Guest Kasamiko Posted August 11, 2004 Report Share Posted August 11, 2004 Guys check this out.another DC to AC power inverter based on 4047 chip with adjustable frequency..What can you say?? worth discussing?? :DRhonn ;) ;) Quote Link to comment Share on other sites More sharing options...
ante Posted August 11, 2004 Report Share Posted August 11, 2004 Hi Rhonn,100 Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 12, 2004 Report Share Posted August 12, 2004 Hi Rhonn,Oh goody, another one!I figure that it will give about 200W. A few comments:1) Those nice 2N3772s cost nearly 3 times as much as ordinary 2N3055s. If you parallel two 2N3055s on each side, the performance will be the same but cost less.2) The BD239s will be running at their limit of 2A. I recommend BD243s which are rated for 6A and cost nearly the same.3) Those 100 ohm resistors will dissipate 1/2W. Use 1W.4) 100 ohms will be fine instead of those 6.8 ohm ones.5) Use a higher current anti-reverse diode. The little one might protect the fuse then it will say "c'mon" to the rest of the circuit. ;D 6) Why have such a wide range of adjustable frequency? A SPST switch can select a fairly accurate 50Hz and 60Hz. Quote Link to comment Share on other sites More sharing options...
Guest Kasamiko Posted August 12, 2004 Report Share Posted August 12, 2004 @AnteThe circuit will guarante 100 watts I think.. ???@AudioguruA swith approach is better I think..Pls tell me how to do it..Here's the whole article: (Please somebody can translate it)"Para reducir o elevar una tensión determinada nada se adapta mejor que un transformador, pero este componente no funciona en corriente continua, que es la disponible en baterías o vehículos. Entonces debemos colocar un oscilador que genere una alternancia en la CC para así tener en la bobina del transformador CA. El circuito integrado (4047) es un oscilador cuyas salidas son una inversa con respecto de la otra. Esto quiere decir que mientras una está en estado alto la otra está bajo y viceversa. Estas señales son demasiado débiles para mover el trasformador así que se implementa un driver formado por tres transistores en cadena. El diodo en paralelo con cada uno de los transistores finales evita que la corriente inversa producida al retirar la corriente del bobinado queme el transistor. El diodo de 5A colocado en paralelo con la línea de alimentación genera un cortocircuito cuando la polaridad es accidentalmente invertida, haciendo que el fisible salte. El preset de 50K permite ajustar la frecuencia del oscilador, que es directamente proporcional con la frecuencia de la CA producida en el trafo. Para que el oscilador trabaje estable se ha dispuesto el resistor de 220 ohms como limitador de corriente y el zener de 9.1v junto con sus capacitores de filtrado. Este conjunto hace que sin importar los cambios en la batería la tensión en el oscilador sea de 9v. El transformador puede ser uno común de los que se emplean para hacer fuentes de alimentación, solo que en este equipo lo usaremos inversamente. En vez de aplicar tensión en el devanado de 220v y retirarla por el de 18v lo que haremos es ingresar la tensión por el devanado de 18v y retirarla por el de 220v. En realidad los cálculos de este elemento dan como necesario un bobinado de 220v y otro de 9.3v+9.3v, pero como no es común este tipo de valores hemos implementado uno de 9+9 que es muy habitual en los comercios. Dado que esto genera algo mas de 220v si quiere puede emplear un transformador de 10+10 (que también está disponible) pero la tensión generada, alimentando el conjunto con 12v será de 204v. Ud. decide. En nuestro caso empleamos el de 9+9. La capacidad del mismo debe ser de 100VA Los transistores de salida deben ser colocados sobre disipador de calor. Respetar las potencias de los resistores en los casos que sea indicado. Comprobar la posición de los diodos y capacitores electrolíticos. Utilizar cables de sección adecuada para la conexión de la batería. Cables demasiado delgados pueden causar caídas de tensión o funcionamiento errático. Una buena alternativa para comprobar el funcionamiento visualmente es colocar un indicador de neón en la salida de 220V. Así, solo cuando el sistema trabaje adecuadamente el indicador brillará. Calibración: Basta con alimentar el sistema y colocar un frecuencímetro ú osciloscopio en la salida del trafo. Girar el preset de 50K ubicado en el 4047 hasta que la frecuencia medida sea de 50Hz. Luego de esto la calibración habrá concluido. Simple. IMPORTANTE: Este equipo genera corriente alterna cuya forma de onda es cuadrada. Esto es así porque los transistores están dispuestos en corte / saturación. Esto no presenta problemas para los equipos resistivos, como soldadores, lámparas o fuentes. Pero equipos de TV o grabadoras de vídeo que empleen como referencia la frecuencia y onda de la red pueden no funcionar correctamente.Nota: Todos los esquemas aquí puestos son en parte recopilados de la red así como otros son de creación propia o bien desde algún aporte. Por tanto no han sido todos testeados y verificados. Si alguien se anima a armar alguno están bienvenidos los resultados y inconvenientes que se tuviesen en el camino así como también los pcb Quote Link to comment Share on other sites More sharing options...
ante Posted August 12, 2004 Report Share Posted August 12, 2004 Rhonn,Yes 100VA is almost 100W but you can Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 13, 2004 Report Share Posted August 13, 2004 Hi Sasi,Where is your schematic?I agree that MOSFETs are better, but are not available everywhere like 2N3055 transistors are.Those MOSFETs cost the same as 2N3055s in Canada, and they are both very cheap.Hi Rhonne,To switch the frequency of the 4047, use a 47K resistor for 100/50Hz, and switch a 220K or 240K in parallel for 120/60Hz. Permanently join one side of the resistors. Long switch wires might pickup interference. Quote Link to comment Share on other sites More sharing options...
ante Posted August 13, 2004 Report Share Posted August 13, 2004 Sasi,Look good, a few points: Why not use a 230VAC changeover relay, this will lower the cost and component count (one transformer, one rectifier, and one cap). At least remove the capacitor to get a quicker changeover? But don Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 13, 2004 Report Share Posted August 13, 2004 Hi Sasi,Congratulations, that's a nice simple circuit and a good PCB layout.What is the relay for? It looks like a relaxation oscillator, where the gate drivers start running then quickly turn-off then start again, etc.What are the paralleled 0.01 ohm resistors for?How much current does the inverter draw from the battery without a load?At full 500W load, don't the MOSFETs get only a little bit warm when mounted on a fairly small heatsink? Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 13, 2004 Report Share Posted August 13, 2004 Ohhhhhh! I get it! Thanks.The relay is for the inverter to give automatic changeover so that it can be a backup AC power supply when the mains fails.Your gamble is that the inverter will be overloaded for a moment if the load is a straining motor or an appliance with a huge filter capacitor, due to the inrush current after the relay delay. Those 50A MOSFETs have a combined rating of 200A(!), so they should be able to handle it and your wiring will limit that huge current anyway.Sasi,I am interested about its current without a load because most heating of the MOSFETs may occur when both sides are turned-on at the same time during their switching. Using a biased LM393 (it is an 8-pin dual) fast-turnoff comparator and pullup resistors on its outputs (it has open-collector output), instead of the slow opamp may help that. You can do a lot of simple timing tricks with a comparator (small delay before gate turn-on, etc.) And the inductance of a transformer that is used backwards may not be enough to make it efficient with light loads. Quote Link to comment Share on other sites More sharing options...
ante Posted August 13, 2004 Report Share Posted August 13, 2004 Audioguru,The man says Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 13, 2004 Report Share Posted August 13, 2004 Hi Ante,Sure, the SG3524 will do a pretty good job here, and in addition to providing an adjustable "dead time", it can provide output voltage regulation and current-limiting, too. But its age shows in its single-ended outputs, designed for driving bipolar transistors, not high-gate-capacitance MOSFETs. Nowadays a maker must have a modern SMPS IC with push-pull outputs.A quick look at IR showed that their 330W IRF2804 MOSFET with its 2.3milliohm on-resistance is coming down in price. A single pair on a small heatsink will easily give an inverter a 500W output. Quote Link to comment Share on other sites More sharing options...
ante Posted August 13, 2004 Report Share Posted August 13, 2004 Audioguru,What Quote Link to comment Share on other sites More sharing options...
Guest Kasamiko Posted August 14, 2004 Report Share Posted August 14, 2004 It looks like a complete UPS system for me.. minus the sine wave output;DI thought I was the only interested in inverter circuit here since BLACK-OUT is part of our daily life. :'(Mosfet is very effecient for this work but very hard to find in my place and to tell you, a good quality of 2N3055 cost about US$1.00 here. A single MOSFET? almost 3X the price of 2N!!!Rhonn ;) ;) Quote Link to comment Share on other sites More sharing options...
MP Posted August 14, 2004 Report Share Posted August 14, 2004 Sasi, nice inverter drawing. Are you using a 24 Volt supply? I see + 12V and - 12V Battery connections on the drawing. Or does this refer to the + and - posts of a single 12 volt battery buss?Here is also something to consider on your inverter projects: Bigger is not better as in what you get from the power grid. From your power company, you have a meter and you will only be charged for what you are drawing. The inverter will always be using the battery power as long as it is connected to the battery. The more watts, the more draw. If you are connected to solar panels, this means a lot since you have to buy more panels to keep the batteries up. A good solution is to use several smaller inverters which are isolated from the battery when nothing is plugged in to them ...or if it is a backup system, isolate the battery drain when the mains are on. It will save you $$.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 14, 2004 Report Share Posted August 14, 2004 The relay will function as a bypass, if there is a presence of AC mains the inverter will not start, it will just bypass the mains line to load.when the mains fails the inverter will start and load will energised, and when the mains come back it stop the inverter automatically. If there is no voltage in oscillator section the mosfets will remain in maximum resistance, will not draw any current from battery.MP,Sasi's backup system circuit has RLY Pole #1 to "isolate the battery drain when the mains are on". Quote Link to comment Share on other sites More sharing options...
MP Posted August 15, 2004 Report Share Posted August 15, 2004 That is not isolation. The circuit is only turned off. If he were to use a DPDT relay, and the second section was where the fuse is, then this would be considered isolation. Isolation means not connected.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 15, 2004 Report Share Posted August 15, 2004 MP,I'm sorry, the circuit is not truly isolated but instead simply reduces battery current drain. Sasi cleverly used a fairly fast-switching low current relay contact to accomplish this, instead of your recommended slow-switching high current relay.I thought that you were concerned that the circuit was running all the time which will drain the battery.With Sasi's relay turning the circuit off, it doesn't need complete isolation since the MOSFETs' uA of leakage current is much less than the self-discharge current of the battery. Quote Link to comment Share on other sites More sharing options...
ante Posted August 15, 2004 Report Share Posted August 15, 2004 Audioguru,I like an inverter for car use and I need it to be sine wave output 230-240VAC 50Hz @ 300W. It would be very nice if it where variable 0 Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 15, 2004 Report Share Posted August 15, 2004 Hi Ante,An inverter with 300W sine-wave output but without much heat?Sure, it's nearly the year 2005, isn't it?I'll look at using a class-D audio amp driver IC (but it will probably be surface-mount) and discrete MOSFETs. An output transformer will step-up the voltage and a low-power volume control will adjust the output voltage. Quote Link to comment Share on other sites More sharing options...
MP Posted August 16, 2004 Report Share Posted August 16, 2004 ...With Sasi's relay turning the circuit off, it doesn't need complete isolation since the MOSFETs' uA of leakage current is much less than the self-discharge current of the battery. No need to be sorry, audioguru.Show the math. Thevinen's theorem would not agree with your statement. This would be like a parasitic draw on the battery. It would also tend to add other problems in the future. All it takes is a simple add on to isolate the battery. Not a big design add-on. Why would anyone not want the added protection? I have to deal with these types of problems in the field all the time. All it takes is a little thinking ahead.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted August 16, 2004 Report Share Posted August 16, 2004 Hi MP,Math? But I can't find my old slide-rule!I'm glad that you agree that Sasi's relay turns-off his circuit, because my new-fangled calculator tells me that a 150A/hr car battery will power the maximum leakage current (25uA each at 55V) of the six MOSFETs for more than 116 years!Maybe Sasi needs backup only for the short time for his hard-drive to park, and uses just a string of 4A/hr D-cell Ni-Cads. They will power that "parasitic" leakage for more than 3 years!I think that the little solar-cell (a real one, not just a label) on my calculator can supply the 150uA of leakage current. ;DHi Sasi,Your PCB layouts are very nice. You should post a complete project. Quote Link to comment Share on other sites More sharing options...
Guest Kasamiko Posted August 16, 2004 Report Share Posted August 16, 2004 Sasi,How about adding some charging circuit whenever there is a main power? I think replacing the transformer that power the relay with bigger one to supply a charging current to the battery is a nice idea.What can you say guys? 8)Rhonn ;) ;) Quote Link to comment Share on other sites More sharing options...
ante Posted August 16, 2004 Report Share Posted August 16, 2004 Audioguru,Sounds good, I am looking forward for a schematic to be posted sometime in the near future. Quote Link to comment Share on other sites More sharing options...
ante Posted August 16, 2004 Report Share Posted August 16, 2004 Audioguru,Oh, I forgot to mention I have a ton of RFG50N05 so it would be nice if they were usable . Its not a must I can get some new if required. Quote Link to comment Share on other sites More sharing options...
ante Posted August 19, 2004 Report Share Posted August 19, 2004 I did not analyse the complete circuit but if the battery is 12V then the LM7812 will not work here. Quote Link to comment Share on other sites More sharing options...
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