P-channel MOSFET biasing - Page 1
Forum section: electronics / beginners

author: dande
author_rank: Contributor
timestamp: April 07, 2014, 08:50:57 pm
content: Hi Guys!I am relatively new to electronics especially when it comes to MOSFETs. I am trying to put a positive bias on the ground pointof Rload. What I tried to do is having a P-channel MOSFET, and put a voltage divider (should be 1.25V) at the Source pin have theDrain pin connected to the Virtual Ground (which should be around 2V above Ground) and the Gate grounded.However the circuit behaves completely different as I expected it (I know I should have expected a big-ol-FAIL at the first try, but I am an optimist).Here is the schematic:When simulated, it shows around 1V voltage drop on Rload, around 10.28V at the drain.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 07, 2014, 09:46:48 pm
content: It's backwards. See the little triangle pointing away from the middle of the three line segments: that triangle represents the internal body diode. Also, gate voltage is relative to that terminal (source).So, you've got a voltage divider with a diode in the middle. The diode drops probably 0.6V, so you get 11.4V across 100 + 1000 ohm resistors, or pretty darn close to 1 and 10V drop, respectively.Tim

author: dande
author_rank: Contributor
timestamp: April 07, 2014, 09:54:43 pm
content: Thanks! So it means, this is not the right way to bias the "ground" for Rload?

author: LukeW
author_rank: Frequent Contributor
timestamp: April 07, 2014, 10:46:19 pm
content: What is the basic thing you are actually trying to accomplish with your circuit?Are you trying to switch on and off the load with the MOSFET?

author: dande
author_rank: Contributor
timestamp: April 08, 2014, 01:27:45 pm
content: Okay, total huge FAIL for me! I wanted to have 2V between Vg and ground, so that Rload's "ground" point is 2V higher than the rest of the circuit's ground.

author: madires
author_rank: Super Contributor
timestamp: April 08, 2014, 01:49:04 pm
content: 3 Si diodes in series are about 2V ;-)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 08, 2014, 05:27:15 pm
content: The gate-source voltage (built in) is probably more than 2V, so even if you ground the gate and use a source follower (the source points up), you won't get 2V. It will also vary with temperature and load (if Rload varies at all).Three diodes in series will do a better job, making probably 1.5 to 2.2V depending on diode type, temperature and load.An emitter follower (using a PNP BJT) is more stable, because BJTs have higher gain than MOSFETs, but the voltage will still vary by around +/- 0.2V, depending on setup. If you need it more precise than this, you will need an op-amp or voltage regulator or something like that.Tim

author: dande
author_rank: Contributor
timestamp: April 08, 2014, 06:57:52 pm
content: Quote from: T3sl4co1l on April 08, 2014, 05:27:15 pmThank you! Could you please tell me why is the higher gain important for voltage stability?I feel so stupid for not realizing that I can simply use three diodes. Anyways, since Rload will probably vary and I try to design the circuit for around 1A and I am not sure that diodes can work with that many current :/

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 08, 2014, 07:05:39 pm
content: Well, what is gain? The gain of a transistor is its transconductance, a change in output current for a change in input voltage. Think about itTim