Explain Miller Effect

Discussion in 'General Electronics Chat' started by bug13, May 31, 2013.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Hi guys

    I came cross this terms many times, and still don't understand it. Can someone explain it to me in a simple way please.

    Please don't link me to Wikipedia as I usually come back with more questions than answers.

    Thanks
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    The Wikipedia article is actually pretty good.

    The basic idea is that if you have a feedback element between the output and input of an inverting amplifier, the element looks smaller by roughly the magnitude of the gain of the amplifier.

    The basic mechanism is pretty simple. You change the input voltage by ΔV. If the other side of the feedback impedance, Z, were at a fixed voltage, then the change in current into the input, ΔI, would be ΔI = ΔV/Z and so the incremental input impedance (due to the feedback element only) would be ΔV/ΔI=Z.

    But the other side of the feedback impedance is NOT at a fixed voltage, but instead it changed by an amount of -A*ΔV. So the change in voltage across the impedance is

    ΔVz = ΔV - -A*ΔV = ΔV(1+A)

    Thus the change in current in the feedback impedance is

    ΔI = ΔVz/Z = ΔV(1+A)/Z

    and the effective impedance seen by the input is

    Zeff = ΔV/ΔI = Z/(1+A)

    If the feedback impedance is a capacitor, then it looks like a capacitor that is (1+A) times as large.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    In a nutshell the capacitance from the input to output (such as the base-collector capacitance) provides negative feedback which is proportional to the gain of the stage and the frequency of the signal (rather like an integrator).
     
    spark8217, absf and bug13 like this.
  4. circuitfella11

    Member

    May 10, 2013
    56
    5
    most inverting amplifiers have this..sometimes it increases the effects on capacitances of base to collector terminal.. its proportional to the gain, that's why it exists..
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,395
    1,607
    Let's forget about capacitors, amplifiers, and let's just look at a one ohm resistor. One terminal is free, the other is grounded.

    If you put 1 volt on this terminal you expect to see 1 amp. That's fine.

    Now say we sneak in a dwarf with a voltmeter and an adjustable supply. He's fast and sneaky, and he's going to measure the voltage you put in, and set his supply to 9 times that, and connect it backwards to that resistor.

    So when you put in that 1 volt, he puts in 9 volts, and now there is 1+9=10 volts across the resistor. If you "assume" the resistor goes to ground to calculate it's resistance, you get 1 volt yields 10 amps, or R=1V/10A = 0.1 ohms.

    So by using the dwarf (who's name is Miller) you get a Miller's effect resistance of 1/(9+1) times the "real" resistor value. It's an equivalent value resistor that gives the same results as the resistor-dwarf-voltmeter-powersupply hookup; the equivalent simplifies the analysis.

    Now just replace the dwarf with a supply by an amplifier, and the resistor with a capacitor, and you get the real Miller effect.
     
    bug13 and PackratKing like this.
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Sounds like a fairy tale.:p
     
  7. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    bug13 likes this.
  8. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Thanks guys for all your helps, I can't say I am fully understand the concept in details, but now know what it is and what it will do to an inverting op amp circuit.
     
  9. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    Have you heard of Maxwell's deamon ?
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    I always thought that sounded like a fairy tale, too. :D
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I was trying to understand why ErnieM had settled on a dwarf as the star of the story. I just noticed the OP is a resident of "Kiwi" country - was there a subtle reference to Lord of the Rings perhaps?
     
  12. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Sorry to bring this thread back up again, I simply want to say thank you all again, it's funny how I look at this again, and thing looks a lot more easier.

    Thanks guys!

    PS: I like how different people explained it from a different angle, that helps a lot.
     
    ErnieM likes this.
  13. analogdesign

    New Member

    Aug 29, 2013
    14
    1
    The key takeaway about the Miller effect is that it is due to GAIN across a gate-drain junction (or base-collector). So to reduce Miller effect you have to reduce the gain across that device? How do you reduce the gain across an individual device but maintain the same gain in the overall amplifier? You use a cascode transistor. Have you heard that term yet? Very, very important in practice.

    Also, make sure you can distinguish between the Miller and Budwiser effects.
     
  14. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Here in Golden, CO, we tend to perfer the Coors effect.
     
  15. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Let me try to give another - very simple - explanation:

    Both ends of a resistor (in our case: the input resistance of an inverting amplifier) are connected to two voltages with different sign.
    Now - the current through this resistor is allocated to one of both voltages only. That means: we imagine that the input current is caused by the input voltage only, because this is in accordance with the definition of an input resistance.
    As a result: The current is larger than the fictitious current caused by Vin only (without the second voltage at the other end).
    That means: The input resistance appears to be decreased.
     
  16. analogdesign

    New Member

    Aug 29, 2013
    14
    1
    Ha! Here in the Bay Area, I suppose we would have to call it the "Snooty Overpriced Microbrew Effect".
     
  17. #12

    Expert

    Nov 30, 2010
    16,346
    6,831
    OK. I'll see if bug relates to my way of speaking.

    Think about an inverting amplifier of your choice, common emitter transistor, vacuum tube, or op-amp.

    For any inverting amplifier, there is a small amount of (parasitic) capacitance between the output and the input. As you apply a voltage to the input of any inverting amplifier, the output responds by changing in the opposite direction (voltage wise). That opposite voltage change sends current through the sneaky, parasitic capacitance back to the input and diminishes the input signal.

    When the inverting amplifier has a gain of more than one, the reversed voltage on the output has more effect than it would if the gain was only one. When the frequency of the signal is fast, the impedance of the parasitic capacitor is less, and so the reversed voltage has more effect on the input than it would if it was slowly changing.

    Now you can go back and look at the math oriented responses and they will make more sense.
     
Loading...