Idea shootout: ideal diodes. - Page 1
Forum section: electronics / projects

author: technix
author_rank: Super Contributor
timestamp: April 17, 2020, 08:15:41 pm
content: I have detected a certain powered USB hub feeding power back to my motherboard, causing me problems in booting. Now it seem to me that I will need some USB 3.0 power diodes to protect my motherboard against that. Using Schottky diodes leads to a 0.3V drop, which can cause problems in some loads...So here is the shootout: an ideal diode circuitry. Specs: 5.5V, 2A.My entry is the following:Ignore the diode and the cap, as this is excerpt from one of my projects. The power comes in from top left and leaves top right, and the bottom rail is ground. Basically the same circuitry taken from the Raspberry Pi 2 schematics, but with some component selection changes.

author: exe
author_rank: Unknown
timestamp: April 17, 2020, 09:15:53 pm
content: "ideal diode controller" , such ashttp://www.ti.com/power-management/power-switches/ideal-diodes-oring-controllers/overview.html.

author: richard.cs
author_rank: Super Contributor
timestamp: April 18, 2020, 12:54:01 pm
content: I've been staring at that circuit trying to understand it and I have to say I am struggling to get my head around how it works, though throwing it into simulation it does seem to behave quite nicely. It doesn't switch on until there's half a volt or so forward across the MOSFET but then goes to about 50 mV drop and turns off before the current reverses though I am not really sure what causes that - is it triggered by the fall in the forward drop as the current falls?It's quite neat anyway, I like it.

author: magic
author_rank: Super Contributor
timestamp: April 18, 2020, 05:53:18 pm
content: Any problem with simply letting the hub power itself?

author: not1xor1
author_rank: Frequent Contributor
timestamp: April 18, 2020, 06:35:13 pm
content: Quote from: richard.cs on April 18, 2020, 12:54:01 pmwhen there is an input voltage the BJT on the right is OFF and so once the capacitor on the right get charged via the MOSFET body diode above the threshold voltage the mosfet goes ON. Once the input voltage is removed, the base of the right BJT is at 0V and the BJT goes ON shorting the MOSFET gate and source.

author: magic
author_rank: Super Contributor
timestamp: April 18, 2020, 06:52:51 pm
content: More elaborately,ignoring base currents, it takes 5:1 ratio of collector currents to equalize collector voltages of the PNPs and pull the FET's gate to Vin-0.7V, which is the threshold region of this part. Coincidentally (or not), it takes about 50mV of Vbe difference in a pair of identical BJTs to produce such ratio. Since the bases are shorted, the FET is gradually turned off as the difference of emitter voltages approaches 50mV. Note that the left PNP is diode-strapped and its collector current is quasi-constant and determined by R10.

author: not1xor1
author_rank: Frequent Contributor
timestamp: April 19, 2020, 06:29:14 am
content: Quote from: magic on April 18, 2020, 06:52:51 pmthanksI had not read the first lines of the original message so completely misunderstood the purpose of the circuit

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 19, 2020, 02:49:13 pm
content: You can find the idea and some explanation in the following article:https://www.powerelectronictips.com/inexpensive-ideal-diode-mosfet-circuit/

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 19, 2020, 04:14:45 pm
content: Quote from: SiliconWizard on April 19, 2020, 02:49:13 pmThat is a good explanation. Additional insight can be gained if the bipolar circuit is viewed as a differential amplifier (acting as a comparator).

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 19, 2020, 04:47:51 pm
content: Hi,This circuit has a flaw that could cause it to fail catastrophically in some circumstances:If the circuit is reverse biased, Vout is positive with respect to Vin Q1 Emitter Base junction is forward biased.Q2 Base emitter is reverse biased. If the voltage exceed the reverse breakdown voltage of Q2, Q2 will be destroyed.In other words this circuit can only be used applications where the maximum reverse voltage is less the reverse breakdown voltage of Q2.Regards,Jay_Diddy_B

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 19, 2020, 05:02:53 pm
content: Quote from: Jay_Diddy_B on April 19, 2020, 04:47:51 pmA solution might be to limit the current by simply putting a resistor between the bases.

author: technix
author_rank: Super Contributor
timestamp: April 28, 2020, 05:00:14 pm
content: Quote from: magic on April 18, 2020, 05:53:18 pmIf I leave the wall wart in there, the hub would feed power back its uplink port (which it should not) and keep some part of my motherboard powered when it is supposed to be powered off (which is bad.) It is also ill-advised to cut VBUS as that makes the port unable to detect the uplink status which can affect enumeration. This is where this ideal diode circuit comes in, as it really should have been integrated in the hub to begin with, but somehow I have to patch it in myself.Quote from: Jay_Diddy_B on April 19, 2020, 04:47:51 pmThis is protecting a 5V rail using transisotrs with a breakdown voltage of no less than 6V. Both sides also have 5.6V Zener clamps.Quote from: Wimberleytech on April 19, 2020, 05:02:53 pmThis requires me having access to that common base, which is not the case of using a transistor pair chip pre-connected for current mirror use. The BCV62 transistor pair chip I used have the bases connected internally (Notice the different color in the schematic - maroon lines are internal to the chip, and green ones are external.)

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 28, 2020, 05:05:46 pm
content: Quote from: Wimberleytech on April 19, 2020, 05:02:53 pmQuoteYup...thats a problem!

author: exe
author_rank: Unknown
timestamp: April 28, 2020, 05:50:42 pm
content: I think there are plenty transistors without bases tighten together, such ashttps://www.tme.eu/Document/b038a6211c62764e189393900f367e18/BC807UE6327.pdf. I'd say look for transistor arrays with 6 leads, chances are parts are separated.PS I don't know if there are any caveats related to them being on the same die. May be they are not completely isolated.

author: magic
author_rank: Super Contributor
timestamp: April 28, 2020, 07:02:19 pm
content: Quote from: technix on April 28, 2020, 05:00:14 pmYes, that's what I meant. I don't think it's going to be a problem, because the hub's PSU permanently drives VBUS to 5V anyway.Quote from: exe on April 28, 2020, 05:50:42 pmUnless you pay extra for so-called "monolithic pair", you will get two separate dice in one plastic blob. Some manufacturers may promise that they came from adjacent locations on the same wafer, some even guarantee rudimentary matching (maybe 3mV Vbe offset, some tens of % on beta). NXP offers some matched duals IIRC.In monolitic pairs, collector to collector isolation is done with reverse biased PN junctions and may break down at excessive voltage, in addition to having some parasitic capacitance.

author: technix
author_rank: Super Contributor
timestamp: April 28, 2020, 08:37:42 pm
content: Quote from: exe on April 28, 2020, 05:50:42 pmThe specific chip combo, AO3401 and BCV62, allows me to layout the whole thing within the top layer and the width of a SOT-23, no vias at all. 6-pin parts can make the physical footprint larger, and I have trouble hand soldering things smaller than SOT-23 chips and 0603 passives.

author: ormaaj
author_rank: Contributor
timestamp: April 29, 2020, 07:12:00 pm
content: I've seen this design. I don't know that an actual comparator wouldn't outperform a "simpler" discrete solution like this.