active collector loads (audio freq)

[Jim Thompson]

No. I'd have asked you directly, here. I didn't think it was so but
on a hunch, I checked and I found it in Hans Camenzind's 7MANUAL.PDF
p. 5-11.

<quote>
As we have noted above, the voltage gain in the bipolar transistor
is anything but linear.
Current gain (hFE), on the other hand, is a naturally linear
parameter. For this reason alone it is
easier to achieve high performance stressing current rather than
voltage amplification.
</quote>

Huh? "Stressing" as in talking about it? The linearity/nonlinearity
of the parameter of choice affects performance? What? the
performance of communicating thoughts?

Then he talks about the swing across the Miller cap - you know this
story already, but to keep things in context:

<quote>
But there is a second reason.
Each junction has a capacitance
(created by the "space-charge
region"). Of particular bother is the
collector-base capacitance. Not
that it is especially large (it isn't),
but it is badly situated. Using the
transistor as a voltage amplifier,
base and collector terminals move
in opposite directions (i.e. they are
180o out of phase). Since the
transistor is capable of a large
voltage gain (especially with a
current source load), the voltage
swing at the collector can be
several hundred to several
thousand times as large as that of the base.
</quote> ... you know the rest.

So he'd have no voltage gain at all?

He goes on with the Miller effect a bit more and then he talks about
the cascode fix for the Miller cap and here's the killer:

<quote>
The cascode stage is only a halfhearted use of current
amplification. A better approach (at least
for high-frequency performance) would be to avoid converting to a
voltage altogether.
</quote>

That's where I was left hanging. Maybe he just means that the large
swing across Mr. Miller kills the highs.

That's where slew-rate and "power-bandwidth" come into play in OpAmp
descriptions.

...Jim Thompson

 

[Active8]

That's where slew-rate and "power-bandwidth" come into play in OpAmp
descriptions.

So IOW pumping up the Miller cap with a stiff current is preferred
to a large voltage with small current capability?

 

[Jim Thompson]

On Thu, 30 Dec 2004 11:22:48 -0700, Jim Thompson wrote:
[snip]

That's where slew-rate and "power-bandwidth" come into play in OpAmp
descriptions.

So IOW pumping up the Miller cap with a stiff current is preferred
to a large voltage with small current capability?

Huh?

...Jim Thompson

 

[Active8]

On Thu, 30 Dec 2004 11:22:48 -0700, Jim Thompson wrote:
[snip]

That's where slew-rate and "power-bandwidth" come into play in OpAmp
descriptions.

So IOW pumping up the Miller cap with a stiff current is preferred
to a large voltage with small current capability?

Huh?

If there's not enough current to pump the miller cap, slew rate
suffers. Could you spell it out a bit more for me please, or am I on
track? He said it's best to not convert to a voltage at all.

 

[John Woodgate]

I read in sci.electronics.design that Active8 <[email protected]>

So IOW pumping up the Miller cap with a stiff current is preferred to a
large voltage with small current capability?

You seem to be well adrift.

For a stage with significant Miller effect present, there MUST be enough
current available to charge the Miller cap at the highest rate demanded
by changes in the signal. If not, gross non-linearity occurs. Normally,
this charging current is derived from the d.c. collector current of the
active device.

Whatever the voltage available, insufficient current remains
insufficient.

 

[Active8]

I read in sci.electronics.design that Active8 <[email protected]>


You seem to be well adrift.

For a stage with significant Miller effect present, there MUST be enough
current available to charge the Miller cap at the highest rate demanded
by changes in the signal. If not, gross non-linearity occurs. Normally,
this charging current is derived from the d.c. collector current of the
active device.

The driving device, right?


"I need current"
Cc
||
| +---||------+
| | | || |
| | | |
| | | |/
| +-----+---------|
| | |>
V | |
| |
Imax |
| |
| |
| |
|
|

^^^
"I have current"

SR = Imax/Cc

Whatever the voltage available, insufficient current remains
insufficient.

Yes. I know you need enough current to charge the cap. Han's (see
quoted text that was snipped) said it's best to not convert to a
voltage at all. That's what I'm confused about. I think he's saying
that the Miller stage has voltage gain (non-cascode) which shows up
across the cap and therefore much current is needed to charge it up
(plus, of course, the cap value is mutiplied by the voltage gain
which exacerbates the situation.) So we'd like to keep the voltage
swing low while amplifying current?

 

[Winfield Hill]

Active8 wrote...

I found it in Hans Camenzind's 7MANUAL.PDF p. 5-11.

He goes on with the Miller effect a bit more and then he talks about
the cascode fix for the Miller cap and here's the killer:

<quote>
The cascode stage is only a halfhearted use of current amplification.
A better approach (at least for high-frequency performance) would be
to avoid converting to a voltage altogether. </quote>

That's where I was left hanging. Maybe he just means that the large
swing across Mr. Miller kills the highs.

No, it's a general comment. One can do better sticking with current-
amplification stages, if one can devise them. Once you start going
down that path, it's surprising what can be accomplished. I've only
played with this a little myself, but I have a few books, some thesis
copies and plenty of papers on the subject. "Analogue IC Design, the
Current-Mode Approach," C. Taumazou, et al, is an oft-quoted example.

Generally one eventually has to convert the signal to a voltage, but
sometimes one can continue with current all the way out of the IC.
For example, many multiplier, DA and DDS chips have current outputs.

 

[John Woodgate]

I read in sci.electronics.design that Active8 <[email protected]>

Yes. I know you need enough current to charge the cap. Han's (see quoted
text that was snipped) said it's best to not convert to a voltage at
all. That's what I'm confused about.

Me, too. I abandoned the concept of 'converting a current to a voltage',
or vice versa, around the age of 14 and I don't really expect to back-
track now.

I think he's saying that the Miller
stage has voltage gain (non-cascode) which shows up across the cap and
therefore much current is needed to charge it up (plus, of course, the
cap value is mutiplied by the voltage gain which exacerbates the
situation.) So we'd like to keep the voltage swing low while amplifying
current?

Yes, if you have a stage that has Miller capacitance, configuring it as
a current amplifier (i.e. making the load impedance low) gives you more
bandwidth. But you now need enough d.c. collector current to supply the
load impedance.

 

[Tim Wescott]

Active8 wrote...



No, it's a general comment. One can do better sticking with current-
amplification stages, if one can devise them. Once you start going
down that path, it's surprising what can be accomplished. I've only
played with this a little myself, but I have a few books, some thesis
copies and plenty of papers on the subject. "Analogue IC Design, the
Current-Mode Approach," C. Taumazou, et al, is an oft-quoted example.

Generally one eventually has to convert the signal to a voltage, but
sometimes one can continue with current all the way out of the IC.
For example, many multiplier, DA and DDS chips have current outputs.

Now that's current thinking!

 

[Active8]

Now that's current thinking!

LOL. Thanks John, Win, Tim, Jim So the trade off is slew rate
over Miller compensation, i.e. using the gain to increase the
effective Cc.

What would be a practical example of effecting compensation without
sacrificing slew rate, i.e., small Cc? I understand the dominant
pole and the concept of pole splitting and that I can put an extra
pole on the output. So what other design(s) would keep Cc low while
still achieving compensation?

Also, IIRC, Win and Jim agreed that lower diff stage gm can increase
slew rate - is that what was said? How can that be if the diff stage
current is lower (lower gm)?

 

[Jim Thompson]

[snip]

Also, IIRC, Win and Jim agreed that lower diff stage gm can increase
slew rate - is that what was said? How can that be if the diff stage
current is lower (lower gm)?

No, not quite. Lower gm allows a smaller pole-splitting capacitor,
so, for the same available current, slew rate increases.

...Jim Thompson

 

[Active8]

[snip]

Also, IIRC, Win and Jim agreed that lower diff stage gm can increase
slew rate - is that what was said? How can that be if the diff stage
current is lower (lower gm)?

No, not quite. Lower gm allows a smaller pole-splitting capacitor,
so, for the same available current, slew rate increases.

I see it, but how do I lower gm without lowering available current,
if for a bjt

gm = 40.Ic ?

 

[Jim Thompson]

[snip]

Also, IIRC, Win and Jim agreed that lower diff stage gm can increase
slew rate - is that what was said? How can that be if the diff stage
current is lower (lower gm)?

No, not quite. Lower gm allows a smaller pole-splitting capacitor,
so, for the same available current, slew rate increases.

I see it, but how do I lower gm without lowering available current,
if for a bjt

gm = 40.Ic ?

Pad resistors in emitters ?

...Jim Thompson

 

[Active8]

On Thu, 30 Dec 2004 16:46:20 -0700, Jim Thompson wrote:

Pad resistors in emitters ?

Cute. You're serious? Hard to tell when you use the positive
emoticon more often than not. And is the question mark indicating a
question or is indicative of your hair style


We are talking the same toplogy, right? The emitter current
source/sink (the available current) provides emitter current equally
for both sides of the diff pair when in balance... The same current
that would go through those resistors... So to lower the gm I have
to lower the current available.

 

[Jim Thompson]

On Thu, 30 Dec 2004 16:46:20 -0700, Jim Thompson wrote:


Cute. You're serious? Hard to tell when you use the positive
emoticon more often than not. And is the question mark indicating a
question or is indicative of your hair style


We are talking the same toplogy, right? The emitter current
source/sink (the available current) provides emitter current equally
for both sides of the diff pair when in balance... The same current
that would go through those resistors... So to lower the gm I have
to lower the current available.

Pondering my quicky smart-ass remark I think it's really like this:
reduce gm by reducing current... but required Cp goes down faster than
the current, so you gain slew-rate... I think... I need to review the
theory myself... usually I just wing it ;-)

...Jim Thompson

 

[Active8]

Pondering my quicky smart-ass remark I think it's really like this:
reduce gm by reducing current... but required Cp goes down faster than
the current, so you gain slew-rate... I think... I need to review the
theory myself... usually I just wing it ;-)

Here's as much as I can help to get started and I don't see where it
helps Call I_1 the current through one branch and Imax the tail
current. At frequencies where Cc dominates, the gain A

A = gm/(s.Cc)

at unity gain freq

w = gm/Cc

SR = Imax/Cc = 2*I_1/Cc = 2*I_1*w/gm

So lowering gm at the same freq (w) appears to allow a smaller Cc,
but doing so reduces I_1 and I'm back to square_1.

This is a good topic considering some PhD or another wrote that the
only way to increase slew rate is to raise the unity gain freq or
increase the saturation voltage of the diff pair, the latter leaving
me cold for now.

 

[Stephan Goldstein]

On Thu, 30 Dec 2004 16:46:20 -0700, Jim Thompson wrote:


Cute. You're serious? Hard to tell when you use the positive
emoticon more often than not. And is the question mark indicating a
question or is indicative of your hair style


We are talking the same toplogy, right? The emitter current
source/sink (the available current) provides emitter current equally
for both sides of the diff pair when in balance... The same current
that would go through those resistors... So to lower the gm I have
to lower the current available.

Nope, Jim's right. If you add small resistors in series with each emitter
and tie the common points together (and to the current source) you'll
reduce the _effective_ gm of the stage. It's called "emitter degeneration"
and is a common trick.

Remember that gm is the ratio between output current and input voltage.
The input voltage appears across the base-emitter junction. Adding
the resistors provides some very local negative feedback that acts to
reduce the signal seen across the b-e for a given differential input voltage.
Hence the differential output current is less although the standing current
is unchanged.

If you construct a DC-level small-signal model of the whole thing and go
through all the math, comparing it with a non-denerated circuit operating
at the same current, you'll see that the stage gm is reduced according
to the ratio of GM*RE to 1 (where GM is the value calculated purely on
the basis of the standing current, and RE is the value of each resistor).

Emitter degeneration isn't the answer to all problems. For one thing, it
adds noise, so is right out if that's your primary concern.

steve

 

[Active8]

Nope, Jim's right. If you add small resistors in series with each emitter
and tie the common points together (and to the current source) you'll
reduce the _effective_ gm of the stage. It's called "emitter degeneration"
and is a common trick.

Thanks. I did think about that - I even thought "effective" or "AC"
gm, but Jim's a guru and I'm still not sure he can't add a little
more to this. Or Win or a few others for that matter. There might
even be a resistorless way or another explaination- like Jim alluded
later on in the thread, to wit:

<quote>
Pondering my quicky smart-ass remark I think it's really like this:
reduce gm by reducing current... but required Cp goes down faster
than
the current, so you gain slew-rate... I think... I need to review
the
theory myself... usually I just wing it ;-)

Remember that gm is the ratio between output current and input voltage.
The input voltage appears across the base-emitter junction. Adding
the resistors provides some very local negative feedback that acts to
reduce the signal seen across the b-e for a given differential input voltage.
Hence the differential output current is less although the standing current
is unchanged.

If you construct a DC-level small-signal model of the whole thing and go
through all the math, comparing it with a non-denerated circuit operating
at the same current, you'll see that the stage gm is reduced according
to the ratio of GM*RE to 1 (where GM is the value calculated purely on
the basis of the standing current, and RE is the value of each resistor).

Emitter degeneration isn't the answer to all problems. For one thing, it
adds noise, so is right out if that's your primary concern.

And reduces distortion.

 

[Stephan Goldstein]

Thanks. I did think about that - I even thought "effective" or "AC"
gm, but Jim's a guru and I'm still not sure he can't add a little
more to this. Or Win or a few others for that matter. There might
even be a resistorless way or another explaination- like Jim alluded
later on in the thread, to wit:

Well, I may lack official "guru" status, but let me try to elaborate...

Think about the small-signal hybrid-pi model of a transistor. In its simplest
incarnation, neglecting ac effects entirely, it's just a resistor between base
and emitter, and a dependent current source from collector to emitter. The
current-source's output is gm*Vbe, where Vbe is the small-signal (incremental)
value appearing across the base-emitter resistor. If the emitter is grounded
then the gm is exactly what you'd calculate from the textbook equation:

gm = (q * Ic)/kT.

[I've assumed here that the input signal is applied to the base, and that
there's some bias network that fixes the emitter current at the desired value
but that doesn't interfere with the small-signal performance. This is easy if
you imagine this transistor is half a differential pair, a little trickier for a
single transistor in isolation].

Now add a resistor between the emitter and ground. With the same input
signal still at the base, the incremental voltage across the base-emitter
junction is clearly less than before. The gm in the small-signal model is
unchanged (it's determined by the DC emitter current, which I've assumed
remains constant) but since the output current is gm*Vbe (remember, the
incremental Vbe is used here) the _effective_ gm is reduced. The math to
prove this is not too bad, some algebra that I won't type out - it's edifying
to do this on your own. But the short answer is that the effective gm of
the circuit becomes:

gm' = gm/(1 + (gm*RE))

where gm is the original value from my first equation.

What's happened is that you've managed to reduce the stage gain but
have not changed the standing current. This allows you to retain the
slew rate and improve stability, all other things being equal.

This is well-explained in many standard textbooks. I learned it from
Gray & Searle (the other Gray), now out of print, but I think you can
find it in Gray & Meyer. There's also a great development in Volume
3 (or 5) of the old SEEC series, but good luck finding them!

HTH.

Steve (@ Analog Devices for nearly 23 years)

 

[Active8]

Well, I may lack official "guru" status, but let me try to elaborate...

^^^^^^^^^^^^^^
"official" guru

Sorry, I didn't mean to imply that. It's just that I've seen many
posts from Jim and others regarding their experience and
accomplishments. Now I know you a little better.<snip>

I've understood emitter degeneration for ages, but thanks. It's just
that when Jim says:

<quote>
reduce gm by reducing current... but required Cp goes down faster
than
the current, so you gain slew-rate... I think... I need to review
the
theory myself... usually I just wing it ;-)
</quote>

I get to thinking there's some non-linear relation he needs to
recall. And his " ? " from the original suggestion left me
wondering.

But it's clear to me without the analysis (which I'll do anyway
because I like to see numbers) that a reduced swing will allow a
smaller Miller cap while the higher DC gm (more available current)
allows a high slew rate... something like that.

gm' = gm/(1 + (gm*RE))

It's always nice to have an answer to shoot for.

Just like A = Rc/(re + RE) and without the eqiv ckt derivation:

Rc
A = ----------- = gm'.Rc
re + RE

Rc
= -------------
1/gm + RE

gm.Rc
= -------------
1 + gm.RE

gm.Rc
A = gm'.Rc = -------
1 + gm.RE

gm
gm' = ------------
1 + gm.RE


That was an ugly hack I'll do it the right way later.<snip>

Thanks Stephan. Happy new year and however you spell [C]Haunika...
*that* holiday.