Zuo Youding, Jiang Zelin, Lu Jianzhao: Wireless Charging System[Copy link]
Wireless Charging SystemMay 21, 2017AbstractThe basic principle of the wireless charging system is to set a coil at both the transmitting end and the receiving end. The transmitting end coil is connected to the wired power supply and generates an electromagnetic signal. The receiving end coil senses this electromagnetic signal and transmits it to the device battery to charge. The transmitting end control chip of this system uses NE555 and IRF640 chips for hardware circuit design, including full-bridge inverter circuit and other protection circuits; in the design of the system, the influence of the transmitting coil, receiving coil and contact distance on the system performance is fully considered. By continuously adjusting the design of the coil, the dynamic adjustment of the voltage and the adjustment of the resonant frequency, the system performance is greatly improved. The maximum transmission power of the system reaches 3W. The disadvantage of electromagnetic induction technology is that the use distance is short. As the distance increases, the power loss will become very large. ContentsAbstract………………………………………………………………1Contents…………………………………………………………………21. System solution……………………………………………………41.1 Overall system design……………………………………………41.2 Justification and selection of design solution………………………41.2.1 Justification and selection of rectifier circuit………………………41.2.2 Justification and selection of drive module……………………………51.2.3 Selection and justification of coil…………………………………72. Theoretical analysis and calculation…………………………………7 2.1 Application principle of NE555………………………………………………7 2.2 Application principle of IRF640……………………………………………………9 2.3 (Power Amplification) TDA2030 Principle…………………………………………9 2.4 Wireless Transmission Principle……………………………………………11 2.5 Calculation Formula…………………………………………………113. Circuit Design……………………………………………………12 3.1 Power Module…………………………………………………12 3.2 Drive Module…………………………………………………12 3.3 Transmission Module……………………………………………13 3.4 Power Amplifier Module………………………………………144. Test Plan and Results………………………………………15 4.1 Test method……………………………………………………15 4.2 Test purpose…………………………………………………15 4.3 Test tools…………………………………………………15 4.4 Problem analysis and conclusion…………………………………15 References System Solution1.1 Overall System Solution DesignElectromagnetic induction wireless charging is the most widely used wireless charging solution. It uses a certain frequency of alternating current in the primary coil and then uses electromagnetic induction to generate a certain current in the secondary coil, thereby transferring energy from the transmitting end to the receiving end. At present, the most common wireless charging solution uses electromagnetic induction. We use hollow coils to make transmitting and receiving coils with a diameter of 5cm; use the signal generating circuit to convert the input DC 5V electricity into alternating current to power the transmitting coil. The coil uses the principle of electromagnetic induction to wirelessly transmit electrical energy to the receiving coil, generating current at the receiving end to meet the wireless charging transmission requirements. 1.2 Demonstration and selection of design solutions1.2.1 Demonstration and selection of rectifier circuitsSolution 1: Half-wave rectificationHalf-wave rectification uses the unidirectional conduction characteristics of diodes. When the input is a standard sine wave, the output obtains the positive half of the sine wave, and the negative half is lost. Solution 2: Bridge RectificationThe reverse voltage borne by each diode in the bridge circuit is equal to the maximum value of the transformer secondary voltage, which is half of that of the full-wave rectifier circuit. Bridge rectification is an improvement on the diode half-wave rectification. Bridge rectifier circuit diagram The bridge rectifier uses four diodes, two by two. When the positive half of the input sine wave is on, two tubes are turned on to obtain a positive output; when the negative half of the input sine wave is on, the other two tubes are turned on. Since these two tubes are reversely connected, the output is still the positive half of the sine wave. The utilization efficiency of the bridge rectifier for the input sine wave is twice as high as that of the half-wave rectifier.
Bridge rectifier circuit We use the power transformer to convert 220V AC into 12V AC, and then use the rectifier diode and 7805 to convert the 12V AC into 5V DC to power the transmitter. By comparing the above schemes one and two, choose scheme two. 1.1.2 Selection and demonstration of drive modules
Scheme one: quartz crystal oscillator Since the waveform generated by the quartz crystal oscillator is very stable and has very good frequency selection characteristics, it has a stable series resonant frequency fs, and the equivalent quality factor Q value is also very high. Only the signal with frequency fs is most likely to pass, while the signal groups of other frequencies will be attenuated by the crystal. The role of the resistors R1 and R2 connected in parallel between the input and output of the two inverters in the figure is to make the inverter work in the current amplification area. The resistance value of R is usually between 0.72k2 for the TTLi gate circuit and between 10-1000Q for the CMOS gate. The capacitor C1 in the circuit is used for coupling between the two inverters; the role of C2 is to suppress high-order harmonics to ensure stable frequency output; the capacitor C2 should be selected so that the RC2 parallel network produces a pole at fs to reduce the loss of the resonant signal; the selection of C1 should make the capacitive reactance of C1 at the frequency of fs negligible. The oscillation frequency of the circuit depends only on the series resonant frequency fs of the quartz crystal, and has nothing to do with the value of RC in the circuit. The function of the third CMOS inverter is to improve the output waveform,Enhance the load capacity. Solution 2: Multivibrator CircuitThe multivibrator circuit is a rectangular wave generating circuit. This circuit can continuously and periodically generate rectangular pulses by itself without an external trigger signal. The pulse is composed of a fundamental wave and multiple harmonics, so it is called a multivibrator circuit. Only simple resistors and capacitors are needed to complete a specific oscillation delay effect. Its delay range is extremely wide, ranging from a few microseconds to several hours. Its operating power supply range is extremely large, and it can be used with logic circuits such as TTL and CMOS, that is, its output level and input trigger level can match the high and low levels of these series of logic circuits. The supply current at its output end is large, and it can directly drive a variety of automatically controlled loads. Moreover, it has high timing accuracy, good temperature stability, and is cheap.
Principle analysis of the multivibrator circuit composed of 555Comparing the above two schemes, Scheme 2 has strong stability and low price, so Scheme 2 is selected. Selection and demonstration of 1, 2, 3 coilsScheme 1: Use multi-layer windingsScheme 2: Use single-layer concentric circle plane windingsSince single-layer concentric circle plane coils are easy to operate and easier to calculate capacitance, Scheme 2 is selected. II. Theoretical analysis and calculation2. 1NE555 application principleThe output voltage of the two comparators controls the state of the RS trigger and the discharge tube. Add voltage between the power supply and the ground. When pin 5 is left floating, the voltage at the non-inverting input of the voltage comparator C1 is 2VCC/3, and the voltage at the inverting input of C2 is VCC/3. If the voltage at the trigger input TR is less than VCC/3, the output of the comparator C2 is 0, and the RS flip-flop can be set to 1, making the output OUT = 1. If the voltage at the threshold input TH is greater than 2VCC/3, and the voltage at the TR terminal is greater than VCC/3, the output of C1 is 0, and the output of C2 is 1. The RS flip-flop can be set to 0, making the output low.
Principle Analysis of the Multivibrator Circuit Composed of 555Working Principle: The circuit has no steady state, only two temporary steady states, and no external trigger signal is required. The power supply VCC is used to charge the capacitor C through R1 and R2, so that uC gradually increases. When it rises to 2VCC/3, uO jumps to a low level, and the discharge terminal D is turned on. At this time, the capacitor C discharges through the resistor R2 and the D terminal, so that uC decreases. When it drops to VCC/3, uO jumps to a high level, and the D terminal is cut off. The power supply VCC charges the capacitor C through R1 and R2 again. In this way, the oscillation continues. The capacitor C charges and discharges between VCC/3 and 2VCC/3, and outputs continuous rectangular pulses. Its waveform is shown in Figure (b). 2.2IRF640 Application Principle IRF640 provides a powerful combination of fast conversion, ruggedness, low on-resistance and high efficiency. IRF640 is widely recognized in the industry. The IRF640 in D2PAK package is suitable for SMD installation. Compared with any other existing SMD package, it has the highest power and the lowest on-resistance. The D2PAK package of IRF640 can adapt to high-intensity current applications. 2.3 TDA2030 application principle [attach]373199 [/attach] TDA2030 is a Hi-Fi amplifier integrated circuit used in many computer active speakers. It is simple to connect and affordable. The rated power is 14W. The power supply voltage is ±6～±18V. The output current is large, and the harmonic distortion and crossover distortion are small (±14V/4 ohms, THD=0.5%). It has excellent short-circuit and overheat protection circuits. TDA2030 is a power amplifier integrated circuit with excellent performance. Its main features are high rise rate and small transient intermodulation distortion. Only a few types of products, including TDA 2030, have specified transient intermodulation distortion indicators. Transient intermodulation distortion is an important factor in determining the quality of the amplifier, and it is an important advantage of this integrated amplifier. Another feature of the TDA2030 integrated circuit is its large output power and relatively perfect protection performance. In the TDA 2030 integrated circuit, a relatively perfect protection circuit is designed. Once the output current is too large or the tube shell is overheated, the integrated block can automatically reduce the current or cut off to protect itself. TDA2030 has a load discharge voltage kickback protection circuit. If the peak voltage of the power supply voltage is 40V, an LC filter must be inserted between pin 5 and the power supply to ensure that the pulse train on pin 5 is maintained within the specified amplitude. Thermal protection: Thermal protection has the following advantages. It can easily withstand output overload (even for a long time) or protect when the ambient temperature exceeds the limit. Compared with ordinary circuits, heat sinks can have a smaller safety factor. In the event that the junction temperature exceeds the limit, there will be no damage to the device. If this happens, Po = (of course, Ptot) and Io will be reduced. When designing the printed circuit board, the decoupling of the ground wire and the output must be considered well, because these lines have large currents passing through them. During assembly, the heat sink does not need to be insulated, the lead length should be as short as possible, and the soldering temperature should not exceed 260 ° C for 12 seconds. Although TDA2030 requires very few components, the selected components must be of guaranteed quality. The VIN signal is sent to the 1st pin (positive phase input terminal) of TDA2030 through the 1U coupling capacitor. After amplification, the high-energy signal with the same input is output at the 4th pin (output terminal), which is provided to the transducer (speaker) to make sound. 4001 is a protection diode. For any reason, the 5th pin of TDA2030 generates high voltage (usually generated by the inductance of the speaker), which makes the voltage of the 5th pin equal to the voltage of the power supply. 2.4 Wireless Transmission Principle Design of the charging control circuit of the receiving end. After the electric energy is received by the coil, the high-frequency AC voltage is full-wave rectified by the IN4007 rectifier tube. The resonant power amplifier of the power amplifier driving circuit is composed of an LC parallel resonant circuit and a switch tube IRF640. The power amplifier driving circuit is mainly used to amplify the oscillation signal generated by the front-stage oscillation circuit, so as to send a higher oscillation signal to the lower-stage high-frequency power amplification circuit. 2.5 Calculation formula The discharge time constants of the multivibrator are: tPH≈ln2×(R1+R2)×C1 tPL≈ln2×R2×C1 The oscillation period T and oscillation frequency f are: T=tPH+tPL≈ln2×(R1+2R2)×C1 f=1/T≈1/[ln2×(R1+2R2)×C1] III. Circuit design 3.1 Power module 373200 3.2 Driving module [attach]373201 [/attach] The multivibrator composed of 555 timer is shown in Figure 3. RA, RB and C are external timing components. [attach]373202 [/attach]In the circuit, the high level trigger terminal (pin 6) and the low level trigger terminal (pin 2) are connected in parallel and then connected to the connection between RB and C, and the discharge terminal (pin 7) is connected to the connection between RA and RB. Since the capacitor C has no time to charge at the moment of power on, the voltage across the capacitor is low, which is less than (1/3) Vcc, so the high level trigger terminal and the low level trigger terminal are both low, the output is high, and the discharge tube V1 is cut off. At this time, the power supply charges the capacitor C through RA and RB, so that the voltage rises exponentially. When it rises to (2/3) Vcc, the output is low level, and the discharge tube V1 is turned on. The voltage rises from (1/3) Vcc to (2/3) Vcc. Since the discharge tube V1 is turned on, the capacitor C discharges through the resistor RB and the discharge tube, and the circuit enters the second quasi-stable state. The length of its maintenance time is related to the discharge time of the capacitor. As C discharges, it drops. When it drops to (1/3) Vcc, the output is high level, the discharge tube V1 is turned off, and Vcc charges the capacitor C again, and the circuit flips to the first quasi-stable state.
555 multivibrator circuit principle analysisAs shown in the figure, changing the value of C1 can change the period. Adjusting Rp can change the duty cycle, or R1 or R2 can be replaced with a variable resistor to adjust the period. 3, 4 Power amplifier module [attach]373204 [/attach] Fourth, test plan and results 4, 1 Test method and steps 1. Calibrate the instrument to check whether the oscilloscope and multimeter are in good condition. 2. Test waveform First, connect the wireless charging transmitter module to the power supply. After it is connected, you can check whether the voltage at both ends of the transmitting coil is a square wave. (Both sets of coils must be tested) 3. Test voltage After the waveform is determined, overlap the receiving module coil with the transmitting coil in the vertical direction. At this time, the indicator light on the receiving module is on, and use a multimeter to measure the input voltage on the receiving module. (Both sets of coils must be tested) 4. Analyze the data and record the measurement results and analyze them. 4.2 Test purpose: 1. Use an oscilloscope to check whether the input of the wireless charging transmitter module is direct current, whether the output voltage waveform is a square wave, and check the stability of the square wave. 2. Test the voltage on the wireless charging transmitter module and the receiving module, and adjust the voltage to the specified value. 4.3 Test tools: Wireless charging transmitter module Wireless charging receiving module Two sets of large and small coils Oscilloscope Multimeter 4.4 Problem analysis and conclusion When winding the large coil, the insulation layer of the coil is damaged, causing the coil to be unusable; If the wireless charging transmitter is not a square wave, it will not work properly; The voltage at the receiving end should be 5 volts, otherwise it cannot provide normal voltage for the audio. From this, we can conclude that: first, we need to change the transmitter circuit so that it emits a square wave; the voltage requirement of the receiving module must reach 5 volts, so we need to change the type of coil to increase the transmission power and increase the voltage at the receiving end; if that doesn't work, use a voltage amplifier to increase the voltage at the receiving end so that the audio can work normally. Music program part/**************************************************************************** [File name] C51 music program (August Osmanthus) [Function] Play music through the microcontroller/******************************************************************************/ #includeTest waveform First, connect the wireless charging transmitter module to the power supply. After it is connected, you can check whether the voltage at both ends of the transmitting coil is a square wave. (Both sets of coils need to be tested) 3. Test voltage After the waveform is determined, the receiving module coil and the transmitting coil are vertically overlapped. At this time, the indicator light on the receiving module is on, and the input voltage on the receiving module is measured with a multimeter. (Both sets of coils need to be tested) 4. Analyze the data and record the measurement results and analyze them. 4.2 Test purpose: 1. Use an oscilloscope to check whether the input of the wireless charging transmitter module is direct current, whether the output voltage waveform is a square wave, and check the stability of the square wave. 2. Test the voltage on the wireless charging transmitter module and the receiving module, and adjust the voltage to the specified value. 4.3 Testing tools: Wireless charging transmitter module Wireless charging receiver module Two sets of large and small coils Oscilloscope Multimeter 4.4 Problem analysis and conclusion When winding the large coil, the insulation layer of the coil was damaged, causing the coil to be unusable; If the wireless charging transmitter is not a square wave, it will not work properly; The voltage at the receiving end should be 5 volts, otherwise it cannot provide normal voltage for the audio. From this, it is concluded that the transmitter circuit must be changed first so that it transmits a square wave; the voltage requirement of the receiving module must reach 5 volts, and the type of coil must be changed first to increase the transmission power to increase the voltage at the receiving end; if not, use a voltage amplifier to increase the voltage at the receiving end for normal operation of the audio. Music program part/************************************************************************ [File name] C51 music program (August Osmanthus) [Function] Play music through the microcontroller/**************************************************************************/ #includeTest waveform First, connect the wireless charging transmitter module to the power supply. After it is connected, you can check whether the voltage at both ends of the transmitting coil is a square wave. (Both sets of coils need to be tested) 3. Test voltage After the waveform is determined, the receiving module coil and the transmitting coil are vertically overlapped. At this time, the indicator light on the receiving module is on, and the input voltage on the receiving module is measured with a multimeter. (Both sets of coils need to be tested) 4. Analyze the data and record the measurement results and analyze them. 4.2 Test purpose: 1. Use an oscilloscope to check whether the input of the wireless charging transmitter module is direct current, whether the output voltage waveform is a square wave, and check the stability of the square wave. 2. Test the voltage on the wireless charging transmitter module and the receiving module, and adjust the voltage to the specified value. 4.3 Testing tools: Wireless charging transmitter module Wireless charging receiver module Two sets of large and small coils Oscilloscope Multimeter 4.4 Problem analysis and conclusion When winding the large coil, the insulation layer of the coil was damaged, causing the coil to be unusable; If the wireless charging transmitter is not a square wave, it will not work properly; The voltage at the receiving end should be 5 volts, otherwise it cannot provide normal voltage for the audio. From this, it is concluded that the transmitter circuit must be changed first so that it transmits a square wave; the voltage requirement of the receiving module must reach 5 volts, and the type of coil must be changed first to increase the transmission power to increase the voltage at the receiving end; if not, use a voltage amplifier to increase the voltage at the receiving end for normal operation of the audio. Music program part/************************************************************************ [File name] C51 music program (August Osmanthus) [Function] Play music through the microcontroller/**************************************************************************/ #include#include //This example uses 89C52, and the crystal oscillator is 12MHZ //As for how to compile music code, it is actually very simple. You can see the following code. //The frequency constant is the pitch in musical terms, and the beat constant is the number of beats in musical terms; //So take out the score and try to compile it! sbit Beep = P1^5; unsigned char n=0; //n is the beat constant variable unsigned char code music_tab[] = { 0x18, 0x30, 0x1C , 0x10, //Format: frequency constant, beat constant, frequency constant, beat constant, 0x20, 0x40, 0x1C , 0x10, 0x18, 0x10, 0x20 , 0x10, 0x1C, 0x10, 0x18 , 0x40, 0x1C, 0x20, 0x20 , 0x20, 0x1C, 0x20, 0x18 , 0x20, 0x20, 0x80, 0xFF , 0x20, 0x30, 0x1C, 0x10 , 0x18, 0x20, 0x15, 0x20 , 0x1C, 0x20, 0x20, 0x20 , 0x26, 0x40, 0x20, 0x20 , 0x2B, 0x20, 0x26, 0x20 , 0x20, 0x20, 0x30, 0x80, 0xFF, 0x20, 0x20, 0x1C , 0x10, 0x18, 0x10, 0x20 , 0x20, 0x26, 0x20, 0x2B , 0x20, 0x30, 0x20, 0x2B , 0x40, 0x20, 0x20, 0x1C , 0x10, 0x18, 0x10, 0x20 , 0x20, 0x26, 0x20, 0x2B , 0x20, 0x30, 0x20, 0x2B , 0x40, 0x20, 0x30, 0x1C , 0x10, 0x18, 0x20, 0x15 , 0x20, 0x1C, 0x20, 0x20 , 0x20, 0x26, 0x40, 0x20 , 0x20, 0x2B, 0x20, 0x26, 0x20, 0x20, 0x20, 0x30 , 0x80, 0x20, 0x30, 0x1C , 0x10, 0x20, 0x10, 0x1C , 0x10, 0x20, 0x20, 0x26 , 0x20, 0x2B, 0x20, 0x30 , 0x20, 0x2B, 0x40, 0x20 , 0x15, 0x1F, 0x05, 0x20, 0x10, 0x1C, 0x10, 0x20 , 0x20, 0x26, 0x20, 0x2B , 0x20, 0x30, 0x20, 0x2B , 0x40, 0x20, 0x30, 0x1C , 0x10, 0x18, 0x20, 0x15 , 0x20, 0x1C, 0x20, 0x20, 0x20, 0x26, 0x40, 0x20 , 0x20, 0x2B, 0x20, 0x26 , 0x20, 0x20, 0x20, 0x30 , 0x30, 0x20, 0x30, 0x1C , 0x10, 0x18, 0x40, 0x1C , 0x20, 0x20, 0x20, 0x26 , 0x40, 0x13, 0x60, 0x18, 0x20, 0x15, 0x40, 0x13, 0x40, 0x18, 0x80,0x00 }; [size=4 ]void int0() interrupt 1 //Use interrupt 0 to control the beat { TH0=0xd8; [size=4 ] TL0=0xef; n--; } [ b] void delay (unsigned char m) //Control frequency delay[size= 4]{ unsigned i=3*m; while(--i); void delayms(unsigned char a) //millisecond delay subroutine { while(- -a); //Use while(--a) Do not use while(a--); You can compile and see the assembly result to know! } void main() [size =4]{ unsigned char p,m; //m is the frequency constant variable unsigned char i=0; TMOD&=0x0f; TMOD|=0x01; TH0=0xd8;TL0=0xef; /b] IE=0x82; play: while (1) { a: p=music_tab; /b] if(p==0x00) { i=0, delayms(1000); goto play;} //If the end character is encountered, delay 1 second and go back to the beginning and play again else if(p==0xff) { i=i+1;delayms(100),TR0=0; goto a;} //If a rest is encountered, delay 100ms and continue to play the next note else {m=music_tab[i++], n=music_tab[i++] ;} //Get frequency constant and beat constant TR0=1; //Open timer 1 [size= 4] while(n!=0) Beep=~Beep,delay(m); //Wait for the beat to finish, output audio through P1 port (multi-channel!) [ size=4] TR0=0; //Turn off timer 1 } } [/ size]Go back to the beginning and do it again else if(p==0xff) { i=i+1;delayms(100),TR0=0; goto a;} //If a rest is encountered, delay 100ms and continue to pick the next note else {m=music_tab[i++], n=music_tab[i++];} //Get the frequency constant and beat constant TR0=1; //Turn on timer 1 while(n!=0) Beep=~Beep,delay(m); //Wait for the beat to complete, and output audio through P1 port (multi-channel is possible!) TR0=0; //Turn off timer 1 } } Go back to the beginning and do it again else if(p==0xff) { i=i+1;delayms(100),TR0=0; goto a;} //If a rest is encountered, delay 100ms and continue to take the next note else {m=music_tab[i++], n=music_tab[i++];} //Get the frequency constant and beat constant TR0=1; //Turn on timer 1 while(n!=0) Beep=~Beep,delay(m); //Wait for the beat to complete, and output audio through P1 port (multi-channel is possible!) TR0=0; //Turn off timer 1 } }
Very well written and very detailed. I believe that if you really understand these contents, you will make great progress. Come on, come on, and post more in the future
提升卡
变色卡
千斤顶
京公网安备 11010802033920号
