply design: At this point, all our circuits are completed: BOOST boost, bidirectional DCDC, ADC Voltage and current collection, two minimum systems of 32, one for boosting and one for bidirectional DCDC. (In fact, the boost is also modified using two-way). 5: System control: According to the diagram above, we collect data first after powering on. If there is an abnormality, it will enter the fault state, directly perform audible and visual alarms and stop outputting PWM. If there is no abnormality, our boost MPPT will start to boost the voltage. When the output voltage is higher than 31V, we will compare Uo with the set value and send it to pid for voltage stabilization. Then when the voltage drops and does not reach 31V, we perform MPPT maximum power tracking. In this way, we directly achieve full-range maximum power tracking. The program is written according to the diagram above. The core program is as follows: This core actually also adjusts the pid. This is true for the two major modules of the system, one boost and one bidirectional. The rest is collected by the ADC. 6: Test adjustment:        When testing, we need to connect the battery first, because the voltage we initially connected is 50V. If there is no battery for charging (it must be in mode 1 at this time, if you don’t understand, you can read the analysis at the beginning), Then we can't stabilize the voltage. Why connect 50V at the beginning? Because our auxiliary power needs to work first.        During the test, the battery will be fully charged. At this time, the system will be unstable and very unstable. At this time, just switch the mode to mode 2, which is the discharge state, and the input voltage can be adjusted to 25V. , at this time the discharge current should be around 1.5A. This test was the result of the final debugging at that time. The efficiency of the two core parts was just over 95. Because two CPUs were used, there were some shortcomings, so this data seemed quite good. Test video of station B: https://www.bilibili.com/video/BV1ma411r7qa/

 

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Some questions about this video: In the video, I adjusted the internal resistance of the photovoltaic using buttons during the test. This is because I have this program in the program. No matter whether your 10Ω resistor is accurate or not, I can adjust it with 2 buttons. This is the core of our MPPT. Suppose it is not 10Ω, but 9,814,13...how to solve it? We need to understand the core meaning of MPPT first. Divide the voltage in half, U1=Rs*Is=1/2Us; U1 and Is are both collected. When we write a program for this, we can just make Rs changeable.
 
In the physical display,
8HbKNqpnJpExPkMAuCxA6yVsEYj1vuVC7hO072C6.png 
the main control we use is STMF103VCT6. MOS, MOS driver, current and voltage acquisition, etc. are all written on it. These things are available on Lichuang Mall. We also bought a lot of things on it when we were preparing for the competition. Our test points are marked on the acrylic board. We used a total of 7 multimeters. These test points are all based on the requirements of the questions. There is a sticker at the beginning.
 
7: Postscript:
This question is indeed more difficult than the question in 2015, but because the preparation time is very long, and then my previous teammates have taken the postgraduate entrance examination, I am the only one left who has nothing to do to find a good job. More than a week before the next game was about to start, two new teammates came in. The epidemic really caused twists and turns in the game. Fortunately, the result was good in the end, and it can be considered a very satisfactory result!
 
Acknowledgments: Teammates who
brought our teachers along for the ride Love to my girlfriend and everyone who supported us…



 
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