In the previous tutorial, a square wave generator was designed having a symmetric output waveform having 50 Hz frequency and 12 V peak to peak voltage. The square wave generator designed using 555 timer IC was meant to provide input square wave in the inverter circuit. Now, it is time to complete the square wave inverter circuit by adding a switching mechanism and step up transformer in the previous designed circuitry.
The square wave generated from the 555 IC based circuit has peak to peak voltage of 12 V. This voltage needs to be stepped up using a step up transformer. So a step up transformer of 12V-0-12V rating is used for this purpose. But not only the voltage need to be simply stepped up, there is also a need for switching mechanism which allows drawing power from the inverter on power failure. The switching mechanism is developed with the help of transistors which are connected as high side switches in the circuitry. As transistors, MOSFET are used as they are most suitable for such AC application.
Components Required –
Fig. 1: List of components required for Switching Mechanism of Square Wave Inverter
Block Diagram –
Fig. 2: Block Diagram of Square Wave Inverter
Circuit Connections –
The square wave inverter has the following building blocks –
1) DC source – The power for delivering to the load is primarily drawn from a DC source. As a DC source, a 12 V lead acid battery is used.
2) Square Wave Generator – This is a 555 IC based astable multivibrator configured to have an output having 50% duty cycle or same period for both positive and negative cycles and an output frequency of 50 Hz.
3) Switching Mechanism – The switching mechanism is developed with the help of transistors. There are two stages of switching transistors used in the circuit. One stage is formed by the transistor Q1 at the base of which the output of the square wave generator is connected. The other stage is formed by transistor Q3 at the base of which the output of the square wave inverter is connected by a logic inverter switching transistor (shown as transistor Q2 in the circuit diagram). The output from the transistor stages is connected to the terminals of the primary winding of the transformer while its centre tape is directly connected to the positive terminal of the battery used as primary source of power. There are also two capacitors (shown as C3 and C4 in the circuit diagram) connected between the centre tape of the step up transformer and the common ground to eliminate noise.
4) Step Up transformer – A output from the transistor stages drive a 12-0-12 V step up transformer which converts the input 12 V AC to 220 V AC level.
Fig. 3: Prototype of Half Bridge Square Wave Inverter
While assembling this circuit, it is important to take care of the following precautions –
1. The current rating of the 12V DC, transformer and the MOSFET should be greater than and equal to the required current at the output. Otherwise, they will be unable to supply the required current at the output.
2. Always use a resistor at the base of switching transistor BC547 and at the gate of MOSFET. This will save the BC547 from any damage and solve the problem of ringing (parasitic oscillations) and voltage spike in the MOSFET.
3. Always use pull-down resistance from gate to source of MOSFET to discharge its parasitic capacitance.
4. Before working on MOSFET and BC547 refer their datasheet to check their breakdown voltage and maximum current limit. Never exceed these values otherwise, it will damage the transistor.
5. For drawing high current at the output, mount heat sink at both the MOSFET to aid cooling and to increase the lifespan of the MOSFET.
Along with heat sink, one can also use heat plaster to add extra cooling to the MOSFET by applying the plaster to both sides of the MOSFET.
6. Always use an electrolyte capacitor in parallel with low-value ceramic at the input supply. This will absorb any voltage spikes and unwanted fluctuations at the input supply.
7. The heat sink is also a conductor so never short the IC pins to the heat sink as this may damage the MOSFET.
8. Loose connections may result in no waveform or abrupt waveform at the output. So, the connections must be double checked before attaching a battery in the circuit.
9. Make sure the filter capacitor should be discharged before working on a DC power supply. For this short the capacitor with a screwdriver wearing insulated gloves.
How the circuit works –
For 12V DC source, a lead acid battery of 12 V is used. When the circuit is powered up by the 12V DC then the 555-timer starts generating a 50 Hz square wave at its output pin (pin 3 of the 555 IC). The generated square wave is of 12V but the AC appliances require 220 V for their operation. So this 12 V square wave needs to be boost up to a 220 V waveform.
For stepping up, first of all, two MOSFET stages are used as switches. For this 30NF10 MOSFET are used. 30NF10 is an N-channel power MOSFET designed for use in DC to DC converter, UPS and motor control circuits. This MOSFET can sustain a drain to source voltage of up to 100 V and a drain current up to 140 A in off state. In ON state, it requires a typical voltage of 3V (which can be maximum 4 V) at its Gate terminal. So the IC is compatible even with the TTL logic and can be driven using regular switching transistors.
So, there are two MOSFET stages in the circuit. One stage is directly connected to the 555 output and another one is connected through a logic inverter designed with the help of switching transistor BC547. A center tap transformer is used to step up the square wave from 12 V to 220 V. At the primary of the transformer, the centre tap wire is directly connected to the 12 V DC source and the remaining two are connected with the drain of both the MOSFET stages.
During the positive half cycle (from the square wave), the gate of the transistor Q1 (30NF10) and Q2 (BC547) gets a positive supply. This turns ON both the transistors Q1 as well as Q2. Due to conduction of the transistor Q2, it grounds the base of the transistor Q3 and this turns OFF the transistor Q3 (which will be in ON state in the next cycle). So, in the positive cycle, the transistors Q1 and Q2 are ON and transistor Q3 is in the OFF state. In this case, the point ‘a’ gets ground through transistor Q1. Thus, the upper half winding of the transformer gets a complete path and current starts flowing from point ‘b’ to ‘a’ (as shown in the image below). The transformer now step-up this 12V to 220V.
Fig. 4: Circuit Diagram showing Working of Half Bridge Square Wave Inverter
In the negative half, the transistor Q1 and Q2 goes in OFF state. Due to non-conduction of transistor Q2, the base of transistor Q3 now gets a positive supply from 12 V DC through resistor R6. This turns ON the transistor Q3. Thus in the negative half cycle, the transformer’s ‘c’ point gets a ground path and this again completes the circuit of the transformer through the lower winding. Again the transformer step-up this 12V square wave to 220V (as shown in the image above) and at the output, a symmetrical 220 V square wave with 50 Hz frequency is obtained.
Therefore in both positive and negative cycles, a continuous square wave of 220 V is obtained at the output. As in this design, at any time only half of the transformer primary winding is used, this topology is known as Half Bridge Inverter.
Fig. 5: Prototype of Switching Mechanism for Square Wave Inverter
The other components connected in the circuit have their own role. In the circuit, the capacitors C3 and C4 are connected between the centre tape of the step up transformer and the common ground to avoid any high voltage spikes in the input supply as it can affect the output load. The resistor R3 and R6 are used to avoid any parasitic oscillations (ringing) in the circuit.
The presence of parasitic capacitance at the gate of MOSFET can cause the ringing effect and this can heat up the MOSFET due to slow turning ON of it. This problem can be solved by connecting a low value resistor at the gate of MOSFET. So a resistor of value 220 E is connected at the gate of the MOSFET. Any resistor having resistance value between 10 E and 500 E can in fact be used in place of it. The resistor R4 is used to limit the current at the base of the transistor Q2 (BC547). For driving the switching transistor Q2 (BC547) in its full conduction mode, an appropriate value of current must be at its base. The value of the required resistance is calculated as follow – Desired base current BC547, Ib = 55 mA (approx.)
Now by ohms law, the value of resistor R4 can be calculated for 55 mA current.
V = I*R
Input Voltage, Vin= 12 V
Desired current, Ib= 55 mA
By putting all the values,
R = 12/0.055
R = 220 E (approx.)
So the resistors R3, R4, and R6 (as shown in the circuit diagram) are chosen of value 220 E.
The resistors R5 and R7 are pull down resistors in the circuit. The MOSFETs have internal parasitic capacitance and due to this during conduction of MOSFET, the MOSFET start accumulating the charge. So this charge must be discharged by some resistor otherwise, the charge starts rising in the MOSFET and can damage it. Due to this, the resistors R5 and R7 are used as pull-down resistors connected between ‘Gate’ to ‘Source’ of the MOSFET. These resistors drain the inherent capacitance from both the MOSFETs.
Testing the circuit –
In this circuit, a 12V step up transformer is used, so by applying a 12 V square wave at its input, it should provide 220 V at the output. That means the transformer should provide an output voltage which is around 18 times the input voltage. But practically the circuit designed in this tutorial does not output 220 V at the secondary of the step up transformer. One reason for this is that as the input of transformer should be 12 V RMS which should convert to 220V. Here, 12V is the peak voltage, not the RMS voltage. Another reason is the iron or core losses in the transformer which reduces the output voltage. On calculating the RMS voltage (Vrms) at the input of the transformer by using the following standard equation –
Vpeak = 1.4*Vrms
Vrms = Vpeak /1.4
Vpeak = 12V (in this case)
Vrms = 12/1.4 = 8.6V (approx.)
The above calculation clears that 8.6V RMS voltage is given to the transformer, not 12 V RMS. Now as the transformer gives an output which is 18 times the input signal. So,
Output voltage at the transformer, Vout (observed) = 18*Vin(RMS)
Vout (observed) = 18*8.6
Vout (observed) = 156V (approx.)
So the desired value of peak voltage (Vpeak) for 12 V RMS is 17 V. So, when the square wave having a peak voltage of 17 V is applied at the input, then the output of the transformer can be 220 V. On observing the output of the circuit on a cathode ray oscilloscope (CRO), the following waveform was observed.
Fig. 6: Graph of Output Waveform of Half Bridge Square Wave Inverter as observed on a CRO
So, practically Input voltage of the transformer, Vin= 12V
When an AC bulb is connected at the output then the output voltage is observed as follow –
Output voltage of transformer with AC bulb/load, VL = 155 V
Current drawn by the AC bulb, IL = 40 mA
Now calculating the efficiency of Square Wave Inverter –
Efficiency % = (output power/Input power) *100
Input power, Pin= Vin*Iin
Input current drawn by the inverter when load is connected, Iin= 1000 mA
On putting the above values,
Pin = 12*1
Pin =12 W (approx.)
Output power or load power, Pout = Vout*Iout
Vout= 155 V
Iout= 40 mA
Putting the above values,
Pout = 155*0.04
Pout= 6.2 W
By putting above calculated values in efficiency equation,
Efficiency in percentage = (6.2 /12)*100
Efficiency in percentage = 52%
The square wave inverter designed in this circuit has 52 percent efficiency. It has other limitations as well.
Fig. 7: Image showing Half Bridge Square Wave Inverter under Test
Like this inverter requires 500 mA for its operation when no load is connected at the output. So the output current is always 500 mA less than the maximum current that can be supplied by the input source. The maximum output voltage is 155 V which is less than the 220 V (standard AC supply voltage). This can be rectified by increasing the input voltage to primary of the transformer up to 18V by using some boosting circuit. The output voltage drops when current demand increases at the output. This is due to the transformer inefficiency, as in transformer there are iron and copper losses. So by using a more efficient transformer, a constant voltage can be maintained. The output waveform of inverter has some voltage spikes. These spikes can be reduced by using a snubber circuit at the input of the transformer.
The square wave inverter designed in this tutorial has high noise at the output. Also, this inverter does not provide an output with the varying magnetic field. So, it cannot be used with audio and mechanical devices like motor which have internal inductor in their circuitry. If this inverter circuit is used with such devices they will have a noise or humming sound. However, this square wave inverter can be used to power appliances which have internal AC to DC converters like bench power supplies, batteries or with such PCs which can manage with high input harmonics. This inverter circuit is cheap and easy to design and can be used with the low sensitive devices like heaters, lighting systems and with motors which does not have windings.
In the next tutorial, a modified sine wave inverter will be designed. The modified sine wave inverters have an output waveform closer to sine wave which improves the efficiency of the inverter.