Timeline for Voltage Divider to protect P-FET gate on a Polarity Inversion Circuit
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2017 at 21:50 | vote | accept | Victor Santos | ||
| Feb 6, 2017 at 13:30 | comment | added | Victor Santos | @Autistic, thank you. But I don't have a motor connected to the circuit. I have a 5v DC-DC buck converter powering an Arduino. The buck converter is connected to 5x 6v batteries in series. I have a fuse on the positive battery wire and sometimes it blows when someone inverts the connection but in very rare cases the buck converter blows. | |
| Feb 6, 2017 at 4:58 | answer | added | Bruce Abbott | timeline score: 3 | |
| Feb 6, 2017 at 4:45 | comment | added | Autistic | It is plausible that the imaginary equipment with the big Cap has say an internal mosfet for inrush protect .BUT we are talking about outrush current .The imaginary internal mosfet could blow too.The diode is therefore more robust so when you place a fet across it to reduce volt drop you can sense the DS volts and shut down the gate to avoid runback . | |
| Feb 6, 2017 at 4:39 | comment | added | user57037 | @Autistic, if there is that much capacitance, then inrush could be a big problem, too. OP would be well advised to consider both cases. | |
| Feb 6, 2017 at 4:26 | comment | added | Autistic | There is a possible way to blow the FET even if you do the gate source zener .Imagine that your circuit is in say a 24V truck .Then say that your protected output has lots of capacitance across it .This is not unreasonable perhaps it is a big car Audio system or maybe an inverter. Now say if the input volts drops quickly due to say starting the engine .The load capacitance will discharge through the P channel because it is still hard on .The spike current is only limited by the total circuit resistance which could be miliohms ,If the mosfet has a small size and a low on resistance then BANG | |
| Feb 6, 2017 at 3:32 | comment | added | AngeloQ | to ground. The resistor should have a value high enough to avoid exceeding the current rating on the zener (100mA in this case), but again, it doesn't need to be that high, so probably 100k would work fine. So why use both (two resistors and a zener)? If there is some fault or soldering problem e.g. in the above circuit where the zener is not soldered properly to Vout, the Vgs would be exceeded. The same would be true if the high resistor of the divider was missing. MOSFETs have zero tolerance to Vgs beyond the max spec. Usually an extra resistor or zener is much less costly than a FET. | |
| Feb 6, 2017 at 3:27 | comment | added | AngeloQ | Actually use all three, two resistors and the zener, just to be safe. Technically either would work. For th resistor divider approach, assuming you don't need this to switch on very fast, high value resistors are fine to minimize current draw, e.g. in the 100k range or higher. You would select resistors to provide the Vgs at a comfortable level, so e.g. since this FET has threshold Vgs of 4V, set it to provide maybe 10V, i.e. 100k between gate and source, and 200k from gate to ground. If you use the zener, the zener clamps the Vgs voltage, while the resistor tries to pull the voltage... | |
| Feb 6, 2017 at 2:57 | history | asked | Victor Santos | CC BY-SA 3.0 |