# Looking for Information on the Miller Effect

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#### Max Bateman

Would someone be willing to help me better understand the Miller Effect? I've come across this multiple times in various textbooks, but it's poorly explained.

Anyone have a simple version to share? Links are appreciated.

Thank you.

- Max

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#### Granger

Promise I won't share the link with you .... BUT the wiki link is actually pretty helpful. I can break it down (might be better).

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 mechanism is simple. 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.

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

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#### Crucshank

Basically, the capacitance from the input to output (or the base-collector capacitance) provides negative feedback. This is proportional to the gain of the stage and the frequency of the signal (like an integrator).

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#### CircuitMan

Most inverting amplifiers have this. Sometimes it increases the effects on capacitances of the base to collector terminal.

It's proportional to the gain... and that is why it exists.

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Just think of the resistor. One terminal is free, the other is grounded.

Now say we put in a voltmeter and an adjustable supply. Set the supply to 9 times that, and connect it back to that resistor.

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.

That voltmeter is 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-power supply hookup; the equivalent simplifies the analysis.

Now just replace with a supply by an amplifier, and the resistor with a capacitor, and you get the real Miller effect.

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#### Max Bateman

Yeah. That's pretty helpful. I really think it's a matter of getting my hands on some real-world issues regarding the miller effect. I'll let you know what I find.

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The main thing to know about the Miller effect is that it is due to GAIN across a gate-drain junction (or base-collector).

So in reducing the 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 worked with cascodes at this point? They are highly important pieces of the puzzle.

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#### Mikael Anders

I always found this most interesting when I was in school — Here's another

Both ends of a resistor — the input resistance of an inverting amplifier — are connected to two voltages with differing signs.

The current through this resistor is allocated to one of both voltages only. That means the input current is caused by the input voltage only because this is in accordance with the definition of input resistance.

The current is larger than the fictitious current caused by Vin. The input resistance then appears to be decreased.

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#### RalphMetzer

Think about an inverting amplifier of any kind, 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.

When you apply a voltage to the input of an 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.

Make sense?

#### npedja

I believe you are interested in Miller effect in MOSFET transistor. The Miller capacitance is a drain-gate capacitance Cgd, with one end fixed at the gate and the other floating with drain terminal, in simple words. When you want to turn the mosfet on, consider the drain-side of Cgd is forsing down to the potential of the source terminal (or ground level). Forsing the capacitor can be done only by current and the only current you have comes from the driver. Therefore, an amount of gate drive current is thrown away for driving Cgd instead of charging the gate. Effectivelly, it slows down turning on.
During turn off, the drain end of Cgd is forced up from the ground to the Vcc. Again, the only current that you have for this comes from the driver. Again, instead discharging of the gate, the driver must drive the Cgd. Effectivelly, this slows down turning off.
This is just a simplified story. More serious analysis will take into account initial charge of the Cgd, but also external parasitics which can even make the current through Cgd larger than the driver current thus pushing the mosfet in the opposite direction (spurious turn-on or turn-off). But, first things first.

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#### CarlKoop

^^^ I second that explanation.

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#### JasonJones13

When the inverting amplifier gains more than one, the reversed voltage on the output will be more effective than it would if the gain was simply just one.

When the frequency of the signal is fast, the impedance of the parasitic capacitor is less.

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#### Mogar

Agreed. You'll always want to lessen the parasitic capacitance at all costs.

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#### JasonJones13

A reversed voltage should have more effect on the input that if it were to slowly change. The math-oriented responses on here might make more sense now. Or even a textbook might be able to showcase it more applicably.

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Yes.

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#### Max Bateman

Thanks, all. This was good.

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