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First time here so i did not find an appropriate category for my Title.
I am trying charge a bank of capacitors BCAP0360 in a 3s2p configuration. The Buck converter U1 (AP3211) sets the output voltage at 8.1v and a bypass N-MOSFET Q1 (SQP60N06) to charge at a higher current.
Questions:
I am just not sure if i have Q1 connected the correct way to function as a bypass transistor for higher current about 50A (Max).
Is the inductor necessary here ? Would the inductor be able to handle that rated current ?
The AP3211 incorporates a measurement of the current passing through it to achieve regulation; use of an external transistor in attempt to increase output current is probably not a good idea for this reason.
Yes, an inductor typically is necessary for most switch mode supplies. If the desired output current is 50A, components in the direct path of current flow should probably be rated for that; this does not appear to be the case based on the component selections in the diagram given.
Thank you rick ! After couple of time since my last post, Would this be efficient. The charge current being 10A i’ve set it to be be PFM. Is that correct ? The inductor was based on the results from RedExpert
There are several problems with this circuit, the most basic of which is that this part will see a supercap as a short circuit if it is not pre-charged, and shut down due to its overcurrent protection circuitry. It may eventually get enough charge into the capacitor bank to begin normal operation, but it would take a very long time.
You should be using an IC designed for charging supercaps, such as the BQ24640RVAT. It will charge at a constant current rate up to a specified voltage and then switch to constant voltage (correction) mode until the caps are fully charged.
We even carry a development board for it to accelerate your design time: BQ24640EVM
Hi David,
There are a couple of things i am trying to figure out,
I have the Capacitors Input Ripple current rating at 4.7uf 35v but not sure how the Iout ripple is calculated. On page 18 the equation shows a symbol before .29 what is it ?
The input capacitor, based on my calculation should be rated at 4.7uf 35v based on my application, should they be placed near the Vcc or the diode ? because diodes generally have a voltage drop. Yet
In several places on your schematic, you have 0.01uF caps rather than 0.1uF caps. Make sure you follow the datasheet recommendations for the values of all components in their example circuit and only change values from the given examples when necessary to get the output voltage and current you need.
Likewise, since your output voltage is the same as that given in the example schematic, you should use the same values for R2 and R1, which are 300K and 105K Ohms, respectively.
Regarding question on symbol just before “0.29” in equation for Icout on page 18, it means “approximately equal to”, so, not exactly, but close enough for the application.
Your values for R5 and R6 are way off. You want Viset to be about 2V to get a 10A output with a 0.01Ohm Rsense resistor. That means the ratio of R5 to R6 should be 0.65. In their example circuit, the sum of the two resistors is about 122K Ohms, so yours should be of roughly the same magnitude. Using values of 64.9K and 100K for R5 and R6, respectively, will give a Viset value of 2V and have similar magnitude total resistance as their example.
Regarding your question about UVLO, it is not adjustable, as it is internally set. If your concern is that it not start charging if the Vcc voltage is less than the supercap voltage, that is not an issue because it will remain in sleep mode if Vcc is less than VSRN (the supercap voltage). If you have another reason for it to not start below 12V, then you could use an external voltage sensing circuit and use the output of that to control the CE (charge enable) pin.
Regarding input capacitor, it should be placed near Vcc. The diode is not on the same voltage node, so they have no relation to each other.
What is your input voltage range for this system? That will determine the ripple current calculations.
So for a 10A charge current how did you determine the values. I just followed the equation in section 7.3.1 on page 12. I mean there are several ways to get the iSet voltage as 2. Moreover the datasheet did not say to be around those limits. Is it default thing to follow from a datasheet. With that, the product of 64.9K and 100K is 1.82 resulting in an amp less of charge current, 9.1A. Again a 130K would precisely set iSet to 2v.
in the Ripple current calculation, using their variables i get a ripple current of 0.05. is that correct ?
There is usually a reason for the values selected by the vendors in their reference designs. They have experience with their chips and know how to get the best results. In the case of setting the Vout value, having the resistors in the >100K range is probably related to the stability of the circuit. In the case of setting Viset, using higher values reduces the load on the Vref pin, at a minimum.
The datasheet stated that Viset should be no greater than 2V (page 6 and 12). Viset is set by the voltage divider circuit using Vref, R5, and R6. So, in your circuit:
Viset = Vref x {R6/(R5+R6)} = 3.3V x {100K/(100K+65K)} = 2.0V
Then, Icharge uses equation 2 on page 12:
Icharge = Viset / (20 x Rsr) = 2.0V / (20 x 0.010 Ohms) = 10A
I still do not know your input voltage, so I don’t know what your ripple currents will be, but using the values in the spreadsheet you show, I get inductor Iripple of:
Iripple = {Vin x D x (1-D)}/(fs x L) = {19 x 0.43 x (1-0.43)}/(600kHz x 6.8uH) = 1.14A
However, in your case you will be using 3.3uH for L and if your input voltage were different, then you would also have a different value for duty cycle “D”.
Hi @David_1528
Thank you for the information. Never knew that <100k would have an effect here. Another datasheet says not to exceed 100k at the lower resistor. I guess that depends by manufacturers.
When they talk about the input voltage Vin, is it the voltage at the Vcc pin or the overall board powering the whole circuit on a PCB, either way it will be constant at 12v @50-100A for my application.
The recommendations are always based on the characteristics of the specific chip. There are a myriad ways of designing the structures within a chip, and they all will have different external interface optimizations.
Vin refers to your external power source.
That’s a lot of power → 600W to 1200W. This circuit will use less than 100W. That is a constant voltage supply, right? Constant current would be a problem, unless it was self-limiting to 12V and, therefore, never reached its constant current mode.
Hi @David_1528
The power supply is from an HP server it has all the bells and whistles on them to protect the server. I work in the IT industry and these are readily available for me. Moreover they are dirt cheap on ebay of one needs to get hold of one. I’ve been using two of these in parallel to power my LiPo Charger iCharger 406 DUO RC Truck.
Is it ok to have the mosfet (circled in green) this far apart, don’t have much board estate and am trying to keep the SMD on the top side.
Hard to say, but you might have some signal integrity and noise issues with this. I would try to follow layout recommendations on pages 21 - 22 as closely as possible. Power loops should be as small as possible, and they recommend keeping the MOSFET gates close to the BQ24640. Long runs there can cause inadvertent triggering of the gate due to noise or ringing.
The BQ24640EVMuser guide shows their layout in the latter section of the document. I also found a link to gerbers for the desgin here from this TI forum post. They used a 4-layer board and placed some of the components on the under side of the board. That allows for a much more compact design while still maintaining good signal integrity.
Hi @David_1528
I’ve rerouted the board again and managed to make the mosfets closer than closer with a Gate net length of 23mm from HIDRV to MOSFET and 9mm from LODRV to MOSFET.
In section 6.5 of the datasheet, on page5 it says Iac (Adapter Supply Current into Vcc), are they the conditions for CE pin to read LOW and HIGH or should Vcc be be limited to 25mA.
That specification is merely stating how much current will be drawn into the Vcc pin under the conditions stated. You do not have to control the current in any way. It is stating that Vcc will typically draw 25mA when it is in charging mode.
To select the inductor’s current rating for my application, i get a Ripple current 1.31A, so 40% higher would be 5.24A. Would it be rated at 5.24A and saturate at 10A ?
Secondly, do i need to worry about the output capacitance and the LC filter, mentioned in 8.2.2.2 and 8.2.2.3 sections.
The inductor saturation current must be greater than the charge current plus 1/2 the ripple current. You don’t want the current to get close to the saturation current of your inductor or the circuit will fail and the chip will be destroyed. Your duty cycle, D= 67.5% (8.1V/12V), so your inductor ripple current would be 1.33A.
Iripple = {Vin x D x (1-D)}/(fs x L) = {12 x 0.675 x (1-0.675)}/(600kHz x 3.3uH) = 1.33A
So the inductor should be rated for greater than 10A + 1/2(1.33A) = 10.67A. The evaluation board for the series uses Vishay IHLP5050EZERxxxM01 series, so I would recommend using the IHLP5050EZER3R3M01, which is rated for a saturation current of 32A, which gives you plenty of headroom.
Section 8.2.2.2 refers to the input capacitor. The ripple current for the input capacitor is going to be about 4.7A. They recommend using about 20uF low ESR type capacitors like ceramic, something like a pair of CGA8P2X7R1E106K250KA would be a good choice.
You do need to worry about the output capacitance and the LC filter as described in section 7.3.6. The LC filter needs to have a resonant frequency between 12kHz and 17kHz for best stability. Table 5 on page 20 recommends 40uF of capacitance along with the 3.3uH inductor, which gives a resonant frequency of about 13.85kHz. They recommend ceramic capacitors for the output as well, so you could use four of the same CGA8P2X7R1E106K250KA there. Ripple current on the output capacitors will be under 1/2A so they will handle that easily and give only a small ripple voltage.
