Oatley Electronics K173 Multi-purpose Inverter Kit ADVICE

[Nigel Goodwin]

You need a true RMS meter to read the output, as it's not a sinewave - a normal meter just reads the average, based on it been a sinewave only.

 

[smanches]

How are you producing the 24vac right now? Did you finish the inverter portion already? As Nigel said, if you're using the inverter, then your meter is going to be off somewhat.

121v will work, but as you said, at 75w it's already dropped to below 100v. I think brownout condition is normally 90v or 80v, not sure. Boosting the voltage up to around 140v or so will give you more room to work with. You might have to be careful of powering small devices though, as that's nearing the max input voltage of many devices, and any sort of spikes could easily go over that.

You could use your light bulbs as a sort of manual voltage regulator, in a sense. If you need to power a light load, put a bulb or two on to drop the voltage back down to 120v or so, then attach your other device. This will use more power, and is not efficient, but it works.

 

[colin55]

Two things to note:

Firstly, when the voltage drops, the wattage consumed by the lamps also decreases considerably.

It looks like your power supply is not able to deliver the wattage (we can also say: current).

The problem is not the transformer.

When you deliver 24v DC to the transformer (via the oscillator circuitry), the output voltage will be less than 90v.

 

[audioguru]

The project does produce a modified sine-wave output because one set of Mosfets makes a pulse then a pause before the other set of Mosfets make a pulse then another pause.
The output must produce the peak voltage that a sine-wave would produce which is 340V for a 240V RMS sine-wave or 170V for a 120V RMS sine-wave.
Therefore the transformer must be 16V or 18V center-tapped so that 12V on the 8V or 9V winding produces a peak voltage of 170V.

 

[colin55]

The transformer works in partial "fly-back" mode. That's why you cannot simply use a "turns-ratio" equation.
The circuit is using the inductance of the transformer to produce a modified "square-wave // sine-wave" output.

 

[audioguru]

I don't think the transformer is used as a fly-back because it is center-tapped.
When one side has a very strong Mosfet conducting then the other side goes up to double the supply voltage due to transformer action and the conducting Mosfet prevents its voltage from going higher.
with a load, the output is rectangular pulses. their average voltage is the same as the RMS voltage of a sine-wave and the peak voltage is the same as a sine-wave.

 

[colin55]

The MOSFETS are only driven for 50% of the time. When the transformer is not being driven, the field collapses and (depending on the load) the output voltage will vary.

 

[teamleader031661]

The 24vac supply that I am using to test my transformers is a toridail that I use for a battery charger circuit. I'm only using it because I have no other 24vac source. As Collin55 has stated, and is correct, it isn't producing the watts I'd like to see. But, I believe that power would be there with the 24v batteries. I haven't started this inverter circuit yet because I was hoping to get a little more guidance from Oatley. Audioguru has a good point, but I believe he is thinking I will be using 12vdc rather than 24vdc. I will experiment with my MOT adding more turns to the primary for a boost. I am somewhat confused about the center tap. The schematic shows v+ to the center tap. If I'm using 24vdc would that mean tx1 and tx2 are getting 12v each? It seems I need more bench work! Thanks for your help guys!

 

[audioguru]

Since your battery is 24V instead of 12V then the transformer must be 32V to 36V center-tapped. The center-tap connects to the positive supply which is +24VDC.
The mosfets will have a peak voltage of about 55V so make certain that they are rated for it.
You will need to reduce the 24V down to 12V with a 7812 voltage regulator IC for the low voltage parts of the circuit.

The battery will be 27.6V for most of its discharge but the Mosfets and transformer have losses that reduce the voltage to about 24VDC.

 

[colin55]

Since your battery is 24V instead of 12V then the transformer must be 32V to 36V center-tapped.

You have got the concept around the wrong way.
A 32v primary winding is an AC value and the peak of a 32v AC input is above 40v.
So, the concept is: for a 24v DC input, the primary should be wound for about 18v.
Since there is a certain amount of "fly-back" introduced via the outputs of the 4017, I don't know what the actual "turns-ratio" will be.

 

[Sceadwian]

Granted flyback conditions would occur in a transformer setup would it really be producing any significant current at those power rating? I would not call it a flyback setup, but only because any flyback current is secondary to main transformer power transfer.

 

[audioguru]

You have got the concept around the wrong way.
A 32v primary winding is an AC value and the peak of a 32V AC input is above 40v.

No.
A square-wave inverter with a 24V battery uses a 48V center-tapped transformer. The peak output voltage is the same as the RMS voltage.

A modified sine-wave inverter uses a 32V to 36V center-tapped transformer so that the 24V pulses into each side produces a peak output voltage that is 1.414 times higher than the RMS output voltage to form a peak voltage that is the same as a sine-wave.

Half of 34V is 17V. When a 24V pulse is fed into the 17V winding then the output voltage has a peak that is 24/17= 1.412 times higher to make the peak.
The transformer must be about 34V center-tapped to 120VAC RMS.

 

[colin55]

Granted fly-back conditions would occur in a transformer setup would it really be producing any significant current at those power rating? I would not call it a fly-back setup, but only because any fly-back current is secondary to main transformer power transfer.

Where do I start?
The 4017 drives one MOSFET for 25% of the time, then turns off. During this time the output goes positive and then starts to go negative.
The amount of negativity will be discussed later.
The energy from the negative excursion comes from the collapsing magnetic flux.
We can call this "fly-back" or "controlled fly-back" or "collapsing magnetic field." At least it is energy being delivered to the output when the input is not being driven.
Now the other MOSFET is turned on the flux in the transformer, (that is already collapsing) will be driven in the same direction to produce a negative output. After another 25% of the cycle-time, this MOSFET will be turned off and the flux will collapse and when it is approximating zero output volts, the other MOSFET will be turned on to produce the positive portion of the waveform.


Your comment

would it really be producing any significant current

is yes, as the waveform at this portion of the cycle has an area under it that represents “power” or “energy.“

Now we come to the “turns-ratio” question.
Since we are putting 24v DC into the primary, we need to have a primary with the correct number of turns for a DC value.
If it is 6 turns per volt, we need 24 x 6 turns.
The secondary will produce one volt for each 6 turns and since you want to produce the same amount of energy a 110v AC supply will deliver, you will need 110 x 6 x 1.4 turns.

 

[Sceadwian]

colin, what is the flyback current in relation to the primary transformer current?

 

[colin55]

The flyback current is determined by the inductance of the transformer (and lots of other factors). The transformer could be as high as 15H. The slope of the (negative) excursion will provide you with the energy it is delivering.

 

[Sceadwian]

What is it? If you can't directly tell me what the amount of current that is produced from flyback vs what is being transferred via standard transformer action then how can you classify it one way or another?

 

[colin55]

This circuit is not standard "AC-in, AC-out."

 

[Sceadwian]

So that means you don't know.

 

[colin55]

How does anyone know? What is the inductance of the transformer. What is the voltage any point in time?

In 40 years of teaching, I have never had to answer so many questions that go around and around in circles.

The fact is this: we are getting energy from the collapsing magnetic field and delivering it to the output, in conjunction with the enegy from the time-interval when the MOSFETS are turned on.
Since the MOSFETS are only turned on for 50% of the time, they are taking a high amount of energy from the battery during these intervals and delivering it over the full 100% of the cycle. This means you are requiring bursts of very high current-demand from the battery.

 

[Sceadwian]

40 years of teaching? What did you teach?