From what I was told, this is a PWM generator. I am not exactly sure how this works. Can someone explain to me how this works? Whats the use of the repeating sequence and sine wave going into the relational operator? etc.
Electrical – Need help understanding this PWM Generator
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The BTS50055 is not good for fast PWM. If you look at the data sheet in particular "load switching capabilities and characteristics" on page 3 it tells you how fast the device can switch. Turn on time is somewhere in the 100us range typically and turn off might be about 60us. These numbers need to be 1% or less for decent efficiency and 100us represents 1% of 100Hz. If your switching speed is pretty low then use it.
How can I design a MOSFET H-bridge motor driver controlling only 1 PWM for a motor?
If I understand correctly you have a single PWM signal and you want to drive a H bridge with this signal. I'd start by looking for a couple of beafy N channel mosfets like the IRF1010 and then pick a topside/lowside mosfet driver like LTC4444. Feed PWM to the inputs via a couple of logic gates that create a deadband (so both FETs never switch on at once) and this gives you a half H bridge.
Repeat to make the other half but remember to invert the PWM signal. I'd look to be monitoring current on the dc side of things and forget about the BTS50055.
Here's a PWM signal fed to a transformer and superimposed on this diagram is the sinewave that the PWM represents: -
The secondary of the transformer usually has an inductor and a capacitor that form a 2nd order low pass filter thus converting the PWM signal into (more-or-less) a fairly decent sinewave.
For instance, if you take the high frequency content of the PWM waveform it looks like a square wave with varying duty cycle and, you can low-pass filter this quite easily to get this: -
On the left is the original square wave. In the middle a little bit of filtering has happened and on the right the filtering is far greater.
Thus, the high frequency edges of the PWM signal can be greatly reduced leaving the low frequency content that represents the sinewave. In effect you get something that typically looks like this: -
You can still see that the waveform has a little bit of the PWM signal but, in the main, it is a sinewave.
If your PWM frequency is 60 kHz and your AC is 60 Hz you could position a filter to have a cut-off of 600 Hz and there would be 2 decades between it and the 60 kHz. A 2nd order filter would attenuate the 60 kHz by 80 dB (40 dB per decade): -
You might note that I mentioned a filter having a cut-off of 600 Hz and wonder why it is position ten times higher than the AC 60Hz. You might ask why not have it at 60 Hz and this would be a good question. The reason it isn't at 60Hz is two-fold: -
- There would be a 3dB attenuation of the AC
- If the filter was extremely resonant it would consume vast amounts of current because, in effect, it is also a series resonant circuit across the line.
It has to be positioned as far away from 60 Hz as possible to avoid large circulating currents in the L and the C of the filter BUT you don't want it up close at 60 kHz because it won't filter out the high frequency content very well. Minimum is 100 Hz I would say and it should be at least 1 decade away from the lowest PWM frequency (generalism alert!).
Best Answer
Sine wave and a repeating wave generates waves of respective nature continuously. They are fed to a relational operator. This relational operator compares both the input values and outputs TRUE(high) if sine wave magnitude is more than the repeating wave magnitude else FALSE(low). The NOT gate generates the complement of these. They are fed to a MUX. Its output is series of TRUE/FALSE (high/low) values based on the input sine and repeating wave values.
Probably the repeating1, 2 and 3 are of different of different frequencies and duty cycle and so PWM waves of different frequencies and duty cycle