Silly power dissipation question

Discussion in 'General Electronics Chat' started by Skeebopstop, Feb 9, 2009.

  1. Skeebopstop

    Thread Starter Active Member

    Jan 9, 2009
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    A capacitor can be thought to be an energy storage mechanism, Q*V and is to be used to drive a circuit for a small amount of time.

    Half the energy of charging a capacitor is lost due to series resistance (or so I just read somewhere). So if I were sizing this 'reservoir' capacitor to turn a MOSFET on and off, say 200x, how would I account for the energy dissipation from the gate resistor when determining how much Q I need?

    Thanks
     
    Last edited: Feb 9, 2009
  2. Skeebopstop

    Thread Starter Active Member

    Jan 9, 2009
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    Any help on this?

    I think I am flawed in trying to view the capacitor as just a charge storage device to drive electrons in and out of the gate capacitance.

    Should I be observing energies more appropriately? Therefore I need to size the cap to not only charge the mosfet gate, i.e. transfer energy into it, but to also overcome those energy losses due to the resistance?

    So for example, lets say the gate capacitance is 1Farad and my Gate resistance is 100 Ohms and I want to charge it up to 15V.

    Energy-gate (required) = 0.5 * 1 * 15^2 = 112.5

    Working under the knowledge that half of the energy is dissipated in any resistance during charging, I should be able to say that my reservoir cap requires 2*112.5 joules.

    225 = 0.5 * Cx * V^2

    assuming V also = 15V here

    450/15^2 = Cx = 2F

    Is it just a coincidence that the capacitance was twice as much? Is that a rule of thumb?

    It actually brings me back to an equation I once saw when charging up the gate of a mosfet, they had a factor of 2*Qg...
     
  3. Ron H

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    You need to provide a schematic of how you propose to connect the capacitor to the gate, and how you propose to charge and discharge the capacitor.
     
  4. Skeebopstop

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    Jan 9, 2009
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    Attached is the schematic. It won't be so intuitive with the description provided as it is a bootstrapping situation.

    C4 is to act as the reservoir cap, please ignore that it must drive circuitry for now and just focus on its charging and discharging of the upper MOSFET. This happens through the gate driver IC.

    The gate charge is 30nC and I'm looking to charge it up in about 25nS or less.

    Cheers
     
  5. Skeebopstop

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    Jan 9, 2009
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    Heya Ron,

    any insight now that the schematic is up?

    Cheers
     
  6. Ron H

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    I have never used a high side driver IC. International Rectifier Design Tip 98-2 has the equation for selecting the bootstrap capacitor.
     
  7. Skeebopstop

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    Jan 9, 2009
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    Aye, that is the documentation I am referencing, however I cannot make complete sense of it.

    Please just ignore the high side drive component for now and remember that it, for the purposes of my question, can be thought of as simply a reservoir cap charging up another cap through a resistor.

    The only real question I have is, do I need to consider energy losses in the resistor when sizing C4 to be able to fully charge the MOSFET gate?
     
  8. Ron H

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    C4 and the gate capacitance make up a capacitive divider. If the number of coulombs stored in C4 greatly exceeds the gate charge, very little voltage will be lost in C4. The gate resistor just contributes to the time constant that determines the rise and fall times. The gate capacitance will be charged and discharged through the series combination of the gate resistor and the ON resistance of the driver. Power will be dissipated in the gate resistor, but it will not reduce the peak-to-peak gate-to-source voltage.
     
  9. Skeebopstop

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    Jan 9, 2009
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    This is how I was originally thinking, however there is a 2*Qg factor when computing the charge required to be stored in C4 to charge the gate, where Qg represents the mosfet gate charge. So I fail to explain the 2*.
     
  10. Ron H

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    Do you want a circuit that works, or do you want to analyze it to death?:D
    I don't know the reason for 2Qg. I do know that if your bootstrap cap holds orders of magnitude more charge than Qg, it doesn't matter.
    I'm an engineer. I like to understand things, but if I don't understand every minute detail, I don't worry about it. Do you have any idea how many engineers and hobbyists are using bootstrap drivers successfully but have never seen these equations?

    Are you a physicist?:D
     
  11. Skeebopstop

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    Jan 9, 2009
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    Naw, just an engineer who hates using things he doesn't understand :)

    Then you get into nasty situations when debugging where something you can't explain happens and much headache ensues. I therefore prefer to invest the time up front, but alas, sometimes a road block is met.

    I've chosen a very conservative value for my bootstrap, so I know it'll be fine, it just seemed almost too convenient that they do a 2*Qg and we know that half the energy used to charge a cap is lost in resistance.
     
  12. Ron H

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    But the energy is only lost (power is dissipated) during charging and discharging the cap. No voltage appears across the resistor during the flat portions of the waveform, and no power is dissipated by it during that time.
     
  13. Skeebopstop

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    I guess it is just easier to think in terms of energy transfer and to size your reservoir cap large enough to be able to transfer 2x the require final potential stored in any dependant capacitors. The 2x coming from the knowledge that half of that will be dissipated energy in the transfer.
     
  14. Ron H

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    Energy is lost. I don't believe any charge is lost.
     
  15. Skeebopstop

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    Agreed. The loss of energy would represent itself as a loss of voltage in the stored cap as potential energy = Q*V.
     
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