MOSFET Question - Page 1
Forum section: electronics / projects

author: MyElectronsFellOut
author_rank: Regular Contributor
timestamp: March 11, 2015, 01:38:09 pm
content: Hi Folks, just a quick question.I'm using this MOSFET ..http://docs-europe.electrocomponents.com/webdocs/126c/0900766b8126ce49.pdfin a circuit as follows. (See attachement)Im tying the gate to Vd (~=9V after D1) through a pull up resistor, and then using another MOS of the same type to turn it off when a seperate 5V SRC is plugged in. The output at the Vs on Q6 is around 1.5-1.7V down on Vd which is at ~=9V. Vg ~= 8.2VIs this;a) Because Vs can never be 9V as this would in effect make Vgs == 0 and thus it would switch off the MOSFETorb) Because of the drop across the body Diode, though I'm sure as the body diode is (Cathode -> +9V), this is unlikely..I'm thinking it's (a) and if i'm right, what are my options for reducing the voltage drop across the MOSFET. My complete circuit design works and the voltage sources feeding the regulator switch over nicely when I plug in / disconnect the 5V source (USB).Any hints or tips welcome. Cheers.

author: Richard Head
author_rank: Frequent Contributor
timestamp: March 11, 2015, 01:49:09 pm
content: It's because the Gate of the upper MOSFET has to be more positive than the Source pin by at least +/-3V, preferably 12V for normal MOSFETs.In order to switch on the upper MOSFET properly you require a gate voltage that's more positive than your positive rail. That's when P- channel MOSFETs are best. With a P-channel device you only have to pull the gate below the Source, which is easy to do and doesn't require an extra bias supply.

author: MyElectronsFellOut
author_rank: Regular Contributor
timestamp: March 11, 2015, 02:11:21 pm
content: The biggest problem I have had is that I needed Q5 to act as the govenor, when I plug in the USB, Q5 turns off Q6. I can't connect Q5 to the 9V rail because I'm not going to get 9V at the source side to switch off a P CH type fet. Tried it alread and when I put 5V on the gate, I get 3.2V at the source, again becuase Vgs has to be positive to conduct.

author: Rog520
author_rank: Regular Contributor
timestamp: March 11, 2015, 05:39:07 pm
content: Quote from: MyElectronsFellOut on March 11, 2015, 02:11:21 pmYou're going to want to flip the P-channel FET around, so that the source is connected to the 9v input. And you won't need Q5 -- applying +5v to the gate of Q6 will turn it off.

author: MyElectronsFellOut
author_rank: Regular Contributor
timestamp: March 11, 2015, 07:58:02 pm
content: Quote from: Rog520 on March 11, 2015, 05:39:07 pmAt best that would give me a Vgs of -3.3 ~ -3.4V...(Diode D1 would still be in the circuit) That is sufficient for most P CH mosfets to be fully on... unless i'm missing something.On a side note, I have changed Q6 to a different type of N CH EN mosfet, one with a much lower Vgsth, so less of a voltage drop there, even if it was a SSSMini sized device and a pain to solder lol. It does have a higher RDSon though, but the gate resistor is now 1Mohm, so when regulator is disabled, or the USB is plugged in, the leakage current to ground from the 9V battery should be in the 10s of uV range.I've also tied the gate of Q6 to +9V directly which gave me a higher Q6 gate voltage by about 0.4V.