Need help in inverter circuit

Thread Starter

Puspal Manna_1493428956

Joined Apr 28, 2017
2
The circuit connection is alright. The transformer used here is 12-0-12 to 230 volts transformer. Here we have used 25 to 1 but don't worry. In practical we have used 12 to 230 transformer. But the problem is I am not getting the required output. Maybe we are making wrong at a simple point but can't identify. Please help me, guys. I am stuck here for many days.The output of the 555 timer is a square wave. The waves of the bases of transistor BC547, Power transistor 2N6292 s are also given in the attachment respectively. There are spikes that are coming from bc547 and the lower power transistor 2n6292(Q3). If you gen any faulty then please let me know.
Thankyou
 

Attachments

EM Fields

Joined Jun 8, 2016
583
The circuit connection is alright. The transformer used here is 12-0-12 to 230 volts transformer. Here we have used 25 to 1 but don't worry. In practical we have used 12 to 230 transformer. But the problem is I am not getting the required output. Maybe we are making wrong at a simple point but can't identify. Please help me, guys. I am stuck here for many days.The output of the 555 timer is a square wave. The waves of the bases of transistor BC547, Power transistor 2N6292 s are also given in the attachment respectively. There are spikes that are coming from bc547 and the lower power transistor 2n6292(Q3). If you gen any faulty then please let me know.
Thankyou
1. If you want +/- 130 volts out of the transformer and the input voltage is 12-0-12volts, then the turns ratio, n, must be equal to Vs / Vp = 130V/12V = 19.17

You show your load as 100 ohms, which means that with 230 volts out of the transformer the current through the load would be +/- 2.3 amperes and the power dissipated by the load would be P = IE = 2.3A X 230V = 529 watts .

If the transformer was 100% efficient, (which it isn't) then the primary would take 529 watts from the 12 volt source, so the current into each half of the primary would be I = P/E = 529W/12V = 44 amperes.

Since the transformer is being driven push-pull, if it were driven with a 50% duty cycle (which it isn't) the average current through each half of the primary would be 22 amperes.

The data sheet for the 2N6292 is available here: http://www.onsemi.com/pub/Collateral/2N6107-D.PDF
and the absolute maximum ratings indicate that the peak collector current allowed is 10 amperes, so the 2N6292 isn't a good choice.

2. In switching applications, in order to assure saturation of the collector-emitter junction, the driver transistors' bases are usually biased with 10% of the collector current. You've shown 15k ohm base bias resistors, which means that instead of 4.4 amperes of base current, with a 12 volt source, less than one milliampere will flow.

3. Configured as shown, the 555 can't get to 50% duty cycle.
 

tcmtech

Joined Nov 4, 2013
2,867
Since the transformer is being driven push-pull, if it were driven with a 50% duty cycle (which it isn't) the average current through each half of the primary would be 22 amperes.
Actually it will still be 44 amps per side. total input watts stay the same but the duty cycle for each side is half and alternates out of phase so they are not sharing the current.

Good observation sir. It was helpful. But in the transformer, with the increase in load in secondary, the secondary voltage drops. What is the reason behind that?
Realistically for a 12 volt input to 230 VAC output you would want a 12 volt center tapped transformer 6 - 0 - 6 and you would regulate your output voltage by varying the duty cycle on each half of the switching circuit which to do that a 555 IC is bad choice. In fact for any power inverter a 555 IC is a bad choice given how many proper inverter driver IC's there are to work with that cost about the same and use similar numbers of additional components but also give you full voltage and current feedback control as well. the old SG3524 or Tl494 are good basic and easy to find inverter driver IC's to work with.

Although to be honest you can buy a compete ready to run sine wave driver board minus the power switching devices for less than each of those IC's circuits cost to make. https://www.wish.com/c/57233fa2a01557014a65aa93?hide_login_modal=true&from_ad=pla4&_display_country_code=US&gclid=CjwKCAjw3KDIBRAzEiwAQ6jiu45pQSW4UQ6RazUftCnsvVc36yeeqCvzu2QY37ClFQjBr3DicwVJahoCcwkQAvD_BwE ~$8 with shipping. Data sheets and manual. http://www.egmicro.com/download/EGS002_manual_en.pdf and can operate in either full or half bridge input mode.

Also for your power switching devices the basic rule of design for a push pull two device design is that each device have a tle as t a 4X voltage rating over the maximum input voltage and a 4x current rating over the peak current the circuit will be switching. Which for a 12 volt system that might be used on a vehicle that would have a ~14.5 - 125 volt output I would not go less than a 60 volt device rating.

Also for whatever output wattage figure at bet 80% efficiency plus the shorter duty cycle being below 50% and whatever average current you come up with at at the 12 volt input side each switching device (or bank of them) would need to be rated for at least 6X that, minimum, which means in a real life working inverter your 2N6292 transistors feeding a center tapped 12 volt transformer would be good for a roughly 20 watt inverter unit provided you could drive their bases at near 3 amps each.:oops:
 

DickCappels

Joined Aug 21, 2008
6,373
You appear to be using an LMC555C (as opposed to the bipolar versions of the NE555). You can get a symmetric square wave by disconnecting everything from pin 7 of the '555 and connecting the R6-R1 in series to pin 3 of the LMC555C. You can't use this trick with the bipolar versions.

I believe there are no exceptions to this: The secondary voltages of all transformers drop when a load is applied. This varies a lot from transformer to transformer.

The turns ratio might seem right, but the transformer core does not see all of those volt-turns. The winding you are using as the primary has resistance and even when the secondary is not loaded there is a lot of current flowing in the primary -this current is used to magnetize the core and is referred to as "magnetizing current". This current can be very high -for example in a small 12V transformer "backwards" as a step-up transformer I found the magnetizing current to be 1.4 amps peak-to-peak. The IR drop in the primary can be considerable. You can measure the magnetizing current with a current probe that has good low frequency response or by placing a small resistor in series with the power supply to the winding.

upload_2017-5-3_11-46-23.png
Above: Blue is secondary voltage yellow is magnetizing current.

The turns ratio is not what you think it should be. A 12 volt winding will usually have extra turns on it so that when the specified
load is connected the voltage drops down to 12 volts instead of down to something lower. The result is that the turns ratio is different from what we would infer from the rated output voltage. These factors vary by individual transformer design and have to be determined experimentally.

I have done what you are trying to do but only to get voltage and very little current. I had to go through several transformers to find one I could make work. The most significant specification I could come across to predict how well a given transformer would work is in this kind of application is the Regulation specification, which is not available for all transformers.
 

EM Fields

Joined Jun 8, 2016
583
EM Fields said:
Since the transformer is being driven push-pull, if it were driven with a 50% duty cycle (which it isn't) the average current through each half of the primary would be 22 amperes.

Actually it will still be 44 amps per side. total input watts stay the same but the duty cycle for each side is half and alternates out of phase so they are not sharing the current.
Each half of the primary draws 44 amperes, but it only draws it for half the time, so the average current through that part of the winding will be 22 amperes.
 
Top