Problem with 12 to 220 inverter

[mohsen 2012]

Hello , i build 12 to 220 inverter based on 555 Timer :



I use a 9.5 v ac transformer and a 6FM 12V battery , but the output of the inverter give 74V ac !!!!!








i think that the problem is the transformer , please help me and thanks.

 

[BradtheRad]

Your schematic looks as though its output is square waves.
However your meter probably is calibrated for sinewaves.

I too got a low reading from my DMM, when checking output from an inexpensive commercial inverter. (It produced either square waves or modified sinewaves.) I recall my reading was between 90 and 100V.

There might be more steps you should take to make your transistors turn on fully.

 

[ydtech]

The half-bridge applies 12 V peak-to-peak, so 6 V peak. The secondary will receive 6*(220/9.5) = 140 V peak. Since the DVM is calibrated for sine wave the 74 V reading is not unreasonable even if a little lower than expected. If you can't get a true RMS meter, half wave rectify and filter the output, this will give you the true peak value.

According to this: https://answers.yahoo.com/question/index?qid=20120202113838AAARBHK , you should be seeing about 140 * .707 = 100 V on the DVM, so something else is going on. Probably not enough drive.

The 555's output is current limited and the TIP4x's are crappy. Add a couple of BC5x8's in darlington configuration to drive them harder.

 

[mohsen 2012]

I add a 10n 400v capacitor in the out , the voltage is down to 10 v !!!!
I thought that the circuit is tested and work 100% !!!

 

[Audioguru]

The circuit is hopeless because its designer forgot to add all its voltage drops.
You measured it with no load.
1) The output of a 555 with a 12V supply and a current of 25mA is 10.5Vp-p. Look at its datasheet to see it. It is 7Vp-p when its output current is 200mA.
2) The 100 ohm resistor has a voltage drop of 5Vp-p when the current is only 25mA then the signal through it to the bases of the output transistors is only 5.5Vp-p.
3) The output transistors cause a voltage drop of 1.8Vp-p when their base current is 25mAp-p and their collector-emitter current is 250mAp-p. Then the 9.5V winding of the transformer gets only 3.7Vp-p.
Then the output voltage of the transformer is 3.7V x 220/9.5= 85.7Vp-p which is 30.3V RMS.

The output power when the output voltage is only 30.3V is 3.7V x 250mA= 0.93Wp-p which is 0.33W RMS.
But you had no load so your output voltage was higher. The lousy circuit has a square wave output and has no output voltage regulation.

 

[mohsen 2012]

can you please give me a correction modified schematic of the inverter.

 

[c_mitra]

Suggest the following tests:

1. Calculate or measure (you DMM can measure frequency, right) the output frequency. Set it, using the potentiometer, to somewhere between 50-100.

2. Replace the base drive resistor of 100R with a smaller one; try 10R or 22R. Smaller is better.

3. The output cap (2200uF) is perhaps having a high impedance at this low frequency. Just remove it. Place a 2-5A fuse there.

 

[mohsen 2012]

OK sir , I will tell you the result....

 

[betwixt]

3. The output cap (2200uF) is perhaps having a high impedance at this low frequency. Just remove it. Place a 2-5A fuse there.

Will burn something out - probably not the fuse!

You could try increasing the value but the circuit is incapable of producing more than a few Watts output anyway. The 555 will not be able to produce enough output current to fully saturate the transistors. It isn't the most efficient of designs by far.

Brian.

 

[ydtech]

At 50 Hz Xc is 1.2 ohms, not counting harmonics. Most power transformers can go to a few hundred Hz before getting excessively lossy.

 

[Audioguru]

How much output power do you want?
Here is a simple 100W inverter that produces a squarewave output. Its voltage is not regulated so it will be too high with a light or no load and will be low with its rated 100W load. A short circuit or overload on its output will destroy it. It does not detect a low battery voltage and shut off so if it runs too long its battery will be ruined. https://www.electronicshub.org/simple-100w-inverter/

Wait a minute! It has wrong parts values, is missing some parts, has un-needed zener diodes and its transformer voltage is not listed.
The 250 ohm pot R10 adjusting the frequency of the CD4047 should be more than 10k ohms so use a fixed 47k resistor then C1 should be a 5% film 0.1uF (100nF) capacitor.
The transformer should be 20V center-tapped and 120VA to 150VA.
I fixed most of its schematic:

 

[mohsen 2012]

I try removing the 2200 cap but the voltage down at 15 !!!
so i want to build other one

 

[Audioguru]

If you remove the capacitor then the low voltage winding of the transformer gets half-wave positive pulses instead of AC. The pulses are for half of each cycle and is 0V for the other half. Then the output voltage is half of when the circuit has the output coupling capacitor. A squarewave inverter is supposed to have an AC output that has a positive pulse for the first half of each cycle then immediately produces a negative pulse for the other half.

Your tiny little transformer and battery might supply only 40W for one hour.

 

[c_mitra]

Can you please check the transisors too? Are they getting very hot? I check from their datasheets that they do not have high gain and need lots of drive.

 

[Audioguru]

Can you please check the transisors too? Are they getting very hot? I check from their datasheets that they do not have high gain and need lots of drive.

How can the transistors get hot when they have no load?

The circuit is not a linear audio amplifier that uses plenty of collector to emitter voltage (4V in the datasheet) so it has a current gain (hFE) of 40. Instead the transistors switch from saturation (fully turned on with a low collector to emitter voltage) to cutoff. Like every other transistor, for it to saturate its base current must be 1/10th its collector current, the TIP41 has a max saturation voltage loss of 1.5V when its collector current is 6A and its base current is 600mA. But the 555 has a maximum allowed output current of only 200mA and the 100 ohm series base resistor needs 60V across it if the base current is 600mA or needs 20V across it if the base current is 200mA. Therefore the 100 ohm resistors have values much too high and the 555 cannot supply enough output current.

 

[ydtech]

What you describe is the common emitter case, and is correct. But in this case the transistors are emitter followers. So the emitters go to whatever voltage the 555/100R combo permits, leaving quite a bit of Vce for the transistors to dissipate. I do suspect there's not all that much current so they're likely not going to be seriously hot.