I want to do some experimentation with a 60v charger connected to 17x Lifepo4 cells in series.

The idea is the 2 resistor voltage divider will be connected across Battery Bank.
The voltage between resistors is Bank Voltage divided by number of cells.

Shown are (16KΩ + 1KΩ) resistor values for 17 cell Bank
So the voltage between resistors is the equivalent of a single cells charge voltage.
The 1 17th voltage will follow voltage changes across the bank.

What would be the best IC or circuit for at least 1/1.5 Amp output.

 

LM675 can deliver 3A. This is intended for power amplifier applications.
For charger applications, a simple pass power transistor is all you would need.

 

LM675 can deliver 3A. This is intended for power amplifier applications.
For charger applications, a simple pass power transistor is all you would need.

That might do the job.
The voltage will be at the maximum specified.
Do you know if this IC would be a drop in for above circuit?

Thanks

 

I want to do some experimentation with a 60v charger connected to 17x Lifepo4 cells in series.

The idea is the 2 resistor voltage divider will be connected across Battery Bank.
The voltage between resistors is Bank Voltage divided by number of cells.

Shown are (16KΩ + 1KΩ) resistor values for 17 cell Bank
So the voltage between resistors is the equivalent of a single cells charge voltage.
The 1 17th voltage will follow voltage changes across the bank.

What would be the best IC or circuit for at least 1/1.5 Amp output.

Your favorite op amp with an NPN Darlington on the output?

 

Your favorite op amp with an NPN Darlington on the output?

Would the output voltage mirror the input, I am not up with the addition of Darlington Driver.

 

Yes, all you need is an opamp and a power transistor.

 

Yes, all you need is an opamp and a power transistor.

View attachment 123332

The 1 17th divider is best left fixed.
So volts in = volts out when R1 and R2 are equal?
It could be advantageous to make adjustment of a few Milli Volt, would you adjust R1 or R2?

To elaborate further on part of experiment:
Diagram showing 2 of 17 Lifepo4 Cells between DPST relays.

Important: The voltage divider is just to show logic without the Voltage Follower.

A 555 is connected to counter connecting 1 of 17 relays for a few minutes each in a cycle.
The 1 17th bank voltage is connected across each cell one after the other in sequence.
The cell is either higher, lower or equal to the 1 17th bank voltage.

Add: The 17 cells will be connected in series to bank charger whilst this is happening.

This information might introduce a few gremlins?

 

Re: So volts in = volts out when R1 and R2 are equal?
Please ignore that question, not a good day for it.

Thanks for the input.

 

Yes, all you need is an opamp and a power transistor.

The voltage feeding the follower will be 2.9v to 3.529v.
The "Transistor?" would charge a Capacitor.
I envisage a diode would be need between Transistor and Capacitor for protection?
If so, is there a mathematical way to pre-compensate for the voltage drop across Diode, not shown below?

50F - 2x 100F Maxwell "BCAP0100 T01" Capacitors in Series:
http://www.mouser.com/ds/2/257/Maxwell_HCSeries_DS_1013793-9-341195.pdf
Rated Current 5A
Max. Current 18A
Max. ESR 30 mΩ

 

The voltage feeding the follower will be 2.9v to 3.529v.
The "Transistor?" would charge a Capacitor.
I envisage a diode would be need between Transistor and Capacitor for protection?
If so, is there a mathematical way to pre-compensate for the voltage drop across Diode, not shown below?

50F - 2x 100F Maxwell "BCAP0100 T01" Capacitors in Series:
http://www.mouser.com/ds/2/257/Maxwell_HCSeries_DS_1013793-9-341195.pdf
Rated Current 5A
Max. Current 18A
Max. ESR 30 mΩ

Why not buy a 100w audio amplifier. Change the input to your voltage divider (clip out the input capacitor(s)).

 

Why not buy a 100w audio amplifier. Change the input to your voltage divider (clip out the input capacitor(s)).

Q1/Why not buy a 100w audio amplifier.
Because I do not yet know if your suggestion will work for application!
What cost are we talking?

S2/Change the input to your voltage divider.
I would like to leave the baseline fixed.
This allows for reasonably know increments in fine tuning!
Ideally the adjustments would be made to follower gain circuit.

S3/clip out the input capacitor(s)
I suspect you are referring to a particular wave form.

I am very interested to know more, particularly in a drawing I can follow.
Does your idea act as accurate voltage follower or some thing you set.
Please understand that what I am doing is very unique, not standard practice.

Still like to hear from others about the automatic follower adjustment for diode voltage drop.
This is the hard way, but I do not want to be playing with these variables whilst testing the idea.

Cheers,
Gary

 

Most 100W audio amplifiers made any time recently (in my experience) would desperately try to servo out that DC offset, and probably self-destruct in the process.

Gazza AU, you don't really need a complementary buffer because your output is single-sided. One of your two driver transistors in the schematic you posted is just going to be twiddling its thumbs while the other does all the work. The original schematic posted by MrChips seems like it would be fine for your purposes...if you don't need any gain, just leave the resistors R1 and R2 out altogether and run the op amp as a unity gain buffer with the transistor inside its feedback loop. It *would* be a good idea to use a Darlington or other higher impedance device to keep too much current from being pulled out of the op amp...you don't need to change anything, because the op amp's feedback will automatically compensate for whatever the voltage drop of the device is...if you wanted to include the diode you mentioned, you could just put it inside the feedback loop, too, and the op amp will compensate for its drop.

 

The original schematic posted by MrChips seems like it would be fine for your purposes...if you don't need any gain, just leave the resistors R1 and R2 out altogether and run the op amp as a unity gain buffer with the transistor inside its feedback loop. It *would* be a good idea to use a Darlington or other higher impedance device to keep too much current from being pulled out of the op amp...you don't need to change anything, because the op amp's feedback will automatically compensate for whatever the voltage drop of the device is...if you wanted to include the diode you mentioned, you could just put it inside the feedback loop, too, and the op amp will compensate for its drop.

Thanks, can I borrow your time a bit further.

The Voltage follower and connected capacitor are one voltage, just to simplify.
Then there is the Charger and connected cells series voltage.
Interaction only occurs between cell and capacitor from this basic position.

The "voltage follower"/ "Capacitor charge transistor" will see lower and higher external voltages,
by design it only needs to work when higher then the external voltage.

What darlington driver would you suggest, and would the diode be still needed?
I cannot figure how the diode in feedback loop would work, hopefully its unneeded.

Cheers
Gary

 

Thanks, can I borrow your time a bit further.


What darlington driver would you suggest, and would the diode be still needed?
I cannot figure how the diode in feedback loop would work, hopefully its unneeded.

Cheers
Gary

Gazza, you only need a single op amp and transistor like post#6, or Darlington pair for higher current, you will get the same results with either, the Darlington will give upto 10amps, a Tip121, Tip126, 2n3055, Mje3055, any will give you 1amp.

 

Gary,

Re-reading your original posts, I'm becoming a little concerned about whether or not we're following your intended application. Are you wanting to charge your cells with this circuit or to buffer the voltage coming from the cells? Lithium cells need to be charged a specific way...they need constant current up to a certain point and constant voltage beyond that point. Consider this thread: https://forum.allaboutcircuits.com/threads/lifepo4-simple-charger-circuit.53192/

Following your stated specification of just needing a voltage follower capable of delivering an amp, DodgyDave's circuit will do exactly that. It's not clear to me what the purpose of the capacitor at the output of the circuit you posted is...what is this circuit ultimately intended to drive?

 

Gary,

Re-reading your original posts, I'm becoming a little concerned about whether or not we're following your intended application. Are you wanting to charge your cells with this circuit or to buffer the voltage coming from the cells? Lithium cells need to be charged a specific way...they need constant current up to a certain point and constant voltage beyond that point. Consider this thread: https://forum.allaboutcircuits.com/threads/lifepo4-simple-charger-circuit.53192/

Following your stated specification of just needing a voltage follower capable of delivering an amp, DodgyDave's circuit will do exactly that. It's not clear to me what the purpose of the capacitor at the output of the circuit you posted is...what is this circuit ultimately intended to drive?

I am good with Lifepo4 characteristics and fundamentals for SOC.
The question is correct, the divider voltage charges the capacitor to Time Constant.
Best not to worry about what it is meant to do.
I am systematic trying to take in a lot of different stuff!

Next I will need to look at how capacitors voltage sag, perhaps look at this again.

If I keep the baseline, there is always an original reference.

 

Hello, I have read your post, I have several questions:

1. Your application is not clear. What do you intend to do? Charge your battery with your follower? Measure its voltage to track its charging state? If you want to charge your battery with this schematics, that means that the more discharged the battery is, the less current you will inject (which could be a good idea since what To3metalcan explained is correct, you will want to charge with current control at the begining and with voltage control at the end) but still this is the best idea. If you just want to measure the voltage, then you don't need 1A output OPA (but I m pretty sure that is not want you want...)
2. The idea proposed here are great, take an OPA with small output current and then make it control a BJT power stage. BUT, remember that BJT have 0.6V Vbe dropout, and Darlington has 2*Vbe dropout=1.2V which can make this stage not work if the OPA output signal is not connected through a serie capacitor (High pass filter) and your Darlington base not correctly biased, so you may want to use a Szilai connection to partially solve it
   
   This has the advantage of reducing the Vbe drop out to only 1Vbe. You have work to do in chosing the base resistor and emitter resistor, and designing the highpass filter + base voltage biasing, but this is doable.

Good luck

 

The idea proposed here are great, take an OPA with small output current and then make it control a BJT power stage. BUT, remember that BJT have 0.6V Vbe dropout, and Darlington has 2*Vbe dropout=1.2V which can make this stage not work if the OPA output signal is not connected through a serie capacitor (High pass filter) and your Darlington base not correctly biased, so you may want to use a Szilai connection to partially

So you are saying the BJT stops switching with voltage source below .6V and Darlington at 1.2v?
If the voltage drops this low, does the device not behave or just not switch on

Note the 2.9v to 3.529v operating range given above!

The Capacitor could theoretically be rated and charged at 50Amps,
But ultimately it comes down to the cost
600 Farad Capacitor XV3585-2R7607-R
2.6 ESR mΩ
33 Amp current rating
320Amp peak

AUD $75 each, makes the decision easier!
So it looks like a mater of shopping around for the best deal.

Lets say I find some affordable capacitors with 30Amp rating!
Two in series would be 15Amp.

So I am looking in the range of 10 to 15 Amps capacity

 

SunsetB, so far as I can tell his entire output is going to be DC, so a coupling capacitor would reduce it to zero. Also, while I love me some Sziklai (seriously, so overlooked in favor of the often-inferior Darlington) the Vbe of the device isn't going to matter...the op amp will just increase its output voltage to compensate for it because it's INSIDE the negative feedback loop.

EDIT: to emphasize this point, so long as the op amp has a high enough power supply to do it, you can ignore any worries about the turn-on voltage of the device, because it is going to compensate for it automatically using the schematic we've been discussing.

GazzaAU, so is the capacitor's charge time important in your application? Because right now it's going to be "very, very fast" regardless of the input. Since your signal is DC and the only resistance is the output resistance of the transistor(s), you will pretty much instantly have the output voltage across the capacitor with no "sag" at all! Your mention of those terms makes me wonder if you're planning to use the capacitor to sample-and-hold the voltage (in fact, noticing that your other post included mention of relays that seem intended to completely isolate the batteries, I feel pretty sure this is what you're up to.) If you are, then your earlier mention of a diode becomes relevant, because the emitter of the transistor and the (-) input of the op amp will present a quite low impedance, and the capacitor will drain quickly between charges (as soon as there's no voltage at the op amp input there will be no voltage across the cap.) A series diode between the transistor and the capacitor would be necessary, then, since it'll allow charging in one direction but stop discharging in the other.

You mentioned compensating for the diode...the quick and dirty way would be to do something like the attached. The diode in the feedback loop raises the output voltage by one drop, and then you lose that drop across the output diode. How precise it is would depend on how similar the two diodes are, but it'll get you pretty darn close. However, this idea is predicated on the input voltage being continuously there. You can't leave the input "floating" with this design because the diode will break the negative feedback loop when the voltage goes too low to turn it on and the op amp will do...something. Probably not something we want it to do. :-o So if the batteries are not connected to the non-inverting input, it needs to be tied to ground or some other reference.

If you do indeed need to hold the cap voltage, the next part of the circuit needs to be something pretty high-impedance, like a FET-input op amp.

Edit: after reading your subsequent post, it sounds like you're planning to use the capacitor for a sudden, high-current discharge. If that's the case, your application is very mysterious. You're going to need a power supply that can charge the cap via the op amp and transistor (therefore one that's somewhat higher than the total voltage of the LiFePo cells) so either a) the batteries don't need to be there, because you can derive your reference voltage from the power supply, and it won't drain over time or b) nothing *but* the batteries needs to be there because you can charge the cap directly from them, no voltage follower needed, disconnect them via the relays and be ready to go. So...???

(P.S. apologies for the sharpie schematic!! I'm about to go to bed and not booting up the computer with the CAD software on it again at this point.)

 

SunsetB, so far as I can tell his entire output is going to be DC, so a coupling capacitor would reduce it to zero. Also, while I love me some Sziklai (seriously, so overlooked in favor of the often-inferior Darlington) the Vbe of the device isn't going to matter...the op amp will just increase its output voltage to compensate for it because it's INSIDE the negative feedback loop.

EDIT: to emphasize this point, so long as the op amp has a high enough power supply to do it, you can ignore any worries about the turn-on voltage of the device, because it is going to compensate for it automatically using the schematic we've been discussing.

GazzaAU, so is the capacitor's charge time important in your application? Because right now it's going to be "very, very fast" regardless of the input. Since your signal is DC and the only resistance is the output resistance of the transistor(s), you will pretty much instantly have the output voltage across the capacitor with no "sag" at all! Your mention of those terms makes me wonder if you're planning to use the capacitor to sample-and-hold the voltage (in fact, noticing that your other post included mention of relays that seem intended to completely isolate the batteries, I feel pretty sure this is what you're up to.) If you are, then your earlier mention of a diode becomes relevant, because the emitter of the transistor and the (-) input of the op amp will present a quite low impedance, and the capacitor will drain quickly between charges (as soon as there's no voltage at the op amp input there will be no voltage across the cap.) A series diode between the transistor and the capacitor would be necessary, then, since it'll allow charging in one direction but stop discharging in the other.

You mentioned compensating for the diode...the quick and dirty way would be to do something like the attached. The diode in the feedback loop raises the output voltage by one drop, and then you lose that drop across the output diode. How precise it is would depend on how similar the two diodes are, but it'll get you pretty darn close. However, this idea is predicated on the input voltage being continuously there. You can't leave the input "floating" with this design because the diode will break the negative feedback loop when the voltage goes too low to turn it on and the op amp will do...something. Probably not something we want it to do. :-o So if the batteries are not connected to the non-inverting input, it needs to be tied to ground or some other reference.

If you do indeed need to hold the cap voltage, the next part of the circuit needs to be something pretty high-impedance, like a FET-input op amp.

(P.S. apologies for the sharpie schematic!! I'm about to go to bed and not booting up the computer with the CAD software on it again at this point.)

You are spot on as far as I can tell. Take me time to go over the reply fully.

I thought up the idea of using second diode but didn't know if it would work, that is excellent.
What would you suggest in diodes, I am not up with their characteristics and could save some time.

I will need to think about the logic conditions of inputs as you mention.
The Capacitor needs to charge fast, then it is paralleled with a cell and losses any voltage above it.
The capacitor is only held at voltage before being paralleled with a cell