Ideal diode circuit question - Page 1
Forum section: electronics / projects

author: perdrix
author_rank: Frequent Contributor
timestamp: February 10, 2022, 11:42:54 am
content: I'm trying to implement a 3 terminal ideal diode replacement for a dual diode like BAT54C operating as a power supply steering diode.I'd like to avoid the need for a ground terminal and the two 10M resistors, but if I omit those I need to leave the BAT54C I'm hoping to replace connected between J1, J2 and J3 - which provides a mechanism to clear the gate charge as the steering circuit switches between battery and power supply.Any suggestions on how best to do this?Thanks, David

author: Peabody
author_rank: Super Contributor
timestamp: February 10, 2022, 04:29:19 pm
content: I don't see how you would eliminate those items without the gate going floating. Something has to pull the battery's mosfet gate low when the external power isn't connected.Does this actually work? When both power sources are present, does the battery mosfet stay off? This is actually a pretty nifty circuit.

author: perdrix
author_rank: Frequent Contributor
timestamp: February 10, 2022, 05:56:01 pm
content: Yes, It works very well indeed with the 10M resistors and the ground connection.David

author: perdrix
author_rank: Frequent Contributor
timestamp: February 10, 2022, 06:23:56 pm
content: Quote from: Peabody on February 10, 2022, 04:29:19 pmThe gate of Q102 is permanently (expect when you change the battery) at just under 3.1V (3.07V ), so when 5V is supplied Q102 goes into conduction, and the gate of Q101 goes to just under 5V (4.95V) so Q101 is turned off hard.When you disconnect the 5V supply, the gate of Q101 discharges quite fast to below 500mV (482mV) which turns on Q101.Technically you don't *need* R2 expect for battery changes under power.I think I'm almost persuaded there's no way to avoid that ground connection.Oh well.D.

author: Peabody
author_rank: Super Contributor
timestamp: February 10, 2022, 07:11:19 pm
content: The only thing that concerns me a bit is that per the datasheet the mosfet's threshold voltage is typically 1.7V, with a max of 2V. But that's where it just begins to conduct. So the 1.93V actual G/S voltage might not turn on the mosfet enough. But you could adjust the value of R102 so the gate voltage is just a bit lower - maybe 0.1V lower so the mosfet would still be off when there's no 5V source, but more fully on when there is. Maybe change R102 to 330K.But looking at it the other way, if you could be sure that nothing would ever be connected to the 5V input that isn't powered up and therefore might sink current, you could eliminate R102 and R2, and just tie that gate to ground. Then 3.1V would be measurable at the 5V pins, but no current would flow. In other words, would it be ok if that mosfet is always on?And actually, you don't need R101. The gate is going to be either at ground or at 5V, so the resistor doesn't really do anything in either case. But you do need those ground connections.I would be interested in whether this circuit could be used with LIPO batteries at 4.2V, and with much higher current and much lower Rds(on). Seems like it could work with the R102/R2 divider adjusted appropriately.

author: Zero999
author_rank: Super Contributor
timestamp: February 10, 2022, 07:24:49 pm
content: The circuit won't work. All you'll get are high currents and a load of magic smoke. I was going to write an explanation, but I just put it into LTSpice instead.It should be obvious, just by looking at the gate voltages.

author: Peabody
author_rank: Super Contributor
timestamp: February 10, 2022, 08:34:29 pm
content: I was completely wrong about having Q102 on all the time. That doesn't protect the gate of Q101 from going high. So the battery would turn itself off.But other than that, I don't understand what doesn't work. The mosfets in your LTSpice model are oriented wrong. The source of both has to be on the load side. Otherwise the body diodes will be the wrong way. They were right in your first schematic.

author: Zero999
author_rank: Super Contributor
timestamp: February 10, 2022, 09:06:30 pm
content: Quote from: Peabody on February 10, 2022, 08:34:29 pmYou're right. It relies on the MOSFETs being connected backwards, so they bypass the body diodes. I mentally reversed them, when analysing the schematic and entering it into LTSpice, because I'm used to P-MOSFETs being connected source to positive.

author: Peabody
author_rank: Super Contributor
timestamp: February 10, 2022, 09:37:24 pm
content: I tried your .asc with the mosfets turned back around, and it looks like it works. But there's about a 0.6V drop across the 5V mosfet. That must be caused by the threshold voltage. The fully-on Rds(on) wouldn't be high enough to cause that.Those mosfets are real stinkers. Low current, high resistance.Edit: I picked another mosfet, the Si1013, and the voltage drop goes away.

author: perdrix
author_rank: Frequent Contributor
timestamp: February 10, 2022, 11:12:55 pm
content: Quote from: Peabody on February 10, 2022, 07:11:19 pmChanging R102 to 330k drops the gate voltage to 3V, so the FET should be on a bit harder...Quote from: Peabody on February 10, 2022, 09:37:24 pmNot too crucial in this particular case as the load only draws 8uA so the voltage drop is close to nothing.David