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1

You are going to get yourself in trouble - but don't let me stop you. As has been commented on, you need what are called high side drivers. In theory, they look something like simulate this circuit – Schematic created using CircuitLab V2 is usually based on a charge pump, which will provide an isolated voltage with relatively low current. Of course, ...


1

boost converter which will boost the 12V to 120V, Transformer will be much more efficient and cheaper than boost converter for that high output/input voltage ratio. Say you need 120V 10A, if it is 100% efficient you will input 12V 100A. If you use a transformer, then you can find low-RdsON low voltage FETs that will handle that current, chop it into a high ...


4

Yes, some very cheap inverters do exactly that: make a high-voltage square wave. That said, to do this directly you need to create a power supply that is more than twice 120V. This is because the AC waveform is actually about 170V peak, and 340V peak-to-peak. You also need to modify the duty cycle of the wave to achieve the same RMS value as a sine: make it ...


4

So-called modified sine wave inverters use a DC rail that is approximately \$\sqrt{2}\$ times the RMS voltage of the sine that is being simulated, so about 170V for 120VAC, and a duty cycle that causes the RMS output to match the sine wave (image from here): Duty cycle should be 50%, so the RMS value and the peak value match that of a sine wave of 120V RMS.


2

This assumes an on-off type H-bridge and not a linear H-bridge used in bridge tied audio amplifiers. How much? If you ignore protection features (required to limit gate-source voltages when the supply exceeds the maximum gate-source voltage), to efficiently drive current through the load requires these gate voltages to be present: - And clearly, the ...


2

No. 120 V AC is the RMS voltage. Peak voltage is \$ 120 \sqrt 2 \ \text V \$. Tip: 'V' for volt.


1

Consider this as the starting scenario: - Q2 is on and Q4 was off, Q3 would be activated (due to Q2) and this forces Q1 to be off. It remains like this for some time. Then, just imagine Q2 was turned off - ask yourself what will cause Q3 to turn off. With Q2 off, Q3's miller capacitance would remain charged and keep Q3 on for some length of time. How long ...


0

Something is wrong with your setup. Higher PWM doesn't influence the working as you describe. Keep in mind that fundamental frequency (in case you use three phase BLDC) would have such impact, not the PWM frequency. Certainly you don't want to install capacitors on the output of H-bridge as you would overload the transistors. The analogy with power supply, ...


0

I see that you reckon that you've found the cause but it is possible for one motor to act as a generator and power the other. Figure 1. When turned by hand the motor will act as a generator and current will flow in through the intrinsic MOSFET diodes (not shown on the schematic symbols) as shown by the red arrows. The reverse polarity MOSFET at (2) is good ...


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There are several issues with your schematic some are listed below. For U5 and U6 a voltage should be applied to pins Tcase and Tj equal to the temperature in Celsius (typically). The A1 inverter should have a statement that indicates high and low voltage. Things like back emf can be modeled with behavioral voltage source. https://www.precisionmicrodrives....


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