Maintaining a Vbe Multiplier's bias value

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

: +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 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?

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

On Mon, 08 Feb 2010 17:39:49 -0800, Jon Kirwan
jonk_at_infinitefactors.org> 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

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

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- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Hang a big capacitor across it.

On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

Hang a big capacitor across it.

Nice try.

Jon

What's a "VAS"?

What exactly are you trying to do?

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

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

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

On Mon, 08 Feb 2010 20:43:03 -0800, Jon Kirwan
jonk_at_infinitefactors.org> wrote:

On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

Hang a big capacitor across it.

Nice try.

Jon

No, seriously, that solves a bunch of problems.

John

On Mon, 08 Feb 2010 20:49:24 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 08 Feb 2010 20:43:03 -0800, Jon Kirwan
jonk_at_infinitefactors.org> wrote:

On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> 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

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.

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.

"Jon Kirwan" <jonk_at_infinitefactors.org> wrote in message
news:okr1n55h5dvjjklg760dllkqq50v7s38ib_at_4ax.com...
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

On Tue, 9 Feb 2010 00:28:27 -0600, "Tim Williams"
tmoranwms_at_charter.net> wrote:

"Jon Kirwan" <jonk_at_infinitefactors.org> wrote in message
news:okr1n55h5dvjjklg760dllkqq50v7s38ib_at_4ax.com...
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.

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
Have you considered making R2 and/or R3 constant current devices