TW201941511A - Relay protection device - Google Patents
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- 239000004065 semiconductor Substances 0.000 claims abstract description 132
- 239000003990 capacitor Substances 0.000 claims description 11
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- 229910044991 metal oxide Inorganic materials 0.000 claims description 5
- 150000004706 metal oxides Chemical class 0.000 claims description 5
- 230000005622 photoelectricity Effects 0.000 claims 1
- 239000012530 fluid Substances 0.000 description 4
- 239000004020 conductor Substances 0.000 description 2
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Abstract
Description
本發明繼電器保護裝置,具有在電路應用過程中負載兩端發生過載或短路之保護功能及其包括有第一半導體、第二半導體、第一電阻器、光電耦合器及繼電器之電子技術領域。 The relay protection device of the present invention has the protection function of overload or short circuit occurring at both ends of a load during a circuit application process, and an electronic technology field including a first semiconductor, a second semiconductor, a first resistor, a photocoupler, and a relay.
如圖1所示,為習知電池保護裝置之電路圖,當開關83閉合時直流電源81電流由直流電源81的正電端經過保險絲82或熱電路斷路器,再經過開關83供電於繼電路85之電磁線圈第一端C1與電磁線圈第二端C2,再回到直流電源81的負電端而完成一繼電器85之供電動作,此時繼電器85的第一接點K1與第二接點K2導通直流電源81供電於負載84,當負載84兩端發生短路時保險絲82燒斷或熱電路斷路器跳開,以保護電池81之損壞,其有以下之缺點: As shown in FIG. 1, it is a circuit diagram of a conventional battery protection device. When the switch 83 is closed, the current of the DC power source 81 passes the positive terminal of the DC power source 81 through the fuse 82 or the thermal circuit breaker, and then supplies power to the relay circuit through the switch 83 The first end C1 of the electromagnetic coil 85 and the second end C2 of the electromagnetic coil return to the negative electrical end of the DC power supply 81 to complete the power supply action of the relay 85. At this time, the first contact K1 and the second contact K2 of the relay 85 The conductive DC power source 81 supplies power to the load 84. When a short circuit occurs across the load 84, the fuse 82 is blown or the thermal circuit breaker is tripped to protect the battery 81 from damage. It has the following disadvantages:
1.當直流電源與保險絲串聯供電於負載,若負載端發生短路時,短路電流達到保險絲之熔點時保險絲熔斷,直流電源與負載端開路,其缺點為保險絲熔斷前其需要一段時間之電流損耗,而且保險絲熔斷瞬間有火花產生,可能造成火災之危險及更換保險絲 需要時間之浪費。 1. When the DC power supply and the fuse are connected in series to the load, if a short circuit occurs at the load end, the fuse will blow when the short-circuit current reaches the melting point of the fuse, and the DC power supply will open circuit with the load. In addition, sparks are generated when the fuse is blown, which may cause a fire hazard and replace the fuse. It takes a waste of time.
2.當直流電源與熱電路斷路器串聯供電於負載,若負載端發生短路時,短路電流達到熱電路斷路器之熱度時,熱電路斷路器接觸點跳開,直流電源與負載端開路,其缺點為熱電路斷路器跳脫前其需要一段時間之電流損耗及熱電路斷路器要恢復原狀態時需要冷却時間或手壓恢復之動作,而且接觸點跳開的温度與復原的温度不同,其熱電路斷路器(Thermal Circuit Breaker)有動作次數之壽命限制。 2. When the DC power supply is connected to the thermal circuit breaker in series to supply power to the load, if a short circuit occurs at the load end, when the short-circuit current reaches the thermal circuit breaker's heat, the contact point of the thermal circuit breaker trips, and the DC power supply opens to the load end. The disadvantages are the current loss of the thermal circuit breaker before it trips, and the cooling time or manual pressure recovery action of the thermal circuit breaker to restore the original state, and the temperature at which the contact point trips is different from the recovery temperature. The Thermal Circuit Breaker has a life limit for the number of operations.
本發明的目的: Purpose of the invention:
本發明應用第一半導體、第二半導體、第一電阻器、光電耦合器及繼電器,能在直流電源供電中負載發生過載或短路時保護直流電源。 The invention applies the first semiconductor, the second semiconductor, the first resistor, the photocoupler and the relay, and can protect the DC power supply when the load is overloaded or short-circuited during the DC power supply.
本發明應用第一半導體、第二半導體、第一電阻器、光電耦合器及繼電器再外加全波整流器,能在交流電源供電中負載發生過載或短路時保護交流電源。 The present invention uses a first semiconductor, a second semiconductor, a first resistor, a photocoupler, and a relay, plus a full-wave rectifier, to protect the AC power source when the load is overloaded or short-circuited during the AC power supply.
當負載發生短路時,本發明應用第一半導體及繼電器能在極短之時間內執行第二半導體之開路動作,達到保護直流電源或交流電源之功能及避免因負載短路所引起之各種災害。 When the load is short-circuited, the invention uses the first semiconductor and the relay to perform the open circuit action of the second semiconductor in a very short time, to achieve the function of protecting the DC power supply or the AC power supply and avoiding various disasters caused by the load short-circuit.
本發明應用光電耦合器,執行本發明重置(Reset)之功能,達到短路原因排除時不必重新再送直流電源或交流電源的動作。 The present invention uses a photocoupler to perform the reset function of the present invention. When the cause of the short circuit is eliminated, it is not necessary to resend the DC power or AC power.
本發明有下列之特徵: The invention has the following features:
1.本發明之繼電器設有一電磁線圈及二個接點,其負責直流電源或交流電源之開路(Off)與導通(On)供電於負載。 1. The relay of the present invention is provided with an electromagnetic coil and two contacts, which are responsible for supplying power to the load by turning off and on the DC power source or the AC power source.
2.本發明之第一半導體,其負責控制繼電器的電磁 線圈開路與導通動作。 2. The first semiconductor of the present invention, which is responsible for controlling the electromagnetic of the relay The coil opens and conducts.
3.本發明之第二半導體,其負責控制第一半導體電路之開路與導通動作。 3. The second semiconductor of the present invention is responsible for controlling the opening and conducting actions of the first semiconductor circuit.
4.本發明之光電耦合器負責重置功能,當負載短路排除時,不必重啟直流電源或交流電源之開路與導通動作。 4. The photocoupler of the present invention is responsible for the reset function. When the load is short-circuited, it is not necessary to restart the open circuit and conduction of the DC power supply or AC power supply.
5.本發明之第一半導體為N通道金屬氧化半導體場效電晶體。 5. The first semiconductor of the present invention is an N-channel metal oxide semiconductor field effect transistor.
6.本發明之第二半導體包括N型電晶體或N通道金屬氧化半導體場效電晶體兩者可以根據需求自行選用。 6. The second semiconductor of the present invention includes an N-type transistor or an N-channel metal oxide semiconductor field effect transistor. Both can be selected according to demand.
7.本發明所應用之光電耦合器有電晶體輸出型與閘流體輸出型,可以根據需求自行選用。 7. The photocoupler used in the present invention has a transistor output type and a thyristor output type, which can be selected according to demand.
8.本發明之時間控制電路選用電阻電容時間常數電路或其他時間控制電路。 8. The time control circuit of the present invention selects a resistance-capacitance time constant circuit or other time control circuits.
11‧‧‧第一半導體 11‧‧‧First Semiconductor
12‧‧‧第二半導體 12‧‧‧Second Semiconductor
13‧‧‧光電耦合器電晶體輸出型 13‧‧‧Photocoupler transistor output type
14‧‧‧繼電器 14‧‧‧ Relay
15‧‧‧第三半導體 15‧‧‧Third Semiconductor
16‧‧‧光電耦合器閘流體輸出型 16‧‧‧Photocoupler brake fluid output type
17‧‧‧全波整流器 17‧‧‧full wave rectifier
21‧‧‧第一電阻器 21‧‧‧first resistor
22‧‧‧第二電阻器 22‧‧‧Second resistor
23‧‧‧第一電容器 23‧‧‧first capacitor
30‧‧‧第一端 30‧‧‧ the first end
40‧‧‧第二端 40‧‧‧ second end
50‧‧‧第三端 50‧‧‧ third end
60‧‧‧第一開關 60‧‧‧First switch
R‧‧‧第一開關的主接點 R‧‧‧ the main contact of the first switch
V‧‧‧第一開關的第一接點 V‧‧‧ the first contact of the first switch
U‧‧‧第一開關的第二接點 U‧‧‧ the second contact of the first switch
T‧‧‧第一開關的第三接點 T‧‧‧ the third contact of the first switch
100‧‧‧本發明繼電器保護裝置 100‧‧‧ Relay protection device of the present invention
200‧‧‧負載 200‧‧‧ load
300‧‧‧直流電源 300‧‧‧DC Power
310‧‧‧第一直流電源 310‧‧‧First DC Power Supply
320‧‧‧第二直流電源 320‧‧‧second DC power supply
400‧‧‧交流電源 400‧‧‧AC power
500‧‧‧交流負載 500‧‧‧AC load
圖1為習知電池保護裝置之電路圖。 FIG. 1 is a circuit diagram of a conventional battery protection device.
圖2為本發明繼電器保護裝置之電路圖。 FIG. 2 is a circuit diagram of a relay protection device according to the present invention.
圖3為本發明繼電器保護裝置第一實施例。 FIG. 3 is a first embodiment of a relay protection device according to the present invention.
圖4為本發明繼電器保護裝置第二實施例。 FIG. 4 is a second embodiment of the relay protection device of the present invention.
圖5為本發明繼電器保護裝置第三實施例。 FIG. 5 is a third embodiment of the relay protection device of the present invention.
如圖2所示,為本發明繼電器保護裝置之電路圖,自圖中可知,本發明繼電器保護裝置100包括有第一半導體11(First Semiconductor,11)、第二半導體12(Second Semiconductor,12)、第一電阻器21 (First Resistor,21)、光電耦合器13(Photocoupler,13)及繼電器14(Relay,14);外端連接有第一開開60(First Switch,60)、時間控制電路70(Time Controll Circuit,70)、負載200(Load,200)、直流電源300(DC Power Source,300);第一半導體11的源極S(Source,S)連接第二半導體12的射極E(Emitter,E)、光電耦合器13的陰極端N(Cathode,N)、繼電器14的第二接點K2及第三端50;第一半導體11的閘極G(Gate,G)連接第二半導體12的集極C(Collector,C)及第一電阻器21的另一端,第一電阻器21的一端連接繼電器14的電磁線圈第一端C1及第一端30,繼電器14的電磁線圈第二端C2連接第一半導體11的汲極D(Drain,D);第二半導體12的基極G(Base,B)連接光電耦合器13的射極E,光電耦合器13的集極C連接繼電器14的第一接點K2及第二端40;光電耦合器13的陽極端P(Anode,P)連接時間控制器70的輸出端72;第一端30連接第一開關60的主接點R及時間控制電路70的輸入端71,第一開關60設有主接點R、第一接點V、第二接點U及第三接點T,第一接點V連接直流電源300的正電端及負載200的一端,負載200的另一端連接第二端40,直流電源300的負電端及時間控制電路70的接地端73連接第三端50;第一半導體11為N通道金屬氧化半導體場效電晶體,第二半導體12為N型電晶體;光電耦合器13為電晶體輸出型(Transistor Output Type),其輸入側為發光二極體(LED),輸出側為電晶體;繼電器14設有電磁線圈(Electric Magnetic Coil)第一端C1、電磁線圈第二端C2、第一接點K1及第二接點K2,其第一 接點K1及第二接點K2構成一組接點以執行電路的開路(Off)與導通(On)動作。 As shown in FIG. 2, it is a circuit diagram of the relay protection device of the present invention. As can be seen from the figure, the relay protection device 100 of the present invention includes a first semiconductor 11 (First Semiconductor, 11), a second semiconductor 12 (Second Semiconductor, 12), First resistor 21 (First Resistor, 21), Photocoupler 13 (Photocoupler, 13) and Relay 14 (Relay, 14); the first end is connected to First Switch 60 (First Switch, 60), Time Control Circuit 70 (Time Controll Circuit, 70), load 200 (Load, 200), DC power source 300 (DC Power Source, 300); the source S (Source, S) of the first semiconductor 11 is connected to the emitter E (Emitter, E) of the second semiconductor 12, The cathode terminal N (Cathode, N) of the photocoupler 13, the second contact K2 and the third terminal 50 of the relay 14; the gate G (Gate, G) of the first semiconductor 11 is connected to the collector C of the second semiconductor 12 (Collector, C) and the other end of the first resistor 21, one end of the first resistor 21 is connected to the first end C1 and the first end 30 of the electromagnetic coil of the relay 14, and the second end C2 of the electromagnetic coil of the relay 14 is connected to the first The drain D (Drain, D) of the semiconductor 11; the base G (Base, B) of the second semiconductor 12 is connected to the emitter E of the photocoupler 13, and the collector C of the photocoupler 13 is connected to the first connection of the relay 14. Point K2 and the second terminal 40; the anode terminal P (Anode, P) of the photocoupler 13 is connected to the output terminal 72 of the time controller 70; the first terminal 30 is connected to the main contact R of the first switch 60 and the time control circuit 70 Lose At the input 71, the first switch 60 is provided with a main contact R, a first contact V, a second contact U, and a third contact T. The first contact V is connected to the positive electrical terminal of the DC power source 300 and the load 200. At one end, the other end of the load 200 is connected to the second end 40, the negative electrical end of the DC power supply 300 and the ground end 73 of the time control circuit 70 are connected to the third end 50; the first semiconductor 11 is an N-channel metal oxide semiconductor field effect transistor. The two semiconductors 12 are N-type transistors; the photocoupler 13 is a transistor output type. The input side is a light emitting diode (LED) and the output side is a transistor. The relay 14 is provided with an electromagnetic coil (Electric Magnetic Coil) First end C1, second end C2 of the electromagnetic coil, first contact K1 and second contact K2, the first The contact K1 and the second contact K2 constitute a group of contacts to perform the Open and On actions of the circuit.
如圖2所示,當第一開關60的主接點R轉向第一接點V,此時直流電源300的正電端供電於第一端30及時間控制電路70的輸入端71,同時直流電源300供電於負載200的一端到負載200的另一端再到第二端40,而直流電源300的負電端連接第三端50,此時第一端30供電於繼電器14電磁線圈的第一端C1及第一電阻器21的一端,經由第一電阻器21的另一端到第一半導體11的閘極G及第二半導體12的集極C,因此第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,直流電源300供電於負載200,而後時間控制電路70的輸出端72供電於光電耦合器13的陽極端P到陰極端N,此時光電耦合器13的集極C與射極E導通,因為繼電器14的第一接點K1與第二接點K2導通,所以第二半導體12的基極B與射極E電壓低,因此第二半導體12的集極C與射極E開路。 As shown in FIG. 2, when the main contact R of the first switch 60 is turned to the first contact V, the positive terminal of the DC power source 300 is supplied to the first terminal 30 and the input terminal 71 of the time control circuit 70 at the same time. The power supply 300 supplies power from one end of the load 200 to the other end of the load 200 to the second end 40, and the negative electrical end of the DC power supply 300 is connected to the third end 50. At this time, the first end 30 supplies power to the first end of the electromagnetic coil of the relay 14. C1 and one end of the first resistor 21 pass through the other end of the first resistor 21 to the gate G of the first semiconductor 11 and the collector C of the second semiconductor 12, so the drain D and the source of the first semiconductor 11 S is turned on, and the solenoid coil of the relay 14 is supplied with voltage. The first contact K1 and the second contact K2 are conducted. The DC power source 300 supplies power to the load 200, and then the output terminal 72 of the time control circuit 70 supplies power to the photocoupler 13. The anode terminal P to the cathode terminal N. At this time, the collector C and the emitter E of the photocoupler 13 are turned on. Because the first contact K1 and the second contact K2 of the relay 14 are turned on, the base B of the second semiconductor 12 is turned on. Since the voltage with the emitter E is low, the collector C and the emitter E of the second semiconductor 12 are open.
如圖2所示,若將第一開關60的主接點R轉向第二接點U時,直流電源300不供電於第一端30,此時第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,直流電源300不供電於負載100,同時光電耦合器13的集極C與射極E開路,第二半導體12的基極B與射極E無電壓。 As shown in FIG. 2, when the main contact R of the first switch 60 is turned to the second contact U, the DC power source 300 does not supply power to the first terminal 30. At this time, the drain D and the source S of the first semiconductor 11 are at this time. Open circuit, the solenoid coil of relay 14 has no voltage supply, its first contact K1 and second contact K2 are open, DC power supply 300 is not supplying power to load 100, and at the same time, collector C and emitter E of photocoupler 13 are open circuit. The base B and the emitter E of the two semiconductors 12 have no voltage.
如圖2所示,當第一端30接有直流電源300時,直流電源300供電於負載200兩端,若將負載200兩端短路,其等同將直流電源300的正電壓供電於 光電耦合器13的集極C,此時因為第二半導體12的集極C與射極E導通及第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,直流電源300不供電於負載200,而達到短路保護第一直流電源300的目的。 As shown in FIG. 2, when a DC power source 300 is connected to the first terminal 30, the DC power source 300 supplies power to both ends of the load 200. If the two ends of the load 200 are short-circuited, it is equivalent to powering the positive voltage of the DC power source 300 to The collector C of the photocoupler 13 is at this time because the collector C and the emitter E of the second semiconductor 12 are turned on, and the gate G and the source S of the first semiconductor 11 are turned on, so the drain of the first semiconductor 11 is turned on. D is open to the source S, and the solenoid coil of the relay 14 has no voltage supply. The first contact K1 and the second contact K2 are open. The DC power source 300 does not supply power to the load 200, and the short-circuit protection of the first DC power source 300 is achieved. purpose.
如圖2所示,當第一端30接有直流電源300時,直流電源300供電於負載200兩端,當負載200發生短路時,先將負載200兩端短路的原因去除,再將第一開關60的主接點R轉向第二接點U,再轉向第一接點V,此時光電耦合器13的集極C與射極E導通,第二半導體12的基極B與射極E兩端電壓低,第二半導體12的集極C與射極E開路,第一半導體11的閘極G受有正電壓,於是第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,直流電源300供電於負載200;若將第一開關60的主接點R轉向第二接點U,直流電源300不供電於第一端30,此時直流電源300不供電於負載200。 As shown in FIG. 2, when a DC power source 300 is connected to the first end 30, the DC power source 300 supplies power to both ends of the load 200. When a short circuit occurs in the load 200, the cause of the short circuit at both ends of the load 200 is removed, and then the first The main contact R of the switch 60 is turned to the second contact U and then to the first contact V. At this time, the collector C of the photocoupler 13 is connected to the emitter E, and the base B and the emitter E of the second semiconductor 12 are turned on. The voltage across the two terminals is low, the collector C and the emitter E of the second semiconductor 12 are open, and the gate G of the first semiconductor 11 receives a positive voltage. Therefore, the drain D and the source S of the first semiconductor 11 are turned on. The electromagnetic coil has a voltage supply, and the first contact K1 and the second contact K2 are turned on, and the DC power source 300 supplies power to the load 200. If the main contact R of the first switch 60 is turned to the second contact U, the DC power source 300 does not Power is supplied to the first terminal 30. At this time, the DC power supply 300 is not powered to the load 200.
如圖2所示,當第一端30接有直流電源300時,直流電源300供電於負載200兩端,若將負載200加大亦就是增大負載200電流量,此時若繼電器14的第一接點K1與第二接點K2二點間之電壓降值大於第二半導體12的基射極導通電壓時,第二半導體12的集極C與射極E導通,第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓 供應,其第一接點K1與第二接點K2開路,直流電源300不供電於負載200,而達到過電流保護直流電源300的目的。 As shown in FIG. 2, when a DC power source 300 is connected to the first terminal 30, the DC power source 300 supplies power to both ends of the load 200. If the load 200 is increased, the load 200 current is increased. When the voltage drop between the two points of a contact K1 and the second contact K2 is greater than the base-emitter on-voltage of the second semiconductor 12, the collector C and the emitter E of the second semiconductor 12 are turned on, and the gate of the first semiconductor 11 is turned on. Both ends of the electrode G and the source S are conducted, so that the drain D of the first semiconductor 11 and the source S are open, and the electromagnetic coil of the relay 14 has no voltage. Supply, its first contact K1 and second contact K2 are open, and the DC power source 300 does not supply power to the load 200, but achieves the purpose of overcurrent protection of the DC power source 300.
如圖3所示,為本發明繼電器保護裝置第一實施例,自圖中可知,時間控制電路70選用電阻電容時間常數電路,其電阻電容時間常數電路中之電阻為第二電阻器22,其電阻電容時間常數電路中之電容為第一電容器23,第二電阻器22之一端連接本發明繼電器保護裝置100中之第一端30,第二電阻器22之另一端連接第一電容器23的正電端及光電耦合器13的陽極端P,第一電容器23的負電端連接第三端50;第一直流電源310(First DC Power Source,310)之正電端連接第一開關60的第一接點V及負載200的一端,負載200的另一端連接第二端40,第一直流電源310之負電端連接第三端50;第二直流電源320(Second DC Power Source,320)之正電端連接第一開關60的第三接點T,第二直流電源320之負電端連接第三端50。 As shown in FIG. 3, this is the first embodiment of the relay protection device of the present invention. As can be seen from the figure, the time control circuit 70 selects a resistance-capacitance time constant circuit. The resistance in the resistance-capacitance time constant circuit is the second resistor 22. The capacitance in the resistance-capacitance time constant circuit is the first capacitor 23, one end of the second resistor 22 is connected to the first terminal 30 in the relay protection device 100 of the present invention, and the other end of the second resistor 22 is connected to the positive terminal of the first capacitor 23. The electrical terminal and the anode terminal P of the photocoupler 13, the negative electrical terminal of the first capacitor 23 is connected to the third terminal 50; the positive electrical terminal of the first DC power source 310 (First DC Power Source, 310) is connected to the first terminal of the first switch 60 A contact V and one end of the load 200, the other end of the load 200 is connected to the second end 40, the negative electrical end of the first DC power source 310 is connected to the third end 50, and the second DC power source 320 (Second DC Power Source, 320) The positive electrical terminal is connected to the third contact T of the first switch 60, and the negative electrical terminal of the second DC power source 320 is connected to the third terminal 50.
如圖3所示,第一直流電源310的正電端供電於第一開關60之主接點R及負載200之一端,再由負載200之另一端連接到第二端40,當第一開關60的主接點R轉向第一接點V,第一直流電源310的正電端供電於第一端30及第二電阻22之一端,正電壓供電於第一電阻器21到第一半導體11的閘極G及第二半導體12的集極C,因此第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,第一直流電源300供電於負載200,而後第二電阻器22之另一端供電於光電耦合器13的陽極端P到陰 極端N,此時光電耦合器13的集極C與射極E導通,第二半導體12的基極B與射極E電壓低,因此第二半導體12的集極C與射極E開路;其中第二電阻器22與第一電容器23構成一時間常數電路(Time Constant Circuit),可以控制光電耦合器13的陽極端P與陰極端N的導通時間,以達到重置時間的控制,其時間常數電路可以用同等功能的時間控制器替代,而不自限。 As shown in FIG. 3, the positive electric terminal of the first DC power source 310 is supplied to the main contact R of the first switch 60 and one end of the load 200, and then the other end of the load 200 is connected to the second end 40. The main contact R of the switch 60 is turned to the first contact V. The positive electrical terminal of the first DC power source 310 is supplied to one of the first terminal 30 and the second resistor 22. The positive voltage is provided to the first resistor 21 to the first. The gate G of the semiconductor 11 and the collector C of the second semiconductor 12 are connected. Therefore, the drain D of the first semiconductor 11 is connected to the source S, and the electromagnetic coil of the relay 14 is supplied with voltage. The first contact K1 is connected to the second contact. The point K2 is turned on, the first DC power source 300 supplies power to the load 200, and then the other end of the second resistor 22 supplies power to the anode terminal P of the photocoupler 13 to the cathode. Extreme N. At this time, the collector C and the emitter E of the photocoupler 13 are turned on, and the voltage between the base B and the emitter E of the second semiconductor 12 is low. Therefore, the collector C and the emitter E of the second semiconductor 12 are open. The second resistor 22 and the first capacitor 23 constitute a time constant circuit, which can control the conduction time between the anode terminal P and the cathode terminal N of the photocoupler 13 to achieve the control of the reset time. The circuit can be replaced by an equivalent function time controller without limitation.
如圖3所示,若將第一開關60的主接點R轉向第二接點U時,第一直流電源310不供電於第一端30,此時第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,光電耦合器13的集極C與射極E開路,第二半導體12的基極B與射極E無電壓。 As shown in FIG. 3, when the main contact R of the first switch 60 is turned to the second contact U, the first DC power source 310 does not supply power to the first terminal 30. At this time, the drain D of the first semiconductor 11 and the The source S is open, and the solenoid coil of the relay 14 has no voltage supply. The first contact K1 and the second contact K2 are open. The first DC power source 310 does not supply power to the load 200. The collector C of the photocoupler 13 and The electrode E is open, and the base B and the emitter E of the second semiconductor 12 have no voltage.
如圖3所示,當第一開關60的主接點R轉向第三接點T,此時第二直流電源320的正電端供電於第一端30及第一直流電源310的正電端供電於負載200到第二端40,而第一直流電源310的負電端及第二直流電源320的負電端連接第三端50,因為有第二直流電源320供電於第一端30,此時第一端30供電於第一電阻器21到第一半導體11的閘極G及第二半導體12的集極C,此時第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,第一直流電源310供電於負載200,而後經由第二電阻器22供電於光電耦合器13的陽極端P與陰極端N,此時光電耦合器13的集極C與射極E導通,第二半導體12的基極B與射極E電壓低,因此第二半 導體12的集極C與射極E開路。 As shown in FIG. 3, when the main contact R of the first switch 60 is turned to the third contact T, the positive terminal of the second DC power source 320 is supplied to the first terminal 30 and the positive terminal of the first DC power source 310. The terminal is powered from the load 200 to the second terminal 40, and the negative terminal of the first DC power source 310 and the negative terminal of the second DC power source 320 are connected to the third terminal 50 because the second DC power source 320 is powered by the first terminal 30. At this time, the first terminal 30 supplies power to the gate G of the first resistor 21 to the first semiconductor 11 and the collector C of the second semiconductor 12. At this time, the drain D of the first semiconductor 11 and the source S are turned on, and the relay 14 The electromagnetic coil has a voltage supply. The first contact K1 and the second contact K2 are turned on. The first DC power source 310 supplies power to the load 200, and then supplies the anode terminal P and the photocoupler 13 via the second resistor 22. The cathode terminal N, at this time, the collector C and the emitter E of the photocoupler 13 are turned on, and the voltage between the base B and the emitter E of the second semiconductor 12 is low, so the second half The collector C and the emitter E of the conductor 12 are open.
如圖3所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,若將負載200兩端短路,其等同將第一直流電源310的正電壓供電於光電耦合器13的集極C,此時第二半導體12的集極C與射極E導通,等同第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,而達到短路保護第一直流電源310的目的。 As shown in FIG. 3, when the first terminal 30 is connected to the second DC power source 320, the first DC power source 310 supplies power to both ends of the load 200. If the two ends of the load 200 are short-circuited, it is equivalent to the first DC power source 310. The positive voltage is supplied to the collector C of the photocoupler 13. At this time, the collector C and the emitter E of the second semiconductor 12 are turned on, which is the same as that the gate G and the source S of the first semiconductor 11 are turned on, so the first The drain D and the source S of the semiconductor 11 are open-circuited, and the electromagnetic coil of the relay 14 has no voltage supply. The first contact K1 and the second contact K2 are open. The first DC power source 310 is not supplied to the load 200, and a short circuit is reached. Purpose of protecting the first DC power source 310.
如圖3所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,當負載200發生短路時,先將負載200兩端短路的原因去除,再將第一開關60的主接點R轉向第二接點U,再轉向第三接點T,此時光電耦合器13的集極C與射極E導通,第二半導體12的基極B與射極E兩端電壓低,第二半導體12的集極C與射極E開路,第一半導體11的閘極G受有正電壓,於是第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,亦就是第一直流電源310供電於負載200;若將第一開關60的主接點R轉向第二接點U,第二直流電源320不供電於第一端30,此時第一直流電源310不供電於負載200。 As shown in FIG. 3, when the first terminal 30 is connected to the second DC power supply 320, the first DC power supply 310 supplies power to both ends of the load 200. When the load 200 is short-circuited, the cause of the short-circuit at both ends of the load 200 is removed first. Then, the main contact R of the first switch 60 is turned to the second contact U, and then to the third contact T. At this time, the collector C of the photocoupler 13 is connected to the emitter E, and the base of the second semiconductor 12 is turned on. The voltage across B and the emitter E is low, the collector C of the second semiconductor 12 is open with the emitter E, and the gate G of the first semiconductor 11 receives a positive voltage, so the drain D and the source S of the first semiconductor 11 When it is turned on, the electromagnetic coil of the relay 14 is supplied with voltage. The first contact K1 and the second contact K2 are conducted, that is, the first DC power source 310 supplies power to the load 200. If the main contact R of the first switch 60 is turned to The second contact U and the second DC power source 320 are not powered by the first terminal 30. At this time, the first DC power source 310 is not powered by the load 200.
如圖3所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,若將負載200加大亦就是增大負載200電流量,此時若繼電器14的第一接點K1與第二接點K2二點間之 電壓降值大於第二半導體12的基射極導通電壓時,第二半導體12的集極C與射極E導通,等同第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,而達到過電流保護第一直流電源310的目的。 As shown in FIG. 3, when the first terminal 30 is connected to the second DC power source 320, the first DC power source 310 supplies power to both ends of the load 200. If the load 200 is increased, the load 200 current is increased. If between the first contact K1 and the second contact K2 of the relay 14 When the voltage drop is greater than the base-emitter on-voltage of the second semiconductor 12, the collector C and the emitter E of the second semiconductor 12 are turned on, which is the same as that the gate G and the source S of the first semiconductor 11 are turned on, so the first The drain D and the source S of the semiconductor 11 are open-circuited, and the electromagnetic coil of the relay 14 is not supplied with voltage. The first contact K1 and the second contact K2 are open. The first DC power source 310 does not supply power to the load 200, and has reached The purpose of the current protection of the first DC power source 310.
如圖4所示,為本發明繼電器保護裝置第二實施例,自圖中可知,將第三半導體15替代圖2的第二半導體12,其第三半導體15的汲極D替代第二半導體12的集極C,第三半導體15的源極S替代第二半導體12的射極E,第三半導體15的閘極G替代第二半導體12的基極B;將光電耦合器16為閘流體輸出型(Thyristor Output Type)替代圖2的光電耦合器13之電晶體輸出型,其光電耦合器16之輸入側為發光二極體(LED),輸出側為閘流體,其閘流體具有第一端A1與第二端A2,其餘電路組成皆與圖2本發明繼電器保護裝置電路100相同,而不贅述。 As shown in FIG. 4, this is a second embodiment of the relay protection device of the present invention. As can be seen from the figure, the third semiconductor 15 is replaced by the second semiconductor 12 of FIG. 2, and the drain D of the third semiconductor 15 is replaced by the second semiconductor 12. The collector C, the source S of the third semiconductor 15 replaces the emitter E of the second semiconductor 12, and the gate G of the third semiconductor 15 replaces the base B of the second semiconductor 12. The photocoupler 16 is used as the gate fluid output. Type (Thyristor Output Type) replaces the phototransistor output type of the photocoupler 13 in FIG. 2. The input side of the photocoupler 16 is a light emitting diode (LED), and the output side is a gate fluid. The gate fluid has a first end. A1 and the second terminal A2, and other circuit components are the same as the relay protection device circuit 100 of the present invention in FIG.
如圖4所示,第一直流電源310的正電端供電於第一開關60之主接點R及負載200之一端,再由負載200之另一端連接到第二端40,當第一開關60的主接點R轉向第一接點V,第一直流電源310的正電端供電於第一端30及第二電阻22之一端,正電壓供電於第一電阻器21到第一半導體11的閘極G及第三半導體15的汲極D,因此第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,第一直流電源300供電於負載200,而後第二電阻器 22之另一端供電於光電耦合器16的陽極端P到陰極端N,此時光電耦合器16的第一端A1與第二端A2導通,第三半導體15的閘極G與源極S電壓低,因此第三半導體15的汲極D與源極S開路;其中第二電阻器22與第一電容器23構成一時間常數電路,可以控制光電耦合器16的陽極端P與陰極端N的導通時間,以達到重置時間的控制,其時間常數電路可以用同等功能的時間控制器替代,而不自限。 As shown in FIG. 4, the positive electric terminal of the first DC power source 310 is supplied to the main contact R of the first switch 60 and one end of the load 200, and then the other end of the load 200 is connected to the second end 40. The main contact R of the switch 60 is turned to the first contact V. The positive electrical terminal of the first DC power source 310 is supplied to one of the first terminal 30 and the second resistor 22. The positive voltage is provided to the first resistor 21 to the first. The gate G of the semiconductor 11 and the drain D of the third semiconductor 15 are connected. Therefore, the drain D of the first semiconductor 11 is connected to the source S. The electromagnetic coil of the relay 14 is supplied with voltage. The first contact K1 is connected to the second contact. Point K2 is turned on, the first DC power source 300 supplies power to the load 200, and then the second resistor The other end of 22 is supplied from the anode terminal P to the cathode terminal N of the photocoupler 16. At this time, the first terminal A1 and the second terminal A2 of the photocoupler 16 are turned on, and the voltage of the gate G and the source S of the third semiconductor 15 is turned on. Low, so the drain D and the source S of the third semiconductor 15 are open; the second resistor 22 and the first capacitor 23 form a time constant circuit, which can control the conduction between the anode terminal P and the cathode terminal N of the photocoupler 16 Time to achieve the reset time control, its time constant circuit can be replaced with an equivalent function of the time controller without self-limiting.
如圖4所示,若將第一開關60的主接點R轉向第二接點U時,第一直流電源310不供電於第一端30,此時第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,光電耦合器16的第一端A1與第二端A2開路,第三半導體15的閘極G與源極S無電壓。 As shown in FIG. 4, when the main contact R of the first switch 60 is turned to the second contact U, the first DC power source 310 does not supply power to the first terminal 30. At this time, the drain D of the first semiconductor 11 and the The source S is open, and the solenoid coil of the relay 14 has no voltage supply. The first contact K1 and the second contact K2 are open. The first DC power source 310 does not supply power to the load 200. The first end A1 of the photocoupler 16 and The second terminal A2 is open, and the gate G and the source S of the third semiconductor 15 have no voltage.
如圖4所示,當第一開關60的主接點R轉向第三接點T,此時第二直流電源320的正電端供電於第一端30及第一直流電源300的正電端供電於負載200到第二端40,而第一直流電源310的負電端及第二直流電源320的負電端連接第三端50,因為有第二直流電源320供電於第一端30,此時第一端30供電於第一電阻器21到第一半導體11的閘極G及第三半導體15的汲極D,此時第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,第一直流電源310供電於負載100,而後經由第二電阻器22供電於光電耦合器16的第一端A1與第二端A2,此時光電耦合器16的第一端A1與第二端A2 導通,第三半導體15的閘極G與源極S電壓低,因此第三半導體15的汲極D與源極S開路。 As shown in FIG. 4, when the main contact R of the first switch 60 is turned to the third contact T, the positive electric terminal of the second DC power source 320 supplies power to the first terminal 30 and the positive electric power of the first DC power source 300. The terminal is powered from the load 200 to the second terminal 40, and the negative terminal of the first DC power source 310 and the negative terminal of the second DC power source 320 are connected to the third terminal 50 because the second DC power source 320 is powered by the first terminal 30. At this time, the first terminal 30 supplies power to the gate G of the first resistor 21 to the first semiconductor 11 and the drain D of the third semiconductor 15. At this time, the drain D of the first semiconductor 11 and the source S are turned on, and the relay 14 The electromagnetic coil has a voltage supply, and its first contact K1 and the second contact K2 are turned on. The first DC power source 310 supplies power to the load 100, and then supplies power to the first end A1 of the photocoupler 16 through the second resistor 22. And the second terminal A2, at this time the first terminal A1 and the second terminal A2 of the photocoupler 16 When the gate G and the source S of the third semiconductor 15 are turned on, the drain D and the source S of the third semiconductor 15 are open.
如圖4所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,若將負載200兩端短路,其等同將第一直流電源310的正電壓供電於光電耦合器16的第一端A1,此時第三半導體15的汲極D與源極S導通,等同第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,而達到短路保護第一直流電源310的目的。 As shown in FIG. 4, when the first terminal 30 is connected to the second DC power source 320, the first DC power source 310 supplies power to both ends of the load 200. If the two ends of the load 200 are short-circuited, it is equivalent to the first DC power source 310. A positive voltage is supplied to the first terminal A1 of the photocoupler 16. At this time, the drain D and the source S of the third semiconductor 15 are turned on, which is the same as that the gate G and the source S of the first semiconductor 11 are turned on. A semiconductor 11 has its drain D and its source S open, and the solenoid coil of the relay 14 has no voltage supply. Its first contact K1 and second contact K2 are open. The first DC power source 310 does not supply power to the load 200, but reaches Purpose of short circuit protection of the first DC power source 310.
如圖4所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,當負載200發生短路時,先將負載200兩端短路的原因去除,再將第一開關60的主接點R轉向第二接點U,再轉向第三接點T,此時光電耦合器16的第一端A1與第二端A2導通,第三半導體15的閘極G與源極S兩端電壓低,第三半導體15的汲極D與源極S開路,第一半導體11的閘極G受有正電壓,於是第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,亦就是第一直流電源310供電於負載200;若將第一開關60的主接點R轉向第二接點U,第二直流電源320不供電於第一端30,此時第一直流電源310不供電於負載200。 As shown in FIG. 4, when the first terminal 30 is connected to the second DC power source 320, the first DC power source 310 supplies power to both ends of the load 200. When a short circuit occurs at the load 200, the cause of the short circuit at both ends of the load 200 is removed first. Then, the main contact R of the first switch 60 is turned to the second contact U and then to the third contact T. At this time, the first terminal A1 and the second terminal A2 of the photocoupler 16 are turned on, and the third semiconductor 15 The voltage across the gate G and the source S is low, the drain D of the third semiconductor 15 is open to the source S, and the gate G of the first semiconductor 11 receives a positive voltage, so the drain D and the source of the first semiconductor 11 are The pole S is turned on, and the electromagnetic coil of the relay 14 is supplied with voltage. The first contact K1 and the second contact K2 are turned on, that is, the first DC power source 310 supplies power to the load 200. If the first contact 60 is the main contact, R turns to the second contact U, and the second DC power source 320 is not powered by the first terminal 30. At this time, the first DC power source 310 is not powered by the load 200.
如圖4所示,當第一端30接有第二直流電源320時,第一直流電源310供電於負載200兩端,若將負載200加大亦就是增大負載200電流量,此時若 繼電器14的第一接點K1與第二接點K2二點間之電壓降值大於第三半導體15的閘源極導通電壓時,第三半導體15的汲極D與源極S導通,等同第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,第一直流電源310不供電於負載200,而達到過電流保護第一直流電源310的目的。 As shown in FIG. 4, when the first terminal 30 is connected to the second DC power source 320, the first DC power source 310 supplies power to both ends of the load 200. If the load 200 is increased, the load 200 current is increased. If When the voltage drop between the two points of the first contact K1 and the second contact K2 of the relay 14 is greater than the gate-source turn-on voltage of the third semiconductor 15, the drain D of the third semiconductor 15 and the source S are turned on, which is equivalent to the first The gate G and the source S of a semiconductor 11 are electrically connected to each other, so the drain D of the first semiconductor 11 and the source S are open, and the electromagnetic coil of the relay 14 has no voltage supply, and the first contact K1 and the second contact thereof K2 is open, and the first DC power source 310 does not supply power to the load 200, but achieves the purpose of protecting the first DC power source 310 from overcurrent.
如圖5所示,為本發明繼電器保護裝置第三實施例,自圖中可知,本發明繼電器保護裝置100中之第一端30連接第一開關60的主接點R及第二電阻器22之一端,第二電阻器22之另一端連接第一電容器23之正電端,第一電容器23之負電端連接第三端50,第一開關60的第三接點T連接第二直流電源320之正電端,第二直流電源320之負電端連接第三端50,第二端40連接全波整流器17的正電端V+,第三端50連接全波整流器17的負電端V-,全波整流器17的第一交流端AC1連接繼電器14的第一接點K1,繼電器14的第一接點K1連接交流負載500的另一端,交流負載500的一端連接交流電源400的一端,交流電源400的另一端連接繼電器14的第二接點K2,繼電器14的第二接點K2連接全波整流器17的第二交流端AC2,以上電路的組合為構成一組交流電源400的過載或短路的保護裝置。 As shown in FIG. 5, it is a third embodiment of the relay protection device of the present invention. As can be seen from the figure, the first terminal 30 in the relay protection device 100 of the present invention is connected to the main contact R of the first switch 60 and the second resistor 22. One end, the other end of the second resistor 22 is connected to the positive electrical end of the first capacitor 23, the negative electrical end of the first capacitor 23 is connected to the third end 50, and the third contact T of the first switch 60 is connected to the second DC power source 320 The positive terminal of the second DC power source 320 is connected to the third terminal 50, the second terminal 40 is connected to the positive terminal V + of the full-wave rectifier 17, and the third terminal 50 is connected to the negative terminal V- of the full-wave rectifier 17, The first AC terminal AC1 of the wave rectifier 17 is connected to the first contact K1 of the relay 14, the first contact K1 of the relay 14 is connected to the other end of the AC load 500, one end of the AC load 500 is connected to one end of the AC power supply 400, and the AC power supply 400 The other end is connected to the second contact K2 of the relay 14 and the second contact K2 of the relay 14 is connected to the second AC terminal AC2 of the full-wave rectifier 17. The combination of the above circuits constitutes a group of AC power 400 for overload or short circuit protection. Device.
如圖5所示,當第一開關60的主接點R轉向第三接點T,此時第二直流電源320的正電端供電於第一端30,而第二直流電源320的負電端連接第三端50,此時第一端30供電於第一電阻器21到第一半 導體11的閘極G,因此第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二接點K2導通,交流電源400供電於交流負載500。 As shown in FIG. 5, when the main contact R of the first switch 60 is turned to the third contact T, the positive electric terminal of the second DC power source 320 is supplied to the first terminal 30 and the negative electric terminal of the second DC power source 320 is supplied. Connect to the third terminal 50. At this time, the first terminal 30 supplies power to the first resistor 21 to the first half. The gate G of the conductor 11 is connected to the drain D of the first semiconductor 11 and the source S. The electromagnetic coil of the relay 14 is supplied with voltage. The first contact K1 and the second contact K2 are conducted. The AC power source 400 is powered by AC load 500.
如圖5所示,若將第一開關60的主接點R轉向第二接點U時,第二直流電源320不供電於第一端30,第一半導體11的閘極G無電壓,此時第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,交流電源400不供電於交流負載500。 As shown in FIG. 5, if the main contact R of the first switch 60 is turned to the second contact U, the second DC power source 320 does not supply power to the first terminal 30, and the gate G of the first semiconductor 11 has no voltage. When the drain D of the first semiconductor 11 and the source S are open, the solenoid coil of the relay 14 has no voltage supply, and the first contact K1 and the second contact K2 are open. The AC power source 400 does not supply power to the AC load 500.
如圖5所示,當第一端30接有第二直流電源320時,交流電源400供電於交流負載500,若將交流負載500兩端短路,其等同將交流電源400經過全波整流器17正電端V+的正電壓直接供電於第二端40,此時第二半導體12的集極C與射極E導通,等同第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,交流電源400不供電於交流負載500,而達到交流負載500短路保護交流電源400的目的。 As shown in FIG. 5, when the first terminal 30 is connected to the second DC power source 320, the AC power source 400 supplies power to the AC load 500. If the two ends of the AC load 500 are short-circuited, it is equivalent to passing the AC power source 400 through the full-wave rectifier 17. The positive voltage of the electric terminal V + is directly supplied to the second terminal 40. At this time, the collector C and the emitter E of the second semiconductor 12 are turned on, which is the same as that the gate G and the source S of the first semiconductor 11 are turned on, so the first The drain D and the source S of the semiconductor 11 are open, and the electromagnetic coil of the relay 14 has no voltage supply. The first contact K1 and the second contact K2 are open. The AC power source 400 is not supplied to the AC load 500 and reaches the AC load 500. Purpose of short circuit protection AC power supply 400.
如圖5所示,當第一端30接有第二直流電源320時,交流電源400供電於交流負載500兩端,當交流負載500發生短路時,先將交流負載500兩端短路的原因去除,再將第一開關60的主接點R轉向第二接點U,再轉向第三接點T,此時第一端30供電於第一電阻器21到第一半導體11的閘極G,因此第一半導體11的汲極D與源極S導通,繼電器14的電磁線圈有電壓供應,其第一接點K1與第二 接點K2導通,交流電源400供電於交流負載500。 As shown in FIG. 5, when the first terminal 30 is connected to the second DC power supply 320, the AC power supply 400 supplies power to both ends of the AC load 500. When the AC load 500 is short-circuited, the cause of the short-circuit at both ends of the AC load 500 is removed first. Then, the main contact R of the first switch 60 is turned to the second contact U, and then to the third contact T. At this time, the first terminal 30 supplies power to the gate G of the first resistor 21 to the first semiconductor 11, Therefore, the drain D and the source S of the first semiconductor 11 are conducted, and the electromagnetic coil of the relay 14 is supplied with a voltage. The first contact K1 and the second contact K1 The contact K2 is turned on, and the AC power source 400 supplies power to the AC load 500.
如圖5所示,當第一端30接有第二直流電源320時,交流電源400供電於交流負載500兩端,若將交流負載500加大亦就是增大交流負載500電流量,此時若繼電器14的第一接點K1與第二接點K2二點間之電壓降值經過全波整流器17正電端V+的正電壓值大於第二半導體12的基射極導通電壓時,第二半導體12的集極C與射極E導通,第一半導體11的閘極G與源極S兩端導通,於是第一半導體11的汲極D與源極S開路,繼電器14的電磁線圈無電壓供應,其第一接點K1與第二接點K2開路,交流電源400不供電於交流負載500,而達到交流負載500過載保護交流電源400的目的。 As shown in FIG. 5, when the first terminal 30 is connected to the second DC power source 320, the AC power source 400 supplies power to both ends of the AC load 500. If the AC load 500 is increased, the AC load 500 current is increased. If the voltage drop between the first contact K1 and the second contact K2 of the relay 14 passes through the full-wave rectifier 17 and the positive voltage of the positive terminal V + is greater than the base-emitter on-voltage of the second semiconductor 12, the second The collector C and the emitter E of the semiconductor 12 are turned on, and the gate G of the first semiconductor 11 and the both ends of the source S are turned on, so that the drain D and the source S of the first semiconductor 11 are open, and the electromagnetic coil of the relay 14 has no voltage. Supply, the first contact K1 and the second contact K2 are open, and the AC power source 400 does not supply power to the AC load 500, but achieves the purpose of protecting the AC power source 400 from overload of the AC load 500.
發明人從事電子科技研究有50多年,本發明所提的實施例皆經過實驗及實作證明其成功,並且可據予實施,以上所述實施例僅是為充分說明本發明所舉的較佳的實施例,本發明的保護範圍不限於此,包括本技術領域的技術人員,在本發明基礎上所作的等同替代或變換,皆在本發明的保護範圍內。本發明的保護範圍以申請專利範圍書為準。 The inventor has been engaged in electronic technology research for more than 50 years. The examples of the present invention have proved their success through experiments and implementations, and can be implemented according to them. The above-mentioned examples are merely to fully illustrate the advantages of the present invention. In the embodiments, the protection scope of the present invention is not limited thereto, and equivalent substitutions or transformations made by those skilled in the art based on the present invention are all within the protection scope of the present invention. The protection scope of the present invention is subject to the scope of patent application.
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