Interested to make your own power inverter with built in charger? A simple circuit design that can be very easily built and optimized has been provided in this article. Read the complete discussion through neat illustrations.A massive 400 watts power inverter with built in charger circuit has been thoroughly explained in this article through circuit schematics. A simple calculation to evaluate the transistor base resistors has also been discussed.
I have discussed the construction of a couple of good inverter circuits through some of my previous articles and am truly excited by the overwhelming response that I am receiving from the readers. Inspired by the popular demand I have designed yet another interesting, more powerful circuit of a power inverter with built in charger. The present circuit though similar in operation, is more interesting and advanced due to the fact that it has got a built-in battery charger and that too fully automatic.
As the name suggests the proposed circuit will produce a massive 400 watts (50 Hz) of power output from a 24 volt truck battery, with an efficiency as high as 78%. As it’s fully automatic, the unit may be permanently connected to the AC mains. As long as the input AC is available, the inverter battery is continuously charged so that it is always kept in a topped up, standby position. As soon as the battery becomes fully charged an internal relay toggles automatically and shifts the battery into the inverter mode and the connected output load is instantly powered through the inverter.
The moment the battery voltage falls below the preset level, the relay toggles and shifts the battery into the charging mode, and the cycle repeats.
Without wasting anymore time let’s straightaway move into the construction procedure.
Parts List for the circuit diagram
You will require the following parts for the construction of the inverter circuit:
All resistors are ¼ watt, CFR 5%, unless otherwise stated.
R1----R6 = To be calculated - Read at the end of the article
R7 = 100K (50Hz), 82K (60Hz)
R8 = 4K7,
I have discussed the construction of a couple of good inverter circuits through some of my previous articles and am truly excited by the overwhelming response that I am receiving from the readers. Inspired by the popular demand I have designed yet another interesting, more powerful circuit of a power inverter with built in charger. The present circuit though similar in operation, is more interesting and advanced due to the fact that it has got a built-in battery charger and that too fully automatic.
As the name suggests the proposed circuit will produce a massive 400 watts (50 Hz) of power output from a 24 volt truck battery, with an efficiency as high as 78%. As it’s fully automatic, the unit may be permanently connected to the AC mains. As long as the input AC is available, the inverter battery is continuously charged so that it is always kept in a topped up, standby position. As soon as the battery becomes fully charged an internal relay toggles automatically and shifts the battery into the inverter mode and the connected output load is instantly powered through the inverter.
The moment the battery voltage falls below the preset level, the relay toggles and shifts the battery into the charging mode, and the cycle repeats.
Without wasting anymore time let’s straightaway move into the construction procedure.
Parts List for the circuit diagram
You will require the following parts for the construction of the inverter circuit:
All resistors are ¼ watt, CFR 5%, unless otherwise stated.
R1----R6 = To be calculated - Read at the end of the article
R7 = 100K (50Hz), 82K (60Hz)
R8 = 4K7,
R9 = 10K,
P1 = 10K,
C1 = 1000µ/50V,
C2 = 10µ/50V,
C3 = 103, CERAMIC,
C4, C5 = 47µ/50V,
T1, 2, 5, 6 = BDY29,
T3, 4 = TIP 127,
T8 = BC547B
D1-----D6 = 1N 5408,
D7, D8 = 1N4007,
RELAY = 24 VOLT, SPDT
IC1 - N1, N2, N3, N4 = 4093,
IC2 = 7812,
INVERTER TRANSFORMER = 20 – 0 – 20 V, 20 AMPS. OUTPUT = 120V (60Hz) OR 230V (50Hz),
CHARGING TRNASFORMER = 0 – 24V, 5 AMPS. INPUT = 120V (60Hz) OR 230V (50Hz) MAINS AC
P1 = 10K,
C1 = 1000µ/50V,
C2 = 10µ/50V,
C3 = 103, CERAMIC,
C4, C5 = 47µ/50V,
T1, 2, 5, 6 = BDY29,
T3, 4 = TIP 127,
T8 = BC547B
D1-----D6 = 1N 5408,
D7, D8 = 1N4007,
RELAY = 24 VOLT, SPDT
IC1 - N1, N2, N3, N4 = 4093,
IC2 = 7812,
INVERTER TRANSFORMER = 20 – 0 – 20 V, 20 AMPS. OUTPUT = 120V (60Hz) OR 230V (50Hz),
CHARGING TRNASFORMER = 0 – 24V, 5 AMPS. INPUT = 120V (60Hz) OR 230V (50Hz) MAINS AC
Circuit Functioning
We already know that an inverter basically consists of an oscillator which drives the subsequent power transistors which in turn switches the secondary of a power transformer alternately from zero to the maximum supply voltage, thus producing a powerful stepped up AC at the primary output of the transformer.
In this circuit IC 4093 forms the main oscillating component. One of its gates N1 is configured as an oscillator, while the other three gates N2, N3, N4 are all connected as buffers.
The oscillating outputs from the buffers are fed to the base of the current amplifier transistors T3 and T4. These are internally configured as Darlington pairs and increase the current to a suitable level.
This current is used to drive the output stage made up of power transistors T1, 2, 5 and 6.
These transistors in response to its alternating base voltage are able to switch the entire supply power into the secondary winding of the transformer to generate an equivalent level of AC output.
The circuit also incorporates a separate automatic battery charger section.
How to Build?
The construction part of this project is pretty straightforward and may be completed through the following easy steps:
Begin the construction by fabricating the heat sinks. Cut two pieces of 12 by 5 inches of aluminum sheets, having a thickness of ½ cm each.
Bend them to form two compact “C” channels. Drill accurately a pair of TO-3 sized holes on each heat sink; fit the power transistors T3---T6 tightly over the heat sinks using screws, nuts and spring washers.
Now you may proceed for the construction of the circuit board with the help of the given circuit schematic. Insert all the components along with the relays, interconnect their leads and solder them together.
Keep transistors T1 and T2 little aloof from the other components so that you may find sufficient space to mount the TO-220 type of heat sinks over them.
Next go on to interconnect the base and emitter of the T3, 4, 5 and T6 to the appropriate points on the circuit board. Also connect the collector of these transistors to the transformer secondary winding using thick gauge copper wires (15 SWG) as per the shown circuit diagram.
Clamp and fix the whole assembly inside a well ventilated strong metallic cabinet. Make the fittings absolutely firm using nuts and bolts.
We already know that an inverter basically consists of an oscillator which drives the subsequent power transistors which in turn switches the secondary of a power transformer alternately from zero to the maximum supply voltage, thus producing a powerful stepped up AC at the primary output of the transformer.
In this circuit IC 4093 forms the main oscillating component. One of its gates N1 is configured as an oscillator, while the other three gates N2, N3, N4 are all connected as buffers.
The oscillating outputs from the buffers are fed to the base of the current amplifier transistors T3 and T4. These are internally configured as Darlington pairs and increase the current to a suitable level.
This current is used to drive the output stage made up of power transistors T1, 2, 5 and 6.
These transistors in response to its alternating base voltage are able to switch the entire supply power into the secondary winding of the transformer to generate an equivalent level of AC output.
The circuit also incorporates a separate automatic battery charger section.
How to Build?
The construction part of this project is pretty straightforward and may be completed through the following easy steps:
Begin the construction by fabricating the heat sinks. Cut two pieces of 12 by 5 inches of aluminum sheets, having a thickness of ½ cm each.
Bend them to form two compact “C” channels. Drill accurately a pair of TO-3 sized holes on each heat sink; fit the power transistors T3---T6 tightly over the heat sinks using screws, nuts and spring washers.
Now you may proceed for the construction of the circuit board with the help of the given circuit schematic. Insert all the components along with the relays, interconnect their leads and solder them together.
Keep transistors T1 and T2 little aloof from the other components so that you may find sufficient space to mount the TO-220 type of heat sinks over them.
Next go on to interconnect the base and emitter of the T3, 4, 5 and T6 to the appropriate points on the circuit board. Also connect the collector of these transistors to the transformer secondary winding using thick gauge copper wires (15 SWG) as per the shown circuit diagram.
Clamp and fix the whole assembly inside a well ventilated strong metallic cabinet. Make the fittings absolutely firm using nuts and bolts.
Finish the unit by fitting the external switches, mains cord, output sockets, battery terminals, fuse etc. over the cabinet.
This concludes the construction of this power inverter with built in charger unit.
How to Calculate Transistor Base Resistor for Inverters
The value of the base resistor for a particular transistor will largely depend on its collector load and the base voltage. The following expression provides a straightforward solution to calculate accurately the base resistor of a transistor.
How to Calculate Transistor Base Resistor for Inverters
The value of the base resistor for a particular transistor will largely depend on its collector load and the base voltage. The following expression provides a straightforward solution to calculate accurately the base resistor of a transistor.
R1= (Ub - 0.6)*Hfe / ILOAD
Here Ub = source voltage to R1,
Hfe = Forward current gain (for TIP 127 it’s more or less 1000, for BDY29 its around 12)
ILOAD = Current required to activate fully the collector load.
So, now calculating the base resistor of the various transistors involved in the present circuit becomes pretty easy. It is best done with the following points.
We start first by calculating the base resistors for the BDY29 transistors.
As per the formula, for this we will need to know ILOAD, which here happens to be the transformer secondary one half winding. Using a digital multimeter, measure the resistance of this portion of the transformer.
Next, with the help of Ohms law, find the current (I) that will pass through this winding (Here U = 24 volts).
R = U/I or I = U/R = 24/R
Divide the answer with two, because the current of each half winding gets divided through the two BDY29s in parallel.
As we know that the supply voltage received from the collector of TIP127 will be 24 volts, we get the base source voltage for BDY29 transistors.
Using all the above data we can now very easily calculate the value of the base resistors for the transistors BDY29.
Once you find the value of the base resistance of BDY29, it will obviously become the collector load for TIP 127 transistor.
Next as above using Ohms law, find the current passing through the above resistor. Once you get it, you may go on to find the value of the base resistor for the TIP 127 transistor simply by using the formula presented at the beginning of the article.
The above explained simple transistor calculation formula may be used to find the value of the base resistor of any transistor involved in any circuit
Here Ub = source voltage to R1,
Hfe = Forward current gain (for TIP 127 it’s more or less 1000, for BDY29 its around 12)
ILOAD = Current required to activate fully the collector load.
So, now calculating the base resistor of the various transistors involved in the present circuit becomes pretty easy. It is best done with the following points.
We start first by calculating the base resistors for the BDY29 transistors.
As per the formula, for this we will need to know ILOAD, which here happens to be the transformer secondary one half winding. Using a digital multimeter, measure the resistance of this portion of the transformer.
Next, with the help of Ohms law, find the current (I) that will pass through this winding (Here U = 24 volts).
R = U/I or I = U/R = 24/R
Divide the answer with two, because the current of each half winding gets divided through the two BDY29s in parallel.
As we know that the supply voltage received from the collector of TIP127 will be 24 volts, we get the base source voltage for BDY29 transistors.
Using all the above data we can now very easily calculate the value of the base resistors for the transistors BDY29.
Once you find the value of the base resistance of BDY29, it will obviously become the collector load for TIP 127 transistor.
Next as above using Ohms law, find the current passing through the above resistor. Once you get it, you may go on to find the value of the base resistor for the TIP 127 transistor simply by using the formula presented at the beginning of the article.
The above explained simple transistor calculation formula may be used to find the value of the base resistor of any transistor involved in any circuit
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