使用UC3842设计的CUK降压电路(无PCB电路板)

1、使用UC3843设计的CUK降压电路第一章 开关电源简介 1.1 开关电源原理分析开关电源是通过脉宽调制或频率调制，控制MOS管导通时间，继而控制电感线圈的磁通量，同时又要保证电感线圈不会达到磁饱和状态，从而控制输出电压的高低。同时通过反馈电路保证负载变化和输入电压变化时，输出电压仍能保证在一定范围内的稳定。1.2、开关电源分类DC/DC变换是将固定的直流电压变换成可变的直流电压，也称为直流斩波。斩波器的工作方式有两种：一是脉宽调制方式Ts不变，改变ton(通用)；二是频率调制方式，ton不变，改变Ts(易产生干扰)。其具体的电路由以下几类：(1) Buck电路降压斩波器，其输出平均电压Uo小

2、于输入电压Ui，极性相同。(2) Boost电路升压斩波器，其输出平均电压Uo大于输入电压Ui，极性相同。(3) Buck-Boost电路降压或升压斩波器，其输出平均电压Uo大于或小于输入电压Ui，极性相反，电感传输。(4) Cuk电路降压或升压斩波器，其输出平均电压Uo 大于或小于输入电压UI,极性相反，电容传输。第二章 3843设计的CUK DC-DC电路2.1、3843性能介绍The 3842A(AM)/43A(AM)/44A(AM)/45A(AM ) are fixed frequency current mode PWM controller. They are specially

3、designed for OFF Line and DC to DC converter applications with a minimal external components. Internally implemented circuits include a trimmed oscillator for precise duty cycle control, a temperature compensated reference, high gain error amplifier, current sensing comparator, and a high current to

4、tempole output ideally suited for driving a power MOSFET. Protection circuitry includes built undervoltage lockout and current limiting. The 3842A(AM) and 3844A(AM) have UVLO thresholds of 16 V (on) and 10 V (off). The corresponding thresholds for the 3843A(AM)/45A(AM) are 8.4V (on) and 7.6V (off) .

5、 The MIK3842A(AM) and MIK3843A(AM) can operate within 100% duty cycle. The 3844A(AM) and UC3845A(AM) can operate within 50% duty cycle.The 384XA(AM) has Start-Up Current 0.17mA (typ).Features Low Start-Up and Operating Current High Current Totem Pole Output Undervoltage Lockout With Hysteresis Opera

6、ting Frequency Up To 300KHz (384XA) 500KHz (384XAM)2.2、引脚定义2.3、由3843设计的CUK降压电路原理图2.4、工作原理介绍当+12V通过D1加到U1的第7脚后，随着电容C2两端电压慢慢升高，当电压超过8.6V时，U1开始启动，第8脚输出+5V50MA稳压电源。同时V1也经过D1和L1给电容C1充电，C1两端的电压达到Vi，且左+右-。第8脚输出的+5V经R1加到U1第4脚，同时为电容C3充电。这时R1和C3组成的RC振荡电路开始工作，为U1提供稳定的工作频率。当RC振荡进入稳态后，U1的第6脚开始输出PWM脉冲，并给R4耦合给Q1的G

7、极。在Ton周期，Q1导通，输入电压Vi通过L1，Q1，为电容C充电，C为负载供电。同时C1中的电能释放，通过L2和电容C沟成回路，对电容C充电，下正上负，同时部分电能转化为磁能存储在L2中。电源管Q1中的电流有两个，一个是L1中的电流，另一个是C1中的电流。在Toff周期Q1截止，输入电压Vi通过L1开始为C1充电，电容C1的电压上正下负。同时L2的电流方向不变，通过二极管D，对电容C充电，磁通转化为电能，存储在电容C中。二极管D中有两种电流，一个是对C1的充电电流，一个L2中的电流。所以无论Q1导通与否，C在Ton期间和Toff期间都有连续的充电电流。（2）、状态指示R7和LED1为指示状

8、态设置，当5V电压输出时，LED1亮起，说明5V电压已经产生。当5V负载过重或短路时，U1保护，第6脚停止输出PWM脉冲，5V电压消失，LED1熄灭。（3）、过流保护电阻R6为过流取样电阻，该电阻有两个作用，一是当该电阻上的分压大于0.8V时，U1过流保护动作，第6脚停止输出PWM脉冲，输出电压5V消失。二是该电阻能够对每一个方波进行检测，当通过的方波开启时间过长时，U1能够强制停止第6脚的输出，防止意外干扰导致Q1长时间导通引起过热损坏。所以R6不能使用导线短接，否则会引起U1长时间检测不到信号，而导致Q1长时间导通而过热损坏。（4）、输出电压调整CUK电路的输出为负压，而UC3843的BF

9、反馈端要求输入的电压为-0.3V5.5V，所以直接使用电阻分压取样是不符合要求。所以需要利用TL431和光耦配合，将负压的反馈信号转为正向电压。R1和VR1组成反馈取样电路，调整VR1的阻值，改变VR1与R1的分压比，把此电压输出到U1的第2脚，U1改变第6脚输出的PWM占空比的大小，从而改变输出电压的高低。2.5、元件的选用与取值对于电感L1、L2与耦合电容C1及输出电容C8的计算，需要假定一些条件和参数不变，如下图所示Fig 2.5When MOSFET Q1 switches on, the right hand side of inductor L1 is shorted to gro

10、und. The current in the inductor ramps according to the equation  where V is the voltage across the inductor (in this case it is equal to the input voltage), L is the inductor value and di/dt is the change in inductor current with time. Thus with a fixed voltage across the inductor and a f

11、ixed inductor value, the change in current with time is constant. When the MOSFET Q1 switches off, the inductor tries to maintain its current flow. It does this by creating a voltage across it where the right hand side tries to fly positive (to push current out of the right hand end) and the left ha

12、nd side flies negative. Since the left hand side of the inductor is clamped to the input voltage, the right hand side of the inductor flies positive to a voltage above Vin in order to maintain current flow. The energy from the inductor flows into capacitor C1 charging it with a positive voltage (whi

13、ch is higher than Vin). The right hand side of C1 is clamped to +0.3V by diode D, but for the sake of convenience we will ignore this voltage drop and assume the right hand side of the capacitor is clamped to 0V. We will work out later exactly what voltage C1 charges to, but for the moment it is suf

14、ficient to assume it charges to a voltage higher than Vin. We will call this voltage Vcap. Since the voltage Vcap is higher than Vin, the voltage across the inductor now has the opposite polarity to before. The inductor discharges according to the equation where V is the voltage across the

15、 inductor, thusIt is interesting to note that the value of di/dt is determined ONLY by the inductance value and the voltage across the inductor. The controller IC has nothing to do with setting the inductor ramp current.When the MOSFET switches on again the voltage on the drain of the MOSFET Q1 goes

16、 from Vcap to 0V. Since the voltage across a capacitor cannot change instantaneously, an equal negative going voltage appears on the anode of diode D so this node transitions from 0V to Vcap. We now have a negative amplitude square wave voltage (at the right hand node of C1) being applied to an LC f

17、ilter (L2 and C2). The LC filter averages out this square wave to produce a flat DC voltage whose amplitude is somewhere between 0V and Vcap. This amplitude is dictated by the duty cycle of the square wave. We are now going to calculate the duty cycle (the ratio of the ON time of the MOSFET Q1

18、to the total switching period) and the voltage (Vcap) on the coupling capacitor C1.  The inductor charge and discharge currents are equal when the circuit is in steady state. Thereforewhere dt1 is the ON time of the MOSFET and dt2 is the OFF time of the MOSFET. Dividing both sides by (dt1+dt2)

19、givesIf the Duty Cycle (DC) can be represented by  thensoHere we can see the Drain voltage going from 0V to Vcap (as yet uncalculated) and the ac coupled drain voltage on the anode of the diode. The capacitor has removed the dc offset and the diode has clamped the positive excursions to roughly

20、 0V.Now, when the circuit is regulating there will be a flat negative dc voltage on the output. Thus, when V(diode) is at 0V there will be a positive voltage from V(diode) to V(out) and the inductor current in L2 will ramp in a positive direction. When V(diode) is negative there will be a negative v

21、oltage from V(diode) to V(out) so the inductor current will ramp to a more negative value.In steady state, when the MOSFET switches ON V(diode) is at Vc and the voltage across inductor L2 is (-Vout-(-Vcap), thus the change in current is represented by  When the MOSFET switches OFF, the voltage

22、across L2 is (0-(-Vout), so the change in current is represented by Equating the values of di gives Dividing both sides by (dt1 + dt2) gives where DC is the duty cycle as defined above. Thus From before we know that  So So  Vout is the magnitude of the output

23、voltage. This is because in the above derivation, we have ignored the slope of di it is positive in L1 when negative in L2, so cannot strictly equate the 2 statements for DC without considering this. The result of knowing Vcap is that we now know that the Drain of the MOSFET is exposed to a vol

24、tage equal to (Vin + |Vout|) and has to be sized accordingly (as does the capacitors working voltage). Knowing thatand We can work out the Duty Cycle in terms of Vout and Vin. Thus Again, Vout is the magnitude of the output voltage. The duty cycle is set by the input and output v

25、oltages only. The inductor value does not feature in setting the duty cycle, nor does the controller IC. The above is true as long as the current in the inductor does not fall to zero. This is called Continuous Conduction Mode (CCM). If the inductor current falls to zero, the duty cycle equatio

26、n above does not hold and the controller enters Discontinuous Conduction Mode (DCM).In CCM, if the load current increases, the duty cycle remains unchanged (in steady state). The circuit reacts to the increase in load current by keeping the duty cycle constant, but the midpoint of the inductor curre

27、nt (its dc offset) increases. The switching frequency and the amplitude of the inductor ripple current remain unchanged. 1）、IC的选择根据设计要求和CUK电路原理计算公式，该电路输入电压12V，输出电压5V，输出电流500MA，所以可以计算出占空比为DC=5/(5+12)=29% < 50%由于UC3843和UC3842的占空比可以达到100%，而UC3844和UC3845的占空比最高为50%。同时UC3842和UC3844的开启电压为16V，而UC3843和UC3

28、845的开启电压为8.4V，所以设计该电路时PWM控制IC可选择为UC3843或UC3845.2）、Inductor Choice电感L1与L2It is good design practice to keep the ripple current in the inductor at 40% of the total current. This is a good trade off between small inductor size and low switching losses. The inductor on the output of a Cuk Converter is c

29、onfigured identically to that of a buck converter. With the buck converter, the average inductor current is equal to the output current. On the input, the Cuk Converter has an inductor configured identically to that of a boost converter and the average inductor current in a boost converter is equal

30、to the average input current. With an output voltage of 5V and a load of 0.5A, this represents an output power of 2.5W. Allowing for an efficiency of 80% for the converter, this means our input power has to be 3.125W. With an input voltage of 12V, this represents an average input current of 260mA.&#

31、160;If the input inductor current ripple is 40%, then the peak inductor current is 260mA x 1.2, or 312mA and the trough inductor current is 260mA x 0.8, or 208mA. The change in current is therefore 104mA. V=12V , dt=(1/100KHZ)*29%=2.9US, L=(12/0.104)*2.9US=33UHTo calculate the output inductor v

32、alue, we go through the same procedure. We know that and we know the voltage on the anode of D in FIG 2.5 is a square wave with amplitude of Vcap, we know that the output inductor has a voltage across it of Vcap Vout (=Vin) when the MOSFET is ON, so for the same ON time our output inductor

33、 should be the same value as the input inductor for the same change in current. The purists would argue that since our output current is different to the input current then keeping both inductor values the same will result in a different ripple percentage in the output inductor, so the output induct

34、or could be sized differently to reflect this, but the resulting change in circuit performance is minimal for most applications. However, it should be noted that the current in the output inductor is considerably higher in this case (since we are stepping down the voltage, so stepping up the cu

35、rrent). Therefore, if the average output inductor current is equal to the output current and we have a ripple current of 104mA, our peak output inductor current will be (1A + 52mA) = 1.052A. So our input inductor needs to have a saturation current rating of at least 312mA and our output inducto

36、r needs to have a saturation current rating of at least 1.052A. It is convenient to select 2 identical inductors (for ease of purchasing), so two 33uH inductors with a saturation current of at least 1.052A are suitable.  If too much current flows in the inductor, the ferrite that the inductor i

37、s wound on saturates and the inductor loses its inductive properties. From the equation  if the inductor value falls, the current ramp increases causing the ferrite to further saturate Therefore must make sure that the inductor never saturates.3）、MOSFET Choice电源管的选择MOSFET的选择根据电源管的工作电压和电流来确定。MOS

38、FET的耐压主要由加在电容C1两端的电压VC1决定。VC1的计算由上面推导的公式计算可得VC1=Vin+Vout=12+5V=17VMOSFET电源管的电流计算由前面的计算可知，MOSFET Q1中的电流至少为1.36A，所以选择的MOSFET管子有很多，本例中选择IRF530，当然也可以选择FQPC2N20C等电源管。当然选择不同的电源管，要考虑与R2的匹配问题。，4）、Output Capacitor Choice输出电容In continuous conduction mode, the capacitor has a continual current flowing into it

39、from the output inductor. Unlike a boost converter, the output capacitor in a buck regulator does not have to hold up the output while the inductor is being charged.The output is made up of 2 components: the ripple current from the output inductor producing a voltage across the effective series resi

40、stance (ESR) of the output capacitor and the ripple current charging the output capacitor according to the equation  Unlike a boost converter where the rectifier diode current jumps from 0A to the peak inductor current as the MOSFET switches off, the ripple in a buck architecture is determ

41、ined by the ripple current amplitude, not the peak inductor current.Recent innovations in ceramic capacitor design mean that very low ESR capacitors are available with high capacitance values. Ceramic capacitors have a typical ESR of 10mOhms.Failing that, low ESR tantalum capacitors are available in

42、 much higher capacitance values with ESR of upwards of 50m Ohms. Of course capacitors can also be paralleled to increase the capacitance and reduce the ESR.In our example the inductor ripple current is 104mA and the ESR is of a typical tantalum capacitor is 70m Ohms, giving an ESR ripple of 16.5mV.

43、To calculate the charging ripple, from the equation above we can see  For convenience the output capacitor ESR has been reduced to 0 Ohms to fully illustrate the effect of discharge ripple. It can be seen that the capacitor current has the same amplitude as the inductor ripple current, but does

44、 not have the dc offset current (of approx. 1A). This is easy to picture, since the output current is equal to the average inductor current (i.e. a straight line drawn through the middle of the inductor current) and any current that does not flow into the load must flow in and out of the capacitor.

45、To obtain the capacitor current, just subtract the output current.Now, we can see that while the capacitor current is positive (above the dotted white line) the output capacitor voltage goes up and while it is negative, the output capacitor voltage goes down. To work out the amplitude of the ripple

46、voltage on the output capacitor, we must calculate the average of the positive part of the capacitor current (above the dotted white line). Since we know the peak to peak ripple current (is equal to the inductor ripple current), the peak ripple current is Iripple/2 and hence the average of this curr

47、ent (since the current is triangular) is Iripple/4. We can now work out the charging ripple.From We can see that dt is equal to half the period, so we can say Since our capacitor current is positive for half the ON time and half the OFF time, the above equation holds true regardless of dut

48、y cycle.Lets assume we want a ripple voltage of 1% (50mV). We already have 16.5mV of ripple as a result of the capacitor ESR, so we now have to have a charging ripple of 33.5mV If our ripple current is 104mA and we are operating at a switching frequency of 100kHz, a capacitor of 3.3uF should su

49、ffice. Comparing this to the circuit in FIG 8, we can immediately see that for the same output current, the Cuk Converter has much less output capacitance. This is due to the fact that the output inductor current continually flows into the load whereas the output capacitor in the single inductor inv

50、erter has to keep the load current alive while the inductor is being charged. 5）、Output Diode Choice续流二极管D2 The output diode needs to have the lowest voltage drop possible to give the lowest power dissipation (and hence the lowest loss). A Schottky diode is an ideal choice. During the inpu

51、t inductor charge phase, the diode is exposed to a reverse voltage of Vcap, which we have determined is equal to Vin + |Vout|, thus the reverse breakdown voltage of the diode should be higher than Vcap.  To calculate the diode current we need to first consider the current in the output inductor

52、. The voltage on the anode of the diode oscillates from 0V (assuming 0V drop across the diode) to Vcap where Vcap is more negative than Vout. To keep a negative voltage on the output capacitor, the average current flowing in the output inductor must flow towards the diode (from right to left through

53、 L2 in FIG 2.5). If the ripple current in the output inductor is low compared to the average current, the current in L2 will not fall to zero so there will always be a current flowing in inductor L2 from Vout towards the diode.  When the MOSFET switches off, the current from the input inductor,

54、 L1, flows into the diode. In addition, the current from the output inductor will also flow through the diode. Therefore the total diode current during the discharge phase of the input inductor is equal to the peak current from both inductors. The diode current rating should be select accordingly. O

55、ur peak current is 1.36A, so the BYD31J is a good choice.6）、D1D1是起保护作用，防止+12V电源接反烧毁U1。由于实验板整机工作电流不大，所以该二极管选用常见的1N4007即可，参数为1.5A1000V。7）、输入电容C2C2为输入电源滤小电容，由于该电路工作在12V，所以选用16V470UF的铝电解电容即可。8）、R5与C4根据上述公式，R5应大于5K，选为10K，f=1.72(R2*C4),，如果UC3843工作在100KHZ，R5*C4=100KHZ/1.72C4=10/1.72=5.8nF9)、R4与VR1R4与VR1构成反

56、馈电压取样电路。因为U1的第2脚为FB输入端，不第2脚不接入电路时，该脚电压为2.5V，该电压与反馈电压进行比较，IC内部放大器将误差信号放大后，继而调整第6脚输出端的占空比大小。R4选用4.7K 1/8W的金属膜电阻，VR1可选用10K的半可变电位器。10） 、C6与R3C6与R3是U1的第1脚补偿输入端，因为该电路中不需要同步，所以不需要接入外接振荡源，使用标准参数接入第2脚即可。R3选用150K 1/8W的金属膜电阻，C6选用0.01UF50V的瓷片电容。11） 、C2、C7C2、C7主要是起消振防自激作用，所以一般使用1000PF50V的瓷片电容即可。12）、C3C3为缓启动电容，用于

57、在加电瞬间，因为第2脚有电压输入，防止U1出现保护。用于在加电瞬间，U1的第2脚输入电压为零。13）、R2与R7R2的作用主要是匹配IC与Q1，该电阻的选值不宜大于100欧，本电路中选用4R7 1/8W的金属膜电阻。R7是防止在Q1截止期间有意外干扰信号导致Q1意外导通而引起Q1损坏，该电阻一般选用10K-20K 1/8W的电阻。该电阻不宜过小，过小会把U1第6脚输出的PWM脉冲拉低，导致Q1不能完全开启而发热量增大。14） 、R1与R8 R1为过流检测电阻，该电阻的大小与输出功率相关，该电阻上的压降应小于1V，所以我们可计算出Q1的电流大小及电路的输出功率大小。因为该电路输出功率较小，可使用

58、0.51欧1W的金属膜电阻。RS=1.0V/(1.052A+312MA)=0.68OHMR8为隔离电阻，用于将R6的取样信号耦合给U1的第3脚，该电阻的取值一般不低于0.5K，不大于2K，防止易导致U1工作不稳定。15） 、R9与LED1该电路的输出为-5V，LED的工作电流设定为10MA，根据计算工式+5V/0.01A可得出R9的阻值应选为500欧。由于LED的工作电流应不超过20MA，最大30MA，所以改变R7大小可改变LED的发光亮度。16） 、RLRL为负载电阻，选用1W50欧的金属膜电阻，主要是防止电源空载时，输出电压持续上升。P=U0（U0/50）=5*（5/50）=0.5W17）、R12因为TL431的灌入电流可以从1MA到100MA，而光耦PC817的输入端的电流为2.530MA，所以可取TL431的工作电流