30W-simple inverter using 6 transistors.

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  Posted by momename - January 8, 2014 at 2:17 pm

This is small inverter on 30 watts, It converts DC voltage from 12V battery to AC 220V-230V at 50Hz which is electricity same use in your house. It can provide 2-3 Air pump or other. You will like them because so cheap and easy to builds.

The Air Pump requirement
Before see the Aquarium Air Pumps are chosen as Figure 1, He said that small,cheap and works great.

1-aquarium-air-pump-is-chosen
Figure 1 Aquarium Air Pump is chosen

2-the-aquarium-air-pump-special-feature
Figure 2 The Aquarium Air Pump special feature.

3-the-internal-structure-of-aquarium-air-pump
Figure 3 The internal structure of Aquarium Air Pump.

Then, what is it required. When look at its bottom will see that a special feature as Figure 2 notice Importance : Use for voltage of 220 volts, Frequency of 50Hz(AC:220V/240V 50Hz) And use AC electrical power of 2.8 watts (power:2.8W)

The internal structure of Aquarium Air Pump.

The air Pump consists of a rubber ball (similar to a pneumatic for bicycle tire) to generated to output. By The driving force between the magnet and electromagnet.

4-block-diagram-of-the-12-volts-battery-to-220v-ac-50hz-circuit
Figure 4 Block diagram of the 12-volts battery to 220V AC 50Hz circuit.

Then, try to apply AC 220V to them will see that Permanent magnet vibrates very fast, so Pumping the air out. Near the same that have a similar device is a transformer coil pack, but with only a single coil. It is the electromagnet there.

The AC electricity cause Electromagnetic field collapse the alternates a pole continually, Magnetic oscillate shorter, Set rubber air pump is working.

Issues to think

In detail of the Air pump above, We can handle issues, as follows :
1.The 50Hz frequency: has very importance, if too low frequency Strong winds are not very and when high frequency the coil will respond before, until stationary to do.
2.The Waveform balance is best perfect, When see in home electricity waveform will see that has peaks to peaks high-low. Which we can use other waveform instead so may will use the square wave.

To design power converter circuit

As above, we have to convert 12V battery to 220V AC 50Hz. by an inverter circuit. Which we can draw the simple block diagram As Figure 4. When apply 12-volts battery to the 50Hz frequency generator on square waveform.

After that, current will flow through amplifier up. To drive the primary coil of transformer and Inductor current to the secondary coil. into the 220V AC 50 Hz as you need ready to use with the air pump.

The frequency generator circuit

In conditions, I select the oscillated generator circuit by dual transistors. As used an 2 led alternating flasher circuit. As Figure 5 is the 2 led alternating flasher circuit.

5-the-2-led-alternating-flasher-circuit-using-two-transistors
Figure 5 is the 2 led alternating flasher circuit using two transistors.

I modified to change the some devices until will be the 50Hz frequency generator circuit as Figure 6
6-the-50hz-frequency-generator-circuit
Figure 6 the 50Hz frequency generator circuit

The selects the parts
1.Q1,Q2 I selected the transistors as number: 2SC1815. Because cheap, suitable for use not over than 20-volts and 0.1A, normally gain under 200 times.

2. R1,R4 are resistors same load of Q1,Q2. When we want to set the waveform is balance (50% duty cycle) both has voltage or no voltage.
In this case I select both R1,R2 = 10K will can cause current flow to lead C-E of Q1,Q2

called “ICQ” by ICQ = (VB – VBE) / R1_______(1)

To set : VB = 12V (battery voltage), VCE = 0.3V (voltage across pin C-E while full conduct current)

R1 = R2 = 10K

Put in

ICQ = (12V – 0.2V) / 10K
= 1.18mA

3. R2,R3 are resistors to bias at B of Q1,Q2 ,they so conduct current. By resistance of R2,R3 have effect to frequency and waveform so should set equal.

By R2 value : (Vb – Vbe)/ Ib

We can set them : VB= 12V, VBE = 0.6V(voltage across BE pin)

But now we not know IB? so do below:
IB = ICQ / hFE

We dertermaned value by ICQ = 1.18 mA, hFE = 20 times.

IB = 1.18 mA / 20
= 0.06 mA

Therefore, we insert them into (1)
R2 = (12V – 0.6V) / 0.06mA
= 190K
But this value not sell in any store, so use 180K instead better.
4. C1,C2 are capacitors that set frequency output, should same value, so calculate only C1 in single. In this case, I choose a 0.1uF because is medium , easy to find.
from: T = 0.7 x R2 x C1
Back new recipes (with move sideways) is
C1 = T/(0.7 x R2) ___________(A)

Try to configure R2 = 180K; But T is unknown, Is looking:

By T is time period associated with the frequencies (F) is:
T = 1/F
Substitute:
T = 1/ 50Hz = 0.02S

Try substituting values in: (A)
C1 = 0.02 S /(0.7 x 180K)
= 0.15uF

But I used 0.1uF, because nearby can be used interchangeably.

Increase current to drive transformer
Next, the current amplifier circuit by increasing signal from the frequency generator rises up before drive the transformer next, this is importance section.

First of all, I think to use 2SD313-transistor only one. Because Can tolerate high current of about 1A. ( at 12V). But has disadvantages are low gain to have add increase current transistor other one.

So is simple current amplifier circuit as Figure 7. When is set as this Darlington form, cause very high gain and have high input impedance, so reduce noise so much.
7-30watts-mini-inverter-circuit
Figure 7 The 30 watts simple inverter version 1

Because we need waveform that has high power both positive and negative. So should use transformer that has center trap and use two amplifier set, that has same electricity feature.

By Each set is alternately working, when Q1 conduct current, result to Q2 not works. So current flow through R4, R6 provide current bias to Q4,Q6 conducts current. Result have current from the battery to transformer coil.
But when Q1 stop cause Q2 works so has current flow through to R1,R5, to provide current trigger to Q3,Q5 conduct current replace Q4,Q6. Thus current from battery so flow through other transformer coil instead. Cause electricity inductance through the transformer switch between positive and negative with high power.

Selecting parts for drive transformer.
1. Q3,Q4 are pre driver amplifier, use many number that is NPN type, I choose 2SC1815 again.
2.Q5,Q6 are high output drive amplifier transistors. I selected a 2SD313 because high current collector more than 1A and cheap. You can use H1061 or TIP41 quality better (high current and voltage)
3. R5,R6 are resistors to provide current to bias the base of Q3,Q5 and Q4,Q6, which they will work alternately as principle above. In this case I so choose them are 4.7K.

4. T1 is transformer increase voltage from 12VAC into 220VAC at frequencies of 50Hz.
First we have to know power to use before. Which is 8.4W as the air pump spec. I set up to 10 watts better. And then power output is 10 watts.
Therefore, power input of the transformer is also 10 watts when use 12-volts battery will use current is : 10W/12V = 0.83A or about 1A.

Real application circuit
In Figure 8 is perfect circuit, to adding efficiency of this project so I add many parts in the perfect circuit.
8-the-perfect-30watts-inverter-circuit-using-6-transistor
Figure 8 The perfect 30watts inverter circuit using 6 transistor.

- D1,D2 are diode protect noise from working of transformer. I choost an 1N4007 (1A-1000V).
- D3 is diode protection if connected in wrong polarity. The negative voltage will flow through D3 instead. Makes circuit safe. But D3 may be short and F1-fuse blown instead.
- F1 is fuse protect over current 1A and in case of wrong polarity F1 also blown.
- LED1 is power on display when have the power : 220V 50Hz completely. The R7 acts to reduce current to LED1. And there is D4 set the polarity to LED1 appropriately

The building and how to use its.
To begin with we have to many parts below.

The parts list.
1. Q1-Q_______2SC1815 or others equivalent transistors ____ = 4 pcs.
2. Q5,Q6______2SD313 or others equivalent transistors _____ = 2 pcs.
3. D1-D5 Diode: 1N4007 or equivalent ___________________ = 5 pcs.
4. LED1-LED as you like________________________________= 1 pcs.
5. C1,C2___0.1uF 100V__Mylar capacitors________________= 2 pcs.
6. C3___0.01uF 100V___Mylar capacitors__________________= 1 pcs.
7. C4______100uF 25V____Electrolytic capacitors__________= 1 pcs.
8. R1,R4____10K 0.25W 5%__ Resistors________________= 2 pcs.
9. R2,R3____180K 0.25W 5%___Resistors_______________= 2 pcs.
10. R5,R6___4.7K 0.25W 5%___Resistors_______________= 2 pcs.
11. R7______100K___”____________”_____________________= 1 pcs.
12. F1______Fuse 1A___________________________________= 1 set.
13. T1______Transformer___0.5A__12V CT 12V_____________= 1 pcs.
14. Heat sink,PCB,and others parts, etc.

Then make PCB as Figure 9 and assemble many parts on PCB as Figure 10
I think many people visiting this site regularly. And Building projects have similar characteristics. I would not advise you how to create this project.

9-actual-size-of-single-sided-copper-pcb-layout
Figure 9 Actual-size of Single-sided Copper PCB layout.
The-components-layout
Figure 10 The components layout.



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