For step-down conversion, use the maximum expected operating voltage for VIN(MAX). If the expected VOUT
operating range (typically from VFBMIN = 2.125V to VFBMAX = 2.75V) includes VIN(MAX)/2, use that value for VOUT.
If the entire operating range is below VIN(MAX)/2, use the value corresponding to VFBMAX = 2.75V. If the entire operating
range is above VIN(MAX)/2, use the value correspondingto VFBMIN = 2.125V.
For step-up conversion, use the maximum output voltage (typically corresponding to pin VFBMAX = 2.75V) for VOUT(MAX).
If the expected VIN operating range includes VOUT(MAX)/2, use that value for VIN.
If the entire input operating range is below VOUT(MAX)/2, use the maximum operating voltage for VIN. If the entire input
operating range is above VOUT(MAX)/2, use the minimum input operating voltage for VIN.
So i would use a 5,6µH inductor with an RMS current of >= 10A and a saturation of 10A*1,25 = 12,5A?Magnetics vendors typically specify inductors with maximum RMS and saturation current ratings.
Select an inductor that has a saturation current rating at or above 1.25 • IMAX, and an RMS rating above IMAX.
In my eyes the datasheet is quite clear.The datasheet is not really clear here and i am a bit confused about the use of IMAX and ILMAX and which one to use in the equations.
= "dynamically adjustable via I_limit_pinwhere IMAX is the programmed inductor current limit.
= peak inductor current limit.ILMAX is
the converter maximum inductor current as programmed by the two sense resistors.
No need to worry. (I ddidn't check your calculations)For me these values seems to be quite a lot to small!? Did i miss something here?
The Eval Board uses a 5,6µH Inductor.
I assume you need to recalculate the values with the above given informations.So i would use a 5,6µH inductor with an RMS current of >= 10A and a saturation of 10A*1,25 = 12,5A?
For me this seems to be completely wrong?
Basically yes.If yes, how to adjust the voltage from an microcontroller?
Maybe a current sink DAC, wich adjusts the feedback node could do the job?
average current here actually mean?"...inductor maximum average current in the DC/DC converter inductor (ILMAX)"
Iin is not known!? - But the datasheet also says:IL ~ Iin*(Vin/Vout)
-> so here i could simply go with the 10A, right?When the converter is stepping down, or operating in buck mode, the inductor current will be roughly
equivalent to the converter output current.
Which is greatest, if the Input is at VinMin and the output is on VoutMax. (Step Up 10V -> 53V)IL ~ Iout*(Vout/Vin)
No need to worry. (I ddidn't check your calculations)
You calculated 2uH minimum value. Now you need to choose an inductor that has at least 2uA worst case.
The inductor value (5,6uH) is the nominal value of the inductor. In reality this nominal value may be smaller due to production tolerance and inductor current..
Thus with 5.6uH you are on the safe side.
We had too low inductance. The formulas on page 23 have an extra 'Imax' term in the denominator, which shouldn't be there.
If you also were using that formula, you might have a too low inductance. Check it out. // Daniel
LMIN = (VOUTx(1–VOUT/VIN(MAX) ) / (fO x deltaIMAX x IMAX))
Basically yes.
You may have a look at this. Just use your DAC instead of PWM (if you like)
Thank you for that link and schematic! This is what i was thinking about.
If i use the full output range from 3V to 53V output with 10 bit i get 50V/1024 ~ 49mV steps.
Is this enough accuarcy for a constant voltage charger?
Above 12 bits the DAC are getting quite expensive, so i would like to avoid this, if this is not really necessary!
Kind Regards
Danie
...
Basically yes.
You may have a look at this. Just use your DAC instead of PWM (if you like)
https://www.edaboard.com/showthread.php?377215-Design-of-PWM-controlled-SMPS-output-voltage
When in shutdown, all charging functions are disabled and input supply current is reduced to 27.5µA.
... RNG/SS voltage can also be manipulated using an active device, such as employing a pull-down transistor to disable charge current or to dynamically servo maximum charging current. Because this pin is internally pulled to ground during fault conditions, active devices with low impedance pull up capability cannot be used.
Your Vin is 10 to 48v
Vout is 3 to 53v
Iout is up to 10A.
At 10Vin, your Iin is 53A max, so this will require paralled converters. I believe the LTC4020 has a transconductance error amplifier so its output can be joined to the output of the other LTC4020 error amp output.
I will look into your other qu's
So does this mean by setting "BAT_FET_DISABLE" to high, the battery FET is completely disconnected? (BGATE pulled high)... RNG/SS voltage can also be manipulated using an active device, such as employing a pull-down transistor to disable charge current or to dynamically servo maximum charging current. Because this pin is internally pulled to ground during fault conditions, active devices with low impedance pull up capability cannot be used.
this question is too general. What in detail do you want to ask?1. Possible that way? I
You try to do this linearely with huge dynamics. Isn't it easier/ more safe to use resistor strings and analog MUX to set for 1, 2, 3 ... cells? I assume you don't have 3.45 cells connected.I am not sure if i understand the battery stack voltage programming part correctly.
For this you need to read the DAC datasheet about allowed pin current.2. Is the series resistance of 8k3 (R306 ... R309) enough to protect the DAC (MCP4728) from overvoltage, while the FBG pin is open (SHDN = low)
We don't know what precision you expect....Or even to high to get enough precision?
You try to do this linearely with huge dynamics. Isn't it easier/ more safe to use resistor strings and analog MUX to set for 1, 2, 3 ... cells? I assume you don't have 3.45 cells connected.
With your circuit, there are crossed out parts. What does that mean? Not decided yet, open, short circuit?
I recommend to focus at one part at a time.
Now I do this at V_FB node.
* Why 100nF from +BATT? I don't see it in the datasheet example circuit.
* You don't care about the FBG_open feature. I recommend to use an analog switch to your DAC to comply with it.
* the DAC output voltage is 0..2.048V...with 8k3 and 126k ( ratio of 15.2) you get a VBatt range of 2.048 x 15.2 = about 31V.
You want it that huge?
Maximum pin current is +/-25mA. Worst case current (FBG open) is while DAC voltage is 0V. This will definitely result in a quite much lower current. (53V/126k+8k3 = ~0,4mA)For this you need to read the DAC datasheet about allowed pin current.
Precision should be high enough to guarantee safe operation in all states.We don't know what precision you expect....
CSOut: What waveform do you expect there?
In time I'm a bit busy, so I'll be back in more than 12h.
Makes no sense.DAC output voltage is 3.3V (Vref=VDD)
Do you expect us to calculate the value(s)?Precision should be high enough to guarantee safe operation in all states.
The battery current ... is it caused by a linear regulator, or ist it caused by a switching regulator...with high switching freqzency ... and high frequency currenr ripple?CSOUT is just an analog voltage representing the battery charge current -> VCSOUT = 0.25 + 20 • (VCSP – VCSN)
VCSP-VCSN is the voltage across the battery charge current sense resistor.
Makes no sense.
Before you talked about 0.1% resistors. This is initial accuracy ... and it has even less drift tgan thise 0.1%.
Then you talk about using VDD as VRef. VDD maybe with an initial accuracy of 5% and drift caused by input voltage, load, temperature and time....all this errors will be 1:1 transferred to the output...
Your system's overall precision will not be better than VDD precision.
Its caused by a switching regulator - My fault :laugh:The battery current ... is it caused by a linear regulator, or ist it caused by a switching regulator...with high switching freqzency ... and high frequency currenr ripple?
I know - but there was just not enough space left :bang:First: standard signal flow at a schematic schould be from left to right.
The first Opamp (right side in your schematic) gets high frequency input. But the Opamp may be too slow to amplify it correctly, thus the output will be distorted and unreliable. The error will not be calculable.To avoid this you should place an RC LPF in front to the Opamp.
That was the reason why i placed the buffer right after the CSOUT pin.Current Sense Amplifier Output and Charge Current Monitor. Connect 100pF capacitor to ground. Pin output impedance is 100kΩ, so any loading for monitors must be high impedance.
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