active collector loads (audio freq)

[[email protected]]

just wondering if there is a reference out there that goes into lotsa
detail about using BJT current sources as collecor loads in analog
amps. why can't i look at the output of an actively loaded gain stage
(open loop) with a scope? if it's a "current output," shouldn't the
scope's input Z turn the current into a voltage? the authors brush
over this in Art of Electronics, though there is plenty of detail about
current sources and mirrors themselves. in the AoE, it is stated that
an active load should not be heavily loaded, or much of the gain will
disappear, but if it's a current output, shouldn't the next stage be
low Z for best current transfer? i'm so confused.

one more thing about differential stages loaded with a current mirror:
with a differential amp, i thought the differential signal was
represented by the collector current difference between the two
transistors (before conversion to a voltage). but the mirror forces
the collector currents to be largely equal. why doesn't this eliminate
the output signal?

if one wants to actively load a differential stage, and have a
differential output at the same time, how does one go about this? the
current mirror restricts you to a single ended output, since the gain
on the diode-connected BJT side is low.

please help! thanks.

Sean B

 

[John Woodgate]

I read in sci.electronics.design that [email protected] wrote

just wondering if there is a reference out there that goes into lotsa
detail about using BJT current sources as collecor loads in analog
amps. why can't i look at the output of an actively loaded gain stage
(open loop) with a scope? if it's a "current output," shouldn't the
scope's input Z turn the current into a voltage? the authors brush
over this in Art of Electronics, though there is plenty of detail about
current sources and mirrors themselves.

No, it's not brushed over. You are missing what I tend to agree is at
first a not-so-easy visualization of what is happening in the circuit.

You *can* look at the output of an actively-loaded gain stage with a
scope, but the input impedance of the scope will change the *voltage*
gain. A very good active load might have an impedance of 100 Mohms, so
even a 10 Mohm scope probe will greatly reduce the *voltage* gain.

in the AoE, it is stated that
an active load should not be heavily loaded, or much of the gain will
disappear, but if it's a current output, shouldn't the next stage be
low Z for best current transfer? i'm so confused.

If your next stage is a low impedance, there is not a lot of point in
using an active load, unless you need to squeeze the last drop of
*current* out of the first stage. If you do, you are not at all
interested in the *voltage* gain of the first stage but its *current*
gain or transconductance, depending on which of the input signal
parameters you are interested in.

 

[Kevin Aylward]

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


No, it's not brushed over. You are missing what I tend to agree is at
first a not-so-easy visualization of what is happening in the circuit.

You *can* look at the output of an actively-loaded gain stage with a
scope, but the input impedance of the scope will change the *voltage*
gain. A very good active load might have an impedance of 100 Mohms,
so even a 10 Mohm scope probe will greatly reduce the *voltage* gain.


If your next stage is a low impedance, there is not a lot of point in
using an active load, unless you need to squeeze the last drop of
*current* out of the first stage. If you do, you are not at all
interested in the *voltage* gain of the first stage but its *current*
gain or transconductance, depending on which of the input signal
parameters you are interested in.

This is why I use a common mode feedback feeding extra emitter followers
in http://www.anasoft.co.uk/Mospoweramp2.jpg. The two stage gain amounts
to the order of 160db gain. The 130db of feedback at low frequencies
gets one rather low distortion

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Active8]

just wondering if there is a reference out there that goes into lotsa
detail about using BJT current sources as collecor loads in analog
amps. why can't i look at the output of an actively loaded gain stage
(open loop) with a scope? if it's a "current output," shouldn't the
scope's input Z turn the current into a voltage? the authors brush
over this in Art of Electronics, though there is plenty of detail about
current sources and mirrors themselves. in the AoE, it is stated that
an active load should not be heavily loaded, or much of the gain will
disappear, but if it's a current output, shouldn't the next stage be
low Z for best current transfer? i'm so confused.

No that loads thing down.

one more thing about differential stages loaded with a current mirror:
with a differential amp, i thought the differential signal was
represented by the collector current difference between the two
transistors (before conversion to a voltage). but the mirror forces
the collector currents to be largely equal. why doesn't this eliminate
the output signal?

Say it's an NPN input stage. When the currents thru the bjts become
unbalanced, the following stage either gets the surplus current from
the active load or gives some up to the controlling bjt.

if one wants to actively load a differential stage, and have a
differential output at the same time, how does one go about this? the
current mirror restricts you to a single ended output, since the gain
on the diode-connected BJT side is low.

You want to mirror an extra transistor off the diode connected
reference.

 

[Active8]

This is why I use a common mode feedback feeding extra emitter followers
in http://www.anasoft.co.uk/Mospoweramp2.jpg. The two stage gain amounts
to the order of 160db gain. The 130db of feedback at low frequencies
gets one rather low distortion

Interesting circuit. You mean the common mode feedback going to the
active loads on the input pair?

I'd like to hear your explaination of the next stage. I'm only
halfway understanding its operation. Looks like emitter followers
with an improved Widlar current mirror at the emitters. Cascode
outputs with all that surrounded by the miller caps. Looks like the
feedback current out the bottom of all that bucks the current in its
respective diff input branch.

 

[Jim Thompson]

Interesting circuit. You mean the common mode feedback going to the
active loads on the input pair?

I'd like to hear your explaination of the next stage. I'm only
halfway understanding its operation. Looks like emitter followers
with an improved Widlar current mirror at the emitters. Cascode
outputs with all that surrounded by the miller caps. Looks like the
feedback current out the bottom of all that bucks the current in its
respective diff input branch.

Looks to me like Kev got overwhelmed with his own BS ;-)

130dB feedback... ROTFLMAO!

...Jim Thompson

 

[Kevin Aylward]

Looks to me like Kev got overwhelmed with his own BS ;-)
Nonsense.


130dB feedback... ROTFLMAO!

It does do 130db of feedback at LF, according to spice. Its two stages
of cascode amplification. 160 db of gain is 2 of 80db. This is 2 of
10,000. A non cascode stage might do Va/Vt = 50/25m = 2000. So, a
cascode can certainly get 10,000 in one stage.

Why don't you actually spice it before opening your mouth and inserting
ones foot in it.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Active8]

It does do 130db of feedback at LF, according to spice. Its two stages
of cascode amplification. 160 db of gain is 2 of 80db. This is 2 of
10,000. A non cascode stage might do Va/Vt = 50/25m = 2000. So, a
cascode can certainly get 10,000 in one stage.

I take back that part about the Wilson mirror. The other half has
the same circuit so it's just 3 emitter followers.

How many MOSFET models in SS are accurate in the sub-threshold
region? Surely none you've just dropped in from the manufacturer. If
it's as bad as Win suggests, I may as well forget expecting any
Spice results to be accurate.

I'd like to find some good models for audio power.

 

[[email protected]]

Say it's an NPN input stage. When the currents thru the bjts become
unbalanced, the following stage either gets the surplus current from
the active load or gives some up to the controlling bjt.

thanks, this is helping. which one is the controlling bjt? the one
that has the diode-connected transistor in it's collector circuit? and
if so, what is the other bjt of the input pair doing if the controlling
bjt and the mirror are doing all the work?

You want to mirror an extra transistor off the diode connected
reference.

can you describe the connections? what do you do with the diode
connected bjt? seems like you'd have to remove it from the collector
circuit of the 1st input bjt to get the gain up. where does it go?

also, generally, what's the clue for when one should think "current
operation" and when one should think "voltage operation"?
Thanks!

Sean B

 

[Active8]

thanks, this is helping. which one is the controlling bjt? the one
that has the diode-connected transistor in it's collector circuit? and
if so, what is the other bjt of the input pair doing if the controlling
bjt and the mirror are doing all the work?

Both input NPNs control current in their own branch. You can cut the
circuit down the middle to simplify it and I was explaining things
in relation to one side only.

active load current source
|
|
|
|
+------- current difference
|
|
|
NPN current, controlled

KCL, dude.
|
|

can you describe the connections? what do you do with the diode
connected bjt?

leave it as a reference

seems like you'd have to remove it from the collector
circuit of the 1st input bjt to get the gain up. where does it go?

off to the side out of the way or in between. Wherever you want to
draw it.

also, generally, what's the clue for when one should think "current
operation" and when one should think "voltage operation"?

Uh, who wants to start the list? Topology is one consideration if
you're talking about analysis. In this case, you can figure out the
currents. You might want to convert to a voltage and hit an emitter
follower with an active load on its emitter. You might not.

Scroll thru 7MANUAL.pdf and designanalogchips.pdf from
www.arraydesign.com for a pile of examples.

 

[John Woodgate]

I read in sci.electronics.design that Kevin Aylward
ueyonder.co.uk>) about 'active collector loads (audio freq)', on Sun, 26
Dec 2004:

The 130db of feedback at low frequencies
gets one rather low distortion

Until the signal reaches the end-stop, of course. x volts out,
distortion 0.00...1%. (x + 0.1) V out, distortion 1%. (x + 1) V out,
distortion 30%. (All voltages r.m.s, of course)

 

[Kevin Aylward]

Interesting circuit. You mean the common mode feedback going to the
active loads on the input pair?

Yes. It keep a low common mode impedance impedance to set the common
mode bias voltage, but a high impedance for differential signals.

I'd like to hear your explaination of the next stage. I'm only
halfway understanding its operation. Looks like emitter followers
with an improved Widlar current mirror at the emitters.

Its just a diff amp with emitter followers that have a current source.
Its not a widlar.

Cascode
outputs with all that surrounded by the miller caps. Looks like the
feedback current out the bottom of all that bucks the current in its
respective diff input branch.

Bog standard diff amp with current mirror load.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

I take back that part about the Wilson mirror. The other half has
the same circuit so it's just 3 emitter followers.

How many MOSFET models in SS are accurate in the sub-threshold
region? Surely none you've just dropped in from the manufacturer. If
it's as bad as Win suggests, I may as well forget expecting any
Spice results to be accurate.

Yes and no. BSim3 models are usually quite accurate in all regions, if
you can get the correct models. However, small signal distortion analyis
is not supported for them in spice.

I'd like to find some good models for audio power.

One cheats.

SS has two set of models for the EC10N20 and EC10P20 range. One is a
kludged BSim3 (EC10N20 and EC10P20) one is a level 1 (EC10N16 and
EC10P16). Both are set up to approximately match each other especially
the capacitances. I do runs with each type to check stability, and use
the level 1 for small signal distortion.

If one is not in subthreshold, the level 1 will give results that are
"reasonable". Who cares if its 100% in error if one is designing say at
0.01%. What one is trying to do is know that the distortion is not 0.1%
if one is after 0.01%. So, if one wants say 0.002%, one tries to get
spice to show its better by a factor of, say two. Engineering is not
exact, but one can still work around this with guestimates.

I can send you the SS files for the amp if you want.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Active8]

Active8 wrote:

One cheats.

SS has two set of models for the EC10N20 and EC10P20 range. One is a
kludged BSim3 (EC10N20 and EC10P20) one is a level 1 (EC10N16 and
EC10P16). Both are set up to approximately match each other especially
the capacitances. I do runs with each type to check stability, and use
the level 1 for small signal distortion.

If one is not in subthreshold, the level 1 will give results that are
"reasonable". Who cares if its 100% in error if one is designing say at
0.01%. What one is trying to do is know that the distortion is not 0.1%
if one is after 0.01%. So, if one wants say 0.002%, one tries to get
spice to show its better by a factor of, say two. Engineering is not
exact, but one can still work around this with guestimates.

I can send you the SS files for the amp if you want.

Thanks. I think I could learn something from that. "mcol" should
bring me up in your address book. If not let me know.

 

[[email protected]]

.. In this case, you can figure out the

currents. You might want to convert to a voltage and hit an emitter
follower with an active load on its emitter. You might not.

how does one convert? thanks for the cool web reference!

SB

 

[Active8]

. In this case, you can figure out the
how does one convert? thanks for the cool web reference!

Personally, I always think of a bjt as a voltage controlled device,
so I don't think in terms of converting a diff stage current to a
voltage, but it's referred to as a current source AFAIK because of
the high impedance of the source.

I read somewhere that it's best to not convert to a voltage at all
and I don't know wtf the author was talking about.

Those pdfs cover diff stages and opamps. One of them covers output
circuits and level shifters. Study that.

What you do is follow the diff stage with a CE stage, sometimes
called the 2nd stage, gain stage, or often, the Miller stage (google
that - also google it in the archives of this group.)

Anyway, the Miller stage does the "current" to voltage conversion
though the output of the diff stage does have a voltage wiggle to
it. The Miller stage gives you lots of voltage gain and a good place
for a compensation cap. At this point, your voltage swing is going
to be closer to one rail than the other and you might level shift it
down prior to the output. The output stage would be some kind of
emitter follower, either 1 bjt and a current source/sink or a
push-pull emitter follower. That gives you extra current drive. Of
course, you'll see from the schems of existing chips that there's
other neat stuff in there.

You can find a few versions of the 741 opamp schem on the web. The
National linear data book has schems for opamps, too.

 

[Jim Thompson]

Personally, I always think of a bjt as a voltage controlled device,

[snip]

Barf ;-)

...Jim Thompson

 

[Active8]

Personally, I always think of a bjt as a voltage controlled device,

[snip]

Barf ;-)

Ok. I learned a lot from the h-param equiv circuit with beta and
all.

Maybe you could comment on my confusion over that author saying it's
best to never convert to a voltage after the diff stage. What could
he have meant?

 

[Winfield Hill]

Active8 wrote...

Maybe you could comment on my confusion over that author saying it's
best to never convert to a voltage after the diff stage. What could
he have meant?

Was that us you're referring to? Where'd we say that?

 

[Active8]

Active8 wrote...

Was that us you're referring to? Where'd we say that?

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.