Solenoid Driving Circuit

Posted on Feb 5, 2014

The solenoid requires a certain amount of current to generate its magnetic field. If the solenoid was a perfect inductor, the DC current would rise above all means and would most likely damage other circuit components. However, solenoids inherently have a significant amount of DC resistance used to limit the current magnitude. Provided you place a

Solenoid Driving Circuit
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bypass capacitor (to absorb high-frequency current pulses induced by changing the current magnitude) between GND (close to the mosfet source) and the 12 V connection solenoid, you do not have to worry about a significant overshoot. Your selected mosfet has breakdown voltage of 100 V, which is certainly an overkill. The mosfet also has a non-zero on-state resistance Rdson (160 mOhm), which will slightly reduce the current through the solenoid. Another implication of Rds is mosfet power dissipation - which is negligible in this case (160 mOhms provided the channel is fully open). 2) One problem I see with your circuit is that the gate voltage will be 3. 3 V but the MOSFETs gate voltage is specified between 2 and 4 V. In practice, it`s fine because even if you get a "bad" part, the MOSFET will still partially close and allow current current to flow through its channel. An implication of low gate voltage is that the switch will work in the linear mode, where its on-state resistance is much higher than the guaranteed value. EDIT The gate threshold voltage is the minimum voltage where the MOSFET starts conducting current; however, the channel current would most likely not be enough to turn on the solenoid. Look at Figure 1 in datasheet, which correlates gate voltage with drain current and drain-source voltage. allowed source peak current - some PWM gate drivers can well support 30 A peak, which (with 10 Ohm gate resistor - R1)...

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