I am presently trying to design a solar panel based system. I want to procure the evaluation board SM-3220-BATT-EV . While trying to procure that, the following questions cropped up.
1) nRF24LE1 micro controller has been used . Do you supply the programmer kit to program the controller ? Or else how do I program it ?
2) In the application note 2121. The controller is PIC rather than the nRF24LE1 . Do you have the same evaluation board using PIC controller. Actually we don't need the RF communication stuff.
3) If we use nRF24LE1 in the evaluation board, and I need to send some debug information to the PC via UART how should I do that. What receiver should I procure ?
4) In my application I need some 18A of load current, if I change the components like, inductor , Mosfet, will the PCB be able to support 18A of current. Or can you please kindly mention us the changes that we need to make.
Please kindly help us with your feedback
Do you measure the voltage at the battery terminal or at the output of the board?
To run at 20A you will mainly need an inductor that can handle the peak current which will probably be around 25A at worst case temperature. Your FETs will need to be properly heatsunk to keep reasonable junction temperature. You might have to choose FETs with smaller ON resistance (Rds ON) and similar gate charge.
Once this is changed, a change in the output current sensing loop will be necessary. A simple way to achieve this is to change the output current sense resistor R10 from 4ohm to 2ohm. This way you will also reduce the heat disipation on the resistor at full current.
Thank you for quick reply. Well I had been measuring the voltage at the battery terminal and not at the output of the board. But now I measured the voltage also at the output of the board. I am getting Vout_board = 13.6v and Vbat = 12.85V, when the current was between 4.5 to 5A. I will once more confirm on this after I do full rigorous test, this week. Why is it that the voltage at output of board is higher that voltage at battery terminal ?
In the present eval board circuitry, how do you actually limit the current to 9A( also low current limit to 500mA ), is it just by the resistance R10 ? Can you kindly share with me the calculation on this. I completely understand the calculations for the buck-boost converter, but I am unable to understand calculation on how the circuit is actually limiting the current to 9A. It will be helpful for me while I am making the circuit, as I will try to make it capable to give to give a huge output current.
In the circuit there is an OP-AMP U11B, whose output is connected to RST, is this OPAMP not being used in the circuit ?As the resistors at the "+ve " pin are not connected.
Please help me with the calculation that you have used in the circuit for current limitation.
Ca you please inform something about the calaculation that had been done in circuit regarding limiting the current to 9A. I understand the buck-boost converter part, but not able to figure out the other extra stuffs.
Well I understood that part on current calculations. Actually I got bit puzzled by the fact that since the circuit already contained a 27uF inductor, where as the requirement was only 2uF ( We have already discussed on this before, in the previous posts ) . So I was thinking that since 27uF might be already be having a high current rating, so why should I be changing the inductor , that is already on board, so there might be some thing else to limit the current. But later i got the idea, that the present the inductor might not be having so much of peak capacity.
Can you kindly inform, what is the current rating for the present inductor on board.
Can you please help me with the three things ;
1) The reason why the current decreases, as the voltage crosses 12.5V. I have already explained the scenario before.
2) The current rating of the inductor present.
3) What is the use of U11B ic, since some of the resistors in the schematic for the ic are not connected.
Kindly, clarify the doubts.
1) Where do you measure the voltage? The system senses the voltage before the board's output terminals. If the cables are long and/or thin, there might be a voltage drop between the output of the converter and the battery. Therefore, the system could get into floating charge when the battery voltage is lower than the set value. As the current drops the voltage at the output of the converter and the battery will equalize towards 13.5V.
2) The peak current rating of the inductor is 13A at 25 degree celsius.
3) U11B performs an overvoltage protection that shuts down the converter if the output voltage crosses above a certain value. The resistors left unconnected perform a hysteresis function (reset at a certain voltage, release reset at a lower voltage) that we do not utilize in the design.
Thank you for your reply. I had measured the voltage both both at the output terminal and the battery terminal. Well there was difference in voltages, which I have already mentioned you earlier. I will surely check on this today and update you by changing the wires.
I had replaced the inductor in the inductor on the board with a 6.52uF and 30A peak inductor. http://www.digikey.com/product-detail/en/HC3-6R0-R
When I placed a multimeter in series and measured the current, the current was reaching barely 1 A. Do you have any idea why it can be so ?
I was making a small experiment for getting more output current, by the replacing the inductor with a 6.52uF and 30A peak inductor. http://www.digikey.com/product-detail/en/HC3-6R0-R.
As I run the circuit, a continuous BEEP sound comes from the inductor. When I placed a multimeter in series and measured the current, the current was reaching barely 1 A. Do you know what can be causing this to happen ?
This inductor will result in high current amplitude due to the decreased inductance. It is possible that the high current ripple affects the sensing circuits, either by the added noise to the current reading or due to increased EMI on some other part of the circuit.
Can you take waveform of the issue? specifically inductor current or at least voltage at the AIOUT pin?
Is the system working well with the original inductor?
The circuit was running fine with the original inductor. Yesterday I had performed a different experiment to test. I had two evaluation boards. I had opened the inductor of one board and connected it in parallel to the inductor of another board. In this way, the inductance gets decreased, but the current capacity increases, I had performed this test just to test whether the circuit works with LOW value of inductor. But again there was no current output.
You have used a 27uF inductor, how did you come up with the value 27uF. None of the calculations in the Application note generate such value. According to my calculation I needed 6uF.
I feel there is something wrong in my side with the mathematics. I will be sharing the calculations . Can you please tell me how you reached the value of 27uF. Also how much value of inductance I should be using to get at least 25A of current .
I feel there is something wrong in my side with the mathematics. I am sharing the calculations . Can you please tell me how you reached the value of 27uF. Also how much value of inductance I should be using to get at least 25A of current .
**************BUCK BOOST CONVERTER DESIGN*****************MAXIMUM_OUT_VOLTAGE : Vout_max = 13.600000 MINIMUM_OUT_VOLTAGE : Vout_min = 12.000000 MAXIMUM_POWER : Pmax = 300.000000 MAXIMUM_V_MPP : Vmpp_max = 45.000000 MINIMUM_V_MPP : Vmpp_min = 15.000000 MAXIMUM_FREQUENCY : Fsw_max = 200000.000000 MINIMUM_FREQUENCY : Fsw_min = 180000.000000 MAXIMUM_OPEN_CKT_VOLTAGE : Vin_max = 50.000000 MAXIMUM_SHORT_CKT_CURRENT : = 11.000000 MAXIMUM_OUT_CURRENT : Iout_max = 30.000000 Delta_I : Delta_I = 9.000000 Delta DVin pp = 2.250000Delta DVout_pp = 0.680000Duty cycle of buck mode : Dbk_min = 0.266667Duty cycle of boost mode : Dbst_max = -0.102941Value for L1 >= Vout_max * ( 1 - Vout_max/Vin_max )/( Fsw_min * Delta_I ) = 0.000006 Value of Ipk = 28.999677
According to the calculations mentioned in the website , I get the value of the inductance to be 6uF. Can you please help with what should be the actual value.
Your formula seems ok but with 9A of ripple current, there will be much more noise generated on the board and its circuits and possibly some issue with the driving of the H-bridge itself. To find the cause of the issue would require checking each sensing chain (amplifiers and filters) and verify that they all perform correctly.
Then, the various auxilliary circuits will need to be tested too.
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