201037479 • 六、發明說明： . 【發明所屬之技術領域】 本發明涉及電子技術領域，特別涉及一種電源模組。 【先前技術】 電源模組已被廣泛應用於各種電子設備，如電視機、 電腦、DVD播放機等中，其作用在於將外部220V交流電源 轉換為直流電源，以提供給負載。 通常電源模組開啟瞬間會在其輸出端產生一過沖電 ® 壓。此過沖電壓會對連接於電源模組輸出端的負載造成損 壞。針對該問題，傳統電源模組設計時通常採用一個光耦 來構成反饋回路。即將光耦的第一輸入端通過分壓電阻連 接電源模組的輸出端，並在光耦的第一輸入端及第二輸入 端之間串接一個取樣電阻，同時光耦的第二輸入端接地。 在電源模組開啟瞬間，分壓電阻及取樣電阻共同對電源模 組輸出端產生的過沖電壓進行分壓。因此，光耦根據其第 ^ 一輸入端及第二輪入端之間的電壓差值，即取樣電阻兩端 分得的電壓值，在其輸出端產生反饋訊號。該反饋訊號被 輸入到電源模組中的控制單元，該控制單元根據反饋訊號 控制電源模組的輸出端輸出較低的電壓。 然而，該反饋訊號的幅值較小，從而使得反饋效果不 理想，並不能有效地抑制電源模組在開啟瞬間產生的過沖 電壓。 【發明内容】 有鑒於此，有必要提供一種防止電壓過沖的電源模 4 201037479 . 組。 . 一種電源模組包括電壓變換單元、輸出整流濾波單元 及反饋單元及放大單元。電壓變換單元連接交流電源，並 對交流電源輸出的交流電進行變換以生成負載交流電。輸 出整流濾波單元對負載交流電進行整流及濾波以生成負 載直流電。反饋單元對負載直流電進行取樣以分別產生第 一取樣電壓及第二取樣電壓，第一取樣電壓與第二取樣電 壓之間的差值為反饋電壓，並根據該反饋電壓產生反饋訊 ® 號，電壓變換單元根據反饋訊號調整負載交流電的大小。 放大單元包括電容及二極體，電容的第一端接收該負載直 流電，第二端接地。二極體的陰極與電容的第一端相連， 陽極與該反饋電壓的低電位端相連。負載直流電對電容進 行充電以產生電壓值缓慢增大的第三取樣電壓，放大單元 用於在第二取樣電壓及第三取樣電壓之間的差值達到二 極體的導通電壓時，使反饋電壓增大。反饋單元根據增大 ^ 的反饋電壓產生增大的反饋訊號，電壓變換單元用於根據 增大的反饋訊號減小負載交流電的大小。 該電源模組，藉由設置一個放大單元，如此當電源模 組開啟瞬間其輸出的電壓過沖時，放大單元可使反饋電壓 增大，反饋單元根據增大的反饋電壓使反饋訊號增大，電 壓變換單元根據增大的反饋訊號減少負載交流電的大 小。從而可相應地降低負載直流電的大小，達到防止電源 模組輸出電壓過沖的目的。 【實施方式】 5 201037479 如圖1所示為一較佳實栋方+201037479 • VI. Description of the invention: 1. Technical field of the invention The present invention relates to the field of electronic technology, and in particular to a power module. [Prior Art] Power modules have been widely used in various electronic devices, such as televisions, computers, DVD players, etc., and their functions are to convert an external 220V AC power source into a DC power source for supply to a load. Usually, when the power module is turned on, an overshoot + voltage is generated at its output. This overshoot voltage will damage the load connected to the output of the power module. To solve this problem, the traditional power module is usually designed with an optocoupler to form a feedback loop. The first input end of the optocoupler is connected to the output end of the power module through a voltage dividing resistor, and a sampling resistor is connected in series between the first input end and the second input end of the optocoupler, and the second input end of the optocoupler is simultaneously connected Ground. When the power module is turned on, the voltage dividing resistor and the sampling resistor jointly divide the overshoot voltage generated at the output of the power module. Therefore, the optocoupler generates a feedback signal at its output according to the voltage difference between the first input terminal and the second wheel input terminal, that is, the voltage value divided across the sampling resistor. The feedback signal is input to a control unit in the power module, and the control unit controls the output of the power module to output a lower voltage according to the feedback signal. However, the amplitude of the feedback signal is small, which makes the feedback effect unsatisfactory, and can not effectively suppress the overshoot voltage generated by the power module at the moment of opening. SUMMARY OF THE INVENTION In view of the above, it is necessary to provide a power supply mode that prevents voltage overshoot 4 201037479 . A power module includes a voltage conversion unit, an output rectification and filtering unit, a feedback unit, and an amplification unit. The voltage conversion unit is connected to the AC power source and converts the AC power output from the AC power source to generate load AC power. The output rectification filtering unit rectifies and filters the load AC to generate a load DC. The feedback unit samples the load DC power to generate a first sampling voltage and a second sampling voltage, respectively. The difference between the first sampling voltage and the second sampling voltage is a feedback voltage, and a feedback signal is generated according to the feedback voltage. The transform unit adjusts the magnitude of the load AC power according to the feedback signal. The amplifying unit comprises a capacitor and a diode, the first end of the capacitor receives the load DC power, and the second end is grounded. The cathode of the diode is connected to the first end of the capacitor, and the anode is connected to the low potential end of the feedback voltage. The load DC charges the capacitor to generate a third sampling voltage whose voltage value is slowly increased, and the amplifying unit is configured to make the feedback voltage when the difference between the second sampling voltage and the third sampling voltage reaches the conduction voltage of the diode Increase. The feedback unit generates an increased feedback signal according to the feedback voltage of increasing ^, and the voltage conversion unit is configured to reduce the magnitude of the load alternating current according to the increased feedback signal. The power module is provided with an amplifying unit, so that when the voltage of the output is overshooted when the power module is turned on, the amplifying unit can increase the feedback voltage, and the feedback unit increases the feedback signal according to the increased feedback voltage. The voltage conversion unit reduces the magnitude of the load AC power according to the increased feedback signal. Therefore, the magnitude of the load DC power can be reduced accordingly, so as to prevent overshoot of the output voltage of the power module. [Embodiment] 5 201037479 As shown in Figure 1, a better real building +
At m 模組圖。t源模組100用:十交^ 1源拉挺100之功 行轉換，以生成負載直流電提供：00輸出的交流電進 包括電壓變換單元,。、輸出整㈡=電源模_。 及放大單元40。 、反饋單元3〇 電屋變換單元10與交流電源2〇〇相連，六 200輸出的交流電進行變換，從而產生負載六又抓黾减 輸出整流濾波單元20用於對負二電。At m module diagram. t source module 100: Ten-crossing ^ 1 source puller 100 power line conversion to generate load DC power supply: 00 output AC power into the voltage conversion unit. , output whole (two) = power mode _. And amplifying unit 40. The feedback unit 3 〇 the electric house conversion unit 10 is connected to the alternating current power source 2〇〇, and the alternating current of the six 200 outputs is converted, thereby generating the load six and subtracting the output rectifying and filtering unit 20 for the negative two electric power.
濾波，以生成負載直流電。 疋订玉机及 進行取樣，以分別產生 第一取樣電壓及第二取 鲭單元3〇根據該反饋電 反饋單元30用於對負載直流電 第一取樣電壓及第二取樣電壓，該 樣電壓之間的差值為反饋電壓。反 壓產生反饋訊號。 放大單元40用於對負載直流電進行取樣以產生第三 取樣電壓，第三取樣電壓從零開始緩慢地上升。在本實施 方式中，放大單元40包括一電容值較大的電容，電容=第 一端用於接收負載直流電，電容的第二端接地，負载直流 電對電容進行充電，從而導致第三取樣電壓，即電容的第 一端電壓值，從零電位缓慢地上升。 放大單元40還用於在第二取樣電壓及第三取樣電壓 之間的差值達到預定電壓值時，使反饋電壓值增大。在本 實施方式中，該放大單元4〇包括二極體，當第二取樣電壓 及第二取樣电壓之間的差值達到二極體的導通電壓時，第 二取樣電壓被拉低至與第三取樣電壓大致相近，即第二取 6 201037479 樣電壓與第二取樣電壓之間相差一個二極體的導通電 壓，從而使反饋電壓值增大。 反饋單元30根據增大的反饋電壓產生增大的反饋訊 號，該增大的反饋訊號被傳送到電壓轉換單元10，如此電 壓轉換單元10根據該增大的反饋訊號降低負載交流電的 幅值大小，從而使輸出整流滤波單元20輸出幅值較小的負 載直流電。 圖2所示為電壓變換單元10之功能模組圖。電壓變換 〇 單元10包括輸入整流濾波單元12、控制單元14及開關變壓 單元16。 輸入整流濾波單元12連接交流電源200，其用於對交 流電源200輸出的交流電進行整流及濾波，以生成初級直 流電。 控制單元14用於接收該初級直流電以作為啟動電 壓，並產生脈衝電壓。 _ 開關變壓單元16用於接收脈衝電壓，並根據該脈衝電 ❹ 壓將初級直流電轉換為負載交流電。 反饋單元3 0所產生的增大的反饋訊號被傳送給控制 單元14，從而控制單元14根據增大的反饋訊號減少脈衝電 壓的占空比，即脈衝電壓的上升沿時間較短，下降沿時間 較長。在本實施方式中，開關變壓單元16包括MOS管及變 壓器，該MOS管的柵極與控制單元14相連，該MOS管的源 極通過電阻接地。該變壓器的初級線圈的一端連接輸入整 流濾波單元12，另一端連接MOS管的汲極。MOS管在脈衝 7 201037479 ’電壓的上升沿導通，在脈衝電壓的下降沿截止。由於M〇S .官Ml導通瞬間，變歷器的初級線圈中無電流流過，而基於 •電感對突變的電流有阻礙作用的特性，因此流過初級線圈 的電流係緩慢地增大的。由KM〇s管的導通時間較短，導 致雙壓為的初級線圈中流過的電流幅值較小，變壓器的次 級線圈感應變化的電流產生的負載交流電的幅值也相應 地降低。輪出整流濾波電路2〇基於幅值降低的負載交流電 減小負載直流電的幅值大小。 〇 當電源模組100在剛開啟瞬間產生過沖電壓時，控制 單元14可藉由減少脈衝電塵的占空比，以降低電源模組 100輸出的負載直流電的幅值大小’從而可防止負載直流 電幅值過大對負載造成損壞。 圖3所示為電源模組100之具體電路圖。輸入整流濾波 單元12與交流電源200相連。控制單元14與輸入整流滤波 單元12相連。開關變壓單元16包括MOS管Ml、下拉電P且 ❹ R7及變壓器T1。MOS管Ml的柵極連接控制單元14，m〇S 管Ml的源極通過下拉電阻R7接地。變壓器τΐ具有初级線 圈L1及次級線圈L2，初級線圈L1的一端連接輸入整流濾波 早兀12 ’另·~~端與MOS管Μ1的〉及極相連。輸出整流滤波 單元20與負載300相連。 反饋單元30包括第一取樣電阻R1、第二取樣電阻r2、 第一分壓電阻R3、第二分壓電阻R4、晶閘管Q1、光耦XJ2。 第一取樣電阻R1的一端連接輸出整流濾波單元2〇，另〆端 連接光耦U2的第一輸入端。第二取樣電阻R2連接於光耦 8 201037479 则第—輸人端及第二輸人端之間。晶閘fQ1的陰極連接 光辆U2的第二輸人端，晶閘fQi的陽㈣地。第—分 =極端連接輸出整流濾波單則 :門極1二分壓電阻R4連接於晶閘⑽的門極及地 ^單元40包括二極體D1、第三分壓電阻以及電容 Ο 〇 :端車Γ壓電阻Μ的一端連接輪出整流濾波單元1另 電容⑽第—端。電容C1的第二端接地。二極體 的陰極連接於第三分壓電阻R5及電容C1的第一端之 ㈣=體D1的陽極連接光抑2的第二輸人端，電容C1 容，。在其他實施方式中，電容C1可為電解電 電解電容的電容值較大。 電療模组100的工作原理如下： *當電源模組咖剛開啟瞬間，控制單元14輸出一個占 i衝=的脈衝電壓’也即脈衝電壓的上升沿時間較長， 的下升沿時間較短。MOS管Ml在脈衝電壓的上 線通’故M〇SfM1的導通時間較長。因此流過初級 流W)的電流（即流過MOS管Ml的汲極及源極之間的電 4也較大，從而使變壓器T4的次級線圈L2產生的負載 、尚大^大，由此輸出整流濾波單元20產生的負载直流電 ^八^〜分壓電阻R3及第二分壓電阻R4對負載直流電進 刀& ’晶閘管Q1具有一基準電壓值。 也浐口為負載直流電較大，第二分壓電阻R4分得的電壓值 也乂大，如此晶閘管Q1的門極電壓值將大於基準電壓值， 9 201037479 晶閘管Q1導通。當電源模組100穩定工作時，負載直流電 . 將較為穩定’此時第二分壓電阻R4分得的電壓值將較小， 使得晶閘管Q1的門極電壓值小於基準電壓值，晶閘管(^ 截止，因此第二取樣電阻R2上無電流流過，光搞U2的第— 輸入端及第二輸入端之間的電壓差為零，光耦U2的輸出端 不產生反饋訊號輸入至控制單元14。 晶閘管Q1導通時’晶閘管Q1的陽極及陰極之間有電济 流過。第一取樣電阻R1及第二取樣電阻R2對負載直流電進 ® 行分壓分別產生第一取樣電壓VI及第二取樣電壓V2,光執 U2的第一輸入端的電壓值即為第一取樣電壓，光執 的第二輸入端的電壓值即為第二取樣電壓V2。 負載直流電通過第三分壓電阻R5給電容C1充電，固為 電容C1的電容值較大，故第三取樣電壓V3，即電容Cl^ 第一端電壓值，由零電位緩慢地上升。 當第二取樣電壓V2與第三取樣電壓乂3之間的電聲差 ❹值達到二極體D1的導通電壓Vc(硅管二極體的導通電枣 0.7V，鍺管二極體的導通電壓為〇 2V)時，二極體〇1導通… 從而使得第二取樣電壓V2被拉低至與第三取樣電壓^^大 致相近’即V2=V3+Vc。因此，光搞奶的第一輪入端 二輸入端之間的電壓差值，即反饋電 V4=Vl-V2=Vl-V3-Vc。 炎 電源模組_剛開啟瞬間其輸出的負載直流 (電壓過沖）時，第三取樣電壓化開始為零電位， 電壓V4為其最大電壓，即V4max=V1_Vc。此後，第三取^ 201037479 • 電壓V3緩慢上升至與第二取樣電壓V2之間的壓差小於Vc 時，二極體D1截止，反饋電壓緩慢下降為V4min=Vl_V2。 相對於習知技術中沒有設置放大單元40的電源模組 100，其光耦U2的第一輸入端及第二輸入端之間的電壓差 值，即反饋電壓V4b=Vl-V2。本發明的電源模組100剛開 啟瞬間，其光耦U2的反饋電壓V4大於V4b。由於光耦U2 的反饋電壓值增大，從而使得流過光耦U2的第一輸入端及 第二輸入端之間的電流增大，即藉由光耦U2的輸出端反饋 〇 到控制單元14的反饋訊號增大。 電源模組100剛開啟瞬間其輸出的負載直流電過大 (電壓過沖）時，控制單元14可根據增大的反饋訊號有效 地將其輸出的脈衝電壓的占空比（脈衝寬度）減少。如此 變壓器T1的次級線圈L2感應初級線圈L1中的變化電流所 產生的負載交流電的幅值將降低，從而使得輸出整流濾波 電路20對負載交流電進行整流及濾波所產生的負載直流 電壓值也降低。如此可有效地防止電源模組100開啟瞬間 其負載直流電過大對外部電子設備造成損壞。 綜上所述，本發明符合發明專利要件，爰依法提出專 利申請。惟，以上所述僅為本發明之較佳實施方式，舉凡 熟悉本案技藝之人士，在援依本案創作精神所作之等效修 飾或變化，皆應包含於以下之申請專利範圍内。 【圖式簡單說明】 圖1為一較佳實施方式的電源模組之功能模組圖。 圖2為圖1中電源模組的電壓變換單元之功能模組圖。 11 201037479 圖3為圖1中電源模組之具體電路圖。 【主要元件符號說明】 電壓變換單元 10 電源模組 100 輸入整流濾波單元12 控制單元 14 開關變壓單元 16 輸出整流濾波單元20 父流電源 200 反饋單元 30 放大單元 40 負載 300 變壓器 T1 初級線圈 L1 次級線圈 L2 MOS管 Ml 電阻 Rl、R2、R3、 光耦 U2 R4 ' R5 ' R6 ' R7 電壓 VI Ύ2 ' V3 ' V4 晶閘管 Q1 二極管 D1 電容 ClFilter to generate a load DC. Sampling the jade machine and sampling to generate a first sampling voltage and a second sampling unit 3, respectively, according to the feedback electric feedback unit 30 for the first sampling voltage and the second sampling voltage of the load DC, between the samples The difference is the feedback voltage. The back pressure produces a feedback signal. The amplifying unit 40 is for sampling the load direct current to generate a third sampling voltage, and the third sampling voltage slowly rises from zero. In this embodiment, the amplifying unit 40 includes a capacitor with a large capacitance value. The capacitor=the first end is used to receive the load DC power, the second end of the capacitor is grounded, and the load DC power charges the capacitor, thereby causing the third sampling voltage. That is, the voltage value of the first terminal of the capacitor rises slowly from the zero potential. The amplifying unit 40 is further configured to increase the feedback voltage value when the difference between the second sampling voltage and the third sampling voltage reaches a predetermined voltage value. In this embodiment, the amplifying unit 4 includes a diode. When the difference between the second sampling voltage and the second sampling voltage reaches the on-voltage of the diode, the second sampling voltage is pulled down to the second The three sampling voltages are approximately similar, that is, the second taking 6 201037479 sample voltage and the second sampling voltage are different from each other by a turn-on voltage of the diode, thereby increasing the feedback voltage value. The feedback unit 30 generates an increased feedback signal according to the increased feedback voltage, and the increased feedback signal is transmitted to the voltage conversion unit 10, so that the voltage conversion unit 10 reduces the magnitude of the load AC power according to the increased feedback signal. Thereby, the output rectification filtering unit 20 outputs a load DC power having a small amplitude. FIG. 2 is a functional block diagram of the voltage conversion unit 10. The voltage conversion unit 10 includes an input rectification filtering unit 12, a control unit 14, and a switching transformer unit 16. The input rectifying and filtering unit 12 is connected to the AC power source 200 for rectifying and filtering the AC power output from the AC power source 200 to generate primary DC power. The control unit 14 is for receiving the primary direct current as a starting voltage and generating a pulse voltage. The switch transformer unit 16 is for receiving a pulse voltage and converting the primary DC power into load AC power according to the pulse voltage. The increased feedback signal generated by the feedback unit 30 is transmitted to the control unit 14, so that the control unit 14 reduces the duty cycle of the pulse voltage according to the increased feedback signal, that is, the rising edge time of the pulse voltage is shorter, and the falling edge time is Longer. In the present embodiment, the switching transformer unit 16 includes a MOS transistor and a transformer. The gate of the MOS transistor is connected to the control unit 14, and the source of the MOS transistor is grounded through a resistor. One end of the primary coil of the transformer is connected to the input rectifying and filtering unit 12, and the other end is connected to the drain of the MOS tube. The MOS transistor is turned on at the rising edge of the pulse 7 201037479 ', and is turned off at the falling edge of the pulse voltage. Since the M〇S. official M1 is turned on instantaneously, no current flows through the primary coil of the manometer, and the current flowing through the primary coil is slowly increased based on the fact that the inductance has a hindrance to the abrupt current. The conduction time of the KM〇s tube is short, and the magnitude of the current flowing in the primary coil caused by the double pressure is small, and the amplitude of the load alternating current generated by the secondary coil of the transformer inducing the change is correspondingly reduced. The round-trip rectification and filtering circuit 2 reduces the amplitude of the load DC by the load AC based on the amplitude reduction. When the power module 100 generates an overshoot voltage immediately after the power is turned on, the control unit 14 can reduce the magnitude of the load DC power output by the power module 100 by reducing the duty ratio of the pulsed dust. Excessive DC amplitude causes damage to the load. FIG. 3 shows a specific circuit diagram of the power module 100. The input rectification filter unit 12 is connected to the AC power source 200. Control unit 14 is coupled to input rectification filtering unit 12. The switching transformer unit 16 includes a MOS transistor M1, pull-down power P and ❹ R7, and a transformer T1. The gate of the MOS transistor M1 is connected to the control unit 14, and the source of the m〇S transistor M1 is grounded through the pull-down resistor R7. The transformer τ ΐ has a primary coil L1 and a secondary coil L2, and one end of the primary coil L1 is connected to the input rectification filter. The first 兀 12 ’ other terminal is connected to the MOS transistor 〉 1 and the pole. The output rectification filtering unit 20 is connected to the load 300. The feedback unit 30 includes a first sampling resistor R1, a second sampling resistor r2, a first voltage dividing resistor R3, a second voltage dividing resistor R4, a thyristor Q1, and an optical coupling XJ2. One end of the first sampling resistor R1 is connected to the output rectifying and filtering unit 2〇, and the other end is connected to the first input end of the optocoupler U2. The second sampling resistor R2 is connected to the optocoupler 8 201037479 between the first input end and the second input end. The cathode of the thyristor fQ1 is connected to the second input end of the light unit U2, and the anode (four) ground of the thyristor fQi. The first minute = the extreme connection output rectification filter single: the gate 1 two-divider resistor R4 is connected to the gate of the thyristor (10) and the ground unit 40 includes the diode D1, the third voltage dividing resistor and the capacitor Ο 〇: One end of the brake resistor Μ is connected to the first end of the rectifying and filtering unit 1 and the other capacitor (10). The second end of the capacitor C1 is grounded. The cathode of the diode is connected to the third voltage dividing resistor R5 and the first end of the capacitor C1. (4) = the anode of the body D1 is connected to the second input end of the light suppressor 2, and the capacitor C1 is charged. In other embodiments, the capacitor C1 may have a larger capacitance value of the electrolytic electrolytic capacitor. The working principle of the electrotherapy module 100 is as follows: * When the power module is just turned on, the control unit 14 outputs a pulse voltage that accounts for i rush = that is, the rising edge of the pulse voltage is longer, and the rising edge time is shorter. . The MOS transistor M1 is connected to the upper line of the pulse voltage, so that the on-time of M〇SfM1 is long. Therefore, the current flowing through the primary stream W) (that is, the electric current flowing between the drain and the source of the MOS transistor M1 is also large, so that the load generated by the secondary coil L2 of the transformer T4 is still large, The output DC power generated by the output rectification and filtering unit 20 has a reference voltage value for the load DC current feed & 'Thyristor Q1. The voltage value of the second voltage dividing resistor R4 is also large, so the gate voltage value of the thyristor Q1 will be greater than the reference voltage value, 9 201037479 thyristor Q1 is turned on. When the power module 100 is stable, the load DC power will be more stable. 'At this time, the voltage value of the second voltage dividing resistor R4 will be smaller, so that the gate voltage value of the thyristor Q1 is smaller than the reference voltage value, and the thyristor (^ is turned off, so no current flows through the second sampling resistor R2. The voltage difference between the first input terminal and the second input terminal of U2 is zero, and the output terminal of the optocoupler U2 does not generate a feedback signal input to the control unit 14. When the thyristor Q1 is turned on, there is a power between the anode and the cathode of the thyristor Q1. Flowing through. A sampling resistor R1 and a second sampling resistor R2 respectively divide the load DC power into a first sampling voltage VI and a second sampling voltage V2, and the voltage value of the first input end of the light holding U2 is the first sampling voltage, and the light The voltage value of the second input terminal is the second sampling voltage V2. The load DC power is charged to the capacitor C1 through the third voltage dividing resistor R5, and the capacitance value of the capacitor C1 is large, so the third sampling voltage V3, that is, the capacitor C1 ^ The first terminal voltage value rises slowly from zero potential. When the electroacoustic difference between the second sampling voltage V2 and the third sampling voltage 乂3 reaches the turn-on voltage Vc of the diode D1 (silicon tube diode) When the conduction current of the jujube is 0.7V and the on-voltage of the diode diode is 〇2V), the diode 〇1 is turned on... thereby causing the second sampling voltage V2 to be pulled down to be substantially similar to the third sampling voltage ^' V2=V3+Vc. Therefore, the voltage difference between the two input terminals of the first round of the light milk, that is, the feedback power V4=Vl-V2=Vl-V3-Vc. When the output of the load is DC (voltage overshoot), the third sampling voltage starts to be zero potential, voltage V4 For its maximum voltage, that is, V4max=V1_Vc. Thereafter, the third is taken as ^201037479 • When the voltage V3 rises slowly and the voltage difference between the second sampling voltage V2 and the second sampling voltage V2 is less than Vc, the diode D1 is turned off, and the feedback voltage is slowly decreased to V4min. =Vl_V2. Compared with the power module 100 in which the amplifying unit 40 is not provided in the prior art, the voltage difference between the first input terminal and the second input terminal of the optocoupler U2, that is, the feedback voltage V4b=Vl-V2. When the power module 100 of the present invention is turned on, the feedback voltage V4 of the optocoupler U2 is greater than V4b. Since the feedback voltage value of the optocoupler U2 is increased, the first input end and the second input end of the optocoupler U2 are caused to flow. The current between the two increases, that is, the feedback signal fed back to the control unit 14 by the output of the optocoupler U2 increases. When the load DC power of the output power module 100 is too high (voltage overshoot), the control unit 14 can effectively reduce the duty ratio (pulse width) of the pulse voltage outputted by the control unit 14 according to the increased feedback signal. Thus, the secondary coil L2 of the transformer T1 senses the magnitude of the load alternating current generated by the varying current in the primary coil L1, so that the output DC voltage value generated by the output rectifying and filtering circuit 20 rectifying and filtering the load alternating current is also lowered. . In this way, the power module 100 can be effectively prevented from being damaged when the load DC power is too large. In summary, the present invention complies with the requirements of the invention patent and submits a patent application according to law. However, the above description is only the preferred embodiment of the present invention. Any person who is familiar with the art of the present invention, equivalent modifications or variations in the spirit of the present invention should be included in the following patent application. BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a functional block diagram of a power module of a preferred embodiment. 2 is a functional block diagram of a voltage conversion unit of the power module of FIG. 1. 11 201037479 Figure 3 is a specific circuit diagram of the power module of Figure 1. [Main component symbol description] Voltage conversion unit 10 Power supply module 100 Input rectification and filtering unit 12 Control unit 14 Switching transformer unit 16 Output rectification and filtering unit 20 Parental power supply 200 Feedback unit 30 Amplification unit 40 Load 300 Transformer T1 Primary coil L1 times Stage coil L2 MOS tube Ml Resistor Rl, R2, R3, optocoupler U2 R4 ' R5 ' R6 ' R7 Voltage VI Ύ 2 ' V3 ' V4 Thyristor Q1 Diode D1 Capacitor Cl
1212