# Explain Miller Effect

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

1. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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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.

Thanks

2. ### WBahn Moderator

Mar 31, 2012
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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
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6,118
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).

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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
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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.

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6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Sounds like a fairy tale.

Apr 5, 2008
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8. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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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
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Have you heard of Maxwell's deamon ?

10. ### WBahn Moderator

Mar 31, 2012
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I always thought that sounded like a fairy tale, too.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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 Senior Member

Feb 13, 2012
1,609
58
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.

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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
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7,379
Here in Golden, CO, we tend to perfer the Coors effect.

15. ### LvW Well-Known Member

Jun 13, 2013
779
105
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
18,076
9,678
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.