Sunday, February 12, 2012

Making a Pure Sine Wave Inverter - Concept Explored


Through a simple concept, the idea of building a pure sine wave inverter has been explained in the article. Read on to know the whole construction details.

A pure sine wave inverter or rather a sine wave inverter circuit is hard to find and the available ones are too complicated for everyone to understand. A simple concept through circuit schematic has been explained here.

A Viable Concept

It’s true that sine wave inverters are not easy to build, due to many different reasons. But it’s probably the most sort after circuit and also pretty difficult to find. For the folks who are desperately looking for such a circuit, perhaps this article can help.

After a lot of thinking, I probably seem to have designed an easier (though not quite efficient) concept of a pure sine wave inverter circuit. Since the circuit has not been tested by me so won’t be able to tell much regarding the exact specifications of the circuit and would like to leave it up to the readers to decide the feasibility of the present circuit.

The idea struck me while reading the circuit description of a MOSFET audio amplifier. We all know that when an audio signal is fed at the input of an amplifier, it produces an amplified output power having exactly the same properties as the input.

That simply implies, in place of an audio signal if a pure AC signal say from a Wien bridge circuit is applied to the input of a power amplifier and an inverter transformer connected to its output (where normally a speaker would be connected), it would certainly produce an amplified replica of the input. And the secondary winding of the connected inverter transformer would definitely produce a sine wave AC power (My assumption).

The only big problem is the loss of a significant amount of battery power in the form of heat through the power devices reducing the overall efficiency of the inverter.

Let’s move on and see how the different stages of the proposed circuitt functions.








The Oscillator Circuit

The simple sine wave generator circuit shown alongside may be used to produce the required sine waves at the input of the power amplifier, let’s study regarding its functioning through the following steps:

Op amp A1 is basically wired as an astable multivibrator,

Resistor R1 and the capacitor C1 define the frequency of oscillation of the astable.

The square wave from A1 is fed to A2 which is configured as a double pole low pass filter and is used to filter out the harmonics from A1.

The output from A2 will be almost a pure sine wave, the peak will obviously be dependent on the supply voltage and on the type of the op amp used.

The frequency of the present circuit has been fixed to approximately 50 Hz. If the values of parts shown in the parenthesis are selected, the frequency will be around 60 Hz.

Parts List

All resistors are 1/8 watts, 1%, MFR

R1 = 14K3 (12K1),

R2, R3, R4, R7, R8 = 1K,

R5, R6 = 2K2 (1K9),

R9 = 20K

C1, C2 = 1µF, TANT.

C3 = 2µF, TANT (TWO 1µF IN PARALLEL)

C4, C6, C7 = 2µ2/25V,

C5 = 100µ/50v,

C8 = 22µF/25V

A1, A2 = TL 072

IC2 = LM3886 (National Semiconductor),

HEATSINK FOR IC2 AS SHOWN IN THE IMAGE,

TRANSFORMER = 0 – 24 V/8 AMPS. OUTPUT – 120/230 V AC

PCB = GENERAL PURPOSE








The Power Amplifier Circuit

In view of keeping the design specifications very simple, and the component count as minimum as possible, a single chip amplifier was the basic requirement. A reasonably powerful amplifier using IC LM3886 (National Semiconductor) was ultimately selected by me for the purpose. The salient features of this power amplifier chip are as follows:

Truly versatile and a high performance IC compared to the other types of hybrid and discrete devices.

Totally internally protected from instantaneous peak temperatures,

Has got a dynamically protected safe area of operation,

The out put is perfectly shielded against a short circuit with the ground or the positive supply through an internal current limiting circuit network.

The output is also protected against output over voltages due to inductive load transients,

Can be operated with voltages as low as 20 volts up to a staggering 94 volts.

Its technical specifications are as follows:

Input sensitivity is 1 Vrms

Output power will be in the vicinity of 100 watts if the transformer primary resistance is around 4 Ohms.

Power bandwidth is a massive 10 Hz to 100 KHz.

Construction Hints

The circuit basically consists of just two ICs as the main active components and a handful of other passive components, so the construction procedure should be very easy. The whole assembly may be simply done over a piece of general purpose board (approximately 4 by 4 inches).

IC2 should be positioned at the edge of the PCB to facilitate easy fitting of the heat sink. The present utilizes two large 24 volt truck batteries. Connect them as shown in the diagram.

A separate battery charger is required to charge the batteries.

28 comments:

  1. Many thanks for your post. I am trying to build a pure sine wave inverter, which converts 110v dc to 110v ac. Can you please suggest me a mosfet power amplifier circuit. I need some mosfets which operates in this voltage range and support 1A current.

    ReplyDelete
    Replies
    1. Presently I do not have 110V input inverter design, but hopefully I'll try to design one and post it for you.....

      Delete
  2. Thank you Mr Majumdar,

    Thank you for your tireless contribution to the likes of me that has limited knowledge in advance electronic design. I am now building this Pure Sine Inverter and need your expertise on how to increase the wattage output to at least 4k watts. What parts do I have to change to meet this project. By the way, I am from Florida USA.

    Ray
    email: lobana51@gmail.com

    ReplyDelete
    Replies
    1. Hello Ray,

      Thank you very much!

      However I would like to inform you that many of the designs including the above is based on my assumptions only, and these have never been tested practically...so you would want to get them verified from some expert.....nevertheless I am always there to help you out in case you have any doubts.


      Regards.

      Delete
  3. Hi Engineer Swagatam,
    My pure sine wave inverter I built is working well, but as a person who would like to be well versed am concerned to know about the following:,
    1 How do i measure or between which two points determine my 200HZ and 500HZ?
    2 My Oscillator output between pin 2 and 7 is 9 v AC when 12.8vdc is applied to the input is this ok? rather what is the maximum output for this case?
    What is the transformer the voltage output (secondary coil) suppose for this particular circuit?

    ReplyDelete
    Replies
    1. Hi Kaluya,

      500Hz can be confirmed at pin2 of IC2 (555) and 200Hz at pin14 of IC 4017

      can you please show me the link of the inverter circuit that you have constructed, it'll help me to suggest correctly.

      Delete
  4. Hello Sir.
    It seems Kaluga never sent you his circuit he built.
    Sir did he leave an email or something to contact him.

    ReplyDelete
    Replies
    1. Hello Michael, no I was not contacted regarding the subject.

      Delete
  5. Sir I intend constructing an inverter for someone using push pull configuration.
    Anyway I want to protect the mosfets from back emfs when powering some indictive loads but
    1)I don't know the exact suitable type of diode to use across the source drain and
    2)I don't know how I will place the diodes...
    3) what will be the current rating of the diodes..
    The inverter will be 24vdc 1500w at 220vac

    ReplyDelete
    Replies
    1. Michael, the diode should be a rectifier diode, it can be connected across drain and source of the mosfets with anode connected to the source (for N channel mosfets).

      the current should be approximately equal to the current of the transformer secondary

      Delete
    2. OK Sir but
      1)what type exactly. A schotkey rectifier or silicon rectifier.?
      At 24dc the max current be 1500w/24= 62.2A
      This is max current the MOSFETs switch at full load.
      2) so will the diode be rated at this current.?

      3) you said for N channel anode is connected to the source and cathode to the drain right?

      4)So I want to know how the diode protects the FETs from back EMF because I know FETs have an inbuilt diode?
      want to have a clear idea
      What happens for knowledge sake.
      Because mosfeta have blown out before.

      Delete
    3. If the frequency of operation of the inverter is over 1kHz then you may require a schotkey diode, otherwise any ordinary rectifier diode will do.

      according to me if there's a load connected at the output of the inverter then the diode can be eliminated but if not then the transformer back emfs might cause damage to the mosfets, although mosfets have built in protection diodes, I am not sure how reliable these built-in diodes may be.

      yes the diode must be rated at 62amps for ultimate security.

      the diode allows the reverse EMF (during OFF time of the frequency) from the transformer primary winding to pass through it and prevents it from passing through the mosfet and in this way it protects the mosfets.

      Delete
  6. Also sir
    Apart from MOSFET protection diode, can i also connect a flywheel diode from the MOSFET drain to transformer centre tap as the transformer is an inductive device.
    I have seen inverter circuits put two diodes from drain to center tap for both sides of the push pull configuration and then the centreof the trafo goes to the positive of the battery.
    What do you think?
    Both MOSFET protection diode and drain to centre tap should provide suitable maximum protection.

    ReplyDelete
    Replies
    1. Michael, the two positions do exactly the same job that is shorting the trafo primary back emfs and preventing it to pass through the mosfet, so any of the positions can be selected, both together is never required.

      Delete
  7. Sir
    Have been reading up on diodes again and so if my anaylysis is right,placing a diode across drain source is like reverse biasing a diode when the FET is off.the characteristics of the diode prevent any flow of current thereby stopping transient voltages or back EMF from flowing into the FET.
    Am I correct in my analysis?

    ReplyDelete
    Replies
    1. yes you are quite right, the diodes allow the reverse current generated by the transformer winding to pass from its anode to cathode thus shorting it out and neutralizes the effect. In the process the diodes make sure that the reverse emf passes through itself and not through the mosfet's source to drain.

      Delete
  8. Sir I wanted to ask if I wound a say 12-0-12/350v trafo. Then when I power my inverter and I should get 350vac then I set it to required 220v thus reducing duty cyle in the process.
    My question is this
    1) does getting required voltage of 220v at a lesser duty cycle say for example 10% duty cycle during idle mode when inverter is not powering anything make the inverter more efficient than getting 220v at say 25% duty cycle during standby mode.?
    Meaming that we should aim for slightly higher trafo voltage winding than required.
    Am I making any sense

    ReplyDelete
    Replies
    1. Michael, having a higher voltage transformer than the battery voltage will allow you to get a constant voltage with a selected PWM regardless of the load (to a specified extent), if you are not using a PWM in that case you can go for a slightly higher rated transformer, which will only make sure that the transformer is little smaller in size. nothing more than.

      So higher voltage transformer will allow you use PWM and get a constant voltage but will result in a bulkier transformer, contrary to this a equally rated transformer will keep the size small but not allow PWMs or guarantee a constant voltage with regard to the load

      Delete
  9. Hello sir
    The inverter I intend to build is modified sine wave 24vdc at 1500w.
    I am using irfp260n as it is popular here with
    RDS= 0.04ohm Vdss=200v Id= 50A power dissipation = 300w.
    My aim is to have the inverter have a surge rating of
    At least twice its output power capacity of 1500w.
    So i want its surge rating to be 3000w for a brief period for loads that require higher starting power.
    How do I go about it.

    ReplyDelete
    Replies
    1. Hello Michael,
      I am afraid that may not be feasible until you actually have a transformer rated at 3000 watt, because a 1500 watt transformer will not be able to produce power above its rated value even for a second

      Delete
  10. Ok let's say I have a trafo rated at 3000w. Then I s there a circuit to help me determine the amount of time it surges to 3000w?
    Then in reality I think I will wind a 2000w trafo or 2500w and the reason is because I like to give enough room so that efficiency can be high.
    So my plan is for a 1500w inverter I use a 2000w or 2500w trafo then if in the future I build a 2000w inverter I use a 3000w trafo.
    So is 2000w good or 2500w good or is the room too much.?
    Anyway how can I use make the inverter surge because I will definitely be using a slightly higher rated trafo than needed.

    ReplyDelete
    Replies
    1. for that you will have to include an overload cut-of circuit with a delayed response, this delay can be set as per the preference.

      the trafo rating must be according to its required maximum surge handling capacity....if its 3kv then it should be rated to handle 3kv

      Delete
  11. OK sir
    An overload cut off circuit with a delayed response.
    This means that the circuit will do Two things
    1) allow a surge more than the inverters maximum 1500w output capacity for a brief period of time.
    So lets say I choose 2000w surge
    2) if it turns out that it surgers more than 2000w then it should cut off immediately
    3) if the inverter surge is at 2000w for a stipulates set time to allow start up then how does allow it it run at 1500w after set stage time has expired
    Am beginning to see comparators at work here.
    Sir do you have any circuit that I can use or similar.
    Mean while am trying to see if I can get any around.

    ReplyDelete
    Replies
    1. first of all a 1500 watt transformer will not produce 2000 wtts of power even for a fraction of a second...so the transformer needs to be rated at the specified surge level.

      we can have a current sensing resistor which will trigger an opto coupler which in turn can be integrated with the shutdown input of the inverter IC for cancelling the effect after a predetermined delay through a capacitor somewhere in the middle for introducing the delay effect as preferred.

      Delete
  12. Also about trafos
    Sir are toroidal trafos better than the normal ones.
    Can torroidals withstand high surge currents better than the normal trafos.
    I have been hearing a lot of noise about torroidals but I don't seem convinced it its ruggedness as I feel they are more prone to damage.

    ReplyDelete
    Replies
    1. torroidal types are better in terms of efficiency at full load....but no trafo in world will generate higher power than what they may be rated at.

      Delete
  13. Hello Sir
    Am thinking of trying out h bridge but I am more comfortable with push pull so I want to ask
    1) can I employ a centre tap transformer in h bridge bit I will just by pass the center tap and use the two ends then I can switch to push pull later.
    I want to have the flexibility of using one transformer to try out both topologies.
    Is it possible.
    Also sir I think eficiency using both topologies are close but with push pull things will be a bit bulkier?
    Is that so?

    ReplyDelete
    Replies
    1. Hello Michael,

      yes that's possible, a center tap trafo is actually a 2-wire trafo with a pulled-out wire from the center of this winding.

      A 2 wire trafo can possibly produce upto 95% efficiency, but a center tap trafo will never reach even 80% efficiency simply because of the inclusion of the two winding at the input, which is like using two transformers for achieving the output equivalent of one.

      efficiency = output power/input power

      Delete

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