Ideal diode circuit confusion - Page 1
Forum section: electronics / beginners

author: kernels
author_rank: Newbie
timestamp: January 20, 2020, 09:42:21 pm
content: Hi guys,I built the circuit below last night as part of an ideal diode experiment and am confused about the result I'm getting. The bit that confuses me is that the circuit works, when I apply 20V to the gates of the two N-Channel mosfets, I get 10V across the load resistor, take away the 20V gate drive and I get 0V across the load resistor.What I don't understand is where the two sources are referenced to, does it have something to do with the gate-source capacitance ? or maybe the body diode?

author: hermitengineer
author_rank: Regular Contributor
timestamp: January 20, 2020, 11:02:33 pm
content: The MOSFETs in your diagram have the gate relative to the source, so that's the reference they use. When your gate voltage is 20V, VGS on the left transistor is +10V (20V-10V). VGS on the right MOSFET is 20V (20V-0V) initially, but once current flows, VGS on the right also drops to +10V since it's passing the left MOSFET's current now.When you turn off the +20V, VGS left is -10V and it turns off. VGS on the right is 0V, so it also turns off.The diodes don't really figure into this circuit because both MOSFETs are either on or off at the same time, and the diodes block each other's current flow when the MOSFETs are off.

author: kernels
author_rank: Newbie
timestamp: January 21, 2020, 12:18:43 am
content: But the 10V is on the drain of the left mosfet, not on the source, the two sources are tied together without any other connection.

author: langwadt
author_rank: Super Contributor
timestamp: January 21, 2020, 12:43:23 am
content: Quote from: kernels on January 21, 2020, 12:18:43 amcosider what happens when you look at the right mosfet as a diode and then when the left mosfet turns on

author: kernels
author_rank: Newbie
timestamp: January 21, 2020, 12:49:35 am
content: Ahhhhhh I see, initially the left side mosfet source just sees ground through a diode (and load), so it can turn on, which then puts 10V on the sources and allows the other one to turn on too, thanks langwadt!

author: Zero999
author_rank: Super Contributor
timestamp: January 21, 2020, 10:53:31 am
content: It's not an ideal diode circuit, which is a bit more complicated.https://www.analog.com/media/en/technical-documentation/tech-articles/lt-journal-article/LTJournal-V26N4-01-df-LTC4357_59-MeilissaLum.pdf

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 21, 2020, 04:28:33 pm
content: Quote from: kernels on January 21, 2020, 12:18:43 am+10 V is on thesourceof the left mosfet. And twodrainsare connected together.And, as it was said, it isn't an ideal diode, it is just an electronic switch.

author: kernels
author_rank: Newbie
timestamp: January 21, 2020, 08:22:27 pm
content: Nope, as I said, they are N-channel Mosfets. And thanks, yeah, I know the circuit does not function as an ideal diode, I was only experimenting with the switch part of the ideal diode circuit since I intend to drive them myself. I note that the ideal diode controllers tend to have a wire to the common source pins, so was trying to figure out why they have that when it appears you can do the switching without it. It may be that you can switch them faster with a connections to the common sources.

author: kernels
author_rank: Newbie
timestamp: January 21, 2020, 08:25:20 pm
content: Sorry ! , I see the confusion, the online schematic package I was using didn't show the body diode as part of the mosfet, so I drew it in manually, but showed the two mosfets themselves reversed!Vovk_z sorry, you were right, I had built the circuit with common sources, but when drawing the schematic to post it here I reversed them.

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 21, 2020, 09:12:50 pm
content: Yes, real dual mosfet switch has sources connected together.