Gigillo74
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The main X9C102 issue in this circuit is max voltage between pin 3-6 that exceded max value. to reduce this voltage and make able pot to feedback on L200 i think a resistor need to be added in series of pin 3.Hi,
I might be wrong, an op amp buffer between L200 Out and the digi-trim-pot may prove helpful, or is the X9C102 high impedance input and enough to not steal current from L200 Out signal?
Ok, thank you.Again, I may be wrong but the suggestion of a resistor, and only there to burn up heat on a sensitive node, causes me marginal distress. Maybe it's fine. An op amp high impedance input buffer that attenuates instead of amplifies is also do-able there.
Yes, it is.* L200 is the (voltage and/or current?) regulator in a charger circuit which has two selectable current limit preset options (15R or 3.3R).
Yes, i would to do it but not dinamically. Only set current charge by dig pot.* L200 will be externally set via code (R/W, etc.) by you at xyz wiper voltage to be able to dynamically select L200 output current limit.
Yes also for this. It is not necessary to force any current. I need only a voltage from pin 5 to shunt resistor to feedback into pin 2 to regulate constant current. The idea is to place a parallel dig pot instead a manual pot (just working with this solution) to regulate charge current. Output voltage is not defined. It adjust itself automatically in funcion of type of battery is connected.* The X9C102 isn't going to use 10mA from the L200 output line at all, in fact it's probable that it will only use a minimal amount of current because it is only a feedback sensing device. It can handle a maximum of +-4mA at either input (H or L), but you have no reason/need to force +-8mA through it for your purposes (much less 10mA). I only read the datasheet quickly but don't understand the need for maximum current at pin 6 (Low) e.g. 1.8k resistor (8V/1.8k = 4.4mA) and no current limiting at pin 3 (High).
The 1,8K resistor is used as voltage divider to make dig pot max voltage low to 5v.* L200 pin 6 (Low) - after the 1.8k resistor is that going to see 0V continuously or what voltage do you expect there?
Pin 3 must have some resistor as pin 6 to complete the voltage divider. Only forgot to put it in sch.* Why does the X9C102 pin 3 (High) need to be powered from the L200 output? Why is pin 6 (Low) not connected to ground? I assume both questions are answered by saying that they are a part of the dynamic feedback set-up for this circuit for it to function correctly, but I am too stupid and lazy to try to undestand what's going on there.
No, any negative voltage. Power supply is 12V* Does your circuit actually use a negative voltage or current anywhere at all? From the schematic I can only see +5V and GND. There is no -5V, or -8V anywhere, neither a power supply nor a reverse signal that could generate a 'negative' value (such as a bi-directional battery charger circuit that has charger current in and load current out sampled on the same path).
If i know how to insert it in the circuit, the reply is yes.* Do you think adding an attenuating op amp buffer between L200 Out and digipot High to subtract those excess 3V is useful or is just adding an unnecessary additional IC where Rdiv alone is actually enough of a solution?
3k & 5k are most interesting solutionAnyway, to save you laborious time answering so many (pointless, and clueless) questions:
Now that I think I understand what I think I understand about your circuit, and I doubt losing a few uA or even mA here or there along the charger out path matters because the charging current (8V/15R = 0.533A; 8V/3.3R = 2.424A) is large in comparison to the maximum digipot current, maybe you are right and a simple resistive divider is more than acceptable - so long as it doesn't distort the feedback to the L200.
And looking at the L200 output impedances you have (15R or 3.3R), even RdivTotal = ~800R for ~10mA + digipot current, guessing that the load to be charged must also be pretty low impedance as an input given the 0.5A or 2.4A, then the 8V>5V divider could be 800R (0.625 x 800R = 500R) at e.g. 500R x 0.01A^2 = 50mW. Rtop = 300, Rbottom = 500. 30mW and 50mW, not so bad, surely? Using 3k and 5k would mean 1mA through the divider and 3mW and 5mW on each resistor. I haven't considered the implications of placing 500R fixed resistor to ground directly in parallel with a variable 1k to ground, though, might get some interesting voltages there that are not 5V precisely and again distort/warp/skew the feedback signal.
Thank you for this sample. I've reply to all question in previous post. If you confirm it as good solution, i'll try to check the ckt building it.Hi again,
You really have +-8V, do you? I ask as I realised (one of) the 'flaw(s)' in my suggestion, which is that a non-inverting amplifier is always Vo = R2/R1 +1, so as that is unworkable mathematically, it only really leaves my toolbox with: single supply solution [buffer + rdiv + buffer] or dual supply solution [inverting amplifier with its helpful Vo = -R2/R1 + another inverting amplifier to turn negative back to positive...]. If you don't already have a negative supply, then the inverting amplifier solution is (maybe necessary but) unappealing just because it means more parts and effort and PCB space.
Here is a quick sim of the inverting amplifier version. It's clunky/would take up more space than you might be happy with, compared to a resistive divider alone. The resistive divider version simulation was unconvincing for me. So, again, for a resistor to drop a few mW dissipating e.g. 3V at what I still assume is negligible current into the digipot High pin, even 10mA across a 3k resistor would only be 0.3W.
Sim results just meant to give idea of what I thought might work as a solution, and as orientative because e.g. L200 is an IC, not a 0 Ohm impedance voltage source, and e.g. Vout OA at 5.001V is amusingly too-perfectly accurate for reality...
View attachment 178915
Hi,If you confirm it as good solution, i'll try to check the ckt building it.
Perfect, i'll do it. I will update you.I'd try the simple 3k-5k divider first, if changing the digipot resistance value alters the 5V coming from the 3k-5k divider, then a buffered approach as shown above might be next-best option.
Hi,Hi,
Thanks for all the information you provided.
Caveat emptor, please, it's only a possible solution - but you still need to check it first as valid, either by doing a little bit of maths and then a few simulations under different circuit conditions (and assessing the credibility of the simulated results) and/or a little bit of maths and then breadboarding it before soldering anything.
I'd try the simple 3k-5k divider first, if changing the digipot resistance value alters the 5V coming from the 3k-5k divider, then a buffered approach as shown above might be next-best option.
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