is the heart of most lower-cost switching power supplies.

I thought you knew what PWM stands for, and how it applies to the context of this thread. My error.

With a pulse-width modulated squarewave signal driving the transformer primary (through saturated switching power transistors), the transformer never "sees" the 50 Hz component of the modulated waveform.

ak

Oops. Looks like I didn't read the PWM bit. I thought you were suggesting to use the transformer to amplify the 50Hz oscillator signal to get the output. My bad, sorry about that. But commercial inverters still seem to have an intermediate DC stage, why do they go that route instead of driving the transformer directly?

I don't know enough electrical engineering to know for sure, but could it be that the 50Hz component still influences the transformer?

 

you should be aware

As my posting history indicates, I have a fair grasp of switching power supply theory and practice.

As for the rest of your post, again you missed the point of my first response. The system you describe has two switching stages, one to boost the battery up to 360 Vdc, and one to chop that into the output transformer, modulated for the output waveform, a 50 Hz sinewave. My point is that you can combine both of those into one stage. Chop the 12 V with the sinewave-modulated PWM waveform that you would use for the output stage in your version. The output transformer can handle boosting the battery voltage up to peak line voltage.

ak

 

Thanks for that very interesting post! So if I understand you correctly, the sine table is a way of numerically storing a sine wave in a microcontroller or ASIC? I.e. just a list of the PWM duty cycles to output at different times? So for example, if the output frequency is 50Hz, the length of one sine wave (360°) is 1/50 = 0.02 seconds. So at, say t=0.007 seconds the controller output would be sin (0.007/0.02 * 360) = sin (126°) = 0.809 so the base PWM duty cycle to the output stage would be 80.9% at that moment. Is that about right (ignoring any trimming due to feedback)?

Your last paragraph was very interesting, in fact that's what I was about to ask. As a mechanical engineer I usually prefer using negative feedback to run things closed-loop rather than trying to model everything precisely and have the system run open-loop. Once the product is out in the wild, you never know what unusual loads the customer is going to throw at it, so it's great that you build some degree self-correction into the system.

Hi,

Well you only work over a half cycle, but i think you have the right idea.

Yes the loads can change so some negative feedback is good.

 

Oops. Looks like I didn't read the PWM bit. I thought you were suggesting to use the transformer to amplify the 50Hz oscillator signal to get the output. My bad, sorry about that. But commercial inverters still seem to have an intermediate DC stage, why do they go that route instead of driving the transformer directly?

I don't know enough electrical engineering to know for sure, but could it be that the 50Hz component still influences the transformer?

They have a DC to DC converter first, then they chop it up.

 

As my posting history indicates, I have a fair grasp of switching power supply theory and practice.

As for the rest of your post, again you missed the point of my first response. The system you describe has two switching stages, one to boost the battery up to 360 Vdc, and one to chop that into the output transformer, modulated for the output waveform, a 50 Hz sinewave. My point is that you can combine both of those into one stage. Chop the 12 V with the sinewave-modulated PWM waveform that you would use for the output stage in your version. The output transformer can handle boosting the battery voltage up to peak line voltage.

ak

I got what you're saying, my question is why don't commercial inverters do it that way and save the extra conversion stage? If you look at the dissection of a commercial invertor here (at 14:44):

and look also at the corresponding datasheet you'll see that it requires a 400VDC supply for the switching transistors of the output, and you'll see that in that inverter implementation it's achieved by a chopper and transformer. Your way sounds more elegant and efficient, but for some reason they opted for an intermediate DC stage. There must be something we're missing?

 

could it be that the 50Hz component still influences the transformer?

Nope.

Because it has a constant amplitude and varying spectrum, PWM is a cousin of FM. Nothing in an FM station's transmitter, feedline, or antenna see any aspect of the audio other than the antenna having enough bandwidth for the FM carrier's sidebands, all of which are way above 87 MHz.

PWM has a different spectrum structure, but basically the same physics.

ak

 

Nope.

Because it has a constant amplitude and varying spectrum, PWM is a cousin of FM. Nothing in an FM station's transmitter, feedline, or antenna see any aspect of the audio other than the antenna having enough bandwidth for the FM carrier's sidebands, all of which are way above 87 MHz.

PWM has a different spectrum structure, but basically the same physics.

ak

I was thinking along the same lines (coincidentally I thought of the analogy the 50Hz being the "audio frequency" and the ~22kHz PWM frequency being like the "carrier frequency").

The big question still remains - why don't cheap real inverters do it this way? In my experience when one discovers a seemingly better way of doing things than the status-quo, there's a reason why no one does it like that. I lack the knowledge to figure out that reason myself. You seem to know more about the topic than myself and also seem to suggest it's possible. Do have the equipment to test the theory and report back the results? Take a microcontroller, program it to output 22kHz PWM to emulate a 50Hz sine wave, use it to power some transistors to drive a 1:20 transformer and measure the output?

Or if not maybe the test can be run in some simulation software?

 

Was looking at this many moons ago and found it a great explanation

http://d1.amobbs.com/bbs_upload782111/files_37/ourdev_625454LS3D3U.pdf

 

My guess is that the output impedance of an almost 30:1 step-up transformer is not low enough to look like a good voltage source. The impedance ratio from primary to secondary would be 1 : 800.

Elegant on paper, not so much in the real world.

ak

 

My guess is that the output impedance of an almost 30:1 step-up transformer is not low enough to look like a good voltage source. The impedance ratio from primary to secondary would be 1 : 800.

Elegant on paper, not so much in the real world.

ak

So that must be it, the output waveform will be too distorted. I suppose that's why they use the transformer to convert to high-voltage DC, and then reconstruct a good quality sine wave at the output using PWM.