Power Supply Circuit Simulator Proteus
So you are saying that from a '12V' battery you will drain '220W/12= 20A'?I dont believe 'IC 4047' can supply '20A'?I think the circuit is wrong to begin with. From 'IC 4047' you only make a sine signal, than you amplify this signal with a transistor in order to use whatever power supply you have (12V accumulator battery, solar panel, etc.). You need to amplify the current so you can drain the maximum from the power supply. Whatever you drain from the power supply (220W) must be supported by the transistors, they should have a voltage rating higher than '12V' and current rating higher than '20A', there will be spikes if you dont take measures so the transistors need to be able to take that.
Atx Power Supply Circuit
I have made inverter circuit in proteus,have used cd4047 to generate 50hz square wave, i have used irfz44 mosfet. What are the transformer rating for proteus simulation, using a 12V battery, output should be 220 watts, i am confused how to set the step up transformer for proteus. SIMULATION OF INVERTER CIRCUIT USING MULTISM AND PROTEUS A Akhikpemelo, P Matsunde, F. Ebenso To cite this version: A Akhikpemelo, P Matsunde, F. SIMULATION OF INVERTER CIRCUIT US-ING MULTISM AND PROTEUS. Continental J. Engineering Sciences, 2016, 11 (2), pp.1-11. All type of electronic devices requires power supply from electric.
Also you will need a radiator for the transistors and preferably use a PWM signal to switch them.The rating of the transformer = 220W on the first side and 220W on the second one. Also the coils need to be 12V primary and 220V secondary. The greatest problem is to supply a large amount of current to the primary coil of '12V'. Your electronics need to be able to do that.EDIT: Do keep in mind that you will be using '+-12V' effective not amplitude value for the transfomer and on the other side you will have '+-230V' effective value not amplitude value. Amplitude value is the maximum value of the wave (also called the peak) and the effective value needs to be calculated from the maximum normally.That means that you will have to supply a larger value on the primary coil to get the effective value that you need on the secondary. So you are saying that from a '12V' battery you will drain '220W/12= 20A'?I dont believe 'IC 4047' can supply '20A'?I think the circuit is wrong to begin with.
From 'IC 4047' you only make a sine signal, than you amplify this signal with a transistor in order to use whatever power supply you have (12V accumulator battery, solar panel, etc.). You need to amplify the current so you can drain the maximum from the power supply. Whatever you drain from the power supply (220W) must be supported by the transistors, they should have a voltage rating higher than '12V' and current rating higher than '20A', there will be spikes if you dont take measures so the transistors need to be able to take that. Also you will need a radiator for the transistors and preferably use a PWM signal to switch them.The rating of the transformer = 220W on the first side and 220W on the second one. Also the coils need to be 12V primary and 220V secondary.
Proteus is not recommended for analog designs.Dont use Arduino.I dont know about the arduino pins, but a PIC microcontroller should have a '1k: resistor in order to limit the current to '5v/1000=5mA'. The maximum current from a PIC microcontroller pin is '25ma', but the recommended one is '5ma-10ma'. Check the datasheet for your Arduino if you intend to use.The principal is this:You generate a sine signal with a low current and '12V' voltage (perhaps from an MCU). This sine signal must be amplified, where your MOS transistors get in. The amplified sine signal goes to the primary coil and is '300W', '12V', '24A', so the secondary coil will transform '300W', '12V', '24A' into '300W', '230V', about '1A'.The MOS transistors are normally voltage controlled, but you need at the output of the transistors a high current and voltage with sine form.
That means you have to apply on the 'gate' of the MOS such a signal that you will get on the output a sine again. The MOS transistors are only to amplify the sine signal, however if you have a sine signal on the gate it has to be with '50Hz', in order to get '50Hz' on the output as well, not to mention that you will have to calculate what needs to be the 'peak to peak' voltage of the gate sine signal, in order to get '12V' on the drain sine signal.The MOS transistors in your first circuit were doing exactly this.
They were producing a sine signal with '12V' peak to peak amplitude and high current. But I dont think they can withstand '300W'.If you want an easy circuit, I will suggest a driver. The driver will produce a sine wave with '12V', '300W' and will require a sine wave signal with less power on its input.Or you can do this: use a thyristor, put a sine wave on the thyristor gate and you will get the same sine signal on the output. If you need a higher wattage, connect more than 1 thyristor in parralel. The thyristors, transistors, or whatever you use, need to be able to bare '300W', that means that in the datasheet it has to be said this element can dissipate '300W'.A 'H bridge' can also be used as an inverter. But with the 'H bridge' 2 transistors take the load and the wattage is halved.
On your first circuit, 1 transistor takes the whole wave.Your MOS transistors will not be able to take '300W' if you dont switch them at a high frequency '10KHz, 20KHz'. This becomes modulation and control of the MOS transistors and it becomes very diffuclt. In other words you need to put a sine wave generator of '300W' as the power supply for the transistors's, but you need to switch them on the gate with a 'PWM' signal of '10KHz'. And that means make the power supply to be a sine wave first, which is not a good idea for such high values. Here is a simple sine wave amplifier with only 1 step of transistors. You need to put 3 steps normally. The first step is 2 transistors working in 'class A' for smaller signals like the '12V, 5mA' from the MCU.
The second step is in class B, and is a driver circuit which will supply enough current for the last step, and the last step is in class 'AB', the one I give you which needs to output '12V, 15A'. The maximum of the '2n3055' transistor is '60V, 15A' and can dissipate '115W' under idealistic conditions, lets say that you can get '10A' at best, that is about '120W'. You also need a proper PNP transistor, '2n3905' is not good, as it can supply a small collector current.
Find a complementary transistor of '2n3055' and make a circuit. After every step, the voltage will drop because the transistors have a '0.6V' losses over the base-emitter junction.You can provide a sine signal with an MCU or analog circuit (including the one in your first post), amplify the voltage to '36' or '48V', than make a bipolar amplifier with at least 3 steps.
'120W/48V = 2A', this is much better for working with. Than make 2 channels of '120W' and you will get what you need. You will never get the full '115W'. But you will need a radiator or even a better system with a fan, maybe water to cool this invertor.If you prefer you can make a power supply with a sine wave '48V, 6A', than use a 'H-bridge' with MOS transistors but switch them at '4KHz, 20KHz' so they dont overheat. You will still need a radiator or a fan. It can work with '12V, 25A' but this is the power supply with a sine form and the cooling system will be very high tech! You need to pick your transistors properly.
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Lets say a bipolar amplifier to get the MCU sine signal to '48V, 2A, 100W', than use this as the power supply for the step up transformer.Summary - MCU sine signal - bipolar amplifier to '100W', step up transformer.MCU impulse signal - MOS transistors H bridge gates - power supply for the MOS transistors H bridge at either '12V, 25A' or '48V, 6A'.I never did get where do you take the VDD of '12V' from? You are a beginner, I understand, probably a student still.If you have time and want to do more, here is the principle:Sine wave - amplified by transistors - given to the primary coil - becomes 230V at the secondary coil.Whatever wattage you give to the primary coil, the same wattage you will get at the secondary coil, only the voltage will be higher and respectively the current smaller to make up for the power being the same.
How you implement the power supply for the primary coil is up to you. Dont go connecting high watages before knowing what you are doing, because like the previous post said, high wattage inverters are tricky. They can have a '2meter' capacitor for power plant inverters. You are doing a micro inverter.The elements (including the transfomer) need a cooler (if the wattage is high a fan).
If you can make the circuits by yourself, you will get what you need. 300W is a lot, start with something smaller, '2W'.
What Is Power Supply Circuit
The circuit is really easy to understand.A resistor of low value (the resistor value will be explained later) is connected in series with the output of the power supply. I've created a separated step for this because this series resistor is the most crucial part of the circuit.
As I've said before, this resistor is connected in series with the power supply output. As current starts to flow through it a small voltage drop will appear on it.You need to choose a resistor that the voltage drop on it is around 0.50.7 volts when the overload current is passing through it. The overload current is the point that the protection circuit actuates and shuts down you power supply output to prevent damage on it.You can choose a resistor by using ohms law: V=R.I. In this case we're going to use: R= V/I.The first thing you need to determine is the overload current of your power supply. In this part I can't help you, you've got to know the maximum current your power supply can supply and therefore dimension your series resistor value.Let's say your power supply can supply 3 amps(The voltage of your power supply does not matter in this case). So, we've got R= 0,6V/3A.
If you calculated the resistor and the result is not a commercial value, don't worry. Just get a commercial value resistor that is near to your calculations results.The next thing you must do is calculate the power dissipation on this resistor, so it does not burn when current is flowing through it. You can calculate the power dissipation by using the formula: P=V.I.If we use our last example we would get: P=0.6V.3A. P=1.8W a 3W or even a 5W resistor would be more than enough.
To build a board like mine, you will need:1 - TL082 (dual op amp)2 - 1N4148 (diode)1 - TIP122 (NPN transistor)1- BC558 (PNP transistor. You can use a BC557, BC556 or equivalent. They are all fine for this application)1 - 2700ohm resistor1 - 1000ohm resistor1 - 10Kohm resistor1 - 22Kohm resistor1 - Series resistor (see previous step)1 - 10Kohm potentiometer1 - 470uf capacitor1 - 1uf capacitor1 - Normally Closed Momentary Switch (see attached picture. Any normally closed momentary switch will work fine)1 - Relay model T74 (This is a very common relay model. Easy to find in eBay. Just try to search for 'G5LA-14' on eBay. There are many coil voltages and contacts amperage.
If you dont want to use this model, make sure to change the PCB layout). I've used Express PCB to design the board and the file I uploaded here is free for you to edit as you like. Edit it as you wish to fit the components you have. A PDF version of the board has been uploaded too if you don't even want to edit it or generate the PDFs yourself.This circuit is not really that big so I've fit it on a 5cm x 5cm board.Note: The board has an optional led (see PCB picture for more details, I've left a note there) so you can know when the protection circuit have disarmed the output of your power supply. If you don't want to use a led, you must short the pins that the led would be connected or else the circuit won't work.
If you want to connect the led, the square pin is the anode and round pin is the cathode. You can connect any LED but high brights ones.I've tested it on a bread board and you should test it too so you can know for sure if the protection circuit will work with our power supply. I used the toner transfer method to transfer the printed circuit to my PCB.Tips for a good tone transfer:- Print the circuit using a good laser jet printer.- Use coated paper ( see: ) to print your circuit.- Make sure your Iron can get as hot as 170ºC200ºC (338ºF392ºF).- Before starting your tone transfer, clean the board using a thin steel wool. Your board will get really shinny and clean.- You can always use YouTube to see new methods and how people make their transfer.By following these tips, you will definitely get a good toner transfer.After the tone transfer you can etch you board using your favorite etching method. I used iron perchlorate.I did not took any picture while drilling the holes to mount the components, sorry about that. A drill with 1mm will do just fine.If the output current of your power supply is higher than 2A, you should reinforce the lines of the series resistor and relay. See the attached PCB picture for more information.
To turn this circuit on, you will need to supply it a voltage that can be from 9V to 15V. See the attached picture for more information about the input.To calibrate the circuit, measure the voltage on the op amp inverting input and turn the potentiometer. As you turn it the voltage will increase or decrease according to the side you are turning it.
Proteus Simulator Free Download
The value you need to adjust this potentiometer is the gain of the input stage times 0.6 Volts (something around 2.25 to 3 volts if your amplification stage is like mine).This procedure takes some time and the best method to calibrate it is trial and fail. You may need to adjust a higher voltage on the potentiometer so the protection does not trigger on peaks. As I've said, it takes some time to calibrate it. Hi Azul, i'm trying to make this circuit but I can't get it to work.
Every time i connect my 12v supply to power the circuit the relay turns on. Even when I turn the potentiometer all the way to the left or right. Can you also explain to me what you mean with those two negative psu connection. I don't know where to connect them.
As op-amp i'm using an ne5532 as transistors an bc557 and bd679. As the reference resistor i'm using aan 0,22 ohm 5 watt resistor. In the pictures you will find my breadboard layout.
Thanks for your help. Thats happening because because you did not connect your psu ground pin to the sensor resistor therefore the opamp input is getting some noise and amplifying it.The ground output of your psu should be connected to your 0.22 ohm resistor. On your third picture is the wire on the right of the resistor.I would also change this bd679. Its a darlington pair and the way the circuit works could change a bit. Try another general use NPN transistor like a bc548 or a TIP122 or even a 2n2222. Just be careful with the pinout if you use my design.Sorry for the delayed answer.