Tracking regulator for battery chemistry replacement? - Page 1
Forum section: electronics / projects

author: bitbanger
author_rank: Regular Contributor
timestamp: December 24, 2017, 04:40:33 pm
content: Hi everyone - looking for some basic input on output tracking based on input voltage.I'm designing a 2x18650 cartridge to replace a series of four NiMH batteries in a device. My goal is a low quiescent current regulator which linearly tracks the 2x18650 series input voltage to "mimick" the NiMH battery span. In this way the original battery meter on the device can still function. For cell protection I plan on using undervoltage lockout circuits for each cell, but may decide to just monitor the series Li-iON voltage and feed this into a regulator EN/SHUTDOWN pin(the lockout circuit however only consumes 4-5uA at shutdown so it's possibly more desirable in that regard).Anyway what's the best approach to this? My first knee-jerk thought was to simply reference the LDOs ADJ pin to the input side, but I don't think this would work (any internal adjustment to the pass transistor conduction won't be "seen", and so this would likely either rail or turn off completely?). Thoughts?Thanks folks.JamiePS the undervoltage lockout is here:http://cds.linear.com/docs/en/lt-journal/LT1389_0699_Mag.pdf

author: David Hess
author_rank: Super Contributor
timestamp: December 24, 2017, 07:33:07 pm
content: In the past I have seen a similar battery replacement which used a low quiescent current inverter to provide a fixed ratio between input and output voltage with little loss. Making the output of a linear regulator track its input instead of its reference can be done with an operational amplifier or in some cases by driving its feedback pin from the input voltage through a voltage divider.

author: bitbanger
author_rank: Regular Contributor
timestamp: December 26, 2017, 05:38:30 am
content: Thanks for the response. Unfortunately I've discovered that the battery fuel meter in the device doesn't behave as suspected - the battery "depleted" and "low batt shutdown" functions work as expected, but the gauge itself doesn't. So, there's less concern about linearity, just hitting the depleted and shutdown points. (This is a personal project by the way.)This is the general mapping of input/output voltages I'm shooting for (sum of cells):6.2V -> 4.3V6.6V -> 4.7V...8.4V -> 6.5VAs you can see simply dropping 1.9V in/out would functionally work. Unfortunately this means over a watt of wasted power (max current is nearly 600mA). Also a restriction is that the Vout can't exceed 7.5 in the case that a charger is connected: the device provides 7.5Vmax charge voltage at the VBATT terminal and so if the LiION exceeds this it will backfeed the charging circuit.Now that I'm typing this all out, even at 7.4Vout, with 8.4Vin (new batteries), power dissipation is over 600mW......... looks like I need to consider a switcher afterall.Best,Jamie

author: SeanB
author_rank: Super Contributor
timestamp: December 26, 2017, 07:53:13 am
content: Single cell with a boost converter and a micropower microcontroller to give you the ability to have an adjustable setpoint will work. Just use a common boost converter to 5V and then use the micro in sleep mode with a slow clock to sample input cell voltage every so often so that as the cell reaches say 25% of capacity it drops the output to low battery level, then at 10% it turns off the output to simulate flat without damaging the lithium cell. The cell protection circuit will stop discharge at a point well after that, but the boost converter will still allow unregulated cell voltage through, less the Schottky diode drop in the converter, till that point, so will keep any RAM alive in the device till the protection drops out the output. Makes charging easier as well, you can use a slightly modified charge circuit and a common 1 cell charger, available off the shelf to add to the cell.Otherwise just go buy a USB power bank and use the board off it, it has every part aside from the voltage variation inside a single small package, and will work. the boards are also available cheap on fleabay, with either USB ports input and output, or with only the board and no connectors for integration into your own equipment, just add a protected cell to it or a non protected cell and a ultra cheap cell protection board, which is also available in 10 packs.

author: ajb
author_rank: Super Contributor
timestamp: December 26, 2017, 03:40:31 pm
content: Some switchers have a separate input to adjust the output voltage, but that's not a common feature. You can adjust the output of any switcher, though, by feeding a control voltage into the feedback divider via a third resistor. This control voltage though, because it goes into the inverting side of the switcher's error amp, so a higher voltage will decrease the output and vice versa. You could make this work with an op-amp inverting and offsetting the battery voltage, or you could use an MCU. There are a couple of 8-pin PICs with DACs that you could use to produce any input->output mapping you wanted, or you could use a digital output or two to switch between two or four discrete settings. I suspect the MCU might be the lower power option, but there may be lower power op-amps that could work.

author: bitbanger
author_rank: Regular Contributor
timestamp: December 26, 2017, 06:50:33 pm
content: The microcontroller option has certainly crossed my mind - the in/out voltage correlation would be much more straightforward! Not to mention it's more inside my comfort zone. However the added complexity nags me. Especially when I feel like I could achieve this with a few opamps (if I were more of an opamp wizard). If this were a one or two feature more complex device I would go for it. Really I just need to work out an opamp circuit to scale/offset, and another as a "limiter" (to effectively limit the regulator output to <7.5V)abj you're spot on with what I've been consider: I've been approaching this with a three terminal regulator in mind, which typically have an ADJ current for the bandgap reference to deal with. However when looking at modern switchers it's simply an error amp input (FB).Considering the LT3621:http://www.linear.com/product/LTC3621I do get a signal back from the device once it's powered on, so the burst mode could be used when not in use; handy.

author: ConKbot
author_rank: Super Contributor
timestamp: December 26, 2017, 08:28:28 pm
content: Put the two 18650s in series, use a buck converter, then hang 2 caps and 2 diodes off the switching node as a negative output charge pump. (http://www.nutsvolts.com/uploads/wygwam/NV_0198_Marston_Figure23.jpgnevermind the ICL7660, but pin2 is a push-pull output like a buck converter with synchronous rectifier would be)This will give you a ~ -(vbat-1v) rail you can reference the feedback divider to. Higher input voltage, lower negative rail, higher output voltage to hold the feedback node in equilibrium at the set point. Resistor values are left as an exercise to the reader () A clamping diode to keep the FB pin from going negative on startup may be a good idea too.

author: ajb
author_rank: Super Contributor
timestamp: December 27, 2017, 05:01:12 am
content: Quote from: bitbanger on December 26, 2017, 06:50:33 pmRight, so you feed a control voltage into the FB node via a third resistor, as in the attached schematic. The control voltage must be inverted (higher control voltage gives a lower output voltage).Forum member Tim Williams has a handy calculator for the resistor values:http://www.seventransistorlabs.com/Calc/ResDiv.html("Biased Divider", the last one).

author: bitbanger
author_rank: Regular Contributor
timestamp: December 28, 2017, 11:33:22 pm
content: That's a mighty handy calculator - thanks for the link!I'm going to have a quick proto spun off and I'll update when it's in/how it turns out.

author: bitbanger
author_rank: Regular Contributor
timestamp: December 30, 2017, 02:10:16 am
content: Say I choose typical 2x programming resistors for the FB such that the switcher provides 7.4V output: 100k/8.82k for R1/R2. If I then want 6.24V out, that means the feedback node would see 0.347V using these same resistors alone. So I need to inject enough current through the 3rd bias resistor (and resistor to ground) to artificially raise this feedback point back to 0.6V (essentially adding 253mV). If that's the case, great, got it.What I'm spinning my wheels on is deciding on or choosing the bias voltage range (or resistor value)? Seems there's no real restriction here: I can use an inverting opamp with DC offset to provide essentially any (inversely) scaled bias. I suppose in this case then, simplicity rules (for example if I can avoid using a fixed DC offset for the inverting amp, that'd be best).Thoughts?Happy Holidays!

author: ajb
author_rank: Super Contributor
timestamp: December 30, 2017, 04:21:54 am
content: Quote from: bitbanger on December 30, 2017, 02:10:16 amYou want to consider the FB voltage to be fixed and calculate your resistors such that with bias voltage A you get 6.5V output and with bias voltage B you get 4.3V output (or whatever). It's negative feedback, so just like analyzing an op amp circuit. On second look, Tim's calculator doesn't give you that configuration as an option, unfortunately.The simplest solution would be to use a zener to establish a reference point somewhere in the middle of your battery voltage and use that as the 'ground' reference of an inverting amp. However the zener will consume some current, so you'd have to check if you could get a sufficiently stable enough reference at an acceptably low current.

author: David Hess
author_rank: Super Contributor
timestamp: December 30, 2017, 05:26:24 am
content: Quote from: ajb on December 30, 2017, 04:21:54 amMost switching regulators with external feedback use a reference to ground unlike a floating regulator where the reference is from the output voltage. If the feedback pin is stuck at 0.6 volts as with the LT3621, why not use that as the reference? I know it seems incestuous since you will be driving a variable current back into that pin but it could for instance be decoupled from the non-inverting input of an operational amplifier with an RC filter for isolation if AC stability would otherwise be compromised.Another way I considered is using a two transistor current mirror to invert the signal from the input to the adjustment pin. Emitter degeneration will be necessary unless matched transistors are used but that still leaves a collector-emitter voltage of 0.3 volts across the output transistor to drive the 0.6 volt adjustment pin of an LT3621 which is plenty. BC548s might be the best for this but 2N3904s should work also.

author: bitbanger
author_rank: Regular Contributor
timestamp: January 09, 2018, 01:27:21 am
content: Hi Everyone -I'm still tinkering with/simulating a continuously-adjusting feedback, but I took a quick lazy stab at a discrete approach:Don't think I need to bother with hysteresis/schmitt because I wouldn't expect the battery to recover after switching to an output resulting in a higher Vdrop (if anything it would sag).