FET circuit not working as expected. - Page 1
Forum section: electronics / beginners

author: Clear as mud
author_rank: Regular Contributor
timestamp: December 24, 2013, 04:57:28 pm
content: I have an on-off switch in a recent circuit design. The actual electro-mechanical switch is only rated for 100 mA, so I used a P-channel MOSFET to switch a higher amount of load current. I'm attaching a schematic of the switching arrangement. +12V is supplied from a "wall wart" type of supply, VCC goes to the input of a 5V and 3.3 volt regulator (L78S05CV and LD1086 linear regulator ICs, respectively). The switch applies either +12V or GND to the gate of the P-channel MOSFET, a IRF9388PbF. Data sheet:http://www.irf.com/product-info/datasheets/data/irf9388pbf.pdfThe problem is that the MOSFET remains always on. I have not connected any load to the circuit except for a power indicator LED that is connected to the output of the 5V regulator. The LED remains lit regardless of the position of the switch. I measured voltages and everything else seems fine. The gate voltage on pin 4 of the IRF9388PbF changes as expected, but the drain and source remain connected. Someone told me Drain-Source shorted is a common failure mode, so I tried taking the MOSFET off the board and installing another one, but it did the same thing. Am I not being careful enough to avoid damaging the FETs, or is there an error in my design?

author: kxenos
author_rank: Frequent Contributor
timestamp: December 24, 2013, 05:05:57 pm
content: I think you have to switch source with drain for it to work as you want it

author: ConKbot
author_rank: Super Contributor
timestamp: December 24, 2013, 05:07:39 pm
content: I'm sure the FET is switching just fine, but your body diode isntswap your drain and source, and it should all be fine.EDIT:if you add a diode symbol to your schematic symbol for the mosfet, it makes it a lot easier to double check for mosfet polarity correctness.

author: Clear as mud
author_rank: Regular Contributor
timestamp: December 25, 2013, 06:35:38 am
content: Well, darn it. I guess that's why I posted this question in the Beginners' section. I know I checked the polarity before I laid out the board, but I must have looked at a wrong example or gotten it confused somehow.

author: sleemanj
author_rank: Super Contributor
timestamp: December 25, 2013, 06:43:58 am
content: Standard mosfet symbols that don't show the body diode should be eradicated if you ask me. I ALWAYS get the polarity of Source and Drain mixed up, if the body diode is on the symbol though it's plainly obvious "which way is up"!

author: Clear as mud
author_rank: Regular Contributor
timestamp: March 12, 2014, 04:53:13 pm
content: I went ahead and transposed the source and drain. Finally, just recently, I sent it off to seeed studio to have another ten boards made, and now I expect it will work correctly.I have another question about the same design. As you can see in the schematic I posted with the first question, I put a 51k resistor tied between the gate and the input voltage. I did that so as not to leave the gate floating during switching the switch, but since I have had more time to think about it, I realized I didn't do anything to guard against switch bounce. As is, the resistor will keep the gate voltage from bouncing when turning the switch OFF, but when turning it ON, every time the switch contacts hit, the gate will go low, and then it will be pulled high through the resistor every time the switch bounces and loses contact. As I mentioned, this feeds a 7805 and a LD1086-3.3, linear regulators. The board has a place for an 0805 capacitor at the input of each regulator, and I specified 0.33 uF in the design, so that will help somewhat with switch bounce, but I am not sure if I am comfortable with this "solution" to the switch bounce.1st suggestion for a better solution: I could add a small circuit with a capacitor and Schottky-input inverter, to debounce the signal before applying it to the gate. I've been using the circuit athttp://www.labbookpages.co.uk/electronics/debounce.htmlin some other circuits I designed.2nd suggestion - an alternative solution: Why don't I just leave the resistor out of the circuit, and let the gate float? For the short time that the switch is between positions, the capacitance of the MOSFET gate will hold it in whatever state it was in. Or, I could make the resistor a much larger value such that its resistance times the gate capacitance is longer than the expected switch bounce time. That way, it would still function to keep the circuit switched off in case of mechanical failure of the switch, but under normal conditions the gate should float and retain most of its charge for the amount of time the switch is between positions or bouncing.

author: Rigby
author_rank: Super Contributor
timestamp: March 12, 2014, 05:25:58 pm
content: Quote from: sleemanj on December 25, 2013, 06:43:58 amThe words source and drain confuse me. I have to think of it as "DRAINing from a source" and "SOURCing to a drain" or I mix it up.

author: free_electron
author_rank: Super Contributor
timestamp: March 12, 2014, 06:21:16 pm
content: source and drain are exatly the seame as emitter and collector.the emitter supplies electrons. the source supplies electrons.the collector collects electrons , the drain lets electrons flow away

author: Rigby
author_rank: Super Contributor
timestamp: March 12, 2014, 06:22:27 pm
content: Quote from: free_electron on March 12, 2014, 06:21:16 pmYes, but I still think of things in terms of conventional current flow, not electron flow. I'll get over this one day.