Thanks for detailed reply,Hi,
Your circuit shows only minor differences to the datssheet schematic Fig. 8-1.
The differences are:
* additional input capacitor --> uncritical
* modified EN voltage --> uncritical
* modified feedback capacitor --> small change, uncritical
* modified feedback resistor (output voltage) --> uncritical
* three inductors in series. Functionally I see no problem. But due to lengthy wiring and bigger switching node and bigger enclosed switching loop area I expect a much higher EMI. So while the SMPS may work it still may fail compliance tests.
Also the PCB layout has big impact no only on EMI, but also on regulation stability and output noise.
Thus I see limited information you can gain from a simulation:
* it´s quite expectable that the circuit will work
* but you can not see the impact of the three inductors in series.
* you can not see the impact of your PCB layout.
* and you can not see EMI results.
And, btw: why three inductors in series at all? The absolutoe value is not that critical and thus you easily should be able to find a one inductor solution.
Klaus
Thank you for your reply,TI has a page for this device on its website.
TPS54302 data sheet, product information and support | TI.com
TI’s TPS54302 is a 4.5-V to 28-V Input, 3-A Output, EMI Friendly Synchronous Step-Down Converter. Find parameters, ordering and quality informationwww.ti.com
They state you can use their design tools including a simulator PSPICE-for-TI.
A buck converter does not output a higher voltage than you input.
However the buck-boost type can accept your specified 8 to 28V, and output 12V inverted, if you can accept inverted polarity.
On a first test I´d use a single inductor with 6.8uH.I had only 6.8 and 3.3 uH inductors
The output voltage is primarily defined by the feedback, not by the inductors.to get an output of 12V
I clearly wrote that I "expect it to work" in the simulation.So by your statement "but you can not see the impact of 3 inductors in series" does it mean its not possible to get 12V output?
If you power the Arduino with 12V you waste 140% of energy (--> heat) compared to powering it with 5V.to power arduino mega 2560
It depends on a lot of things: Mainly - as already written - on the PCB layout.In that perspective can i achieve this using current circuitry or is to necessary to use a single 15uH inductor?
Thank you for your response sir,Hi,
On a first test I´d use a single inductor with 6.8uH.
And if it does not work I´d use rather 2 x 6.8uH than 3 inductors.
Btw: the error to 15uH is less with 2x6.8uH than with your 3 inductor solution.
2x6.8uH gives just 10% error to 15uH. And often 15uH inductors have a tolerance of +/-20%.
***
Final circuit vs simulation:
I´m confused if you want to do the final circuit with one 15uH or several other inductors.
For me it´s more important to know how the final circuit performs than a ciruit with parts laying around.
What information do you gain on the 3 inductor solution vs a one inductor solution?
The output voltage is primarily defined by the feedback, not by the inductors.
I clearly wrote that I "expect it to work" in the simulation.
But your "schematic only simulation" (without placement, signal routing, coupling effects) will not show how the circuit perform as a real circuit. Especially regariding loop stability and EMI.
So my doubt is: What information do you get from a simulation that most likely shows that a circuit (schematic) works, while it´s quite questionable whether the real circuit performs correctly.
If you power the Arduino with 12V you waste 140% of energy (--> heat) compared to powering it with 5V.
What´s the benefit of powering it with 12V?
It depends on a lot of things: Mainly - as already written - on the PCB layout.
It it will be much harder to comply with EMI regulations when you use inductors in series.
I don´t know if yu are experienced with SMPS design and EMI compliant design. A good designer will have no troubles.
An amateur has way bigger troubles with 3 inductors instead of using just 1 inductor. SMPS design is not a simple task.
Did you consider to use a ready to buy module?
Klaus
Added: did you do the math given in the datasheet? You talk about 1A output current ... but use inductors with a saturation current of just 1A. Did you ignore the ripple current?
Ok will make sure that external devices consuming more than 100mA is not connected.Hi,
O.K. well done reading the document.
You are in the safe side if you keep on the document.
However, it says as long as you be careful, using the 5V input is O.K.
Precautions are
* do not connect USB at the same time when using external 5V supply. Because 5V may flow back to the computer when the external 5V is highervthan the USB_ 5V. This is explicitely the case when the compurt is powered down. The (back) current is not limited, thus may be destructive.
* do not connect anything to V_in. ... to avoid any current flow back from the (5V) voltage regulator.
* the same is true for the barrel jack
To protect against V_in and barrel jack problem there is a simple solution: connect a schottky diode from 5V (anode) to V_in (cathode). This is the usual method to prevent voltage regulator damage from reverse currents.
So it's fine to use the external power supply. If I had to use it, I'd rather go with the minimum input voltage plus headroom.
For your regulated power supply you don't need much headroom, since the voltage is reliable, predictable and stable.
The problem is the power dissipation.
With 8V input the power dissipation in the 5V regulator is 3W per ampere.
With 12V input the power dissipation in the 5V regulator is 7W per ampere.
Without heatsink the voltage regulator can stand 0.5W or maybe 1W. It is likely to overheat. (Sadly the document does not tell about this)
It is safe if you run the Arduino only from this power supply, since it will draw less than 100mA. But as soon as you connect external devices, like LED display or anything other that draws considerable current it may cause problems.
Klaus
Simple math:Ok will make sure that external devices consuming more than 100mA is not connected.
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