Miller effect.

[caps]

Ok heres another question guys and girls...


A grid resistor is connected to a capacitor in effect by virtue of the capcitance between the grid and the plate. So it seems then we have a RC network. Hence, it would roll off High frequency response at a certain frequency .

Is this actually part of whats called the miller effect ??

OR...

is the Miller effect confined to the fact that high frequency output signals can be capacitively coupled back to the grid (x u), with the result of possible oscillation and reduced gain?

 

[dale116dot7]

The miller effect says that that capacitance is going to be amplified by a factor of (1+gain). So if you have a 10 pF capacitor connected from grid to anode of a tube with a gain of 25, that capacitor appears as a 250 pF capacitor since you need to charge up the capacitor not only with the changing signal but also with the change in plate voltage.

 

[caps]

Thanks Dale.

When the capcitance charges, is it drawing current on its way from the cathode to the plate ? Is this how it works? Hence the reduced gain with the miller effect?

 

[dale116dot7]

That's right. But you can look at it in many ways. Basically, the voltage of the capacitor needs to charge to the change in grid voltage plus the change in plate voltage. Effectively, you can look at this as a capacitive feedback network, or as a capacitor to ground with a value that's C*(1+gain).

 

[Guest]

[quote author="dale116dot7"]you can look at it in many ways. [/quote]

So called Miller theorem, maybe that Miller, who wrote "Science
of Musical Sound" I don t know, someone here maybe know???

I do not remember exactly, hehe...
Given any linear general network with two nodes A,B, and ground
and having between nodes gain so that Vb=Au*Va.
If capacitor (or it may be resistor or inductor, only must be computed
dually, like admittance) is connected between A and B,
then you see between A and ground capacitance

Cseen=Cab*(1-Au)

and when Au is large and negative, capacitance seen is larger than C real.

But the some with conductance.

But if Au is unity (like in cathode follower), capacitance seen can be 0.


"mistakes reserved!"
xvlk

 

[PRR]

> maybe that Miller, who wrote "Science of Musical Sound"

Or maybe Glenn "In The Mood" Miller, King Of Swing?

The phrase "Miller Effect" may not be well known outside the US, and he may not have been the first man in the world to understand it, but it is a good paper. Here's a scan:

http://web.mit.edu/klund/www/papers/jmiller.pdf

When a tube is cold, it has a capacitance (everything has capacitance).

When hot and working as an amplifier, the capacitance increases. In fact the input capacitance changes with plate load, very annoying. Why? How much?

John Miller's paper is heavy reading, try the light version:

http://en.wikipedia.org/wiki/Miller_effect

 

[skipwave]

I have nothing to offer, just wanted to say thanks to caps for asking specific questions. I've learned much from the threads you start here in the Drawing Board. Danka! :sam:

 

[Larrchild]

It is sort of like reading Carl and Jerry from a 1955 Popular Electronics, aint it.
Bravo.

 

[Guest]

[quote author="PRR"]Or maybe Glenn "In The Mood" Miller, King Of Swing?
[/quote]
PRR, thanx for a paper,
I love swing, epoque of direct disc cutting and ribbon microphones
I heard lot of shellaq disc when I was child while spending hollidays
with my grandparents.

xvlk

 

[PRR]

> When the capcitance charges, is it drawing current on its way from the cathode to the plate ? Is this how it works? Hence the reduced gain with the miller effect?

I don't understand the question.

If you have a ZERO impedance source, you do not worry about input capacitance, static or Miller.

The input of a triode working with gain may be 5pFd grid to cathode (assumed grounded) and 150pFd of Miller effect, plus stray capacitance in the socket and wiring: round up to 200pFd.

If we drive that with a 50Ω source, such as a strong signal generator, the -3dB point is about 20MHz.

But if we have a 50Ω source, we could always transform it to a higher impedance and get "free" voltage gain. We normally do that with tube mike inputs. If the mike is transformed to 50KΩ, the -3dB point is now 20KHz. We also couple from the output of one tube to the grid of another: if the first stage's output impedance is 50KΩ (pretty typical) then 200pFd input to the next tube is again -3dB at 20KHz.

What bothered Miller so bad was not audio. It would be nice to tune-up the coils in a radio with the tubes off (pre-tune them in the coil factory). But adding the tube de-tunes the coil: first when you connect the cold tube, and again when the tube is working. Miller wanted to know the reason and the amount of this extra capacitance so it could be designed around. (The answer to HIS problem lay a few years in his future: good tetrode tubes have really low Miller effect.)

 

[caps]

Thanks PRR, that was indeed a badly phrased question.

I guess what I was trying to ask is...

on the DIScharge of this capacitance between grid/anode, where does the dischrged current go Seems maybe back through the grid? Though that cant be good.... ?

 

[dale116dot7]

Back into the signal source.

One thing to note is that this effect provides a substantial amount of negative feedback in a condensor microphone circuit. Since the source is a capacitance, and the feedback element is a capacitance, the feedback is pretty close to flat in terms of frequency response.

 

[clintrubber]

[quote author="dale116dot7"] Since the source is a capacitance, and the feedback element is a capacitance, the feedback is pretty close to flat in terms of frequency response.[/quote]
And more of that: pad realized by means of a switchable cap. :thumb:

 

[caps]

[quote author="dale116dot7"]Back into the signal source.

One thing to note is that this effect provides a substantial amount of negative feedback in a condensor microphone circuit. Since the source is a capacitance, and the feedback element is a capacitance, the feedback is pretty close to flat in terms of frequency response.[/quote]


So the miller effect is less for a condensor microphone, as opposed to a dynamic mic ?

 

[dale116dot7]

Not really..... we're talking about the tube 'head amplifier' located right at the condensor capsule, not about a preamp or anything like that. A dynamic mic does not have a tube preamp right in the mic body.

Many signal sources are basically effectively a resistor and a voltage source. A dynamic mic works this way, so does a condensor mic AFTER the head amplifier. So the mic preamp in your rack or console, if it uses a tube (or for that matter, any amplification), the Miller effect diddles with the amplifier's frequency response.

In the condensor microphone head amplifier, the signal source is capacitive (after all, a condensor microphone capsule is a capacitor), so its impedance varies with frequency. If you make a feedback element out of a capacitor, its impedance varies with frequency in exactly the same manner, so you get a pretty flat frequency response despite the fact that the impedances in the circuit vary substantially with frequency.

The Miller effect is still there in a condensor mic, BUT it does not cause a high-frequency rolloff the same way that it would in 'most' amplifier circuits.

In RF circuits, the Miller effect is still there, BUT the effect is not so much a rolloff of high frequencies, but rather, a shift in the tuning frequency of the LC filters used in radio receivers and transmitters, as was mentioned earlier. It gets quite complicated as in RF circuits, you often have tuned circuits on both the input and output of the amplifier, so there can be some interaction because of this.

So this Miller effect is not so much concerned with the end application effect, but the magnitude of this effective capacitance.

 

[caps]

Thanks Dale!

PRR and others who have contributed.

Unfortunately, I still cant get my head completely round the miller effect. I know what it can result in, I.e - loss of gain, high frequency roll off etc., but Im trying to form a mental picture of how it actually works.

I went back to review how capcitors work but this hasnt helped. I will continue trying.

 

[NewYorkDave]

It's really not that difficult. Between any two electrodes, in this case grid and plate, there exists some finite amount of capacitance that is measurable even when the tube is unplugged and sitting on a shelf. The two electrodes act as the plates of a capacitor, with the vacuum as the dielectric. It's important to remember that capacitance exists between any physical bodies, and it become more significant as the distance between those bodies is decreased, or the surface area of those bodies is increased, or both. There's even capacitance between the earth and the moon--something on the order of a few microfarads.

Back to the triode. On the grid side of our vacuum capacitor is the input signal; on the plate side is the input signal, multiplied by the gain of the tube, in opposite polarity. Since this adds to the net voltage difference across the capacitor, the inter-electrode capacitance that already exists is effectively multiplied by a factor equal to the gain of the tube.

 

[caps]

Thanks dave, I can get my head round that, thanks for taking the time.

One last question before I go back into my hole though...

What is the actual result of (by way of the miller effect), having grid current flowing back into the grid from the capcitance discharging ? How does this react with the input coming into the grid?

Current into the grid coming up against current coming out of the "capacitor", whats the result? thanks again.

 

[CJ]

Because of the way a tube is constructed there is capacitance between the plate and grid. Because the tube inverts the signal the capacitance appears to be much bigger than it actually is. Your grid signal is swinging the miller cap one way, but the plate is swinging it in the opposite direction big time.

Since charge on a cap equals volts time capacitance, Q = CV, increasing the volts on this miller cap means you increase the charge, or current. This current has to come from somewhere, and since the grid circuit is high impedance, the current does not come from there very easily. It comes from the source, which means signal degradation .


This has the effect of increasing the input capacitance.

 

[caps]

[quote author="CJ"]Because of the way a tube is constructed there is capacitance between the plate and grid. Because the tube inverts the signal the capacitance appears to be much bigger than it actually is. Your grid signal is swinging the miller cap one way, but the plate is swinging it in the opposite direction big time.

Since charge on a cap equals volts time capacitance, Q = CV, increasing the volts on this miller cap means you increase the charge, or current. This current has to come from somewhere, and since the grid circuit is high impedance, the current does not come from there very easily. It comes from the source, which means signal degradation .


This has the effect of increasing the input capacitance.[/quote]


CJ,

you mean....


The current has to come from the source, so whatever the amplifier is connected to ? As in the case of a pre, a mic for instance ?