Explain Miller Effect

Thread Starter

bug13

Joined Feb 13, 2012
2,002
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
 

WBahn

Joined Mar 31, 2012
29,976
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.
 

crutschow

Joined Mar 14, 2008
34,280
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).
 
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..
 

ErnieM

Joined Apr 24, 2011
8,377
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.
 

Thread Starter

bug13

Joined Feb 13, 2012
2,002
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.
 

t_n_k

Joined Mar 6, 2009
5,455
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?
 

Thread Starter

bug13

Joined Feb 13, 2012
2,002
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.
 
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.
 

WBahn

Joined Mar 31, 2012
29,976
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.
Here in Golden, CO, we tend to perfer the Coors effect.
 

LvW

Joined Jun 13, 2013
1,752
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.
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.
 

#12

Joined Nov 30, 2010
18,224
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.
 
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