How to get a constant voltage drop of 1.1V ? - Page 1
Forum section: electronics / projects

author: ballen
author_rank: Regular Contributor
timestamp: May 17, 2025, 11:29:42 am
content: I have a large number of battery-operated thermostats, powered by 2 AA alkaline batteries in series. For many of those, I'd like to use a standard Li Ion 18650 battery (3.5 - 4.2V) instead. The thermostats draw a few hundred mA when closing the valve, so we can assume a current < 1 Amp.The simple solution would be a 3V (or 3.3V) 1A LDO regulator. The issue is that the thermostat will then report that the battery is 100% charged, until the battery is completely discharged.So, I'd like a solution where, when the battery reaches 3.65V, the output is around 2.55V. That's the point where the battery has about 20% capacity remaining, and where the thermostat will also report 20% remaining (because it assumes 2 x Alkaline AA batteries). So if there is a two-terminal device that give me a fixed 1.1V drop, that would be a good solution.Yes, I could use a Si diode and a Schottky diode in series, that would be about right. But is there a simple way for me to produce a 1.1V drop that is fairly current-independent up to about 1A, in a two-terminal device, with say 50mV of accuracy?A three terminal solution is also OK, provided that the current being drawn for the purposes of regulation is less than about 30 microamps. Over a year that won't use more than about 250mAh.It's my impression after doing some experiments that the thermostat measures the voltage when the internal motor is running, so when current is being drawn. So the two diode solution may be OK, but if there is a simple better way, I'd like to know.

author: Zero999
author_rank: Super Contributor
timestamp: May 17, 2025, 03:55:26 pm
content: Why not use three AAA NiMH cells and a silicon diode?

author: ballen
author_rank: Regular Contributor
timestamp: May 17, 2025, 04:05:28 pm
content: IMO NiMh is obsolete technology. I prefer Li-ion cells for their energy density, lack of memory, very low self-discharge, and wide availability at low cost.So, the question remains open, is there a simple way to get a constant 1.1V drop at up to 1 amp with either zero or low quiescent current?

author: TimFox
author_rank: Super Contributor
timestamp: May 17, 2025, 05:17:52 pm
content: A pass transistor (either BJT or MOS) in series, with a simple op-amp circuit to stabilize the voltage across the transistor (making a three-terminal circuit)?

author: bdunham7
author_rank: Super Contributor
timestamp: May 17, 2025, 05:28:20 pm
content: Quote from: ballen on May 17, 2025, 04:05:28 pmIt's not a terribly uncommon issue and AFAIK there is no simple, perfect solution at that low voltage. You might want to actually measure that operating current so you can perhaps decide on a better compromise. I'm also a little unsure about your assertion that 3.65V would represent 20% SOC on your Li-ion cell, especially if tested under any significant load. If your operating current is say 300mA, then a single standard 1N4007 would probably get you pretty close to where you want to be. If you can live with only charging the Li-ion cells to 4.0V, then your maximum voltage to the thermostat would only be about 3.3, other wise it will be 3.4-3.5V after you install a charged cell.

author: Zero999
author_rank: Super Contributor
timestamp: May 17, 2025, 06:40:33 pm
content: Quote from: ballen on May 17, 2025, 04:05:28 pmThe problem is, you're losing a fair amount of that energy by dropping 1.1V, so you might as well use a lower battery voltage and waste less.

author: ballen
author_rank: Regular Contributor
timestamp: May 17, 2025, 06:47:42 pm
content: Quote from: bdunham7 on May 17, 2025, 05:28:20 pmBother. I was hoping that there was some simple circuit incorporating a FET that would do this.The operating current is the issue. This is a Thread device, so a 2.4GHz transmitter (think Bluetooth LE with a mesh network) designed for IOT use to run on batteries for a year or two. So it sleeps for a fraction of a minute at a few microamps. Then it wakes up, briefly swallows some tens of mA to measure the ambient temperature, transmit and receive its status, then back to sleep again. Then, from time to time, it runs the motor, which I have measured up to 500mA.Regarding the assertion that 3.65V would represent 20% remaining capacity on a Li-ion 18650 cell, just look at a typical discharge curve at 0.5C, which would typically be around 500mA. So my goal is to subtract 1.1V. The fact that this leaves 3.3V at full charge (4.2V) is no problem. I have already tested the device running on 3.3V, and it's perfectly happy with that.My reading of the 1N4007 puts it around 0.8V at 400mA. Hence my ugly solution of that in series with a Schottky to drop another 0.3V.But perhaps there is a clever person here with a better solution. I am still hoping that there is a way to combine a BJT and a FET, using the base-emitter junction of the BJT as a "voltage reference" and the FET to then "buffer" this to make it current independent, with a couple of resistors to somehow shift the 0.7V to 1.1V. I wish Jim Williams was still around, he would probably know how.PS: before anyone suggests just using a straight LiFePO4 cell, unfortunately if I charge it initially to 3.4V, by the time it gets to 2.55V there's hardly anything left, not 20%.

author: ballen
author_rank: Regular Contributor
timestamp: May 17, 2025, 06:52:29 pm
content: Quote from: Zero999 on May 17, 2025, 06:40:33 pmMy goal is NOT to extract all of the energy (eg with a buck converter). I'm happy to get 3/4 of the energy and have the other 1/4 turn into heat in the 1.1V drop "diode". My goal is to have an output voltage of 2.55V when there is 20% charge remaining on the cell, and between 3 and 3.3V when the Li-ion cell is 100% charged.Why? I want the device (pre-programmed for 2 x AA alkalines) to report "20% charge remaining" when my Li-ion cell has about 20% remaining.

author: pqass
author_rank: Super Contributor
timestamp: May 17, 2025, 08:03:42 pm
content: VBEmultiplier may work for you. See attached schematic andsimulation here.Others, more knowledgeable than I, can advise on how sloppy it can get with regards to changes in: current, temperature, HFE, etc.The circuit is explained.

author: bdunham7
author_rank: Super Contributor
timestamp: May 17, 2025, 08:21:52 pm
content: Quote from: ballen on May 17, 2025, 06:47:42 pmNot saying it can't be done, but if it can I'd like to know how. The trick is you don't want any significant quiescent current.QuoteI have looked at many such charts and even measured a few myself and I've never seen any where 3.65V => 20% SOC even OCV. At low discharge rates or for very low impedance batteries, the curve may be very flat and around 3.65V for the middle of that curve, but even under those circumstances your 20% indication could trigger at a SOC of 50% or more with most cells. I'd suggest you look at data--and preferrably run your own test--for the exact specific cell you intend to use. Your job might be a lot easier if you were less rigid about the 20% thing, having the indication come on when the true capacity is only 15% likely wouldn't ruin anyone's day especially since your cells will have more capacity. They will have more capacity than 2 NiMH AA cells, right? Trying to guesstimate the SOC of an Li-ion cell in the middle of its range using only voltage when you have temperature, component variability, etc causing relatively large errors is probably not practical. At 15% SOC or less the voltage curve steepens up a lot.Here are a whole bunch of discharge curves to look at and compare:https://lygte-info.dk/review/batteries2012/Common18650comparator.php

author: bdunham7
author_rank: Super Contributor
timestamp: May 17, 2025, 09:14:10 pm
content: Quote from: pqass on May 17, 2025, 08:03:42 pmI won't claim to be more knowledgeable, but I'll point out that at low current, before the "knee", the circuit is essentially passing full voltage thru the resistors. IDK if that works for the OP, but it certainly won't be a constant voltage drop over the operating range of the thermostat device. It is a true two-terminal solution though.

author: pqass
author_rank: Super Contributor
timestamp: May 17, 2025, 11:27:37 pm
content: The trend for the VBEmultiplier circuit is: lower current -> lower voltage drop.And with very low currents, 100uA, there'sonly a 115mV drop. It may be acceptable if the load can stand a 4V supply (on a full battery).Just tune the voltage drop at the current the thermostat senses the battery level (when motor is running at 500mA).Maybe OP is better off using an LDO to regulate the load and have a battery low-voltage monitoring solution (TL431-based?) light an LED.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 17, 2025, 11:36:00 pm
content: Quote from: ballen on May 17, 2025, 11:29:42 amSimple bipolar parts will be very temperature dependent.You could use a voltage reference, high side relative to drop the positive rail, or a voltage reference, GND relative to drop the -ve rail.A suitable Hi-Side VREF is NJM2823, 1136mV 0.4% 15ppm/'C typLow side parts with 0.6V ref would be TS4436 and ZXRE160, and the NCP100 has 700mV low side ref.Low dropout current boost is not simple...

author: Someone
author_rank: Super Contributor
timestamp: May 18, 2025, 12:35:59 am
content: modify the devices voltage divider.....

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 18, 2025, 03:16:18 am
content: Further to #12, this is what a TS4436 looks like, as a 3 terminal 1.10V shunt, up to 1A.The TS3326 can sink 20mA to < 140mV, so can drive a PNP emitter follower clamp with HFE above 50 to 1A.The PNP selected needs to be large, good enough for ~1 Watt, with HFE high to >> 1A

author: amyk
author_rank: Super Contributor
timestamp: May 18, 2025, 05:10:19 am
content: How about a buck converter with its feedback adjusted by the input voltage?

author: mariush
author_rank: Super Contributor
timestamp: May 18, 2025, 05:19:07 am
content: Could you use a programmable linear regulator with a zener diode / tl431 voltage reference or some voltage divider + schottky inverter ?See for example AP7179D which lets you configure the output voltage with 5 pins :https://www.digikey.com/en/products/detail/diodes-incorporated/AP7179DSKZ20-13/18674707Datasheet :https://www.diodes.com/assets/Datasheets/AP7179D.pdf2.55v : GND GND Open Open Open GND2.95v : GND GND Open GND Open GND3.15v : GND GND GND GND Open GNDSo it would just be a matter of pulling that 4th pin to ground while the battery voltage is higher than 3.65v, to keep the output voltage at 2.95v

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 08:47:50 am
content: Quote from: pqass on May 17, 2025, 08:03:42 pmThanks a lot for this!Unfortunately, this won't work for my application. Since I need up to about 1 amp of current, if the transistor has a beta of 50, I'll need to provide 20mA of base current. This means that the resistor from collector to base has to be no more than about 100 ohms, and the resistor from base to emitter needs a similar value to get a 1.1V drop. Thus, when the device is sleeping and needs just a few microamps of current, the voltage drop across this circuit won't be 1.1V, but rather will be around 1mV. This will expose the device to the full battery voltage (up to 4.2V) which is too much, it is designed for 3V and I am only comfortable exceeding that by 10% (3.3V).If there is an FET-based version of this circuit that has megohm resistors, that might work. But I don't see how to do it.Thanks again, this was a very useful idea, and while it does not solve my problem, I am sure that someone else will benefit from it.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:04:13 am
content: Quote from: bdunham7 on May 17, 2025, 08:21:52 pmYes, that's right.Quote from: bdunham7 on May 17, 2025, 08:21:52 pmWow, thanks, you have looked into this a lot more carefully than I have! My typical cells are Samsung 18650, scavenged from old laptop packs. So let me use your Samsung ICR18650-26H 2600mAh (Pink) curve at 0.5A discharge.Looking more closely at my requirements, my thermostats show 20% remaining at 2.55V and 0% remaining at 2.40V. So if I look at your curve I get 3.55V at 1.87Ah and 3.40V at 2.4Ah. These differ by 22%. So I need to shift the voltage by 1.0V (say at 400mA) and not by 1.1V.Anyway, I will do some experiments with some 1N4007 and Shottky diodes to see how close I get to this.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:07:03 am
content: Quote from: pqass on May 17, 2025, 11:27:37 pmNo way!I have forty of these thermostats spread around the house. One of the reason I am replacing the current 17-year-old electronic ones is because they beep when the battery is low. I can't even figure out which one is beeping! The new ones let me use an App on the iPhone to monitor battery level, and warn me at 20%.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:08:53 am
content: Quote from: Someone on May 18, 2025, 12:35:59 amNot an option. These are tightly integrated small form factor SMD devices integrated into a case which can not be opened without destroying the plastic tabs that hold it together. And there are forty of them. I want to just build a small add-on for providing power.

author: voltsandjolts
author_rank: Unknown
timestamp: May 18, 2025, 09:10:32 am
content: Are you sure the thermostats report battery state in the way you expect, i.e. directly from battery voltage?You might want to try powering one of the from a lab supply and varying voltage to see see effect on reported battery state.Show us the plot.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:15:03 am
content: Quote from: PCB.Wiz on May 18, 2025, 03:16:18 amI'm not quite sure if I understand your circuit diagram, specifically where the battery and load are located. But from looking at the TS4436 specs, the typical operating current is 150uA with max 200uA. That is too much: over a year (8760h) it adds up to 1.3Ah to 1.8Ah.BUT, could I do something similar using a TLV40x1 device? It is a comparator with a precision internal voltage reference and the operating current is just 2uA.https://www.ti.com/lit/ds/symlink/tlv4021.pdf?ts=1747547084664

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:17:28 am
content: Quote from: voltsandjolts on May 18, 2025, 09:10:32 amI don't have a plot but do have a table, which I think is good enough. The data was taken about 24h ago, before I posted my first message starting this thread.The device is an "Eve Thermo". Using a lab power supply, I made a table of input battery voltage (for both in series) and corresponding behavior. The radiator valves seem to monitor the voltage when the motor runs (which can draw up to about 500mA when closing fully) and assume that the voltage will only go down, not up. So if you increase the voltage for testing, you must remove and reapply the voltage, to simulate a battery change. This is not needed if you decrease the voltage.The test protocol is as follows:(1) Set power supply current limit to 2A(2) Power off the valve(3) Set voltage to desired value(4) Power up the valve (which runs the motor)(5) Read battery voltage percentage from App(6) Return to step 2Results are:2.40V 0%2.45V 5%2.50V 10%2.53V 15%2.55V 20%2.58V 25%2.60V 30%2.65V 35%2.67V 40%2.70V 45%2.75V 50%2.77V 55%2.80V 65%2.85V 70%2.90V 75%2.95V 85%3.00V 95%3.05V 100%The valve continues to operate down to 1.8V, although at some point the motor stops operating. At 1.80V the valve starts to make a periodic "chirp" sound and to flash "lo" on the display. So, in effect, when the battery voltage drops below around 2.40V, the device prioritizes its communication/networking.

author: Hydron
author_rank: Super Contributor
timestamp: May 18, 2025, 09:18:55 am
content: Have you opened one up to check what the parts inside are specced for? One would think that something designed specifically for battery use with 2*AA alkaline (a bit more than 3V when fresh and unloaded) might also be designed for a LiIon cell voltage.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:25:14 am
content: Quote from: Hydron on May 18, 2025, 09:18:55 amI have two devices that were damaged and hence can be opened (case tabs already split). While I have not taken out my microscope to study the parts, I have read reports that others have tried this and that the devices did not operate on and/or survive 4.2V.Following my comment from a few minutes ago, I think that this device plus an external transistor or FET might provide a good solution:https://www.ti.com/lit/ds/symlink/tlv4021.pdf?ts=1747547084664

author: shapirus
author_rank: Super Contributor
timestamp: May 18, 2025, 09:26:28 am
content: It looks like anything that uses only basic elements like diodes and transistors will inevitably have a rather soft "knee" without a sharp enough transition, in other words, the voltage drop dependence on the current will be too high, or the quiescent current will be too high.Some op amp based solutions were mentioned before, do you consider them to be too complex for this application? Adding an op amp and a voltage reference to what already has to contain a pass transistor and a few resistors probably won't increase the cost and effort too much.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:27:48 am
content: Quote from: amyk on May 18, 2025, 05:10:19 amThe problem with this is that the device has a low-energy 2.4GHz transmitter/receiver inside. I'm worried that without a lot of care the buck converter is going to make a whole mess of harmonics which interfere with the reception. I'd like to avoid having to do spectrum analysis to get a working design.

author: RoGeorge
author_rank: Super Contributor
timestamp: May 18, 2025, 09:30:38 am
content: Just so you know, I have a few 3V devices connected directly at a Li-ion cell. Some are working for years like that. So far they didn't get any damage, but they are devices I do not care much anyway (weather station, bathroom scale, etc., so nothing critical).Can not guarantee about your thermostats, but I guess they will work, too, if powered directly from a 4.2V Li-ion cell instead of 3V from 2xAAA. A typical Li-ion during discharging has a plateau at 3.7V, so the 4.2V won't stay for long. Or, you can charge the Li-ion cell at only 3.8V, not much charge is stored above 3.9V anyway (probably less than 10-20% from the total battery capacity).

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:32:42 am
content: Quote from: mariush on May 18, 2025, 05:19:07 amIf all I wanted was a FIXED voltage output, I could just use any garden variety 1A low drop out 3.0 or 3.3V regulator.My goal is a VARIABLE voltage output, so that when the input voltage is 4.2 volts, the output is around 3 to 3.3V. When the input voltage is 3.55V, the output should be 2.55V. When the input voltage is 3.4V, the output should be 2.40V. This will correctly interact with the "low battery" system in the device.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:36:36 am
content: Quote from: shapirus on May 18, 2025, 09:26:28 amYesQuoteThat's fine, as long as the active system draws only a few microamps. Cost is not an issue, I'll make about fifty of these. I think that this device plus an external transistor or FET might provide a good solution:https://www.ti.com/lit/ds/symlink/tlv4021.pdf?ts=1747547084664Do you see a way? I'd like to subtract 1V from the input voltage, over an input voltage range from 3.3 to 4.2V, providing a current of up to 1A, with a quiescent/operating current of microamps when there is a small load (megohm).

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 09:38:44 am
content: Quote from: RoGeorge on May 18, 2025, 09:30:38 amAgain, my issue is NOT to correctly power the devices. I can do that using any low quiescent current LDO 3.0 or 3.3V 1A regulator, these are a dime a dozen. The issue is to get a correct low battery indication around 20% remaining energy in the Li-Ion battery.To summarize the requirements:Battery Input 4.20V Output 3.1 - 3.3VBattery Input 3.55V Output 2.55VBattery Input 3.40V Output 2.40VProvide current up to 1AWhen output (load) impedance high (megohm) the current drawn from the battery should be < 30uA.

author: shapirus
author_rank: Super Contributor
timestamp: May 18, 2025, 09:48:16 am
content: Quote from: ballen on May 18, 2025, 09:36:36 amThis is a comparator, which is, if we put it this way, an on/off type device, which isn't really what you need. What you need is a "tracking" device, which would be an op amp. So, you need a micropower voltage reference (not necessarily exactly 1.1V) and a micropower rail to rail op amp which you would connect in a"voltage subtractor"arrangement (with feedback taken from after the pass transistor rather than the op amp's output). One input would be your Lithium battery voltage, another would be the voltage reference's one, and the output will be the difference between them (subject to the ratios set by the resistor values), thus achieving the "battery voltage minus a fixed 1.1V" goal.p.s. the op amp doesn't even have to be a rail-to-rail one, I think (or at least not a rail to rail input -- r2r output may be necessary). Can't tell for sure as I can't grasp it wholly in my head at the moment, need to simulate it and see how it's going to work.

author: mariush
author_rank: Super Contributor
timestamp: May 18, 2025, 09:59:23 am
content: Quote from: ballen on May 18, 2025, 09:32:42 amI understand, the point I was trying to make is that you can change the output voltage of the regulator on the fly, by pulling a pin down to ground. (instead of changing resistor value or doing something else)As I said, the chip has 6 pins that set the desired output voltage, by setting those pins to ground you change the output voltage from a default of 0.8v :2.55v : GND GND OpenOpenOpen GND2.95v : GND GND OpenGNDOpen GNDThe difference between 2.95v and 2.55v is a single pin pulled to ground. So what if you have for example a n-channel mosfet between that pin and ground... and now all you have to do is turn on the mosfet when the voltage goes below 3.65v or whatever voltage you have.First thought that comes to mind is a schmitt trigger inverter (a not gate)... as long as the input voltage is above some threshold, it won't output anything, and when the input goes below some threshold you'll enable the mosfetA SSM3K56 mosfet will trigger with very low Vgs and will have a low rds(on) that shouldn't affect things, it's under 0.3 ohm with 2.5v Vgs :https://www.digikey.com/en/products/detail/toshiba-semiconductor-and-storage/SSM3K56FS-LF/3863603An inverter with schmitt trigger input like MC74HC1G14 would probably work :https://www.digikey.com/en/products/detail/onsemi/MC74HC1G14DFT1G-Q/22161786The negative threshold is 1/2 of the input voltage, but you could add a small trim potentiometer to tweak the voltage divider so that when the lithium battery voltage is 3.65v, you'll get half the voltage on the input and trigger the inverter, which would turn on the output and power the mosfet which will adjust the output voltage of the regulator to 2.55vEDIT: Sorry, I just realized, you don't need to turn on a mosfet, you need to turn off a mosfet, you need to disconnect the pin from ground to get the regulator down to 2.55v.So I guess an AND gate with schmitt trigger input or some other means to turn off the mosfet when battery voltage goes below 3.65v would be correct.

author: nali
author_rank: Frequent Contributor
timestamp: May 18, 2025, 10:06:22 am
content: If you can live with 1.2v then use a LM138Or - if you're confident that the voltage is always measured when the motor is running, and you know the load current, just use a series resistor to drop the 1.1v at that current. If the current isn't constant over the voltage range then choose the set point at which you want it to be most accurate.

author: shapirus
author_rank: Super Contributor
timestamp: May 18, 2025, 10:26:12 am
content: Quote from: nali on May 18, 2025, 10:06:22 amProblem is, that voltage has to be dropped when the motor is NOT running, too. So it has to be maintained across the entire range from microamps to 1A.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 10:34:22 am
content: Quote from: nali on May 18, 2025, 10:06:22 amWon't work, requires too much operating current.QuoteWon't work, when the device is sleeping this will feed 4.2V to the device, above my max 3.3V.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 10:40:51 am
content: Would the following be a simple solution? Suppose that I have a 1.00V LDO voltage regulator with low quiescent current, which can both source and sink current. I use the Li-ion battery as input, and connect my load between battery positive and LDO output. Thus, the load always sees battery voltage minus one volt. Do such LDO voltage regulators exist?Another way of saying this: I want to generate a 1.0V virtual ground, with low quiescent current. Is there a conventional part that can do this? It should sink up to 1 amp. I think a 1V virtual ground with low standby current should be possible using:REF35 1.024V as voltage reference (1 microamp)TLV8801 or similar as an opamp (1 microamp quiescent current)plus a pass transistor or FET to handle the current.Here is the idea:Is there a low quiescent current opamp (30 microamps) that could work here? If the pass transistor is a 2N2955 or similar it will need up to 50mA base current, which I think is unlikely from such a low power opamp. Can I use an FET as the pass device?

author: Zero999
author_rank: Super Contributor
timestamp: May 18, 2025, 11:46:00 am
content: My point is that you lose more energy with a Li-ion cell dropping 1.1V, than NiMH because the voltage is higher, so you gain very little in terms of energy density. Yes, there's less energy capacity in two NiMH cells, compared to a similar sized Li-ion, but at least you get to use it.Heck, if the thermostat is properly designed, it should work down to 2V, so should be fine with two NiMH cells and no other circuitry.Failing that, you can control the voltage of a variable LDO, by switching one of the voltage setting resistors. If you choose comparator with an open collector output, then you can reduce the voltage of the LDO by adding a resistor in parallel with the lower voltage setting resistor, once the supply voltage drops below a set threshold, giving you the low battery warning.

author: ledtester
author_rank: Super Contributor
timestamp: May 18, 2025, 02:24:04 pm
content: You could take a look at the Xtar 1.5V Lithium Ion AA Batteries..They are designed to drop to 1.1V when they becomed discharged...Xtar USB Type C 1.5V Lithium Ion AA Batteries -- Juian Iletthttps://youtu.be/O-dfkikwDvg

author: SiliconWizard
author_rank: Super Contributor
timestamp: May 18, 2025, 02:38:15 pm
content: To get the "most" (minus efficiency) of a Li-ion battery in this case, just use a switching buck converter.If you really want to emulate the voltage drop of a pair of alkaline 1.5V batteries, you'll have to control the set output voltage, but that's going to be a rabbit hole. Possibly a fun project if you're into that. Estimating the SoC of a Li-ion battery from its cell voltage alone, especially when loaded, in short, "doesn't work" without a lot of "trickery". Some fuel gauge ICs can do this reasonably well with "sophisticated" internal algorithms.If the SoC estimation is not critical and just very roughly indicative, you could adjust the output voltage linearly from the battery cell voltage. This is going to take some design work, especially since you'll typically have to use a control voltage that changes in the opposite direction. That's typically done placing a resistor in series with the feedback node of the buck converter to apply the control voltage. Doing so, you'll probably need to use at least one opamp to generate the control voltage appropriately.For something more complex (in terms of "system") and a bit more expensive, but probably easier to design if you're more of a software guy and significantly more accurate, if you don't mind using a small MCU, is to control the buck converter using a small, low-power MCU and a fuel gauge IC such as the MAX17048 or similar. Of course, that makes a project that's a bit more that just a single active component, but it should be reasonably robust, accurate and very flexible. Should cost only a few bucks in parts.

author: bdunham7
author_rank: Super Contributor
timestamp: May 18, 2025, 02:45:27 pm
content: Quote from: ballen on May 18, 2025, 09:04:13 amThat's not my website, it belongs to userHKJwho has done all the hard work. And you've picked the least-flat discharge curve of any 18650 I've seen yet!QuoteIf you are using recycled batteries there may be significant variance in their performance. I would recommend experimenting with simpler solutions first. I've done a fair number of battery retrofits using various types and I've almost always been able to adapt very simple solutions to do well enough. Many pre-Li-ion OEM solutions were pretty crude. Even float charging Li-ion can be done reasonably well with very simple methods.

author: pqass
author_rank: Super Contributor
timestamp: May 18, 2025, 03:01:39 pm
content: Quote from: ballen on May 18, 2025, 10:40:51 amIt seems to workin the simulator.Note theSziklai pairis used to take the load off the op amp.

author: David Hess
author_rank: Super Contributor
timestamp: May 18, 2025, 03:21:59 pm
content: Quote from: ballen on May 17, 2025, 06:52:29 pmThat does not mean a buck converter will not work.Use a micropower buck converter, but with the input voltage used as a reference so the output follows the input and as the input decreases, so does the output.

author: Peabody
author_rank: Super Contributor
timestamp: May 18, 2025, 04:11:33 pm
content: Late to the thread, so sorry if this is already covered. If you connect the 18650 to the collector and base of an NPN transistor, and the emitter to your circuit, it should drop about 0.6V, and the idle current would be zero. Or connect the base through a diode, possibly a schottky, to provide an additional 0.3V to 0.6V drop. The drop would vary somewhat with current, but I think even at near zero current, you would still get most of the drop, and a high gain transistor would minimize the increased drop at high current.Along a similar line, I keep thinking of an LM317 configured as a current limiter, where nothing is referenced to ground, but without the limiting resistor. I wonder how much voltage it would drop.They make "tracking" LDOs like the TPS7B4255Q, that might work if the reference voltage input is connected to the input voltage. But it's only 70mA.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: May 18, 2025, 04:13:09 pm
content: How about this

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 04:46:58 pm
content: Quote from: ledtester on May 18, 2025, 02:24:04 pmAlready tested these. Two known issues. First, the internal buck converter generates noise that interferes with the radio communications between devices. Second, the drop to 1.1 does not work for me, I need a drop from 1.5 to 1.25 or similar. At 2.2V the device is in "communications only" mode and no longer moves the valve.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 04:49:19 pm
content: Quote from: SiliconWizard on May 18, 2025, 02:38:15 pmSee post seven above.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 05:00:06 pm
content: Quote from: pqass on May 18, 2025, 03:01:39 pmThis is super cool! I'd like to play around with the simulator. Is it possible for me to copy your configuration and modify it on my own? If so, could you please give me some basic instructions or point me to the same?QuoteFunny, I thought about using a Darlington pair but two junctions drops is too much. I didn't think about a complementary pair. Also did not know it was called a "Sziklai-pair".Thanks a lot for your help with this, I will figure out what parts to use, make a prototype, and test it.(Thanks also to all the others who contributed to this discussion!)

author: pqass
author_rank: Super Contributor
timestamp: May 18, 2025, 06:21:28 pm
content: Just click on my simulator link, change the schematic, then from the top-left menu, File->Save As|Export as Link|Text.The simulator takes a bit of getting use to. You can use the Draw menu but also right-click on the background to Add a component. You'll learn to use the key shortcuts: w (wire), r (resistor), n (npn), N (NMOS), etc. then click+drag to place. Escape takes you out of Add mode. Moving wires is a bit of a pain to hover+click+drag and hover+cntrl+click+drag on a wire end to change length+orientation.Once placed, you can change some parameters to the components by right-clicking then Edit. I changed the HFEon the PNP,NPN to 100,50. Though I've never tried it, you can use Create New Model in the Edit popup to better characterize the component. I wasn't successful with using a MOSFET as all the ones I know have high VGS(th)but you can give it a try (see its Edit popup to change VGS(th)and "Beta" (note the link)).The simulator isn't perfect but at least it hasn't ruled-out the solution. It predicts that you'll need at most 211uA from the op amp; the TLV8801 should be able to deliver (ISC=4.7mA)? At this stage, I think it's worth picking your real-world parts and prototyping.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 18, 2025, 06:58:04 pm
content: Quote from: SiliconWizard on May 18, 2025, 02:38:15 pm"Just use" sounds too simple, keep in mind this is a very low quiescent current application where total efficiency is not dominated by the occasional 1A load, but the quiescent current of the regulator. Traditionally, totally the application for linear regulator, although today we do have very low quiescent current switchers for battery-operated IoT sensor stuff. But be careful choosing.Otherwise I agree and would not worry about buck converter destroying radio communications with switching noise - those radios are highly selective so having a strong emission peak at the exact poor frequency would be extremely poor luck. High frequency switcher can be made small and noise contained within the circuit.QuoteMaybe true as a generic comment, but in a project like this, you are going to use the cell voltage alone, nothing else, and no trickery at all (unless averaging (e.g. RC) counts as trickery), period. The "trickery" you refer to is specifically coulomb counting and trying to keep it tracked with hints from voltage. But this is all relevant for high average discharge current applications, where filtered down voltage alone is too primitive. But for something which runs for months, you are not going to coulomb count anything (even the internal battery leakage becomes significant, and can't be measured!), instead what you need to do is to measure voltage under minimum load and take a direct look-up voltage->SoC%. A decent "fuel gauge" IC would do the same, recognizing the practically no-load situation and just reset the coulomb counter during that and apply SoC% ~ voltage from LUT.In other words, average voltage with a very long time constant will work fine.One alternative pops to my mind though which could be more valuable than true SoC% - that is minimum voltage under load (motor running). When that is getting too low to risk causing brown-out reset in near future, then step down the output voltage to the level which will trigger the low battery warning. But that would require some kind of minimum storage sample & hold circuit, or a small MCU.Maybe easiest is to pick a suitable regulator - linear or very low Iq switcher - and pick a micropower opamp to inject a bit of current into the feedback pin, inversely relative to the input voltage. Exactly desired linear input voltage -> output voltage mapping can be had that way. Shouldn't be too complicated.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 07:03:52 pm
content: Quote from: pqass on May 18, 2025, 06:21:28 pmAgreed.The design is simple enough that there is not much to go wrong. I might replace the voltage source REF35 by a 1.6V part plus a voltage divider, so I can tune the offset. I'll play around with the simulator but at this point just putting one together is probably quicker and simpler.

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 07:12:10 pm
content: Quote from: Siwastaja on May 18, 2025, 06:58:04 pmThat's exactly how it appears to work: the thermostatic valve records the lowest voltage measured when the motor is running. This is typically when the motor briefly stalls closing the valve, when it draws about 0.5A for a fraction of a second.QuoteI'm going to first build and test the "low Iq virtual ground" solution. Only if that doesn't work well will I try something more complex.

author: inse
author_rank: Super Contributor
timestamp: May 18, 2025, 07:27:32 pm
content: There are Lithium batteries in AA form factor which have a converter to 1.5V integrated and a USB charging connector.I read that they also drop the output voltage to allow low battery detection.Would that be an alternative?

author: ballen
author_rank: Regular Contributor
timestamp: May 18, 2025, 07:30:24 pm
content: Quote from: inse on May 18, 2025, 07:27:32 pmsee post #46 above.

author: Zero999
author_rank: Super Contributor
timestamp: May 18, 2025, 07:40:21 pm
content: It sounds like a case of a poorly designed product. I'd be looking for something better, rolling my own or modifying it.Anyway, here's a rough design to show how a comparator can be used to lower the voltage of a switched mode controller, when the battery voltage drops. The power on output spike can probably be removed by playing with the values of C1 and C4, but I didn't have time. I just got the circuit from the LTC3441 example circuit and changed the voltage divider values. The same idea will work with a LDO regulator.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 18, 2025, 09:06:17 pm
content: Quote from: ballen on May 18, 2025, 10:40:51 amWell done, you have cloned the circuit I gave back in #14.The 1.024V ref is pricey, you can scale it up or down, with 2 resistors, and you can also choose a low Iq LDO.You might choose a XC6504A111MR 1.10V 0.6uA IQ and CAP not essential.Addit: or maybe TPS7A0510PDBVR 1V and 1.3uA max at Io =0mA, with 1% tolerance.Or TPS7A03/02 - very low Iq (100nA), and better PSRR than A05, & better wide tempco, but they do not have 1V in SOT25, only in DFN. They do have a 0.9V TPS7A0309PDBVRQuote from: ballen on May 18, 2025, 10:40:51 amEasy: Don't use such and old and feeble PNP.There are many good PNP's these days with HFE > 200, eg You can buy a '772 type device spec'd for 200-400 at 1A for under 10c

author: amyk
author_rank: Super Contributor
timestamp: May 18, 2025, 10:12:12 pm
content: A simple solution that occurred to me upon reading your problem statement again: isn't your desired input vs. output curve satisfied by a 3.3V linear regulator with a suitable dropout voltage of around 1V? You'll get close to 3.3V with a 4.2V input, and it'll stop regulating and track the input voltage with the desired offset as it drops below that.

author: pqass
author_rank: Super Contributor
timestamp: May 19, 2025, 03:02:43 am
content: Quote from: amyk on May 18, 2025, 10:12:12 pmJust for giggles, I wanted to see what the dropout would be for a (handy)bog-standard 3.3V LDO regulatorunder load.It's not a fancy low quiescent current regulator (5mA) so it isn't useful for OPs battery powered application.Notice that there is a 500mV drop difference (0.9V to 1.39V) between no-load and 850mA load.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 19, 2025, 03:10:48 am
content: Quote from: ballen on May 18, 2025, 10:40:51 amPossibly.The FET needs to be NMOS open drain, source = gnd, then you can drive gate higher than the 1V clamping.That does give you excess gain, but it looks like it can be made stable.Something like this would be a starting point. Choose a low threshold, low gate charge FET of enough power.QuoteHow long does that impulse take ?

author: Zero999
author_rank: Super Contributor
timestamp: May 19, 2025, 07:24:45 am
content: Quote from: ballen on May 17, 2025, 11:29:42 amWhy not simply modify the thermostat? It probably senses the battery voltage with a potential divider. Cut the connection between the top resistor in the potential divider and battery and connect it to the input of the LDO regulator, via another resistor.EDIT:An op-amp and suitable voltage reference can be used to subtract a fixed 1.1V off any input voltage. Rather than trying to power the whole circuit, just inject it into the thermostat's voltage sensing circuit.Alternatively, if you must drop 1.1V, add a pass transistor and use a suitable reference for V2.

author: ballen
author_rank: Regular Contributor
timestamp: May 19, 2025, 08:13:33 am
content: Quote from: PCB.Wiz on May 18, 2025, 09:06:17 pmI see! In #22 I replied that I was confused by #14, pointed out that the quiescent current used was way too high, and asked for clarification about the different elements in the schematic. I wish you had replied. Anyway, a belated thank you!QuoteScale down I understand. Scale up is via a voltage divider to the inverting input of the op amp?I'll have a look at the parts that you suggested. Note that the B772 is specd for 3A but only 0.5W. So here, with 1A x 1W that is not quite enough. I was thinking a Sziklai pair with 2N3904 for the NPN driver then TIP42 for the PNP output. The TO-220 package is nice for DIY.

author: ballen
author_rank: Regular Contributor
timestamp: May 19, 2025, 08:14:54 am
content: Quote from: amyk on May 18, 2025, 10:12:12 pmThat was the first thing I looked at. Problem is that all of the voltage regulators with low standby current Iq are also low dropout. But if you can suggest a part, that would be a great solution!

author: ballen
author_rank: Regular Contributor
timestamp: May 19, 2025, 08:21:57 am
content: Quote from: PCB.Wiz on May 19, 2025, 03:10:48 amI am old: I learned electronics before FETs and power FETs were available and am not good with them. Would IRLML6344 be a good choice? If not, could you suggest a specific part for me to play with? Drain to load, source to ground, correct?QuoteIt's a mechanical device (motor) so time scales of tens to hundreds of ms. The current is 200mA for ten seconds and then spikes to 500mA for a fraction of a second, say 100ms.

author: ballen
author_rank: Regular Contributor
timestamp: May 19, 2025, 08:26:04 am
content: Quote from: Zero999 on May 19, 2025, 07:24:45 amSee post #20QuoteYour circuit draws far too much quiescent current. But a low Iq version of this is already being discussed, see posts #14 and #37.

author: Zero999
author_rank: Super Contributor
timestamp: May 19, 2025, 08:34:29 am
content: Quote from: ballen on May 19, 2025, 08:26:04 amNo it isn't. The circuits discussed in those posts are shunt regulators. I'm talking about using a pass regulator. Put a suitable transistor between U1's output and R3 and power your circuit from it. As far as quiescent current is concerned: obvious, just increase the resistor values and use a suitable op-amp. This is just a guide, not a complete solution. No one is going to do all the work for you for free.

author: ballen
author_rank: Regular Contributor
timestamp: May 19, 2025, 09:00:44 am
content: Quote from: Zero999 on May 19, 2025, 08:34:29 amI see, it is only a question of regulating the current flowing into the device, versus the currently flowing out.In your solution, Iq is needed both to feed the voltage reference and to feed the voltage divider on the non-inverting opamp input. Whereas on the shunt regulator we only need to feed Iq to the voltage reference. Doesn't that give an advantage to the shunt approach?QuoteNot expecting that, but it does take me time to understand, because I am not as experienced as you and many others here.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 19, 2025, 09:51:27 am
content: Quote from: ballen on May 19, 2025, 08:13:33 amThey come in different packages, I was looking at TO252 parts.Quote from: ballen on May 19, 2025, 08:21:57 amThe IRLML6344 I can find come in SOT23, which is a bit teensy, but the die specs are good enough.QuoteThat's not too bad, as it can thermally average that.So I'd suggest a part with some metal cooling, so not SOT23, but with a metal tab, like SOT89 / SOT223 or even better TO252 on a good copper pad.

author: Zero999
author_rank: Super Contributor
timestamp: May 19, 2025, 12:35:29 pm
content: Quote from: ballen on May 19, 2025, 09:00:44 amThat's true if the reference and op-amp power supply is in series with the load, otherwise it doesn't make any difference.Each has its advantages and disadvantages.The shunt regulator will have a minimum load current requirement. Below that, it will not regulate properly. The entire circuit must also run down to 1.1V, which can be challenging. It's also more difficult to stabilise, since the output is just a common emitter/source amplifier connected across the output, although at these low voltages, this will apply to a LDO pass regulator.QuoteYou did come across as being ungrateful when you just say refer to posts x & y, especially when it appeared as though you hadn't taken the time to read what I'd written and look at the schematic properly.Note the circuits I've posted are differential amplifiers. The ratio of the resistors determine the gain, which in this case is below 1, since the reference voltage is higher than the 1.1V you want to subtract.https://www.electronics-tutorials.ws/opamp/opamp_5.htmlIt is a challenging problem. It involves micropower design and low-dropout regulator techniques. This is why I wouldn't try to solve it. I would look at changing the thermostats for ones which suit your requirements.EDIT:I had a quick look at low references and op-amps. Here's another idea. I've moved the reference to the positive rail, to avoid having to use a differential amplifier. You might need to put a large capacitor on the output. I haven't done any stability analysis on the simulation, partly because I don't trust the models.https://www.analog.com/media/en/technical-documentation/data-sheets/1389fa.pdfhttps://www.analog.com/media/en/technical-documentation/data-sheets/600345fd.pdf

author: Someone
author_rank: Super Contributor
timestamp: May 19, 2025, 11:29:03 pm
content: Quote from: Zero999 on May 19, 2025, 12:35:29 pmIt makes no sense to keep the sealing of the original product in place when any addition that is under discussion would need to sit in/with the batteries anyway. Most people I can think of who would be in this situation would find out the divider structure from the opened unit and then come up with a way to access that easily (possibly drill an access hole in just the right spot).Or like anyone who relies on battery powered devices, have a battery change schedule that occurs well before empty and do not rely on last minute warnings.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 20, 2025, 06:01:57 am
content: Quote from: Someone on May 19, 2025, 11:29:03 pmAssuming they draw even roughly similar power (i.e., hopefully no outliers which draw a lot more than others), probably simplest procedure is to swap all the batteries when one of them dies - even if it fails to give a warning in advance. Then it's smooth sailing for a long time. Probably less stressful than getting "low battery" warnings every week and replacing them all the time one-by-one.

author: shapirus
author_rank: Super Contributor
timestamp: May 20, 2025, 07:43:57 am
content: Quote from: Someone on May 19, 2025, 11:29:03 pmI think it was mentioned that the devices in question do not survive when powered from 4.2V directly.So it's not only a matter of adjusting the battery low indicator threshold.

author: Someone
author_rank: Super Contributor
timestamp: May 20, 2025, 11:54:54 pm
content: Quote from: shapirus on May 20, 2025, 07:43:57 amThats more of the OPs added constraints from their estimate/rule of thumb, no evidence for it so far. A teardown and reverse engineering of the device in question would likely be productive so the constraints can be clearly defined.

author: shapirus
author_rank: Super Contributor
timestamp: May 21, 2025, 08:34:42 am
content: Either way, both micropower voltage references (including 1.1V I think to avoid adding resistors) and op amps are available, so some of the proposed circuits should be feasible. Simulation works fine too (I tried some on my own). Time to test it in hardware at this point.

author: Zero999
author_rank: Super Contributor
timestamp: May 21, 2025, 12:19:51 pm
content: Quote from: shapirus on May 20, 2025, 07:43:57 amI don't think anyone was suggested powering it directly from 4.2V I certainly wasn't. The idea was to feed it from a LDO regulator and hack the battery monitoring circuit.Quote from: shapirus on May 21, 2025, 08:34:42 amI haven't seen any 1.1V references, micropower or otherwise. If you find one, please post a link to the datasheet and ideally a distributor who sells it.

author: shapirus
author_rank: Super Contributor
timestamp: May 21, 2025, 12:51:22 pm
content: Quote from: Zero999 on May 21, 2025, 12:19:51 pmYeah I may be wrong about this one. Some resistors may be required then, either in the feedback loop or on the input. Now, these resistors are going to waste some power, too. And to make this loss acceptable, they will have to be quite high value, which, I think, means that we want a FET-input single-supply micropower op amp. Not exactly a jellybean part, not sure if such things actually exist. A quick search shows some likeADA4092-4but this specific one requires (typically) 165 μA, which may or may not be too much.ISL28133might work too. They don't say it's FET-input, but it does have a very low input bias current, which, I think, is what we're after (am I wrong on this?).In any case, at this stage it's a task of finding suitable parts.

author: shapirus
author_rank: Super Contributor
timestamp: May 21, 2025, 05:34:44 pm
content: Quote from: Terry Bites on May 21, 2025, 05:09:35 pm...except that it's not exactly micropowerWe're talking tens of microamps of quiescent current at most, I think.This is actually an interesting little problem on its own, regardless if it can be solved by using a different thermostat or whatever. There may be more applications for it, so thinking of ways how to solve it is not pointless at all.

author: SiliconWizard
author_rank: Super Contributor
timestamp: May 21, 2025, 07:32:35 pm
content: Quote from: Siwastaja on May 18, 2025, 06:58:04 pmYeah, "just use" didn't mean "just use any". You still have to do your homework and pick something adapted to the requirements, both in terms of quiescent current/efficiency at the actual current the device draws in various modes of operation.There are ample offering of low-Iq buck converters these days. Still look carefully at the efficiency at various loads, and the actual current profile of the device. If you don't know about it, you're shooting in the dark.One thing that can go in favor of a buck converter rather than linear regulator is precisely the ability (if properly selected) to draw very little current (in the µA range) when the load is very light while still being able to deliver high peak currents (in the hundreds of mA to a few A), something that can be much harder to find with a linear regulator. Finding a linear reg with a Iq of a few µA but able to deliver say up to 1 A without going berserk may not be as easy as it seems.Only by knowing the actual current profile of the device could we properly choose an "optimal" approach though. A linear reg could yield lower average current draw. Can't be determined without more info.Quote from: Siwastaja on May 18, 2025, 06:58:04 pmHave at it, and let us know. I almost guarantee you that it won't reflect the true SoC in any kind of accurate way, except probably when the battery is nearly fully discharged, by which time, you won't care.The MAX fuel gauge ICs I mentioned do not use "coulomb counting" either. They only use cell voltage, you connect them solely in parallel with the battery. They monitor both short-term voltage variations and more long-term voltage and estimate SoC via models of batteries. That's the kind of thing that can take several years to develop to be just remotely accurate. They work fairly well and have been a better option than "manual" handling of SoC in most cases I've witnessed. On Semi have similar ICs that are also pretty good and possibly cheaper (but last I checked, I think they were only available as very small BGAs).Quote from: Siwastaja on May 18, 2025, 06:58:04 pmYes, indeed, if only a "low battery" condition is required to be detected, you don't need accurate SoC. Still, the benefit of the fuel gauge IC I mentioned is that it would give this info, even if it looks overkill for the application, while drawing only about 3 µA and while taking the peak currents into account, without having to directly measure current or otherwise detect if the device you power is suddenly drawing a high current.Quote from: Siwastaja on May 18, 2025, 06:58:04 pmYes, which is what I suggested as well in my earlier post. But I added that it could be a rabbit hole, and the follow-up of posts of people trying to suggest actual implementations increasingly show that. Something that looks simple enough until you actually implement it. Also, a simple threshold will end up leading to oscillations, which won't look pretty, so you'll want some hysteresis, some filtering and at that point maybe you'll be at your third iteration and you'll see that it all increases power consumption.Which is why I thought and suggested that using something like a MAX17048 coupled to a very small and low-power MCU (that could wake up only to inspect the current estimated SoC and determine a threshold) controlling a low-Iq buck converter - or linear reg, depending on what yields lower average consumption in this particular case - (two output voltage levels would be enough given what the OP said) could end up being the solution that would require the least amount of tweaking, even if it looks overkill, could draw less than 10 µA total on average. Would also be a reusable solution. But certainly more complex that a single active component. Not sure it's really worth doing anyway for a one-shot application like this, I feel like we are mostly seeing this as a challenge rather than as something really useful.

author: Zero999
author_rank: Super Contributor
timestamp: May 21, 2025, 08:25:35 pm
content: Quote from: shapirus on May 21, 2025, 12:51:22 pmOr the LT6003 I showed in my schematic?

author: Peabody
author_rank: Super Contributor
timestamp: May 21, 2025, 09:25:04 pm
content: In replies 44 and 45, @mikeselectricstuff and I suggested using an NPN transistor with a diode in the base line to drop the voltage. I haven't seen any response. Did I miss it, or perhaps had someone already suggested that? It just seems that with the gain of the transistor, the current through the diode would be pretty stable, as would the voltage across it. Just wondering why this wouldn't work.

author: Zero999
author_rank: Super Contributor
timestamp: May 21, 2025, 09:34:18 pm
content: Quote from: Peabody on May 21, 2025, 09:25:04 pmTemperature sensitivity might be a problem.

author: wraper
author_rank: Unknown
timestamp: May 21, 2025, 09:34:28 pm
content: Quote from: ballen on May 17, 2025, 04:05:28 pmAmong those which you listed the only real advantage over low self-discharge NiMH is energy density (which you intend to butcher anyway). Fool's errand.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 21, 2025, 09:50:38 pm
content: Quote from: Peabody on May 21, 2025, 09:25:04 pmIt 'works' but has poor temperature performance.You need to add temperature correction, here is an example of a two terminal 1.1V dropper, with adjustable temp correction.You would thermally couple the dual diode and NPN, and isolate from the heat of the output transistor.Two resistors set the temperature null, so you would need to select one vendor / one reel and then find the values for those.It has a finite slope, which gets worse as you lower the bias current, because the reflected base current becomes a significant % change on Q4.Here that is ~ 63mV from 25uA to 1AIt also has a minimum load current, with these values R6,R5 need ~ 21uAMaybe that is good enough for this application ?

author: shapirus
author_rank: Super Contributor
timestamp: May 21, 2025, 10:12:36 pm
content: Quote from: Zero999 on May 21, 2025, 08:25:35 pmYes, this one looks perfect. Almost too good to be true, I'd say.

author: Peabody
author_rank: Super Contributor
timestamp: May 21, 2025, 10:39:30 pm
content: Quote from: PCB.Wiz on May 21, 2025, 09:50:38 pmWell, it's a thermostat. So if it works, there's unlikely to be much ambient variation. But of course current would cause heating. Does current vary a lot?It just seems that it might work well enough to get the job done. If so, it would be two parts, and there would be no idle current.What about doing the same thing with an N-channel mosfet with a threshold voltage of about 1V? No diode.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 22, 2025, 04:12:37 am
content: Quote from: Peabody on May 21, 2025, 10:39:30 pmThe problem is that'about 1V'- the Vth of a MOSFET is poorly defined with large margins MIN - MAXExample typical curve of a low threshold NMOS. The op wants < 100uA to 1A peaks. It will vary 500mV-1V,beforeyou add temperature and part to part variations.

author: Zero999
author_rank: Super Contributor
timestamp: May 22, 2025, 11:05:45 am
content: Quote from: shapirus on May 21, 2025, 10:12:36 pmNot quite perfect, it's a slow op-amp, same with any micropower device. A big low leakage output capacitor will probably be needed to make up for that.Quote from: Terry Bites on May 22, 2025, 10:23:35 amAs in the circuit I posed above, which is the same as yours but has micropower parts and huge value resistors.https://www.eevblog.com/forum/projects/how-to-get-a-constant-voltage-drop-of-1-1v/msg5919696/#msg5919696I wouldn't bother with any of this. If anything, I'd be more tempted to redesign the thermostat, or at least the control board, if I couldn't find one which works directly from a lithium cell.

author: ledtester
author_rank: Super Contributor
timestamp: May 22, 2025, 12:13:38 pm
content: Maybe consider sodium ion battery tech.Nominal voltage is 3V and chatgpt says they work down to 2V.Similar charging procedure but safer - less likely to catch fire.There been around for awhile now... here's a video from a year ago:Na-ion battery test (Seekwatt 3V 1300mAh 18650) -- DiodeGoneWildhttps://youtu.be/L4UBiNhhG6k

author: Terry Bites
author_rank: Super Contributor
timestamp: May 22, 2025, 03:45:10 pm
content: I had been thinking along similar lines to the emitter follower solutions but as much as it looks fine at DC it is likely to become unstable as the load is applied and removed.So I built it up and it didnt work in standby the transistors are essentially off so the opamp is open loop.Adding a 10k bias resistor fixes that.55uA on standby, full volts-1.1 into a 3R load. The ZTX751 got a bit warm but not dangerously so. A small heatsink is necessary.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: May 23, 2025, 12:05:49 am
content: A variant on the 2-terminal design of #82, this one is a bit more esoteric, and uses recently available parts.It uses a low threshold mosfet, so may need supplier and device selection to ensure it can set 1A at < 900mV VGS.PMOS parts seem to come lower Vgs than NMOS and Vishay and Toshiba have suitable parts.To improve loop gain a current mirror load is used, which slightly increases the min load to 30uA.Looks well behaved in spice, with very low dynamic resistance.Possible PMOS