100W Inverter

Here is a 100 Watt inverter circuit using minimum number of components. I think it is quite difficult to make a decent one like this with further less components.Here we use CD 4047 IC from Texas Instruments for generating the 100 Hz pulses and four 2N3055 transistors for driving the load. The IC1 Cd4047 wired as an astable multivibrator produces two 180 degree out of phase 100 Hz pulse trains.

These pulse trains are preamplified by the two TIP122 transistors.The out puts of the TIP 122 transistors are amplified by four 2N3055 transistors (two transistors for each half cycle) to drive the inverter transformer.The 220V AC will be available at the secondary of the transformer. Nothing complex just the elementary inverter principle and the circuit works great for small loads like a few bulbs or fans.If you need just a low cost inverter in the region of 100 W, then this is the best.



Circuit diagram:

 

100 Watt Inverter Circuit Diagram

Parts:

P1 = 250K
R1 = 4.7K
R2 = 4.7K
R3 = 0.1R-5W
R4 = 0.1R-5W
R5 = 0.1R-5W
R6 = 0.1R-5W
C1 = 0.022uF
C2 = 220uF-25V
D1 = BY127
D2 = 9.1V Zener
Q1 = TIP122
Q2 = TIP122
Q3 = 2N3055
Q4 = 2N3055
Q5 = 2N3055
Q6 = 2N3055
F1 = 10A Fuse
IC1 = CD4047
T1 = 12-0-12V
Transformr Connected in Reverse

Notes:

• A 12 V car battery can be used as the 12V source.
• Use the POT R1 to set the output frequency to50Hz.
• For the transformer get a 12-0-12 V , 10A step down transformer.But here the 12-
• 0-12 V winding will be the primary and 220V winding will be the secondary.
• If you could not get a 10A rated transformer , don’t worry a 5A one will be just
• enough. But the allowed out put power will be reduced to 60W.
• Use a 10 A fuse in series with the battery as shown in circuit.
• Mount the IC on a IC holder.
• Remember,this circuit is nothing when compared to advanced PWM
• inverters.This is a low cost circuit meant for low scale applications.

Design tips:

1. The maximum allowed output power of an inverter depends on two factors.The
2. maximum current rating of the transformer primary and the current rating of the driving
3. transistors.
4. For example ,to get a 100 Watt output using 12 V car battery the primary current will be
5. ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
6. Also here ,each final driver transistors must be rated above 4A. Here two will be
7. conducting parallel in each half cycle, so I=8/2 = 4A .
8. These are only rough calculations and enough for this circuit.

Here is a 100 Watt inverter circuit using minimum number of components. I think it is quite difficult to make a decent one like this with further less components.Here we use CD 4047 IC from Texas Instruments for generating the 100 Hz pulses and four 2N3055 transistors for driving the load. The IC1 Cd4047 wired as an astable multivibrator produces two 180 degree out of phase 100 Hz pulse trains.

These pulse trains are preamplified by the two TIP122 transistors.The out puts of the TIP 122 transistors are amplified by four 2N3055 transistors (two transistors for each half cycle) to drive the inverter transformer.The 220V AC will be available at the secondary of the transformer. Nothing complex just the elementary inverter principle and the circuit works great for small loads like a few bulbs or fans.If you need just a low cost inverter in the region of 100 W, then this is the best.



Circuit diagram:

 

100 Watt Inverter Circuit Diagram

Parts:

P1 = 250K
R1 = 4.7K
R2 = 4.7K
R3 = 0.1R-5W
R4 = 0.1R-5W
R5 = 0.1R-5W
R6 = 0.1R-5W
C1 = 0.022uF
C2 = 220uF-25V
D1 = BY127
D2 = 9.1V Zener
Q1 = TIP122
Q2 = TIP122
Q3 = 2N3055
Q4 = 2N3055
Q5 = 2N3055
Q6 = 2N3055
F1 = 10A Fuse
IC1 = CD4047
T1 = 12-0-12V
Transformr Connected in Reverse

 

Notes:

• A 12 V car battery can be used as the 12V source.
• Use the POT R1 to set the output frequency to50Hz.
• For the transformer get a 12-0-12 V , 10A step down transformer.But here the 12-
• 0-12 V winding will be the primary and 220V winding will be the secondary.
• If you could not get a 10A rated transformer , don’t worry a 5A one will be just
• enough. But the allowed out put power will be reduced to 60W.
• Use a 10 A fuse in series with the battery as shown in circuit.
• Mount the IC on a IC holder.
• Remember,this circuit is nothing when compared to advanced PWM
• inverters.This is a low cost circuit meant for low scale applications.

Design tips:

1. The maximum allowed output power of an inverter depends on two factors.The
2. maximum current rating of the transformer primary and the current rating of the driving
3. transistors.
4. For example ,to get a 100 Watt output using 12 V car battery the primary current will be
5. ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
6. Also here ,each final driver transistors must be rated above 4A. Here two will be
7. conducting parallel in each half cycle, so I=8/2 = 4A .
8. These are only rough calculations and enough for this circuit

Cheap 12V to 220V Inverter

Even though today’s electrical appliances are increasingly often self-powered, especially the portable ones you carry around when camping or holidaying in summer, you do still sometimes need a source of 230 V AC - and while we’re about it, why not at a frequency close to that of the mains? As long as the power required from such a source remains relatively low - here we’ve chosen 30 VA - it’s very easy to build an inverter with simple, cheap components that many electronics hobbyists may even already have.

Though it is possible to build a more powerful circuit, the complexity caused by the very heavy currents to be handled on the low-voltage side leads to circuits that would be out of place in this summer issue. Let’s not forget, for example, that just to get a meager 1 amp at 230 VAC, the battery primary side would have to handle more than 20 ADC!. The circuit diagram of our project is easy to follow. A classic 555 timer chip, identified as IC1, is configured as an astable multivibrator at a frequency close to 100 Hz, which can be adjusted accurately by means of potentiometer P1.

Circuit diagram:

 

Cheap 12V to 220V Inverter Circuit Diagram

As the mark/space ratio (duty factor) of the 555 output is a long way from being 1:1 (50%), it is used to drive a D-type flip-flop produced using a CMOS type 4013 IC. This produces perfect complementary square-wave signals (i.e. in antiphase) on its Q and Q outputs suitable for driving the output power transistors. As the output current available from the CMOS 4013 is very small, Darlington power transistors are used to arrive at the necessary output current. We have chosen MJ3001s from the now defunct Motorola (only as a semi-conductor manufacturer, of course!) which are cheap and readily available, but any equivalent power Darlington could be used.

These drive a 230 V to 2 × 9 V center-tapped transformer used ‘backwards’ to produce the 230 V output. The presence of the 230 VAC voltage is indicated by a neon light, while a VDR (voltage dependent resistor) type S10K250 or S07K250 clips off the spikes and surges that may appear at the transistor switching points. The output signal this circuit produces is approximately a square wave; only approximately, since it is somewhat distorted by passing through the transformer. Fortunately, it is suitable for the majority of electrical devices it is capable of supplying, whether they be light bulbs, small motors, or power supplies for electronic devices.

PCB layout:

 

PCB Layout For Cheap 12V to 220V Inverter Circuit Diagram

 
COMPONENTS LIST
Resistors
R1 = 18k?
R2 = 3k3
R3 = 1k
R4,R5 = 1k?5
R6 = VDR S10K250 (or S07K250)
P1 = 100 k potentiometer
Capacitors
C1 = 330nF
C2 = 1000 µF 25V
Semiconductor
T1,T2 = MJ3001
IC1 = 555
IC2 = 4013
Miscellaneous
LA1 = neon light 230 V
F1 = fuse, 5A
TR1 = mains transformer, 2x9V 40VA (see text)
4 solder pins

Note that, even though the circuit is intended and designed for powering by a car battery, i.e. from 12 V, the transformer is specified with a 9 V primary. But at full power you need to allow for a voltage drop of around 3 V between the collector and emitter of the power transistors. This relatively high saturation voltage is in fact a ‘shortcoming’ common to all devices in Darlington configuration, which actually consists of two transistors in one case. We’re suggesting a PCB design to make it easy to construct this project; as the component overlay shows, the PCB only carries the low-power, low-voltage components.

The Darlington transistors should be fitted onto a finned anodized aluminum heat-sink using the standard insulating accessories of mica washers and shouldered washers, as their collectors are connected to the metal cans and would otherwise be short-circuited. An output power of 30 VA implies a current consumption of the order of 3 A from the 12 V battery at the ‘primary side’. So the wires connecting the collectors of the MJ3001s [1] T1 and T2 to the transformer primary, the emitters of T1 and T2 to the battery negative terminal, and the battery positive terminal to the transformer primary will need to have a minimum cross-sectional area of 2 mm2 so as to minimize voltage drop.

The transformer can be any 230 V to 2 × 9 V type, with an E/I iron core or toroidal, rated at around 40 VA. Properly constructed on the board shown here, the circuit should work at once, the only adjustment being to set the output to a frequency of 50 Hz with P1. You should keep in minds that the frequency stability of the 555 is fairly poor by today’s standards, so you shouldn’t rely on it to drive your radio-alarm correctly – but is such a device very useful or indeed desirable to have on holiday anyway? Watch out too for the fact that the output voltage of this inverter is just as dangerous as the mains from your domestic power sockets.

So you need to apply just the same safety rules! Also, the project should be enclosed in a sturdy ABS or diecast so no parts can be touched while in operation. The circuit should not be too difficult to adapt to other mains voltages or frequencies, for example 110 V, 115 V or 127 V, 60 Hz. The AC voltage requires a transformer with a different primary voltage (which here becomes the secondary), and the frequency, some adjusting of P1 and possibly minor changes to the values of timing components R1 and C1 on the 555.

12V Flourescent Lamp Inverter

Fluorescent tubes use far less energy than incandescent lamps and fluorescent tubes last a great deal longer as well. Other advantages are diffuse, glare-free lighting and low heat output. For these reasons, fluorescent lighting is the natural choice in commercial and retail buildings, workshops and factories. For battery-powered lighting, fluorescent lights are also the first choice because of their high efficiency. The main drawback with running fluorescent lights from battery power is that an inverter is required to drive the tubes.

Circuit diagram:

 

12V Fluorescent Lamp Inverter Circuit Diagram

Fig.1: two switch-mode circuits are involved here: the DC-DC inverter involving IC1, Q1 & Q2 and the fluoro tube driver which converts high voltage DC to AC via IC3 and Q3 & Q4 in a totem-pole circuit.

Inverter efficiency then becomes the major issue. There are many commercial 12V-operated fluorescent lamps available which use 15W and 20W tubes. However, it is rare to see one which drives them to full brilliance. For example, a typical commercial dual 20W fluorescent lamp operating from 12V draws 980mA or 11.8W. Ignoring losses in the fluorescent tube driver itself, it means that each tube is only supplied with 5.9W of power which is considerably less than their 20W rating. So while the lamps do use 20W tubes, the light output is well below par.

Warning:
This circuit generates in excess of 300V DC which could be lethal. Construction should only be attempted by those experimenced with mains-level voltages and safety procedures

Voltage Inverter Using Switch-Mode Regulator

This circuit uses a step-up switch-mode regulator, which is usually used to produce a positive supply, to generate a regulated negative output voltage. The device used here is the MIC4680 from Micrel (www.micrel.com), but the idea would of course work with similar regulators from other manufacturers. Because of coil L1, which performs the voltage conversion by the intermediate storage of energy in the form of a magnetic field, the output is effectively isolated from the input. We can therefore connect the right-hand side of L1 to ground rather than to the positive output without causing a large current to flow.


Then we connect the ground pin of the regulator IC and all the components connected to it as the negative voltage output, isolated from ground. The components on the output side of the regulator are connected as usual: flywheel diode D1, coil L1 and the voltage divider formed by R1 and R2. These last two components set the output voltage, according to a formula given in the data sheet. Example component values for the MIC4680 used here are given in the table. The input voltage should lie within the permitted range for the regulator used, and must in any case be at least as great in magnitude as the desired output voltage (here +5 V or +12 V), so that the step-down regulation technique can work.


It is important to take care when building this circuit to mount the regulator using an insulator, since generally the GND pin of the device is connected to the heatsink tab. Also, the ON/OFF control input cannot be driven using a normal logic signal, since the regulator’s ground reference is the output voltage rather than ground itself. If the ON/OFF function is required, a level shifter or optocoupler must be used.

Discrete Voltage Inverter

The circuit in the diagram enables a negative voltage to be derived without the use of integrated circuits. Instead, it uses five n-p-n transistors that are driven by a 1 kHz (approx) TTL clock. When the clock input is high, transistors T1 and T2 link capacitor C1 to the supply voltage, UIN, which typically is 5 V. During this process, transistor T5 conducts so that T3 and T4 are off. When the clock input is low, T5 is cut off, whereupon transistors T3 and T4 are switched on via pull-up resistor R6 and either R4 or R5.

This results in the charge on C1 being shared between this capacitor and C2 Since the +ve terminal of C2 is at ground potential, its –ve terminal must become negative w.r.t. earth. The high level at the clock input must be of the same order as the positive input voltage, UIN, otherwise T1 cannot be switched on. The clock frequency should be around 1 kHz to ensure a duty cycle ratio of 1:1. Altering the ratio results in a different level of negative output voltage, but this is always smaller than that with a ratio of 1:1.

500W Low Cost 12V to 220V Inverter

Using this circuit you can convert the 12V dc in to the 220V Ac. 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.

How to calculate transformer rating
The basic formula is P=VI and between input output of the transformer we have Power input = Power output
For example if we want a 220W output at 220V then we need 1A at the output. Then at the input we must have at least 18.3V at 12V because: 12V*18.3 = 220v*1
So you have to wind the step up transformer 12v to 220v but input winding must be capable to bear 20A.

Inverter 12VDC to 220AC

Berikut ini adalah skema rangkaian sederhana 12VDC to 220VAC Inverter 100 Watt. Rangkaian ini murah dan sederhana cocok dirakit di sebuah veroboard akan tetapi anda harus hati-hati karena keluara (out-put) invereter ini mencapai 220VAV yang berbaya karena bisa menyengat anda (nyetrum).

Inverter adalah perangkat elektrika yang digunakan untuk mengubah arus listrik searah (DC) menjadi arus listrik bolak balik (AC). Inverter mengkonversi DC dari perangkat seperti batere, accu, panel surya / solar cell menjadi AC.

Gbr.1 Skema Rangkain Inverter 

12VDC to 220AC Sederhana

IC CD 4047 merupakan multivibrator CMOS astabil / monostable berdaya rendah. Di IC inilah akan di bangkitkan rentetan dua pulsa 50HZ (0,01s) yang saling berlawanan 180 derajat yang selanjutnya akan di kuatkan arusnya oleh Mosfet IRF 540. Pin 10 dihubungkan ke pin gerbang Q1 dan pin 11 dihubungkan ke pin gerbang Q2 mosfet melalui resistor R3 dan R4. Resistor R3 dan R4 ini berfungsi untuk mencegah IC 4047 agar tidak terbebani oleh mosfet. Ketika pin 10 yang terhubung ke Q1 tinggi maka arus akan mengalir melalui bagian atas primer transformator yang menyumbang setengah tegangan positif dari output AC. Dan ketika pin 11 yang terhubung ke Q2 yang tinggi maka arus akan mengalir melalui bagian bawah dari transformator utama dengan arah yang berlawanan dan ini akan menyumbang setengah tegangan negatif dari output AC sehingga dengan begitu akan terbentuklah tegangan AC. untuk transformatornya anda bisa menggunakan jenis step-down biasa (220V to 12VCT)/10A yang di balik fungsinya menjadi step-up.

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