使用UC3842設(shè)計(jì)的CUK降壓電路(無PCB電路板)

1、使用UC3843設(shè)計(jì)的CUK降壓電路第一章 開關(guān)電源簡介 1.1 開關(guān)電源原理分析開關(guān)電源是通過脈寬調(diào)制或頻率調(diào)制，控制MOS管導(dǎo)通時(shí)間，繼而控制電感線圈的磁通量，同時(shí)又要保證電感線圈不會(huì)達(dá)到磁飽和狀態(tài)，從而控制輸出電壓的高低。同時(shí)通過反饋電路保證負(fù)載變化和輸入電壓變化時(shí)，輸出電壓仍能保證在一定范圍內(nèi)的穩(wěn)定。1.2、開關(guān)電源分類DC/DC變換是將固定的直流電壓變換成可變的直流電壓，也稱為直流斬波。斬波器的工作方式有兩種：一是脈寬調(diào)制方式Ts不變，改變ton(通用)；二是頻率調(diào)制方式，ton不變，改變Ts(易產(chǎn)生干擾)。其具體的電路由以下幾類：(1) Buck電路降壓斬波器，其輸出平均電壓Uo小

2、于輸入電壓Ui，極性相同。(2) Boost電路升壓斬波器，其輸出平均電壓Uo大于輸入電壓Ui，極性相同。(3) Buck-Boost電路降壓或升壓斬波器，其輸出平均電壓Uo大于或小于輸入電壓Ui，極性相反，電感傳輸。(4) Cuk電路降壓或升壓斬波器，其輸出平均電壓Uo 大于或小于輸入電壓UI,極性相反，電容傳輸。第二章 3843設(shè)計(jì)的CUK DC-DC電路2.1、3843性能介紹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、引腳定義2.3、由3843設(shè)計(jì)的CUK降壓電路原理圖2.4、工作原理介紹當(dāng)+12V通過D1加到U1的第7腳后，隨著電容C2兩端電壓慢慢升高，當(dāng)電壓超過8.6V時(shí)，U1開始啟動(dòng)，第8腳輸出+5V50MA穩(wěn)壓電源。同時(shí)V1也經(jīng)過D1和L1給電容C1充電，C1兩端的電壓達(dá)到Vi，且左+右-。第8腳輸出的+5V經(jīng)R1加到U1第4腳，同時(shí)為電容C3充電。這時(shí)R1和C3組成的RC振蕩電路開始工作，為U1提供穩(wěn)定的工作頻率。當(dāng)RC振蕩進(jìn)入穩(wěn)態(tài)后，U1的第6腳開始輸出PWM脈沖，并給R4耦合給Q1的G

7、極。在Ton周期，Q1導(dǎo)通，輸入電壓Vi通過L1，Q1，為電容C充電，C為負(fù)載供電。同時(shí)C1中的電能釋放，通過L2和電容C溝成回路，對電容C充電，下正上負(fù)，同時(shí)部分電能轉(zhuǎn)化為磁能存儲(chǔ)在L2中。電源管Q1中的電流有兩個(gè)，一個(gè)是L1中的電流，另一個(gè)是C1中的電流。在Toff周期Q1截止，輸入電壓Vi通過L1開始為C1充電，電容C1的電壓上正下負(fù)。同時(shí)L2的電流方向不變，通過二極管D，對電容C充電，磁通轉(zhuǎn)化為電能，存儲(chǔ)在電容C中。二極管D中有兩種電流，一個(gè)是對C1的充電電流，一個(gè)L2中的電流。所以無論Q1導(dǎo)通與否，C在Ton期間和Toff期間都有連續(xù)的充電電流。（2）、狀態(tài)指示R7和LED1為指示狀

8、態(tài)設(shè)置，當(dāng)5V電壓輸出時(shí)，LED1亮起，說明5V電壓已經(jīng)產(chǎn)生。當(dāng)5V負(fù)載過重或短路時(shí)，U1保護(hù)，第6腳停止輸出PWM脈沖，5V電壓消失，LED1熄滅。（3）、過流保護(hù)電阻R6為過流取樣電阻，該電阻有兩個(gè)作用，一是當(dāng)該電阻上的分壓大于0.8V時(shí)，U1過流保護(hù)動(dòng)作，第6腳停止輸出PWM脈沖，輸出電壓5V消失。二是該電阻能夠?qū)γ恳粋€(gè)方波進(jìn)行檢測，當(dāng)通過的方波開啟時(shí)間過長時(shí)，U1能夠強(qiáng)制停止第6腳的輸出，防止意外干擾導(dǎo)致Q1長時(shí)間導(dǎo)通引起過熱損壞。所以R6不能使用導(dǎo)線短接，否則會(huì)引起U1長時(shí)間檢測不到信號(hào)，而導(dǎo)致Q1長時(shí)間導(dǎo)通而過熱損壞。（4）、輸出電壓調(diào)整CUK電路的輸出為負(fù)壓，而UC3843的BF

9、反饋端要求輸入的電壓為-0.3V5.5V，所以直接使用電阻分壓取樣是不符合要求。所以需要利用TL431和光耦配合，將負(fù)壓的反饋信號(hào)轉(zhuǎn)為正向電壓。R1和VR1組成反饋取樣電路，調(diào)整VR1的阻值，改變VR1與R1的分壓比，把此電壓輸出到U1的第2腳，U1改變第6腳輸出的PWM占空比的大小，從而改變輸出電壓的高低。2.5、元件的選用與取值對于電感L1、L2與耦合電容C1及輸出電容C8的計(jì)算，需要假定一些條件和參數(shù)不變，如下圖所示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 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 inductor and a fixed inducto

11、r 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 hand side flie

12、s 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 (which is higher

13、 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 sufficient to a

14、ssume 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 the inductor, thusIt is int

15、eresting 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 from Vcap to 0V. Since

16、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 filter (L2 and C2). The L

17、C 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 the total switching period)

18、 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 Duty Cycle (DC) can be r

19、epresented 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 circuit is regulating th

20、ere 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(diode) to V(out) so the indu

21、ctor 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), so the change in curren

22、t 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 slope of di it is positive

23、 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 capacitors working voltage).

24、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 IC.The above is true as l

25、ong 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 current increases, the duty

26、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 remain unchanged. 1）、IC的選擇根據(jù)設(shè)

27、計(jì)要求和CUK電路原理計(jì)算公式，該電路輸入電壓12V，輸出電壓5V，輸出電流500MA，所以可以計(jì)算出占空比為DC=5/(5+12)=29% 50%由于UC3843和UC3842的占空比可以達(dá)到100%，而UC3844和UC3845的占空比最高為50%。同時(shí)UC3842和UC3844的開啟電壓為16V，而UC3843和UC3845的開啟電壓為8.4V，所以設(shè)計(jì)該電路時(shí)PWM控制IC可選擇為UC3843或UC3845.2）、Inductor Choice電感L1與L2It is good design practice to keep the ripple current in the indu

28、ctor 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 equal to the output current.

29、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 of 2.5W. Allowing for an e

30、fficiency 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 trough inductor current is 260

31、mA 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 with amplitude of Vcap, we k

32、now 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 c

33、urrent then keeping both inductor values the same will result in a different ripple percentage in the output inductor, so the output inductor 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

34、the current in the output inductor is considerably higher in this case (since we are stepping down the voltage, so stepping up the 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 b

35、e (1A + 52mA) = 1.052A.So our input inductor needs to have a saturation current rating of at least 312mA and our output inductor 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 satur

36、ation current of at least 1.052A are suitable. If too much current flows in the inductor, the ferrite that the inductor is 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 satu

37、rate Therefore must make sure that the inductor never saturates.3）、MOSFET Choice電源管的選擇MOSFET的選擇根據(jù)電源管的工作電壓和電流來確定。MOSFET的耐壓主要由加在電容C1兩端的電壓VC1決定。VC1的計(jì)算由上面推導(dǎo)的公式計(jì)算可得VC1=Vin+Vout=12+5V=17VMOSFET電源管的電流計(jì)算由前面的計(jì)算可知，MOSFET Q1中的電流至少為1.36A，所以選擇的MOSFET管子有很多，本例中選擇IRF530，當(dāng)然也可以選擇FQPC2N20C等電源管。當(dāng)然選擇不同的電源管，要考慮與R2的匹配問題。，

38、4）、Output Capacitor Choice輸出電容In continuous conduction mode, the capacitor has a continual current flowing into it 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 u

39、p of 2 components: the ripple current from the output inductor producing a voltage across the effective series resistance (ESR) of the output capacitor and the ripple current charging the output capacitor according to the equationUnlike a boost converter where the rectifier diode current jumps from

40、0A to the peak inductor current as the MOSFET switches off, the ripple in a buck architecture is determined 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. Cera

41、mic capacitors have a typical ESR of 10mOhms.Failing that, low ESR tantalum capacitors are available in 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 curren

42、t is 104mA and the ESR is of a typical tantalum capacitor is 70m Ohms, giving an ESR ripple of 16.5mV. 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 b

43、e seen that the capacitor current has the same amplitude as the inductor ripple current, but does 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

44、current) and any current that does not flow into the load must flow in and out of the capacitor. 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

45、 it is negative, the output capacitor voltage goes down. To work out the amplitude of the ripple 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

46、inductor ripple current), the peak ripple current is Iripple/2 and hence the average of this current (since the current is triangular) is Iripple/4. We can now work out the charging ripple.FromWe can see that dt is equal to half the period, so we can saySince our capacitor current is positive for ha

47、lf the ON time and half the OFF time, the above equation holds true regardless of duty 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 33.5mVIf our ripple current is 104mA and we a

48、re operating at a switching frequency of 100kHz, a capacitor of 3.3uF should suffice. 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 continua

49、lly flows into the load whereas the output capacitor in the single inductor inverter has to keep the load current alive while the inductor is being charged.5）、Output Diode Choice續(xù)流二極管D2The output diode needs to have the lowest voltage drop possible to give the lowest power dissipation (and hence the

50、 lowest loss). A Schottky diode is an ideal choice. During the input 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

51、 we need to first consider the current in the output inductor. 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 ind

52、uctor must flow towards the diode (from right to left through 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 MO

53、SFET switches off, the current from the input inductor, 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. T

54、he diode current rating should be select accordingly. Our peak current is 1.36A, so the BYD31J is a good choice.6）、D1D1是起保護(hù)作用，防止+12V電源接反燒毀U1。由于實(shí)驗(yàn)板整機(jī)工作電流不大，所以該二極管選用常見的1N4007即可，參數(shù)為1.5A1000V。7）、輸入電容C2C2為輸入電源濾小電容，由于該電路工作在12V，所以選用16V470UF的鋁電解電容即可。8）、R5與C4根據(jù)上述公式，R5應(yīng)大于5K，選為10K，f=1.72(R2*C4),，如果UC3843工作在100

55、KHZ，R5*C4=100KHZ/1.72C4=10/1.72=5.8nF9)、R4與VR1R4與VR1構(gòu)成反饋電壓取樣電路。因?yàn)閁1的第2腳為FB輸入端，不第2腳不接入電路時(shí)，該腳電壓為2.5V，該電壓與反饋電壓進(jìn)行比較，IC內(nèi)部放大器將誤差信號(hào)放大后，繼而調(diào)整第6腳輸出端的占空比大小。R4選用4.7K 1/8W的金屬膜電阻，VR1可選用10K的半可變電位器。10） 、C6與R3C6與R3是U1的第1腳補(bǔ)償輸入端，因?yàn)樵撾娐分胁恍枰?，所以不需要接入外接振蕩源，使用?biāo)準(zhǔn)參數(shù)接入第2腳即可。R3選用150K 1/8W的金屬膜電阻，C6選用0.01UF50V的瓷片電容。11） 、C2、C7C2

56、、C7主要是起消振防自激作用，所以一般使用1000PF50V的瓷片電容即可。12）、C3C3為緩啟動(dòng)電容，用于在加電瞬間，因?yàn)榈?腳有電壓輸入，防止U1出現(xiàn)保護(hù)。用于在加電瞬間，U1的第2腳輸入電壓為零。13）、R2與R7R2的作用主要是匹配IC與Q1，該電阻的選值不宜大于100歐，本電路中選用4R7 1/8W的金屬膜電阻。R7是防止在Q1截止期間有意外干擾信號(hào)導(dǎo)致Q1意外導(dǎo)通而引起Q1損壞，該電阻一般選用10K-20K 1/8W的電阻。該電阻不宜過小，過小會(huì)把U1第6腳輸出的PWM脈沖拉低，導(dǎo)致Q1不能完全開啟而發(fā)熱量增大。14） 、R1與R8 R1為過流檢測電阻，該電阻的大小與輸出功率相關(guān)

57、，該電阻上的壓降應(yīng)小于1V，所以我們可計(jì)算出Q1的電流大小及電路的輸出功率大小。因?yàn)樵撾娐份敵龉β瘦^小，可使用0.51歐1W的金屬膜電阻。RS=1.0V/(1.052A+312MA)=0.68OHMR8為隔離電阻，用于將R6的取樣信號(hào)耦合給U1的第3腳，該電阻的取值一般不低于0.5K，不大于2K，防止易導(dǎo)致U1工作不穩(wěn)定。15） 、R9與LED1該電路的輸出為-5V，LED的工作電流設(shè)定為10MA，根據(jù)計(jì)算工式+5V/0.01A可得出R9的阻值應(yīng)選為500歐。由于LED的工作電流應(yīng)不超過20MA，最大30MA，所以改變R7大小可改變LED的發(fā)光亮度。16） 、RLRL為負(fù)載電阻，選用1W50歐的金屬膜電阻，主要是防止電源空載時(shí)，輸出電壓持續(xù)上升。P=U0（U0/50）=5*（5/50）=0.5W17）、R12因?yàn)門L431的灌入電流可以從1MA到100MA，而光耦PC817的輸入端的電流為2.530MA，所以可取TL431的工作電流20MA，這時(shí)R12的取值為R12=5/20MA=250OHM.18)、TL431與光耦PC817的使用因?yàn)镃UK電路輸出的電壓是負(fù)電壓，而