solved: something similar to a "normally on" BJT conduct at <200mV? - Page 1
Forum section: electronics / projects

author: incf
author_rank: Frequent Contributor
timestamp: July 10, 2025, 12:13:44 am
content: Academic question,Very specific thing, not necessarily aiming to solve any practical problem.Suppose there was a high impedance circuit on a super low power battery operated product. In this case the VDD supply rail of a PCB containing a STM32 MCU in deep sleep mode. The things we care about that are connected to this rail are the RTC backup domain registers / power on reset detection circuitry within the MCU. They are only guaranteed to clear when the supply voltage drops below 200mV.When power is removed, the VDD supply rail with it's decoupling capacitors will sit at 600mV for a "very long time" and the RTC registers never clear. This is undesirable behavior. The supply voltage needs to drop to less than 200mV in less than a second when power is removed. This problem can be easily solved in a wide variety of very easy and practical ways.However, I am not interested in a practical solution, but rather a very specific academic solution. (ie. I'm looking to fill a hole in my knowledge of analog circuitry)I'm interested in discharging 1mF of supply rail capacitance from 600mV to roughly 150mV or less in roughly 100ms to 1sec (ie. 3mA average discharge current) but only when VDD is below roughly 1.2V. And when VDD is above 1.8V the leakage of the discharge circuitry should be less than 1uA.Placing a resistor on VDD to ground would "solve" the problem - but would either be too slow, or draw way more than 1uA above 1.8V.Having a switched transistor seems like the way to do it, but there is a problem, BJTs need >600mV on the base to conduct and there is no power available to supply current to the base. A BJT is "normally off" but what this requires is a "normally on" switch. (PNP vs NPN makes no difference here, both require current to flow through the base)Another way to look at it is that the problem could be solved by something like a zener diode where the behavior is reversed (ie. conduct below the zener voltage, and stop conducting above).Are there any clever analog ways to solve this problem?It seems that maybe JFETs and other less common active devices might not be suitable. But maybe I've overlooked somethingAgain, I'm not necessarily interested in the easy or practical solutions along the lines of "change the problem to be easier / more sensible." I'm interested in the academic solution that narrowly fits within the prescribed bounds

author: thm_w
author_rank: Super Contributor
timestamp: July 10, 2025, 01:08:08 am
content: You can get N-fet with Vgs th of 0.4V:https://assets.nexperia.com/documents/data-sheet/BSH103.pdfAnyway MCU can detect power loss, send voltage to the gate of a FET (with cap and high value pulldown resistor), which is clamped across the voltage rail via load resistor. If it has to be all analog, it may be possible to use a comparator as the trigger. They are available with Iq of ~2uA.I'm sure you can figure out a circuit for that to test in simulation.

author: incf
author_rank: Frequent Contributor
timestamp: July 10, 2025, 02:01:34 am
content: When I look at the datasheet for a well-stocked JFET it almost looks like maybe they could do the job after all. I wasn't sure about being able to turn it off enough to get a low enough cutoff current. (In the past I seemed to recall the threshold voltages being a lot higher than what these devices posses)https://www.onsemi.com/pdf/datasheet/tf412s-d.pdfhttps://www.onsemi.com/pdf/datasheet/mmbfj113-d.pdf

author: David Hess
author_rank: Super Contributor
timestamp: July 10, 2025, 04:31:54 am
content: A JFET will do exactly what you want. A depletion mode MOSFET could also be used.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 10, 2025, 05:00:20 am
content: Quote from: incf on July 10, 2025, 12:13:44 amOne academic solution would be a Depletion Mode MOSFET and a Photovoltaic optocoupler.The optocoupler can work current fed from just over 1V, and an active coupler holds off the depletion mode mosfet.An alternative, cheaper but with lower gain, would be a charge pump like TPS60403DBVR, or even 7660, paired with a depletion mode FET.An active charge pump holds off the mosfet

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 10, 2025, 11:09:45 am
content: Various FET's have already been mentioned, but it can also be done with a BJT.When you drive the BE junction of a BJT into saturation, the Collector - Emitter will draw current down to a quite low voltage, often <100mV, or even <50mV. In some (line level) audio circuits this is used to mute the audio during power up / down. It's one of those peculiar things of BJT's that I have not fully investigated yet.

author: David Hess
author_rank: Super Contributor
timestamp: July 10, 2025, 04:59:11 pm
content: Quote from: Doctorandus_P on July 10, 2025, 11:09:45 amThe problem is that there is no other source to provide bias to the base-emitter junction and keep the transistor on after power is removed.A capacitor could be used to hold the bipolar transistor on for long enough after power is removed.

author: Zero999
author_rank: Super Contributor
timestamp: July 10, 2025, 05:31:30 pm
content: Quote from: thm_w on July 10, 2025, 01:08:08 amThat's the minimum threshold voltage with a drain current of 1mA. The maximum threshold voltage will be much higher.It's theoretically possible to design a MOSFET with a maximum threshold voltage of say 200mV but the problem is, it wouldn't necessarily fully turn off at 0V. It would need to be operated in depletion mode to turn fully turn off, making it a depletion mode device.

author: incf
author_rank: Frequent Contributor
timestamp: July 10, 2025, 06:08:27 pm
content: Quote from: Zero999 on July 10, 2025, 05:31:30 pmI have a question about the TG412 . It looks like that datasheet says the guaranteed max threshold voltage is 1.5V - but this only applies at Vds = 10 V, Id = 1 uA , and Tj = 25C.This looks like a "good guarantee to me" on this specific part.Asking specifically about the TG412 is it possible that the threshold voltage might be above 2.0V under certain conditions? (They say 1.5V max)https://www.onsemi.com/pdf/datasheet/tf412s-d.pdf

author: Marco
author_rank: Super Contributor
timestamp: July 10, 2025, 06:15:58 pm
content: Quote from: incf on July 10, 2025, 12:13:44 am600 mV can pull current from a higher voltage, a higher voltage such a separate capacitor which was fed through a Shottky diode from the supply rails. So then you can pull current through the PNP, which can push current through the base of a NPN.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 10, 2025, 08:02:25 pm
content: Quote from: incf on July 10, 2025, 06:08:27 pmIt will never exceed -1.5V, ie you need to drive the gate negative to reduce to 1uA, but note the 0V current can be as low as 1.2mAAnother approach is to diode charge a gate cap of a conventional low threshold fet as Vcc falls.A reset chip can do that at a specific voltage.You do not need this current for much time.

author: Zero999
author_rank: Super Contributor
timestamp: July 10, 2025, 08:04:06 pm
content: Quote from: incf on July 10, 2025, 06:08:27 pmTo be pedantic. It's a depletion mode device. All of the threshold voltages are negative. Actually cut-off voltage is a more accurate term, since the drain current is specified to be close to zero, when the gate voltage is below or equal to the cut-off voltage of -1.5V. The cut-off current is measured to be 1µA when VDS= 10V and TJ= 25°C

author: thm_w
author_rank: Super Contributor
timestamp: July 10, 2025, 09:43:23 pm
content: Quote from: PCB.Wiz on July 10, 2025, 08:02:25 pmyes that is what I said above. cap can hold a voltage high long after the rail is gone, so you just need some mechanism to trigger and discharge the cap. Some regulator will already have this, a Power Good output. No need for exotic parts.

author: NiHaoMike
author_rank: Super Contributor
timestamp: July 11, 2025, 04:24:30 am
content: How about a pair of small MOSFETs, with one being held off by the other and pulled up by a high value resistor from a capacitor that's kept charged using a diode from the supply rail? A tiny ceramic capacitor will keep the MOSFET turned on for a very long time, certainly long enough for it to perform its function.Have you considered a firmware fix by checking for a "brown out reset" and clearing the registers in that case?Thinking laterally, how about a switch that is held open by the battery being installed?

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 11, 2025, 06:00:52 am
content: Quote from: NiHaoMike on July 11, 2025, 04:24:30 amThat's simple but misses the OP's request forvoltage monitoringQuoteThat needs a voltage monitor, with separate power pin, something like Gainsil GS8021 is close - that has RRIO Comp 13us 390nA with 1.2V Ref, so it can do the under-voltage sense at < 1uA.

author: NiHaoMike
author_rank: Super Contributor
timestamp: July 12, 2025, 04:03:07 am
content: Quote from: PCB.Wiz on July 11, 2025, 06:00:52 amSelect a MOSFET with a threshold voltage such that the voltage needed to hold the other MOSFET off is between 1.2V and 1.8V.