active collector loads (audio freq)

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!

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.

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.

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:

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

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:

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.

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

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

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.

The driving device, right?

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

V | | | | Imax | | | | | | | | |

^^^ "I have current"

SR = Imax/Cc

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?

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

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

Pad resistors in emitters ?:-)

...Jim Thompson

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)?

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

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

gm = 40.Ic ?

I read in sci.electronics.design that Active8 wrote (in ) about 'active collector loads (audio freq)', on Thu, 30 Dec 2004:

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.

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.

I read in sci.electronics.design that Active8 wrote (in ) about 'active collector loads (audio freq)', on Thu, 30 Dec 2004:

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.

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.

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.

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:

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 ;-)

And reduces distortion.

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

^^^^^^^^^^^^^^ "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.

I've understood emitter degenerati 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 ;-)

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.

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.

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

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.

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

HTH.

Steve (@ Analog Devices for nearly 23 years)