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Is the 1K resistor at the gate of the low side switch necessary? I've put it there to limit the current but thinking about it the MCU current is in the uA range
The current is around 30mA according to the manufacturer. The voltage is not mentioned in the datasheet so I am not sure that's why I went for a low side switch to be safe.
I assume I can also connect directly from the MCU output pins (high imepdance by default and logic 0 to switch) to terminal 3 and 4 instead of using the low side switches?
Terminal 8 is unconnected by default, it needs to be connected to the 24V line for it to switch internally. You are saying the top part of the potential divider is connected to 24V which means not terminal 8 and it does not make sense. Again, a diagram might make it clearer. I have already...
I was not sure what you meant by aux. it is not possible to connect terminal 7-10 directly to the to the MCU as they have different voltages (3.3 and 24v). In addition, current will be too high for the MCU input pins.
I have solved this issue by using an optocoupler.
the AG-020-B electric actuator is controlled from a microcontroller. Terminal 1 is connected to the DC PSU GND, Terminal 2 is connected to the +24VDC via a high side Power MOSFET(TPS22810) as shown below. This is for power saving.
Low side MOSFET (AUIPS1021) Q5A and Q5B are used to control...
From the text book notes attached, the loop gain is given which makes sense but I don't see how I can incorporate the closed loop gain and bandwidth given using this transfer function to satisfy the requirements.
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