Hello friends, hope you all are fine and having fun. In today’s post we are gonna have a look at LM317 Voltage Regulator in Proteus. In the previous post, we have seen how to design a 5V Power Supply in Proteus ISIS, which I have designed using IC regulator 7805. Today I am going to share How to design LM317 Voltage Regulator Circuit in Proteus. In this circuit 4047 is use to generate the square wave of 50Hz and amplify the current and then amplify the voltage by using the step transformer. 7: The inverter circuit output waveform in proteus 8.0 As can be seen from Figure 7, it is observe that the output wave form is a square wave.

• Build Power Supply Circuit
• Lm7805 Proteus Library Download

When you launch Prteus click on New Project or press CTRL+N. A black colored window would appear that is PCB layout tab, you can switch the tab to schematic captureClick on P button it stands for place, a new window would appear, In this window you can search for the components by typing exact names like 1n4007(diode) or simply type diode but if you want to make PCB you have to check that your selected component has a PCB design which is shown in PCB preview. For this tutorial I designed a Variable Power Supply. Make sure you selected the little diode button. select any component as I clicked 1n4007 from devices section by clicking them once, then click in the area provided for drawing the circuit now the component is available to be place anyzhere, to place click again. If you want another 1N4007 just click again as many times as you want. Similarly place all the components.

to join the components with each other click on one end of a component, a line would start, you can click on the other components where you want it to join like in the picture. Similarly complete your circuit diagram. To change the values of components(resistors, capacitors) double click on the component and change it.

Similarly change the value of VSINE by double clicking it. As it is AC you need to add frequency and voltage value as mentioned at output terminals of transformer. To add a voltmeter click on the meter icon and select DC VOLTMETER then click where ever you want to place it connect it by wires(lines) if necessary. To check the circuit click on the play icon on bottom left of the proteus. I have attached last two pictures by changing the value of variable resistor check the output voltage.

At this step we have simulated our circuit and verified it on Proteus you need to save this file for later purposes. Board Edge (Border or Size).

Click on the PCB layout tab then zoom in to upper left corner as shown in pictures this block is 1x1 cm in area. Now from the left toolbar click on the Square button and from the bottom toolbar select Board Edge.

Draw a rectangle of any size by clicking anywhere inside the blue area of the PCB window, as you can see I have drawn a 2 x 2 cm² block. This is the size of your PCB you can change it by hoovering your mouse pointer at any corner off this Block. As I have also changed it after completing the design. Now click on the small diode icon which indicate that component lists. Select the component and then click inside the yellow Square to place it. When you would be placing a component you will see green lines which indicates the connection of one point to the other point and the Yellow arrow head which indicates that this component is to join with this terminal of the other component.

Place the components wherever you like keep in mind that you choose simple and easy path which is indicated by the green lines. For etching you need to print the PCB layout on a glossy sheet from a laser printerFor printing the PCB layout there are two methods. Go to output click on export graphics then click on export Adobe PDF file in this method you will save the PCB layout as a PDF file and dialog box will open from this box only select the bottom copper and the Board Edge and un-check everything else and click OK. You need to check the actual size printing option while printing from a adobe or any other PDF viewer.

Go to output click on print layout, a dialog box will appear only check bottom copper and board Edge adjust the PCB as shown in the print preview you can print it anywhere on the page and click OK.

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. 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. Saints row 4 music list 2017. 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.

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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. 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.

Lm7805 Proteus Library Download

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'.