Yesterday, I recommend Simple working principle of the inverters. Which you have understood well. Today we see Operation of 200 watt inverter diagram continuously.
In the block diagram of Figure 1. Starting from the frequency generator circuit or the oscillators that works with IC2 numbers SG3526 Which is IC pulse width modulation as figure 2 shows the internal structure. Served to genered the rectangular signal on frequency 50Hz output to Output A and B.
Which both these output signals have phase difference of 180 degrees all the time. That is if a Status Output A High, Output B will be low. Or If Output A is Low, Output B will be the High instead. These two signals are sent to control the output transistors Q1 and Q2.
Figure 1 is 200watt inverters block diagram
Both transistors Q1 and Q2 are Enhancement MOSFET type of N channel So not conduct current simultaneously. Cause the current that flow in the primary coil of transformer T1 flow switch a way all time. Electromagnetic fieldSo swelled and collapsed, inductor to the secondary set. get voltage 220 volts frequency of 50Hz to output.
After the circuit work will cause heat up the transistor. power output (Q1 and Q2) If the heat is too high. It may be dangerous to the transistor. Especially when it requires more power to the load. Therefore, So need to check the temperature of the cooling pad or heat sink for both transistors are overheating or not? This is a device that functions by PTC (Positive Temperature Control) If the temperature rises abnormally, the PTC will send the check to the IC1.
Which is op amp IC1 Reset command is sent to IC2. to IC2 frequency generator to stop. The end result is that the output transistors are both 2 stops prevent damage to the transistor.
We see circuit real use. As shown below. Which has working principle of the circuit is divided into three sections are a oscillator, output and protection section.
But today I’m talking the oscillator circuit before.
Starting when push switch-S1 which is switch turn on voltage +12V will be send to diode-D3 provided the IC2-SG3526N at pin 14. To pass through resistor-R12 of 33 ohm, there are C1 as decoupling backup current to IC2-SG3526N.
The bring a positive voltage. Which is the supply voltage IC2-SG3526N to via the diode-D3 before to want to make IC2 be get the supply voltage constant over time. Because the source voltage of the battery will be sent to the output circuit directly.
The function of output has characteristics of switches. Which when switch is in the range beginning to state on. May cause supply voltage source drop down. Which this drop voltage will cause IC2 malfunction. Must to insert diode-D3 across the voltage source and supply voltage of ICs.
When the supply voltage source has tumbled With starting to conduct current of the output circuit has cause voltage at anode (A) of D3-1N4002 lower than pin cathode (K), The diode-D3-1N4002 has the voltage across of state Reverse bias. They can not conduct current. Makes the output circuit can not draw on the accumulated charge of C1 is correct. As a result, the power supply voltage IC2 not tumbled like a voltage source.
The diode-D3 In addition to maintaining a constant supply voltage to IC2.Can also prevent damage of IC2. Which is the oscillator circuit. Because the connection wires from wrong battery terminals. If the wires alternating positive to negative and negative to positive. The diode-D3 conditions in Reverse bias status. Can not conduct current. Circuit will stop working without damage.
When has supply voltage IC2 so Starting frequency. By the frequency is determined with the value of R11 and C4 connected to pin 9 (RT) and 10 (CT) respectively. This frequency is 50Hz Which corresponds to the frequency of home power. Frequency that IC2 built this, as rectangular signal. By the output pin 13 (OUT A) and pin 16 (OUT B).
The output power MOSFET
As show figure below When IC2-SG3526 can generate the square wave frequency of 50Hz output at pin 13 and pin 16. Both signal that has Phase difference is 180 degrees will send to R13 and R15, to bias pin gate of both transistors Q1 and Q2 number IRFP054. Which are mosfet type N channel, that has diode damper Diode is within, makes both of transistors alternately conduct current. If any one has a bias. signal at gate pin as high will can conduct current(ON). But if it get signal as low will stop conduct current(OFF).
The conduct currents of the two transistors. Cause current to flow in primary coil of transformer T1 into drain out to the source pin of Q1 and Q2, and to fully circuit at ground by pass to R8 value 0.01 ohm. Which is resistor monitoring the amount of current in the output circuit, which is discussed next.
The working of Q1 and Q2 are characterized the switches. The spike voltage will cause the occurs at pin drain of both transistors. Which occurs incunabulum It is changing state from ON to OFF. This spike voltage is the voltage is very high. May be higher than the supply voltage, many times. Which is extremely dangerous to the power output transistors. Therefore, it is necessary to eliminate this spike voltage out. The circuit that performs this function is called a snubber circuit.
The snubber of this circuit consists of D4, D5(BY299), D6, R14, C9 and C10. For D5 and D6 are diode works faster or called Fast Revovery Diode number BY299.
Under normal conditions, D5 and D6 will do not conduct current. Which it will conduct current when voltage at pin drain of Q1 or Q2 rise. Until have value greater power supply (+12 volts), The current will flowing through the both diode into charge to C9 and C10, and if voltage at pin drain of Q1(IRFP054) and Q2(IRFP054) have higher 18 volts, will makes the zener diode D4. Which have value 18 volts conduct current through the R14 of 18 ohm.
Which operation of all the equipment as a result to lower the spike voltage.
Likely that both power transistors will be damaged less.
GET UPDATE VIA EMAIL
I always try to make Electronics Learning Easy.