capacitor multiplier question

Discussion in 'General Electronics Chat' started by coinmaster, Feb 27, 2016.

  1. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    Capacitor multipliers are supposed to multiply a capacitor's ability to reduce ripple
    [​IMG]
    The only thing I see here is an RC filter that is attached to an emitter follower.
    Wouldn't the emitter follower just copy the ripple instead of reducing it? LT spice seems to agree with me, the wave forms look exactly the same on the base and emitter.
     
  2. AnalogKid

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    Aug 1, 2013
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    Yes, but not on the base and the collector. The emitter puts out a current-gain version of what is on the base, and the base has Vin *after* it has been through a lowpass filter. Hanging a capacitor directly on Vin can do the same thing, but the source impedance likely is way lower than R1, so the capacitor would have to be that much larger for the same effect. It is fairly easy for a CM circuit to reduce ripple by 99%, especially if the transistor is a darlington.

    ak
     
  3. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    I still don't get it. You basically told me that the emitter gets a current gain version of the ripple at the base but what we want is ripple attenuation which is a function of voltage not current.
     
  4. ian field

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    The ripple on the base is much smaller than on the collector - and you don't need such a big capacitor to do that because of R1.
     
  5. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    How is that any different then just using an RC filter by itself? The ripple on the base is the result of the RC filter. Perhaps you're hinting at a trait of transistors I'm not grasping?
     
  6. AnalogKid

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    Let's assume there is 5 V peak-to-peak, 120 Hz ripple at Vin, the residual ripple on the big power supply filter capacitors farther upstream. Without the circuit, Vin = Vout, and the ripple has to be dealt with farther downstream. Now, if R1 = 113 ohms and C1 = 470 uF, then the corner frequency is 3 Hz. This is 1/40th of the ripple frequency, so the attenuation factor is 32 dB, or 97.5%. So the amplitude of the ripple at the base, and hence the emitter and hence Vout, is only 0.125 Vp-p. This is the theoretical limit; there are other factors that reduce the circuit's effectiveness in real life, but not by much.

    You asked this while I was posting - the difference with and without the transistor is that the main circuit power goes through the transistor, not the resistor. If the resistor had to handle the full output current, possibly tens of amps, it would need to be so much smaller that the corner frequency would be too high to be effective, and you're back to the same as no circuit at all. Because the transistor bypasses the resistor, the resistor is sized based only on the frequency of interest and the transistor's base current requirement for the full output current. This is why a darlington is better - for a 10 A output current you need less than 10 mA of base current, so R1 can be big enough to be a good filter without attenuating the DC voltage at the base enough to affect the output voltage very much.

    ak
     
    Last edited: Feb 27, 2016
    atferrari likes this.
  7. coinmaster

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    Dec 24, 2015
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    There must be something big I'm missing because what you just told me was how an RC filter works. Yes an RC filter with a 113 ohm resistor and a 470uf capacitor will have a cutoff frequency of 2.9HZ but what is the point of the transistor? It doesn't do anything but copy what the RC filter is already doing. I might as well just take the output at C1.
     
  8. ian field

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    The emitter follower draws less current from the RC filter than if you just stuck the load on there.
    A couple of people have pointed out the trait of the transistor, if you still don't get it - take up woodwork.
     
    #12 likes this.
  9. coinmaster

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    Dec 24, 2015
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    Does ripple increase with current draw?
     
    Last edited: Feb 27, 2016
  10. ian field

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    Its the other way round - maybe even woodwork isn't for you..........................
     
  11. coinmaster

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    Dec 24, 2015
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    Sorry forgot the "with" :p
    So in other words a normal RC filter doesn't actually perform to spec.
     
  12. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    BING BING BING.

    Now tell us which is larger in that transistor, the current into the base or the current into the collector?

    Is this a great circuit? No it is more of a hack that anything useful.
     
  13. coinmaster

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    Dec 24, 2015
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    Oh looks like AnalogKid answered my question a while back in an edit, missed that.
    Wouldn't a mosfet be better than a transistor since they don't require gate current?
     
  14. AnalogKid

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    You asked that once already and I answered it. To eliminate the transistor, try this:

    For a analog system power supply that has 24 V DC plus 4 V p-p ripple on the filter caps, and a load of 5 amps that can work with no more than 200 mVpp ripple but a decrease in the DC value of no more than 2 volts, use a single pole R-C filter.

    1. Calculate the value of the resistor that meets the voltage drop requirement.

    2. Calculate the value of the capacitor that meets the filtering requirement.

    ak
     
  15. ian field

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    A MOSFET needs its gate a few volts above its source - that comes out of your voltage headroom, unless you arrange a voltage doubler to provide a gate bias supply.
     
  16. coinmaster

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    Dec 24, 2015
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    What about a Jfet? They produce less noise than the other 2 and have a high input impedance and as far as I understand they don't need extra voltage on the gate?

    Although LTspice is showing all the current flowing through the gate. Since it's a source follower it is not forward biased, how is current flowing through the gate?
     
    Last edited: Mar 2, 2016
  17. ian field

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    JFETS that can handle enough current for a series pass regulator are few and far between. They pretty much act as constant current elements, the TO92 etc types are typically from about 2mA to about 20mA.

    JFETs have pretty wide parameter spreads, especially for IDss and Vgs cut off. You can set the channel current fairly precisely (or at least repeatably) by grounding the gate to the bottom of a source resistor.
     
  18. coinmaster

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  19. ian field

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  20. coinmaster

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