Maintaining a Vbe Multiplier's bias value

I think this fits in sci.electronics.design, not .basics.

I'd like to consider the Vbe multiplier often used in audio
amplifiers to maintain a bias voltage for the output stage.
The purpose is to better mitigate against ripple in the
unregulated power supply rails and against the the VAS
voltage output resulting from amplified signal voltages.

(The only active device under consideration is a BJT, though.
No JFETs or MOSFETS or opamps or other ICs.)

The basic starting form for a Vbe multiplier is shown in Fig.
1 and the bias voltage output is indicated there. Assume Q1
is thermally coupled in some magic way, for now, in just the
right way so that if the current through the Vbe multiplier
were perfectly stable, that the bias voltage would track just
as needed (The 'Eg' of Q1 is exactly what's needed for the
output stage's temperature tracking in some nice way and the
values of R1 and R2 are set correctly and the thermal
coupling and location is somehow where it needs to be.) The
focus is on the Vbe multiplier's variation of bias in the
face of changes in sourcing current at the top of Fig. 1.

>: +V
>: |
>: resistor or
>: current source
>: |
>: ,---+---,
>: | |
>: \ |
>: / R2 |
>: \ +-----> upper quadrant
>: / | ^
>: | | |
>: | |/c Q1 BIAS
>: +-----| VOLTAGE
>: | |>e |
>: \ | |
>: / R1 | v
>: \ +-----> lower quadrant
>: / |
>: | |
>: '---+---'
>: |
>: VAS ---'
>:
>: FIGURE 1

If I use a resistor as the load for the VAS, it's obvious to
me that the Vbe multiplier will need to cope with varying
currents. But even if I use a BJT (or two) to make a current
source sitting above the Vbe multiplier, it's still not going
to hold entirely still with +V ripple and with varying VAS
drive voltages. That variation will ultimately manifest
itself in a varying Vbe bias voltage. That will change the
operating point for the output stage.

If it is class-A, I suppose it doesn't matter that much. But
I don't want to be forced into class-A operation. Nor do I
want to be forced into regulated rails. So it becomes a
little more important, I think, to get this nailed down
better.

There's the problem, anyway.

To quantify how bad all this really is, I tried my hand at
figuring out the small signal analysis of the Vbe multiplier.
If I got a first approximation about right, it is based
squarely upon the small-re of the BJT. The very familiar
value for (kT/q)/Ic.

There is also the value of R2 shown in Fig. 1, but since its
effect is only affected by the change in base current, I
believe it's contribution is divided by Q1's beta. So the
actual equation is something like:

R_ac = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta

For a 2X multiplier where R2 is about R1, this is:

R_ac = (2/Ic)*(kT/q) + R2/beta

The Vbe multipler value is:

V_bias = Vbe*(1+(R2/R1)) + R2*Ic/beta

(The latter term being a correction for base current.)

Ignoring base current for now and assuming I had Ic set
around 5mA and placed R1=R2=1k for the 2X factor, this R_ac
value works out to about 15.4 ohms.

A variation of half an mA in Ic yields about 7.7mV change in
the bias point.

I decided to see if the Early effect made much of a
difference. The adjustment appears to be something like
this:

R_early = dV/dI = -Ic/VA*R^2

If I'm interpreting it right, it really does show as negative
resistance added to R_ac. The fuller equation, then,
including the Early effect, would be:

R_ac = (2/Ic)*(kT/q) + R2/beta - Ic/VA*R^2

(Which requires a quadratic solution to solve for R.)

If R_ac is 15.4 ohms and Ic is around 5mA, a VA of 100V would
suggest about R_early=-10mOhms. Which is roughly a factor of
1500 less than 15.4 Ohms. Since it now appears to be on the
order of 0.1% or so for typical Ic, VA, and, R_ac values, I
think I can ignore it for these considerations.

So drop it, I will.

I had scouted around a few weeks back (not for this reason)
and found what is shown in Fig. 2. I remembered it, but
didn't understand it then.

>: +V
>: |
>: resistor or
>: current source
>: |
>: ,---+---, <-- node A
>: | |
>: | \
>: | / R3
>: \ \
>: / R2 /
>: \ |
>: / +-----> upper quadrant
>: | | ^
>: | |/c Q1 |
>: +-----| BIAS
>: | |>e VOLTAGE
>: \ | |
>: / R1 | v
>: \ +-----> lower quadrant
>: / |
>: | |
>: '---+---'
>: |
>: VAS ---'
>:
>: FIGURE 2

I think I now understand why R3 was there. Changes in Ic
create changes in Q1's collector voltage, per Ic*R3. The
result is that dV=dI*R3. If R3 is on the order of the above
computed R_ac, then variations at node A caused by changing
currents through the Vbe multipler (most of which are seen as
Ic changes) will be neatly compensated for the change in the
voltage drop caused by R3.

However, that can only be set for some assumed Ic. Nearby
changes will work pretty well. But further deviations will
start to show problems again. Also, the Fig. 2 version will
use a slightly higher multiplier value to get node A up high
enough for the R3 drop to hit the right place required to
bias the output stage. That higher multiplier means that
while, let's say, the two (or four, if that's it) output
BJT's Vbe values vary over temp and the thermally coupled Q1
above also varies it's own Vbe value, the multiplier other
than 2 (or 4) will mean the variation of the bias will match
at only one place -- if it ever did more than one spot. How
important that is, I've not considered yet.

I'm wondering about additional topology changes to improve
the performance still more. Obviously, if they are crazy and
wild, I'm probably going to live with the above and be done
with it. But I think there's got to be something still
better. Another BJT as a bypass route across Q1 and R3?

Getting this nailed down should help mitigate against both
unreg supply ripple (on one side, anyway) putting hum into
the output and also against large scale changes in the VAS
amplified signal voltage (which means distortion.)

Jon

John Larkin wrote:

> On Mon, 08 Feb 2010 17:39:49 -0800, Jon Kirwan
> <(E-Mail Removed)> wrote:
>
>
>>I think this fits in sci.electronics.design, not .basics.
>>
>>I'd like to consider the Vbe multiplier often used in audio
>>amplifiers to maintain a bias voltage for the output stage.
>>The purpose is to better mitigate against ripple in the
>>unregulated power supply rails and against the the VAS
>>voltage output resulting from amplified signal voltages.
>>
>>(The only active device under consideration is a BJT, though.
>>No JFETs or MOSFETS or opamps or other ICs.)
>>
>>The basic starting form for a Vbe multiplier is shown in Fig.
>>1 and the bias voltage output is indicated there. Assume Q1
>>is thermally coupled in some magic way, for now, in just the
>>right way so that if the current through the Vbe multiplier
>>were perfectly stable, that the bias voltage would track just
>>as needed (The 'Eg' of Q1 is exactly what's needed for the
>>output stage's temperature tracking in some nice way and the
>>values of R1 and R2 are set correctly and the thermal
>>coupling and location is somehow where it needs to be.) The
>>focus is on the Vbe multiplier's variation of bias in the
>>face of changes in sourcing current at the top of Fig. 1.
>>
>>
>>>: +V
>>>: |
>>>: resistor or
>>>: current source
>>>: |
>>>: ,---+---,
>>>: | |
>>>: \ |
>>>: / R2 |
>>>: \ +-----> upper quadrant
>>>: / | ^
>>>: | | |
>>>: | |/c Q1 BIAS
>>>: +-----| VOLTAGE
>>>: | |>e |
>>>: \ | |
>>>: / R1 | v
>>>: \ +-----> lower quadrant
>>>: / |
>>>: | |
>>>: '---+---'
>>>: |
>>>: VAS ---'
>>>:
>>>: FIGURE 1
>>
>>If I use a resistor as the load for the VAS, it's obvious to
>>me that the Vbe multiplier will need to cope with varying
>>currents. But even if I use a BJT (or two) to make a current
>>source sitting above the Vbe multiplier, it's still not going
>>to hold entirely still with +V ripple and with varying VAS
>>drive voltages. That variation will ultimately manifest
>>itself in a varying Vbe bias voltage. That will change the
>>operating point for the output stage.
>>
>>If it is class-A, I suppose it doesn't matter that much. But
>>I don't want to be forced into class-A operation. Nor do I
>>want to be forced into regulated rails. So it becomes a
>>little more important, I think, to get this nailed down
>>better.
>>
>>There's the problem, anyway.
>>
>>To quantify how bad all this really is, I tried my hand at
>>figuring out the small signal analysis of the Vbe multiplier.
>>If I got a first approximation about right, it is based
>>squarely upon the small-re of the BJT. The very familiar
>>value for (kT/q)/Ic.
>>
>>There is also the value of R2 shown in Fig. 1, but since its
>>effect is only affected by the change in base current, I
>>believe it's contribution is divided by Q1's beta. So the
>>actual equation is something like:
>>
>> R_ac = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta
>>
>>For a 2X multiplier where R2 is about R1, this is:
>>
>> R_ac = (2/Ic)*(kT/q) + R2/beta
>>
>>The Vbe multipler value is:
>>
>> V_bias = Vbe*(1+(R2/R1)) + R2*Ic/beta
>>
>>(The latter term being a correction for base current.)
>>
>>Ignoring base current for now and assuming I had Ic set
>>around 5mA and placed R1=R2=1k for the 2X factor, this R_ac
>>value works out to about 15.4 ohms.
>>
>>A variation of half an mA in Ic yields about 7.7mV change in
>>the bias point.
>>
>>I decided to see if the Early effect made much of a
>>difference. The adjustment appears to be something like
>>this:
>>
>> R_early = dV/dI = -Ic/VA*R^2
>>
>>If I'm interpreting it right, it really does show as negative
>>resistance added to R_ac. The fuller equation, then,
>>including the Early effect, would be:
>>
>> R_ac = (2/Ic)*(kT/q) + R2/beta - Ic/VA*R^2
>>
>>(Which requires a quadratic solution to solve for R.)
>>
>>If R_ac is 15.4 ohms and Ic is around 5mA, a VA of 100V would
>>suggest about R_early=-10mOhms. Which is roughly a factor of
>>1500 less than 15.4 Ohms. Since it now appears to be on the
>>order of 0.1% or so for typical Ic, VA, and, R_ac values, I
>>think I can ignore it for these considerations.
>>
>>So drop it, I will.
>>
>>I had scouted around a few weeks back (not for this reason)
>>and found what is shown in Fig. 2. I remembered it, but
>>didn't understand it then.
>>
>>
>>>: +V
>>>: |
>>>: resistor or
>>>: current source
>>>: |
>>>: ,---+---, <-- node A
>>>: | |
>>>: | \
>>>: | / R3
>>>: \ \
>>>: / R2 /
>>>: \ |
>>>: / +-----> upper quadrant
>>>: | | ^
>>>: | |/c Q1 |
>>>: +-----| BIAS
>>>: | |>e VOLTAGE
>>>: \ | |
>>>: / R1 | v
>>>: \ +-----> lower quadrant
>>>: / |
>>>: | |
>>>: '---+---'
>>>: |
>>>: VAS ---'
>>>:
>>>: FIGURE 2
>>
>>I think I now understand why R3 was there. Changes in Ic
>>create changes in Q1's collector voltage, per Ic*R3. The
>>result is that dV=dI*R3. If R3 is on the order of the above
>>computed R_ac, then variations at node A caused by changing
>>currents through the Vbe multipler (most of which are seen as
>>Ic changes) will be neatly compensated for the change in the
>>voltage drop caused by R3.
>>
>>However, that can only be set for some assumed Ic. Nearby
>>changes will work pretty well. But further deviations will
>>start to show problems again. Also, the Fig. 2 version will
>>use a slightly higher multiplier value to get node A up high
>>enough for the R3 drop to hit the right place required to
>>bias the output stage. That higher multiplier means that
>>while, let's say, the two (or four, if that's it) output
>>BJT's Vbe values vary over temp and the thermally coupled Q1
>>above also varies it's own Vbe value, the multiplier other
>>than 2 (or 4) will mean the variation of the bias will match
>>at only one place -- if it ever did more than one spot. How
>>important that is, I've not considered yet.
>>
>>I'm wondering about additional topology changes to improve
>>the performance still more. Obviously, if they are crazy and
>>wild, I'm probably going to live with the above and be done
>>with it. But I think there's got to be something still
>>better. Another BJT as a bypass route across Q1 and R3?
>
>
> Hang a big capacitor across it.
>
> John
>
actually, I was going to suggest a diode in the base circuit to VAS to
help with thermo issues with that type of circuit..

oh well.

On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin
<(E-Mail Removed)> wrote:

>Hang a big capacitor across it.

Nice try.

Jon

On Mon, 08 Feb 2010 19:09:28 -0700, Jim Thompson
<To-Email-Use-The-Envelope-(E-Mail Removed)> wrote:

>What's a "VAS"?

Sorry. I read it somewhere regarding audio amplifiers and
the term stuck in my mind, I suppose. It's short-hand for
Voltage Amplifier Stage. It's almost so simple that no one
would bother creating a term for it, except that it seems as
though someone did and folks have used it in places where
I've been reading.

By the way, if you look at the semi-conceptual schematic at
the top of this page:

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

You will see Q3 acting as the VAS. Together with R6 it
converts the beta multiplied current into drive voltage.

(The Vbe/Ic transfer nasties this up, but I think it may be
survivable. Everything is important, but I'm leaving
worrying about this till later.)

That schematic isn't entirely realistic, either. R3/R4 are
better replaced with a mirror, regular, Wilson, or otherwise.
R5 is often itself a current source or sink (depending on
which way you flip the schematic polarities) and may be a BJT
and diodes or two BJTs, etc.

>What exactly are you trying to do?

If you look again at the schematic mentioned above, note the
function of D1 and D2. They stack to create a bias voltage.
That's used to set the point of operation for the output
stage (two-quadrant emitter follower -- which may be just two
BJTs as in that picture, or more.) Often, this is replaced
with an adjustable BJT configured as a Vbe multiplier. That's
what I'm trying to do. Except that I'd like to have the +V
and -V supply rails (ground is also present in the system) be
unregulated.

Part of the function of the Vbe multiplier is to also track
the Vbe requirements for the output stage as it heats up and
cools down. The variation of Vbe is quite large, as you
know, where the controlling Eg term in the Is(T) equation
overwhelms the otherwise oppositely-signed dV/dT of the
Shockley equation. Above -2mV/K. And with the exponential
dependance of Ic on Vbe... well, it serves that function as
well. So the Vbe value needs to track temperature in just
such a way that it maintains the design operating point for
the output stage, over temperature, while also ignoring
variations in the current that sources through it.

I'm trying to keep my options open, regarding the amplifier's
class. If it were operating class-A all the time, my limited
understanding suggests that some variation across the Vbe
multiplier isn't nearly as important as it clearly would be
for, say, class-B operation. I'm not exactly sure where I
want to wind up biasing things.

So I am slowly learning this stuff and, assuming the Vbe
multiplier has some part within it thermally coupled as
appropriate to some well-chosen part of the output stage,
trying to gather how I'd: (1) stabilize the voltage at some
fixed temperature T against variations in the current flowing
through it, and (2) calibrate it's Vbe multiplication factor
in just the right way so that it tracks well with the
effective Eg found in the Is(T) function of the output stage
needed to hold the operating point steady vs temperature.

My question here was regarding (1), not (2). I'm not far
enough along on that one to even begin on that one, yet. To
be honest, I just started learning about audio amplifier
design, including terms like VAS, starting around the 26th
last month. So I may be far off the mark in a few places.

I'm finding it a very interesting education, though, and I'm
glad I started down the road a small bit. But "being exact"
about what I want remains part of the learning process,
itself. So what you see here is as far as I've gotten to.

>My nickname, as a kid engineer at Motorola (48 years ago), was "Vbe"
>Thompson, because I could pull so much magic with Vbe compensation
>methods ;-)

Well, I can believe it. And I mean that as a sincere
compliment. If you can suggest something still better than
what I've already posted, I'd like to look at it.

>(Vbe multipliers generally are used just to create a smaller dead-band
>that is temperature stable.

In this case, I want it to track the output stage so I'm
going to have to couple it thermally in some useful way. What
I'm considering, right now, is how to make it immune to
unregulated supply variations and VAS output voltage swings.

>Class AB bias is an art form of which I
>am expert, but cannot divulge publicly at this time :-)

Well, I want to examine class-AB at some point. It may be
where I want to settle, though class-B would be quite fine
for my needs.

If you can't help with class-AB, then you can't. I will have
to struggle along. However, anywhere else you can send me a
clue I'd certainly appreciate it.

There is no interest other than personal. Certainly nothing
commercial in mind. I'm just a hobbyist trying to learn.

Jon

On Mon, 08 Feb 2010 20:49:24 -0800, John Larkin
<(E-Mail Removed)> wrote:

>On Mon, 08 Feb 2010 20:43:03 -0800, Jon Kirwan
><(E-Mail Removed)> wrote:
>
>>On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin
>><(E-Mail Removed)> wrote:
>>
>>>Hang a big capacitor across it.
>>
>>Nice try.
>>
>>Jon
>
>No, seriously, that solves a bunch of problems.
>
>John

Which problems does a slew-dependent, C*dV/dt bypass current
solve?

Jon

"Jon Kirwan" <(E-Mail Removed)> wrote in message
news(E-Mail Removed)...
> Part of the function of the Vbe multiplier is to also track
> the Vbe requirements for the output stage as it heats up and
> cools down.

The general idea is to put the Vbe transistor on the same heatsink as the
outputs, if not glued to a transistor directly.

Unfortunately, for widely mismatched current densities, this doesn't work.
http://webpages.charter.net/dawill/Images/Ampere.gif
In this boringly typical circuit, the 2N3904 Vbe mult. doesn't have enough
tempco to compensate the far beefier (= lower current density??) output
darlingtons.

I was thinking of adding another CCS so a constant voltage drop appears on
the Vbe's base divider resistor. Algebraically subtracting a fairly stable
voltage results in the effective tempco (percentwise) increasing. The base
divider ratio has to be changed to compensate.

> In this case, I want it to track the output stage so I'm
> going to have to couple it thermally in some useful way. What
> I'm considering, right now, is how to make it immune to
> unregulated supply variations and VAS output voltage swings.

Don't worry about stability -- as John said, bypass and forget about it.
Most of the dynamic VAS/CCS current flows into the output stage, since
that's what it's there for anyway. The capacitor helps turn on the N side /
turn off the P side for rising edges and vice versa.

As for PSRR, the CCS's and gobs of feedback keep that in check. Of course,
in principle you need something to start the CCS's. ICs do this with a JFET
(i.e. current regulating diode) or bandgap reference (e.g., TL431), or
sometimes both, to set a master current, from which everything else is
mirrored. Most discrete circuits just use a resistor, which is "0%" PSRR,
but it's not all that bad because the currents are balanced (*on average*,
which means you'll see IMD products when it's moving).

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

On Mon, 08 Feb 2010 22:35:05 -0800, John Larkin
<(E-Mail Removed)> wrote:

>On Tue, 9 Feb 2010 00:28:27 -0600, "Tim Williams"
><(E-Mail Removed)> wrote:
>
>>"Jon Kirwan" <(E-Mail Removed)> wrote in message
>>news(E-Mail Removed). ..
>>> Part of the function of the Vbe multiplier is to also track
>>> the Vbe requirements for the output stage as it heats up and
>>> cools down.
>>
>>The general idea is to put the Vbe transistor on the same heatsink as the
>>outputs, if not glued to a transistor directly.
>>
>>Unfortunately, for widely mismatched current densities, this doesn't work.
>>http://webpages.charter.net/dawill/Images/Ampere.gif
>>In this boringly typical circuit, the 2N3904 Vbe mult. doesn't have enough
>>tempco to compensate the far beefier (= lower current density??) output
>>darlingtons.
>>
>>I was thinking of adding another CCS so a constant voltage drop appears on
>>the Vbe's base divider resistor. Algebraically subtracting a fairly stable
>>voltage results in the effective tempco (percentwise) increasing. The base
>>divider ratio has to be changed to compensate.
>>
>>> In this case, I want it to track the output stage so I'm
>>> going to have to couple it thermally in some useful way. What
>>> I'm considering, right now, is how to make it immune to
>>> unregulated supply variations and VAS output voltage swings.
>>
>>Don't worry about stability -- as John said, bypass and forget about it.
>>Most of the dynamic VAS/CCS current flows into the output stage, since
>>that's what it's there for anyway. The capacitor helps turn on the N side /
>>turn off the P side for rising edges and vice versa.
>>
>>As for PSRR, the CCS's and gobs of feedback keep that in check. Of course,
>>in principle you need something to start the CCS's. ICs do this with a JFET
>>(i.e. current regulating diode) or bandgap reference (e.g., TL431), or
>>sometimes both, to set a master current, from which everything else is
>>mirrored. Most discrete circuits just use a resistor, which is "0%" PSRR,
>>but it's not all that bad because the currents are balanced (*on average*,
>>which means you'll see IMD products when it's moving).
>>
>>Tim
>
>This topology, thermally coupled Vbe multiplier, was mediocre 50 years
>ago. And still is.

A trek of a thousand miles starts with but the first step.

Jon

"Jon Kirwan" <(E-Mail Removed)> schrieb im Newsbeitrag
news:(E-Mail Removed)...
>I think this fits in sci.electronics.design, not .basics.
>
> I'd like to consider the Vbe multiplier often used in audio
> amplifiers to maintain a bias voltage for the output stage.
> The purpose is to better mitigate against ripple in the
> unregulated power supply rails and against the the VAS
> voltage output resulting from amplified signal voltages.
>
No this is not the purpose of this stage. It is used as an adjustable Zener
and is used to create and temperature compensate the bias voltage of the
output stage. A pur DC function. Since the power Transistors draw quite a
bit of quiescent current the ripple of the power supply will be higher then
without. The supply ripple is reduced mainly by negative feedback from
output to the input stage.

snip>
>>: +V
>>: |
>>: resistor or
>>: current source
>>: |
>>: ,---+---,
>>: | |
>>: \ |
>>: / R2 |
>>: \ +-----> upper quadrant
>>: / | ^
>>: | | |
>>: | |/c Q1 BIAS
>>: +-----| VOLTAGE
>>: | |>e |
>>: \ | |
>>: / R1 | v
>>: \ +-----> lower quadrant
>>: / |
>>: | |
>>: '---+---'
>>: |
>>: VAS ---'
>>:
>>: FIGURE 1
>
> If I use a resistor as the load for the VAS, it's obvious to
> me that the Vbe multiplier will need to cope with varying
> currents. But even if I use a BJT (or two) to make a current
> source sitting above the Vbe multiplier, it's still not going
> to hold entirely still with +V ripple and with varying VAS
> drive voltages. That variation will ultimately manifest
> itself in a varying Vbe bias voltage. That will change the
> operating point for the output stage.
>
> If it is class-A, I suppose it doesn't matter that much. But
> I don't want to be forced into class-A operation. Nor do I
> want to be forced into regulated rails. So it becomes a
> little more important, I think, to get this nailed down
> better.
>

In class A you do not use this kind of bias generator

snip

I think you should understand before calculating. It is not much of a use to
view this stage in isolation without the output stage and the associated NFB
paths.

ciao Ban

On Mon, 8 Feb 2010 23:47:24 -0800 (PST), "(E-Mail Removed)"
<(E-Mail Removed)> wrote:

>Have you read Randy Slone's power amplifier book? This stuff really
>isn't rocket science. Nor is AB. ;-)

I haven't. It's not rocket science. But it is interesting
at my level.

><http://www.amazon.com/High-Power-Aud...uction-Manual/
>dp/0071599258/ref=dp_ob_title_bk>

I'll look, but the title appears more on the contruction
side. I am using this to educate myself a little better.

>The black art is all in assembly, protection circuitry, and making
>sure it starts up cleanly. [Most engineers never look at start up, so
>you get designs that thump when you power them. I have lots of gear
>with power-on thumps.]

I recall reading of a recommendation suggesting that all
electronic devices use less than 1W when on, but not
performing their intended application. This would also seem
to require a little added effort to achieve, as well. But I
take your point.

>I'd pick one of his MOS designs. Bipolar designs often have good
>intentions, but ring like a bell. MOS is mushy, but predictably mushy.

I'm still learning about BJTs. In fact, that's what this is
about for me. MOS later. ICs later. BJTs now.

Jon

On Mon, 08 Feb 2010 23:32:00 -0800, Robert Baer
<(E-Mail Removed)> wrote:

><snip>
> Have you considered making R2 and/or R3 constant current devices
>(depletion FETs are good here)?

I've not. Could you be more specific?

Jon