使用UC3842设计的CUK降压电路无PCB电路板

1、百度文库-让每个人平等地捉升口我使用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，同时乂要保证电感线圈不会达到磁饱和状态，从而控制输出电圧的高低。同 时通过反馈电路保证负载变化和输入电圧变化时，输出电圧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圧变换成可变的直流电圧，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

2、路一一降压斩波器，其输出平均电压Uo小于输入电压Ui,极性相同。Boost电路一一升压斩波器，其输出平均电压Uo大于输入电压Ui,极性相同。(3) Buck-Boost电路一一降压或升圧斩波器，其输出平均电压Uo大丁或小于输入电 压Ui,极性相反，电感传输。(4) Cuk电路一一降圧或升圧斩波器，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。Boost Buck Mode SMPS#百度文库-让每个人平等地捉升口我第二章 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

3、 frequency current mode PWM controller. They are specially desig ned for OFF- Line and DC to DC conv er ter applicati ons with a minimal exter nal comp onents. I nternally implemented circuits in elude a trimmed oscillator for precise duty cycle control, a temperature compensated referenee, high gai

4、n error amplifier, current sensing comparator, and a high current totempole output ideally suited for drivi ng a power MOSFET Protect! on circuitry in eludes built un dervoltage lockout and current limiting. The 3842A(AM) and 3844A(AM) have UVLO thresholds of 16 V (on) and 10 V (off). The correspond

5、ing thresholds for the 3843A(AM)/45A(AM) are 8.4V (on) and 7.6V (off) 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 a nd Operati ng Current Hi

6、gh Current Totem Pole Output Undervoltage Lockout With Hysteresis Operating Frequency Up To 300KHz (384XA) 500KHz (384XAM)2.2. 引脚定义NFunctionDescription1COMPThis pn is me Error Amplifier output and is made for kx>o compensation.2VfbThis is the inverting in pul of tiie Error Amplifier. It is normal

7、ly connected to the switching power supply output throixjh a resistor divider.3I SENSEA TOltas proportional to inductor current is connected to this input The PWM uses this informetion to terminate the output s亦 1ch conduction.4Rt/CtThe oscillator frequency and maxmum Output duty cycle aro programme

8、d Dy connocting resistor RT to V“ and capacitor Cr to o<ound.5GROUNDThis pin is the combined control circuitry and povr qround6OUTPUTThis output directly drives the gate of a power MOSFET. Peak currents up to 1A are sourced and sink by7VccThis pin is the positive supply of the inteqrated circuiL8

9、JThis is the reference outojt. it oroviccs charoma current for caoaator C r throuah resistor Rt 2.3s由3843设计的CUK降压电路原理图52.4、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，当电圧超过 8.6V时，U1开始启动，第8脚输出+5V50MA稳压电源。同时VI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

10、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极。在Ton周期，Q1导通，输入电圧Vi通过LI, 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 4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，输入电圧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Q

11、1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为指示状态设置，当一5V电压输出时，LED1亮起，说明一5V电压已经 产生。当一5V负载过重或短路时，U1保护，第6脚停止输出PWM脉冲，一5V电圧消 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圧大于0.8V时, U1过流保护动作，第6脚停止输出PWM脉冲，输出电压一5V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 百度文阵-让每个人平專

12、地捉升口我使用导线短接，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电路的输出为负圧，而UC3843的BF反馈端耍求输入的电压为-0.3V-5.5V,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，将负圧的反 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 圧输出到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。25、元件的选用与取值对于电感LI、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L$TW

13、hen MOSFET QI switches onz the right hand side of inductor LI is shorted to ground. The current in the inductor ramps according to the equation#百度文库-让每个人平等地捉升口我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

14、in inductor current with time. Thus with a fixed voltage across the inductor and a fixed inductor value, the change in current with time is constant.Whe n the MOSFET QI switches off, the in ductor tries to maintain its curre nt flow It does this by creating a voltage across it where the right hand s

15、ide tries to fly positive (to push current out of the right hand end) and the left hand side flies negative Since the left hand side of the in ductor is clamped to the input voltage, the right hand side of the in ductor flies positive to a voltage above Vin in order to maintain current flow The ener

16、gy from the in ductor flows into capacitor Cl charging it with a positive voltage (which is higher than Vin). The right hand side of Cl 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

17、 will work out later exactly what voltage Cl charges to, but for the moment it is sufficient to assume it charges to a voltage higher than Vin. We will call this voltage Vcap.Since the voltage Vcap is higher than Vinz the voltage across the inductor now has the opposite polarity to before The in duc

18、tor discharges accordi ng to the equatio nwhere V is the voltage across the in ductor, 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 in ductor ramp current When

19、 the MOSFET switches on again the voltage on the drain of the MOSFET QI goes 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 OV to -Vcap We now have a negative amplitude

20、square wave voltage (at the right hand node of Cl) being applied to an LC filter (L2 and C2). The LC filter averages out this square wave to produce a flat DC voltage whose amplitude is somewhere between OV and -Vcap This amplitude is dictated by the duty cycle of the square wave17We are now going t

21、o calculate the duty cycle (the ratio of the ON time of the MOSFET QI to the total switching period) and the voltage (Vcap) on the coupling capacitor Cl.The in ductor charge and discharge currents are equal when the circuit is in steady state. Thereforewhere dti is the ON time of the MOSFET and dt2

22、is the OFF time of the MOSFET.Dividing both sides by (dti+dt2)givesIf the Duty Cycle (DC) can be represented bythenHere we can see the Drain voltage going from OV to Vcap (as yet uncalculated) and the ac coupled drain voltage on the anode of the diode The capacitor has removed the de offset and the

23、diode has clamped the positive excursions to roughly OV.Now, when the circuit is regulating there will be a flat negative de voltage on the output Thus, when V(diode) is at OV there will be a positive voltage from V(diode) to V(out) and the inductor curre nt in L2 will ramp in a positive di recti on

24、. When V(diode) is negative there will be a negative voltage from V(diode) to V(out) so the inductor current will ramp to a more negative valueIn 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 represent

25、ed byWhen the MOSFET switches OFF, the voltage across L2 is (0-(-Vout)z so the change in current is represented byEquating the values of di givesDividing both sides by (dti + dt2)giveswhere DC is the duty cycle as defined above.ThusFrom before we know thatSoSoVout is the magnitude of the output volt

26、age This is because in the above derivation, we have ignored the slope of di 一 it is positive in LI 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 voltage equa

27、l to (Vin + |Vout|) and has to be sized accordingly (as does the capacitor's working voltage)Knowing thatand百度文库-让每个人平等地捉升口我We can work out the Duty Cycle in terms of Vout and Vin. ThusAgain, Vout is the magnitude of the output voltageThe duty cycle is set by the input and output voltages only.

28、The inductor value does not feature in setting the duty cycle, nor does the controller IC.The above is true as long as the curre nt in the in ductor does not fall to zero. This is called Continuous Conduction Mode (CCM). If the inductor current falls to zero, the duty cycle equation above does not h

29、old and the controller enters Discontinuous Conduction Mode (DCM).In CCM, if the load current in creases, the duty cycle remains un cha nged (in steady state). The circuit reacts to the increase in load current by keeping the duty cycle constant, but the midpoint of the in ductor curre nt (its de of

30、fset) in creases The 16百度文库-让每个人平等地捉升口我switching frequency and the amplitude of the inductor ripple current remain unchangedI) 、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的开启电圧为16V,而UC384

31、3和UC3845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 curre nt. This is a good trade off betwee n small in ductor size and low switching losses The inductor on the output of a Cuk C

32、onverter is con figured 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 Con verter has an in ductor configured ide ntically to that of a boost con verter and the average in ductor current in a boost c

33、onverter is equal to the average input current.With an output voltage of 5V and a load of 0.5AZ 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 3i 25W With an in put voltage of 12VZ this represe nts an average input cu

34、rrent of 260mAIf the in put in ductor curre nt ripple is 40%, the n the peak in ductor 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 104mAJ空 StL dt也V=12V, dt=(l/100KHZ)*29%=2.9USzL=(12/0.104)*2.9US=33UHTo calculate the ou

35、tput inductor value, we go through the same procedureWe know that23and 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 in ductor has a voltage across it of Vcap 一 Vout (=Vin) when the MOSFET is ON, so for the same ON time our output i

36、nductor should be the same value as the input in ductor for the same change in curre nt. 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 perce ntage in the output inductor, so the out

37、put in ductor could be sized differently to reflect this, but the resulting change in circuit performanee 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 t

38、he current) 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 in ductor needs to have a saturation current rati ng of at least 312mA and our output inducto

39、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 suitableIf too much current flows in the inductor, the ferrite that the inductor is wound o

40、n saturates and the in ductor loses its in ductive properties From the equationif the inductor value falls, the current ramp increases causing the ferrite to further saturate. Therefore must make sure that the inductor never saturates3) x MOSFET Choice电源管的选择MOSFET的选择根据电源管的匸作电压和电流来确定。MOSFET的耐压主要由加在电容

41、 C1两端的电压VC1决定。计算可得VC1的计算由上而推导的公式VC1= Vin+Vout=12+5V=17VMOSFET电源管的电流计算由前面的计算可知，M0SFETQ1中的电流至少为1.36A,所以选择的MOSFET管了有 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) 、OutDUt CaDacitor Choice 输出电容In continuous conduct!on mode, the capacitor has a continual current flowing into it from the

42、 output inductor Unlike a boost converter, the output capacitor in a buck regulator does not have to hold up the output while the in ductor is being charged The output is made up of 2 comp on ents: the ripple curre nt from the output inductor producing a voltage across the effective series resistanc

43、e (ESR) of the output capacitor and the ripple current charging the output capacitor according to the equationUn like a boost converter where the rectifier diode current jumps from OA to the peak inductor current as the MOSFET switches o仟，the ripple in a buck architecture is determined by the ripple

44、 curre nt amplitude, not the peak in ductor 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 lOmOhms.Failing that, low ESR tantalum capacitors are available in much higher capa

45、citance values with ESR of upwards of 50m Ohms Of course capacitors can also be paralleled to increase the capacitance and reduce the ESRIn our example the in ductor ripple current is 104mA and the ESR is of a typical tantalum capacitor is 70m Ohms, giving an ESR ripple of 16.5mVTo calculate the cha

46、rging ripple, from the equation above we can seeFor 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 not have the de offset curr

47、ent (of approx 1A) This is easy to picture, since the output current is equal to the average in ductor curre nt (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 To obtain the capacitor curren

48、t, 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 voltage on the output capacito

49、r, 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 in ductor ripple current), the peak ripple current is Iripple/2 and hence the average of this current (since the current is tria

50、ngular) is Iripple/4. We can now work out the charging rippleFromWe can see that dt is equal to half the period, so we can say百度文库-让每个人平等地捉升口我Since our capacitor current is positive for half the ON time and half the OFF time, the above equation holds true regardless of duty cycleLet's assume we

51、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 33.5mVIf our ripple current is 104mA and we are operating at a switching frequency of 100kHz, a capacitor of 3.3uF should suffice Comparing this to the c

52、ircuit in FIG 8Z 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 in ductor current continually flows into the load whereas the output capacitor in the single in ductor inverter has to keep the load cu

53、rrent alive while the inductor is being charged 5) 、Output Diode Choice 续流二极管 D2The 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 Duri ng the in put inductor charge phase, the diode is

54、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 VcapTo calculate the diode current we need to first consider the current in the output in ductor The voltage on the anode of the diode oscill

55、ates from 0V (assumi ng 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 L2 in FIG 23百度文库-让每个人平等地捉升口我25) If the ripp

56、le 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 diodeWhe n the MOSFET switches off, the curre nt from the input in ductor, Llz flows into the diode In addi

57、tion, the current from the output in ductor will also flow through the diode Therefore the total diode current during the discharge phase of the input in ductor is equal to the peak curre nt from both inductors The diode current rati ng should be select accordi ngly. Our peak current is 1.36A, so the BYD31J is a good choice6) 、DIDI是起保护作用，防止+12V电源接反烧毁Ul。由丁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1.5A1000Vo7) 、输入电容C2C2为输入电源滤小电容,由丁该电路匸作在12V,所以选用16V470UF的铝电解电容即可。8)