Author Topic: Is this the miller effect at work?  (Read 5894 times)

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Offline EvertonTopic starter

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Is this the miller effect at work?
« on: September 29, 2014, 03:50:13 am »
Hi,

I am trying to make an amplifier that will take a 0 to 1V signal and amplify it to +/-10V.  The signal bandwidth is 20Mhz.

I slapped together a simple commen emitter amplifer as shown in the attachment (ignore the push pull on the right as this was meant to be the final stage), but it seems to roll of around 1 to 2 megs no matter what transistor I put in there.  Is this the Miller effect at work?  If so, what would be the best solution?  Would I be best to use a cascode or differential amplifier?

Thanks

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Online T3sl4co1l

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Re: Is this the miller effect at work?
« Reply #1 on: September 29, 2014, 03:57:05 am »
Yes, and yes.

Note that measuring a real circuit puts >10pF of probe capacitance on the circuit, worsening things even more.

Another way to think of it, 30k isn't very much to drive a nice low impedance signal (~100pF of load capacitance is already 79 ohms at 20MHz!), so if you increase the bias current significantly, you can easily push that up.

Tim
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Offline katzohki

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Re: Is this the miller effect at work?
« Reply #2 on: September 29, 2014, 06:44:47 pm »
If you want to be amplifying a signal at 20MHz, you might consider using an op-amp. Or, if you don't expect to be amplifying anything that high, such as audio, then just live with it.
 

Online T3sl4co1l

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Re: Is this the miller effect at work?
« Reply #3 on: September 29, 2014, 11:16:28 pm »
10V at 20MHz is a lot to ask from an op-amp.  Yes, there are plenty that go well beyond that, but why would you want to?

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Offline vk6zgo

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Re: Is this the miller effect at work?
« Reply #4 on: September 30, 2014, 01:21:58 am »
Hi,

I am trying to make an amplifier that will take a 0 to 1V signal and amplify it to +/-10V.  The signal bandwidth is 20Mhz.

I slapped together a simple commen emitter amplifer as shown in the attachment (ignore the push pull on the right as this was meant to be the final stage), but it seems to roll of around 1 to 2 megs no matter what transistor I put in there.  Is this the Miller effect at work?  If so, what would be the best solution?  Would I be best to use a cascode or differential amplifier?

Thanks

Yes,you have slapped together a nice audio amplifier!

In  small part,your rolloff is caused by Miller Effect,but I would suggest the major cause would be the fact that your collector load consists of 30 kOhm,in parallel with 500kOhm,in turn parallel with the Collector circuit stray capacitance.

At audio,the reactance of this stray capacitance is very high & can be neglected,but as the frequency is increased,it becomes much smaller,& starts to shunt the collector load,reducing stage gain.

Back in the tube days we learned a "rule of thumb" for the rise time of a similar amplifier.
tr= 2.2 Cs RL
Where tr is the rise time,Cs is the stray capacitance, & RL was the (then) anode load.

Then,knowing tr,we could derive the "3dB down point" from  0.35/ tr.
« Last Edit: October 01, 2014, 09:33:16 am by vk6zgo »
 

Online T3sl4co1l

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Re: Is this the miller effect at work?
« Reply #5 on: September 30, 2014, 02:51:03 am »
Or you can just cut right to the quick and go...
f_0 = 1 / (2 * pi * R * C)

It should be unsurprising that,
2.2 / 0.35 ~= 2pi

2.2 = 11/5,
0.35 = 7/20
and,
(11/5) * (20/7) = 44/7 = 2 * 22/7, which of course is double the common approxicant.

That's old tech knowledge for 'ya -- passed down, rarely scrutinized, but usually good, and for a reason.

Tim
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Offline EvertonTopic starter

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Re: Is this the miller effect at work?
« Reply #6 on: September 30, 2014, 03:15:39 am »
Thanks guys.  I was wondering why increasing quiescent current gave me more bandwidth, but Vk6zgo's explanation of stray capacitance makes sense.

So what would be a good way to build an amplifier with a gain of 20 and a bandwidth of say 20Mhz?  Should I build multiple stages at lower gains, or is there a better way to do this?
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Online T3sl4co1l

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Re: Is this the miller effect at work?
« Reply #7 on: September 30, 2014, 07:00:22 am »
You're looking for a GBW of 400MHz, which isn't going to be easy to achieve with 2N3904s (fT ~ 300MHz -- note, by the way, that fT is GBW in terms of current, not voltage; it's still possible to get real voltage or power gain at frequencies above fT!).  A single stage certainly won't, but several rubbed together might get the job done, or you can look at higher frequency devices.

One thing is true: you can always go for very marginal gain per stage, maximizing bandwidth as much as possible; then, it's only a matter of chaining enough stages to get the gain required (which goes exponential in the number of stages, after all!).  Obviously, a gain of 1 takes infinite stages, so you can't go too low; supposedly, the ideal gain per stage is something bizarre like sqrt(pi) (~1.77) or the like.  Practically speaking, you don't want to go to so much bother, copy-and-pasting stages together, so you'd shoot for a modest amount, like 5-10 per stage.  This also helps with noise, which is additive per stage, so if your gain is very small, noise will quickly dominate (the noise, accumulated from the first couple stages, only grows exponentially (along with gain) through later stages!).  A good, low noise, modest gain, front end helps magnify your signal without adding too much noise, allowing subsequent stages to be fairly noisy with little consequence.

Here's an example:

http://seventransistorlabs.com/Images/WidebandAmp.png

MMBTH10 is something like a shrunken, peppy 2N3904, with fT a bit over 800MHz, or something like that.  The V/I ratings are less than half a 2N3904's, so the scaling analogy probably isn't far off.

This amplifier is intended for 50 ohm service, so the input and output impedances are fairly low.  Hence, a common base input stage was chosen.  This still isn't very good, because shunt feedback to the base (necessary to set gain, which increases bandwidth and decreases distortion) increases that impedance, I think into the 1kohm range (which looks capacitive at high frequencies where gain starts to roll off, so it momentarily becomes a good match, then it goes bad again).

To address stray capacitance -- not so much of the stage itself (that would be hard), but of the things attached to it (like the feedback caps and the next stage), a gyrator (tube enthusiasts will recognize it as a "mu follower") load is used.  This has an emitter follower output characteristic (when taken from the emitter, obviously..), nice and low impedance, while somewhat isolating the load from the active gain node (the bottom transistor collector).  The active top load delivers about double the bias current of the first transistor, excess being sunk through the 680 ohm resistor.  I think... this was necessary to get the voltages to line up correctly, while also maintaining linearity (you don't want the transistors swinging into cutoff half the time) and extra pull-down oomph.

The second stage is the same thing, except the 680 pull-down goes through two diode drops and a small resistor (huh, it's not labeled -- seems I used 27 ohms in the real thing), which provides DC bias for the emitter follower output (which could've been more MMBTH10/81 action, but I figured the output impedance of the emitter follower was good enough to drive somewhat 'fatter' transistors, better suited to driving a low impedance; I don't think I noted much difference in the simulation*).

*But then, the simulation was claiming a 300MHz bandwidth, which is optimistic at best.  I forget how I measured 100MHz, think it was step response (against a 350MHz scope).

Anyway, the second stage is arranged as a shunt feedback amplifier, much as you'd build an inverting op-amp stage: the 470 ohm coming from the previous stage, into the transistor base, is balanced by the 2.2k coming from the output, for a gain of about 4 on this stage.  (Similarly, the 6.2k and 2.2k resistors on the base of the first stage set noninverting gain around 4.  Total gain is around 16, but between source padding and load resistance, it comes down to about 10 total, or 20dB.)

Whether this type of design is applicable to your present problem, I don't know.  For example, the "raising input impedance" trick might come in handy.  There are ways of bootstrapping the input impedance away to infinity (and back to negative impedances, for that matter), handy for very weak signals.  Or, if you don't need wideband, but only narrowband use (RF amps), transformers and inductors can do an order of magnitude better than my poorly optimized circuit.  A 2N3904 might even achieve the requisite 26dB at 20MHz, if it's tuned to a limited bandwidth (say +/-10%).

Tim
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Offline Laertes

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Re: Is this the miller effect at work?
« Reply #8 on: September 30, 2014, 04:01:14 pm »
If you want to eliminate the miller effect, you could also build a Cascode setup(see http://en.wikipedia.org/wiki/Cascode for a primer). It should be possible to pull off 1:10 gain at 20MHz with that and you could do it with just two transistors...
 

Offline IconicPCB

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Re: Is this the miller effect at work?
« Reply #9 on: September 30, 2014, 09:49:48 pm »
There are various two transistor topologies whihc will allow control over input output impedance, gain, bandwidth, group delay....

What is it You wish to achieve?
 

Offline GK

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Re: Is this the miller effect at work?
« Reply #10 on: October 01, 2014, 05:31:25 am »
Yes, and yes.


No and no.

He is driving Q4's base with a source impedance of 1 ohm. That (ball park calculation - ~10pF Cob / ~25Av gain) pushes out the ~Miller effect to ~600 MHz.

The high 30k collector resistor in conjunction with the effective capacitance to ground is the dominant bandwidth killer, but that has nothing to do with the Miller effect.


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Online T3sl4co1l

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Re: Is this the miller effect at work?
« Reply #11 on: October 01, 2014, 05:43:07 am »
Oh come on, at least say something serious like "no and yes". ;D

Miller effect is C-B capacitance multiplied; the standard analysis looks at input-referred capacitance, which magnifies Ccb by ~G.  Or in other words, instead of Vin across Ccb (collector grounded), you have Vin on one side, -Vin*G on the other, for a total of (1 + G) times.  This is capacitance referred to ground.  Since... you don't need to evaluate an "effect" if you were analyzing it with the feedback capacitance to begin with.

The effect still works, output-referred... though it's a (1 + 1/G) effect, because Vin is a small fraction of Vout.  It's still higher than Ccb alone, although in most cases, not significantly so.

So, in a technical sense, I'm still right: it's effectively increased.  Practically speaking, I'm not correct, because it's by a negligible amount, and who cares.

Tim
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Offline GK

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Re: Is this the miller effect at work?
« Reply #12 on: October 01, 2014, 05:54:06 am »
So, in a technical sense, I'm still right: it's effectively increased.  Practically speaking, I'm not correct, because it's by a negligible amount, and who cares.


 ::) The OP posted a Bode plot and asked if the of ~6MHz bandwidth limitation shown is due to the Miller effect. It isn't.

But yes, a real life circuit will unlikely be driven from a 1 ohms source, so some variant of a cascode amplifier would be required for a 20MHz bandwidth or more unless the net gain is distributed across a number of cascaded amplifier sections.
 


 
« Last Edit: October 01, 2014, 06:19:53 am by GK »
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Offline XFDDesign

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Re: Is this the miller effect at work?
« Reply #13 on: October 01, 2014, 02:06:40 pm »
To the OP, to get a better feel for what everyone is banging on about, pick up a copy of "Experimental Methods in RF Design." It's $35 US from Amazon, and it covers pretty much everything mentioned here in enough detail to explain, without really demanding a serious engineering background (though I use it often as a serious engineer).
 

Offline EvertonTopic starter

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Re: Is this the miller effect at work?
« Reply #14 on: October 02, 2014, 03:32:21 am »
It's now all falling into place.

~10pF  @ Av 20 referred to the input gives Cb =  210pF.  With a source resistance of 1 ohm that gives a 3dB point (f=1/2*pi*R*C) at ~600Mhz.  This makes it clear it is not the miller effect killing my gain.

On the output side, the equivalent load capacitance is CL = ~10pF (with or without miller), but since R= 30k, the 3dB point is at 530Khz!

I was originally asking myself why increasing the quiescent current would give me more bandwidth, but I now understand that it is simply because with a higher Ic, RC becomes smaller moving the cut off freq to the right with it.

I think that all I may end up doing is setting up 2 stages running at higher quiescent current to get me where I want to be.

Thanks!
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