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Original question:

LM3435: RGB output cap singing

LM3435: Okay to switch to tantalum output capacitor?

Eric Steimle
Expert 2290 points
Part Number: LM3435

We followed your advice and tried several different low acoustic noise ceramic capacitors for our output cap. This lowered the noise but we could still hear it. So we switched to a small tantalum in C47 location below and the problem almost completely went away. We can still hear it very faintly if your ear is right up next to it.  

So I just have to double check if this is okay since the datasheet clearly suggests MLCC in the output.  Is the tantalum cap okay?  The cap we picked is an AVC TLNN476M010 10V  / 47uF / 6000mOhm ESR / 100Khz RMS current of 82mA.  Seems like that might be a little low for this application.

Thank you

  • 0
    TI__Mastermind 28655 points
    Hello Eric,

    From the datasheet: "OUTPUT CAPACITORS SELECTION
    Two output capacitors are required with LM3435 configuration, one for VOUT to Ground, COUT2 and one for decoupling
    the LED current ripple, COUT1. The LM3435 operates at frequencies high enough to allow the use of
    MLCC capacitors without compromising transient response. Low ESR characteristic of the MLCC allow higher
    inductor ripple without significant increase of the output ripple. The capacitance recommended for COUT1 is 10μF
    and COUT2 is 22μF. Again, high quality MLCC capacitors with X5R and X7R dielectrics are recommended. For
    certain conditions, acoustic problem may be encountered with using MLCC, Low Acoustic Noise Type capacitors
    are strongly recommended for all output capacitors. Alternatively, the acoustic noise can also be lowered by
    using smaller size capacitors in parallel to achieve the required capacitance."

    6000 mohm is 6 ohms, this is much too high to maintain low ripple voltage at the output. You can calculate ripple voltage knowing the ripple current and switching frequency using ESR and Xc. I would compare the ceramic to anything that would replace it.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Expert 2290 points
    Thanks. I've read the datasheet of course. Are you referring to the inductor ripple current?
  • 0 in reply to Irwin Nederbragt
    Expert 2290 points

    I didn't see anything in the datasheet specing ESR for the Cout2 or how to calculate the ripple current for the capacitor.

  • 0 in reply to Eric Steimle
    TI__Mastermind 28655 points
    Hello,

    The inductor ripple current will turn into ripple voltage by the two capacitors. Ceramic capacitors have very low ESR. There are graphs showing impedance if you look at ceramic capacitor data sheets. XC = 1/(2*pi*f*C) will be dominate at lower frequencies, one part I just looked at is pure capacitance at 100 KHz and nearly pure capacitance at 1 MHz. This puts the impedance below 100 mohm at 100 KHz and below 10 mohm at 1 MHz. 6 ohms is much too high. This means if there is 2A of ripple there will be less than 200 mV of ripple at 100 KHz and less than 20 mV of ripple at 1 MHz. In this case the 0.1 uF ceramic will dominate above 300 KHz. At 6 ohms you can see it won't work.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Expert 2290 points
    Okay I follow you here. How do I calculate my inductor ripple current? I'm trying to figure that out by re-reading the section on inductor selection but I'm not getting it (also the formulas are very blurry in the datasheet for some reason).
  • 0 in reply to Eric Steimle
    TI__Mastermind 28655 points
    Hello,

    This is part of the whole design process. The output current(s), Voltage(s), duty cycle need to be known as well as input voltage, inductor value and switching frequency. This can be calculated from the datasheet equations or for a simple rough example:

    For example, if the maximum output is 3V and 1A and the input is 5V assuming the inductance value is high enough to keep the ripple reasonable (can assume squarewave). This is just a rough calculation (the equations in the datasheet will give more detail). Pout is 3W, assume Pin is 3W. duty cycle will be 37.5% for the input, the output 62.5%. This means the output current square wave will have an amplitude of 1A/0.625 = 1.6A. The output average current is 1A so charging current is 1.6A - 1A or 0.6A for 62.5% of the time. If the switching frequency is 500 KHz the off-time is 1.25 us. I = C * (dv/dt) shows the voltage excursion with a 22 uF capacitor to be 34 mV (dv = (0.6A/22 uF)*1.25 us), this will be peak to peak ripple. This is because ceramic is still capacitive at these frequencies.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Expert 2290 points
    Alright let me try. My system input voltage is 42V (it's a battery so it could range from 4.2V - 3.0V). I have a 2.2uH inductor, I've used RT to set Tsw to 2uS just like in the datasheet. I measured the LED voltages when the display is running and I see VLED is 3.8V.

    Given the formula in the datasheet
    D = Vled / (Vin + Vled)
    D = 3.8 / (4.2+3.8)
    D = 0.475

    Another formula defines the Inductor Iripple
    Iripple (max) - Iripple (min) = Vin/L * Tsw * D
    Iripple = 4.2V/2.2uH * 0.2uS * 0.475
    Iripple = 181mA

    So given a 6ohm ESR tantalum that would be a 1V ripple?
  • 0 in reply to Eric Steimle
    TI__Mastermind 28655 points
    Hello,

    Sorry, no, a flyback is discontinuous. You need to know the output current/power. You are near 50% duty cycle. If the output current is 1A then the square wave of current is approximately 2A peak to zero (assuming 50% duty cycle). Inductor charges when MOSFET is on (no current goes to the LEDs), inductor provides current to the LEDs when the MOSFET is off. At 50% duty cycle it needs to be 2A to average 1A to the LEDs. The charging of the capacitor is 2A - 1A (current from the inductor minus the output current) = 1A. During the capacitor charging the voltage would have to be 6V higher than the LED voltage. Since the ripple is a small percentage of the 2A square wave it can be somewhat ignored for this though it will add some slope to the 2A square wave. This math works for low ripple voltage, at high ripple voltage the math gets more complicated. Perhaps try simulating an inductor charging then discharging into a capacitor with a fixed load to see the voltages and currents.

    Flybacks, buck-boosts, boosts, have discontinuous LED current, bucks are continuous current for the LEDs.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Expert 2290 points

    Okay I'm back.  I did a basic spice simulation and I read the flyback section of my Pressman power supply design book.  So assuming a 50% duty cycle for a minute, my understanding now is the inductor is only connected to the load 50% of the time. The other 50% of the time it is charging up, while the capacitor provides current to the load. So assuming my load is 500mA, then the cap must supply 500mA for half of the cycle and the inductor must supply 1A for half the cycle, because the inductor has to charge the capacitor back up, and provide the power to the load.

    If I'm using a ceramic capacitor my voltage excursion would be:

    dv = I / C * t

    dv = 500mA / 47uF * 1.25us

    dv = 13.3mV

    But I decided to pick a tantalum capacitor with an ESR of 6 Ohms. When that 500mA flows into that capacitor I could expect a voltage excursion of 500mA * 6 Ohms = 3V.  Thinking about it further I guess that means that poor cap would have to endure 1.5W of power dissipation as well.  All that makes this cap a very poor choice.

    If instead I picked a tantalum with an SR of 0.1Ohms, then I would get 500mA * 0.1Ohms = 50mV (maybe plus the 13.3mV).  With power dissipation at 25mW.

    Am I getting closer here?

    I'm guessing 50% duty cycle is the worst case for these calculations because the capacitor can't be asked to supply more than the loads maximum current.

  • 0 in reply to Eric Steimle
    TI__Mastermind 28655 points
    Hello,

    Yes, you are starting to see how this works. Also note that when the capacitor is supplying the current to the LED the voltage on the capacitor portion has to be much higher than the LED because that same 6 ohms is in series (basically it won't work). Worst may not be 50% duty cycle, it depends on what you are looking at. If they duty cycle were higher (not really possible in your design), say 75%, then the inductor only provides current for 25% of the time to the output meaning it would need to provide 2A, 1.5A to charge the cap back up and 0.5A to the LED. Note that the capacitor has to have an amp second balance which the calculations will show, if it doesn't the voltage on the capacitor will either increase or decrease over time.

    As for the total voltage, they are not the sum since the phase angle is 90 degrees between XC and R (capacitance and resistance) so the total ripple would be the square root of the sum of the squares or about 52 mV.

    I haven't used them but you may want to look at the aluminum polymer capacitors, you would have to research them to see it they would work but from what I've read you don't have the tantalum issues especially if operated at higher ambient temperatures.

    Best Regards,