使用UC3842设计的CUK降压电路无PCB电路板.docx

1、使用UC3842设计的CUK降压电路无PCB电路板使用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：(

2、1) Buck电路降压斩波器，其输出平均电压Uo小于输入电压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 contro

3、ller. They are specially 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 comparat

4、or, and a high current totempole 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

5、.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 and Operating Current High Current Totem Pole Output Undervoltage Lock

6、out With Hysteresis Operating 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

7、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

8、的充电电流。（2）、状态指示R7和LED1为指示状态设置，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

9、压调整CUK电路的输出为负压，而UC3843的BF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 indu

10、ctor L1 is shorted to ground. The current in the inductor ramps according to the equationwhere 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 ind

11、uctor and a fixed 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) an

12、d the left hand 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

13、 voltage (which 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 mom

14、ent 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 Vin, the voltage across the inductor now has the opposite polarity to before. The inductor discharges according to the equationwhere V is the voltage across th

15、e 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 goe

16、s 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

17、filter (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 to th

18、e 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) givesIf the

19、 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 0V.Now, when the

20、 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 voltage from V(dio

21、de) 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 across L2 is (0-(-Vout)

22、, so the change in current is represented byEquating the values of di givesDividing both sides by (dt1 + dt2) giveswhere DC is the duty cycle as defined above.ThusFrom before we know that SoSo Vout is the magnitude of the output voltage. This is because in the above derivation, we have ignored the s

23、lope 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 voltage equal to (Vin + |Vout|) and has to be sized accordingly (as does the cap

24、acitors working voltage).Knowing thatandWe can work out the Duty Cycle in terms of Vout and Vin. ThusAgain, Vout is the magnitude of the output voltage.The duty cycle is set by the input and output voltages only. The inductor value does not feature in setting the duty cycle, nor does the controller

25、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 equation above does not hold and the controller enters Discontinuous Conduction Mode (DCM).In CCM, if the load cur

26、rent 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 current (its dc offset) increases. The switching frequency and the amplitude of the inductor ripple current rema

27、in 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 r

28、ipple 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 configured identically to that of a buck converter. With the buck converter, the average inductor current is equa

29、l 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 to the average input current. With an output voltage of 5V and a load of 0.5A, this represents an output power o

30、f 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.If the input inductor current ripple is 40%, then the peak inductor current is 260mA x 1.2, or 312mA and the troug

31、h 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 value, we go through the same procedure.We know thatand we know the voltage on the anode of D in FIG 2.5 is a square wave wit

32、h 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 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 diffe