Up to now, and quite deliberately, I had
practically ignored the
transmit path in favour of the receive path.
Now let's look closer. Just in case.
The transmit signal at 2nd IF is generated by the DSP, emerging from
the CODEC DAC. It is switched between feeding the speaker amplifier on
Rx and the modulator on Tx. On Tx the complex IF signal is padded
a uniform modulated output signal by using a SOT resistor (RX) with the
published setup procedure. Then it is buffered in a voltage follower
consisting of a TL072 (IC606a) driving complementary bipolar output
transistors (TR604, TR606) that are in the opamp feedback loop.
The high-power output configuration is presumably to allow driving 50R,
which the opamp cannot do alone. The buffered signal then couples via
C632 (220n) and the bus switch IC604, feeding to the 'LF arm' of the
diplexer and hence to the 2nd mixer 'LF-PORT' (MixAdaptor, not shown)
up-conversion to 10.7MHz.
Buffer Output Loading
The signal from the buffer is at low impedance, not the 50R that one
might expect. The source impedance at moderate frequencies is very
close to zero because of the 100% overall negative feedback. However,
at 15kHz the series capacitor C632 (220nF)
fortuitously has Xc of 48R, so maybe there really is a fortuitous 50R
source impedance (but only at 15kHz). The
filter output capacitor is 100nF; it is
driven by the buffer with the wanted 15kHz.
At 15kHz, the 100nF has Zc of 106R, quite low but not awful. The
impedance at the images is much lower. At the first image (33kHz) the
Zc is 48R, at the higher images it will be correspondingly less. At the
same time as Xc of this capacitor reduces, so does Xc of C632, placing
a strenuous load on the buffer at the high frequencies. I think that
reason that the
buffer doesn't oscillate parasitically with the large capacitive load
is that the bus switch series on-resistance fortuitously buffers it
about 8 to 10R). This is not really a satisfactory situation.
Another way of looking at this: The output of the buffer
IC606A+TR604+TR605 feeds through C632 (220nF) and C631 (220nF) in
series, i.e. 110nF. This then feeds to C651 (100nF). Ignoring the LC
arm of the filter for a moment, the path above is a capacitive
attenuator. Only half the signal appears at C651, frequency
independently for a capacitive divider. The load is actually the
resonant complex filter, so it is not this simple, but you probably get
the idea; signal is being lost because of the unnecessarily low value
coupling capacitors, whilst there is no formal impedance control
(although providing it is not difficult).
The low values are not, as we might think, an attempt to shape the
audio response. The incremental amplitude change at 3kHz each side of
15kHz is quite small, but also subject to selected sideband (one way it
boosts at 3kHz, the other way it cuts) and the near-15kHz point in any
corresponds to an audio modulation frequency of zero - the small effect
is therefore inconsistent with any audio frequency shaping we might
contemplate. This shaping is done in the DSP software, quite rightly.
This capacitive behaviour occurs because the filter is being driven at
which is highly capacitive. The 'L' style
of filter is not reciprocal, so driving it 'backwards' at the
output may not give the expected results. In the Rx direction, the
input is (should be) an inductor and the output a capacitor; hf
products are (should be) blocked by the inductor and not drive the
capacitor (but see my earlier ramblings about the shunt cap on the
inductor). The HF arm of the diplexer is intended to be the 10nF+47R
at the mixer port. If the diplexer has been rebuilt as per my
recommendations above, this situation will prevail.
The transmit buffer directly drives 100nF with the synthesised 15kHz
Tx IF waveform. The CODEC should have removed the image frequencies,
and even if some remain then the roofer would remove them. But the
amplifier stability with a capacitive load is questionable, so some
changes may be desired. An inductive element would be better.
Fortunately, I have already proposed this in conjunction with the
What I think is more than useful is to increase C632 (220nF) so that it
is no longer providing the 50R source impedance at 15kHz. As it is, any
residual HF noise and
the Xc of 48R at 15kHz is attenuating the very signal that we need. We
would prefer a bigger capacitor value and a 'real' resistor,
which will stay 50R even at higher frequencies. Both could mount in
series at an angle off the existing C632 pad. Of course, the image
frequencies at the buffer output would then no longer drive into a low
impedance capacitive load, but the new resistor,
so it kills 2 birds with one mod.
For a revised C632,
>2u2 as before is a minimum value, with Xc(15kHz) of 5R. Bigger
is better; I used one of my 10uF ceramics. If it is electrolytic, the
should connect to the emitters of TR604, TR606. But remember to fit 47R
Whilst pondering on this, it struck me that the charging path for this
capacitor is into the FST3125 pin. At switch-on the path is configured
as 'receive', so the switch is off. The normal d.c. state for the
buffer output, asssuming exact components and no power-on glitches, is
half the +10V rail (due to R628, R629), so there is a 5v transfer to
the FST. The absolute maximum input voltage is +7v; there seems to be
no normal rail clamp diode, so the situation is just about OK.
After agonising for a while on whether to make any changes at all, I
gave in. Although it is obvious that many Stars have been built using
the original design and without reported problems, this in itself does
not mean that everything is correct.
As I looked more deeply into the original design, it crossed my mind
that not making changes is not really an option anyway.
It really seemed as if there should be
an inductive arm for the
Transmit opamp buffer to see, so that at high frequencies it wouldn't
be seeing a capacitive load. I know that putting a 10u and 47R in
series at the buffer output, to replace C632, will let the source be
all frequencies, and isolate any capacitive load due to the filter, but
I quite wanted to at least convey a vestige of symmetry to the
The extra rolloff pole might be useful, too.
The entire Transmit 2nd IF path is modelled below, also but Elsie has
to model the 'T' filter in both transmit & receive directions. In
the plots I have set the inductor Q to 50, which has required a 10%
larger value for the capacitor (now 220nF) than with the higher
original Q of 250. If this remains at 200nF, the input impedance rises;
this is mainly visible in the reverse (transmit) direction when it
rises from 50R to 58R at 15kHz. This is unlikely to be a real problem
practice, so I have used two 100nF in parallel as before. In any case,
the lower value keeps the gain up at slightly higher frequencies, which
is not a bad thing.
The two plots below show the
response in forward & reverse direction through the revised, 'T'
The schematic above shows the reverse direction; swap the
inductors for forward path.
Tx '2nd IF' response with 'T' filter and 50R source
Here is the schematic. I used a
substitute opamp from the LT simulator.
The response over a wide range,
100Hz to 10MHz.
The point at which the HF arm of the diplexer comes in is
visible (at about 1MHz)
The response covering 10kHz to
60kHz, to show the rolloff on the first image at 33kHz (about 10dB down
This response is all we might hope for.
Surely this is OK? After all, it has been built by many STAR
constructors over the years. Nevertheless, we need to look into it even
if the outcome is 'no change'.
If you look at the circuit of the
above, you will see that the bipolar complementary
output stage of the buffer amp is surprisingly
with a 470R
the NPN side and 3k3 pulldown on the PNP.
The TL072 amplifier is only specified to drive a load in excess of
which is clearly not happening here. It is driving much less,
the static state with +5v on the opamp "+" input from the potential
divider R628 & R629 (both 10k). The output on the emitters will
at +5V also, by follower action. What our circuit is doing is drawing
around 10mA through the 470R and pushing 1.5mA down the 3k3. The
balance of 8.5mA must
into the opamp output. Statically no significant current, maybe a few
flows in the bases or into the load because it is A.C. coupled and at
The load spec of >2k for the amplifier relates to its ability to
source & sink current. This is limited by on-chip resistors at the
outputs; 64R in series with each emitter and 128R from their junction
to the output pin (from TI Datasheet) which amounts to an output
resistance of 192R for the chip, as well as any active current-limiting
(short-circuit protection). This is the resistive part
through which the surplus 8.5mA flows, dropping 1.6V and dissipating
14mW. The remainder is dissipated in the PNP chip output transistor,
3.4V at 8.5mA being 30mW.
If the input level varies (due to signal from the CODEC), the output
must change also. But in this configuration, the opamp's internal PNP
all the work; the internal NPN never
passes any current.
In addition, the
current in the PNP shifts the internal operating point of the opamp, by
requiring it to have a substantial 'no-signal' output of -8.5mA.
I have no idea why the asymmetry is present in this circuit
is cloned from DSP-10). It pulls more current through the
Vbe-offsetting diode D605, so will increase the forward drop and is
more likely to over-compensate the Vbe of TR604. If static current
flows through TR604, either it must also flow through TR606 or the 100%
-ve feedback loop will rupture, since it cannot alter the diode current
without shifting opamp output voltage and cannot shift opamp output
without also shifting the emitters voltage away from mid-point, which
is required by the -ve feedback.
The original clone source, the DSP-10, has somewhat the same circuitry:
However, you will notice two rather significant differences in the
These differences are important
- The output NPN/PNP pair is powered from +5V, rather than STAR at
- The op-amp is an LM358M; STAR uses a TL072
The reduced (5V) supply produces only
half as much current in the 470R (R28 above) for the op-amp to sink.
Also the LM358 internal circuitry does not
have any resistance in
series with the PNP device, so it is better able to sink that current
without voltage drop.
The danger of cloning circuitry is revealed again! In this case, a
handy implementation was almost cloned, but the items that were
inadvertently altered have seriously affected the operation.
The Buffer is Bad after all!
Having decided to leave the buffer alone, on the basis that others have
used it apparently with no problem, I actually took a 'scope probe and
waveforms from the DSP CODEC output (using the CW key to produce a
steady signal) and had a big surprise.
The waveform from the CODEC is a perfect sinewave at 14.2514kHz.
This is the waveform present at the C657 end of the setup resistor RX
(see Glenn's schematics or the scrap
A sinewave at 14.2514kHz and amplitude about 2.5Vpp from the CODEC.
This is with the signal level set at 100% for this band (actually 40m).
The signal is attenuated by my 'Set-on-test' resistor Rx, but still
looks sinusoidal, so I haven't shown it here.
However, the op-amp output clips the -ve peaks
badly (actually it clips
them very well
, but it
And the emitter followers TR604, TR606 look like this:
op-amp output and emitters have virtually identical waveforms.
Here is a composite image with the
input and output waveforms
superimposed. The distorted trace mean-value is shifted vertically due
to the a.c.
coupling of the 'scope, so I aligned it:
The emitter followers are within the 100% feedback path (voltage
follower) of the buffer. This is highly significant: if the PNP emitter
follower (TR606) were running out of steam and could not pull the load
negative enough, thereby clipping the waveform, the op-amp output would
not do the same, but would shoot more negative in an attempt at making
the emitter output (and thereby the opamp inverting input) go more
negative. But it doesn't - it sticks.
So it is actually the op-amp output
that is unable to go more negative.
This may not be surprising, since in order to go negative it has to
sink even more current from the 470R (R604). For every volt it goes
below mid-rail, it needs to sink another 2.13mA from R604. Even at
mid-rail (quiescent state) it is sinking 8.5mA (see sums above), so it
is already way beyond the TL072 device spec, which says that with a 2k
load (the smallest load indicated) at 25degC it will drive an output of
just above 3Vpk with split 10V (+-5V) supply, typically. Even then this
is not part of the
main spec, but it is seen on the graph (Figure 6 in the datasheet) for
output swing vs frequency. This output of 3.5Vp into 2k is, of course,
We blow 8.5mA into it even when quiescent; no wonder it can't manage
even more to take the bases of the buffer transistors much lower than
So the -ve peak clipping is categorically not due to transistor
limiting in TR606, but is entirely due to output overload of the op-amp
IC606A because of the 470R R604, whose value and purpose I have already
wondered at before.
I once said "if it works then leave well alone."
But it doesn't work. Not properly, or well.
So what can we
do to make the buffer work without clipping?
First, it would not be a bad idea to look at the architecture.
The overall buffer is a 100% feedback non-inverting amplifier (voltage
follower). We do not need to give it gain, so this is fine. The
overall 100% feedback around the whole circuit is good, because it
makes the op-amp work to correct any shortcomings (such as crossover
distortion) as best it can.
We should try hard to remove the resistive asymmetry of R604, R615
(470R, 3k3). If these can be the same high value, then when quiescent
no current will flow into the opamp (except unbalance current) so the
present need to sink considerable current at all times will be removed
op-amp can work in the 'usual' manner. If we can make the resistors
R615 both at least 4k then the combined resistive load would be 2k,
which is within the device specification.
We can take a look now to see how small in value those resistors
We are driving the output into notionally 100R (50R source resistance,
a physical resistor for my mod as described, or Xc of a 220nF which is
48R at 15kHz, plus a load which we would expect to be 50R - the
reconstruction filter). In fact, both the original filter/diplexer and
my modification can present less than 50R, so let us call that part 0R
(it can't be much less!). So our buffer should drive 50R, assuming that
my modification is done, or a capacitive load if not. Even the
capacitive load of the original filter is buffered somewhat by the
on-resistance of the Bus Switch, one section having a resistance of at
least 4R (considerably more with the original switch bias).
To drive 50R with the emitter followers, and assuming a waveform of
2.5Vpp as in the CODEC output 'scope-trace above (CODEC maximum DAC
swing is actually 2.83Vpp), then each transistor
must pull or push the peak voltage (1.25V) into 50R. This must be
comfortable, with no overload or distortion.
The peak load current is therefore (Vpk / Rload) = 1.25V / 50R =
This is the emitter current of a bipolar transistor, 2N3904 or 2N3906,
as emitter follower. Current gains are:
2N3904: min hFE
= 40 at Ic = 0.1mA;
70 at 1mA; 60 at 50mA, for Vce = 1V.
2N3906: min hFE
= 60 at Ic = 0.1mA,
80 at 1mA; 60 at 50mA, for Vce = 1V.
Since we seek resistor symmetry, let's assume both transistors are as
bad as the
NPN, and that they have hFE
(because maximum current gain is
needed when the output voltage and current is highest in magnitude).
load current is 25mA, so using the 50mA figure for hFE
is falling, is reasonable. This
is our minimum specified
The base resistor is the only source of current into the base, since
the diodes prevent the op-amp driving base current (the op-amp can
only 'steal' current).
If the op-amp output is at mid-rail (quiescent), then due to the
symmetry equal current
flows in each base resistor, none into the op-amp and none into the
bases (or only a tiny value). As the signal input appears, it wants
the emitters to slew to 1.25V away from their mid-level, in turn as the
signal goes from +ve to -ve peak. The op-amp output will do what is
within its power
to make the emitters achieve this, by changing
voltage as necessary. Such is the effect of negative feedback.
For either transistor emitter to drive the necessary 25mA into the load
on a peak, it will need a base current of (25mA / hFE
for convenience. In
fact it will be fractionally less, because the 0.417mA base current
also flows into the load, but we can ignore this as it is a small
proportion of 25mA.
Now the million dollar question is: can the 0.4mA be made available by
a realistic base resistor, which we have said would be well chosen at
Well, can it?
If the signal is moving by 1.25V, then both resistors change current
since the diodes node voltage must also change. Assume a -ve peak (just
because...). The emitter junction must move negative by 1.25V, to keep
the feedback loop intact. Each base will also move negative by a tad
more than 1.25V, but it is OK to call it 1.25V for now. The difference
is due to the internal resistances of the transistor base &
emitters, including intrinsic resistance and diode resistance. So the
diode junction also moves -ve by 1.25V. The current in R604 rises by a
bit, and that in R615 falls by approximately the same bit.
current in R615 enough to drive 0.4mA into the base of TR606? Well, the
base is 0.6V (approx, Vbe) less positive than the emitter for the PNP
transistor, and the emitter need to be 1.25V below quiescient,
which is mid-rail 5V. The voltage across R615 will therefore be (5V
-1.25V -0.6V) which is 3.15V at the negative peak, and because we have
guessed it at 4k the current in it will be (3.15V / 4k) = 0.79mA. We
want to take 0.4mA into the base, which leaves the opamp and R604
between them to mop up the remaining 0.4mA (OK, I rounded again, it's
0.39mA really!). At least there is some to mop up, so we win.
Having done this, we can easily see that 4k is very reasonable. If it
were 8k, we would be on the verge of running out of base current, but
as it is, we have 0.4mA in hand, perfectly fine.
The similar argument applies on the +ve peak, but the directions of all
voltages and currents invert.
This all sounds rather convincing, but what about all this current that
needs 'mopping up'? All I said was that "the other half and the
op-amp will take care of it". But how do they do this? What happens to
the spare 0.4mA?
Because the op-amp output nominally follows the output voltage
(assuming that the transistor Vbe's are compensated exactly by the
diode forward drops), then the current in the 'other' resistor will
change also. None will flow into the other transistor base; that
transistor is not contributing to the peak output current. Instead of
that resistor passing it's quiescent current of (5V - 0.6V(diode)) / 4k
(for a 4k resistor) it will now pass (5V - 0.6V(diode) + 1.25V(signal))
/4k which is 1.413mA. The other resistor is only passing 0.4mA into
its diode, the rest goes into the transistor base as we saw.
of the 1.413mA can pass straight into the opposite resistor, leaving
1.013mA unaccounted for. Where does it go? The op-amp kindly mops it
his only desire is that the whole circuit should balance in this
condition, short-lived though it may be.
Because the op-amp in this simple design is at worst driving two
resistors each of 4k in parallel, it is within the manufacturers
specification. We could slightly reduce the resistors, maybe, since the
diodes D605 & D606 reduce their effect by reducing their load
current. In fact, everything seems perfect!
What our simple design has not done (yet) is to put a resistor in each
transistor emitter, the junction being the output of the circuit. These
resistors pad the ever-changing intrinsic
the emitter (and hence collector) current does not tend to 'run away'
to the point where the transistor is destroyed, in the event of a
slight residual base current being passed if the base diode happens to
have a larger forward voltage than the transistor Vbe. Even our op-amp
includes these - see the 64R in the equivalent circuit below:
Do we need these emitter resistors? And do we need a more sophisticated
anti-crossover bias scheme than the present two diodes?
be good design practice, and would avoid needing to hope that no
quiescent current flows in the transistors that might cause a thermal
runaway. Also, do we need distortion?
But it is more difficult to implement than simply changing the values
of two resistors. There are no spare pads for the emitter resistors, so
they would need fitting into the emitter legs of each transistor. Any
changes to the simple 2-diode biassing might need extra components
either in place of the diodes, or in their leads.
we have sanitised the op-amp output requirement enough, we can leave
some crossover to be taken care of by the 100% feedback loop, hoping
that the voltage 'gap' will not be excessive and that the op-amp output
can slew rapidly over it. After all, we are only interested in 15kHz
2kHz for our SSB '2nd IF', anything else will be removed by filters
(the CODEC spec, our nice new reconstruction filter and the
roofing filter). So maybe
crossover distortion is not the same unpleasant problem that it is in a
wideband audio amplifier? We could try
, (it is no worse than
the original design for this) or we can add lots of components as
described above to avoid any problems.
I'm going to suck it & see.
The first buffer modification
We have gone a long way round, in order to do a proper design on this
circuit. All that has actually happened is that we have ended up with
different values for two resistors. So what is the change?
Simples! We replace both R604 (470R) and R615 (3k3) with 4k or
thereabouts. 3k9 sounds good; 4k7 would probably be OK but we may run
short of base current. The op-amp no longer has to sink 8.5mA
quiescent, nor operate way outside it's spec, nor is half of it's
output stage inactivated by the current it always sinks. Everything is
balanced and the op-amp can do it's job as I presume was originally
If we wanted to put emitter resistors and better quiescent biassing
parts in place, it would be rather more difficult to implement,
although distortion would reduce somewhat and we would be protected
from thermal runaway.
But let's try it first without these; after all, being inside a (now)
fully functional 100% negative feedback loop will reduce distortion
greatly, and so long as the biassing ensures quiescent non-conduction
of the transistors then we will not get thermal runaway. Fingers
It works well, with no clipping of -ve peaks using the same setup as
above. There is, however, flattening of the +ve peaks. This was
actually there before, but the horrific -ve peak clipping overshadowed
Positive Peak Clipping
The +ve peak clipping looks like this at the output pin (pin 1) of the
op-amp IC606A. It is not explained by any figures calculated above. Are
my sums wrong, or does it come from somewhere else (I am prejudiced
into thinking it must be somewhere else, of course). You
can also just see the crossover points of the output stage, at the
graticule centre X-axis. I have kept
the range knobs in view for clarity.
This is the waveform at the
output of the bus switch, IC604 pin 8. It shows the waveform clipping
is happening before the
buffer amp. Sorry about the visible flyback above the trace.
No clipping at the DAC input to the bus switch IC504 pin 9,
from the bias network R630, R631 (both 10k).
Just for the record, this is again the input to the bus switch
but d.c. coupled. The 0V origin is set on the bottom graticule line.
The clipped peak is 7.4 divisions up at 0.5V/division, i.e. +3.7V for a
measured Vcc on IC604 of 5.031V.
Yes, but "my"
IF doesn't do that!
You may think I'm chasing something
peculiar to my own STAR. Well, it's a reasonable first reaction and not
exactly an uncommon one. The fact that I can show by calculation that
there will be trouble doesn't seem to stop this reaction, for some
Here's the waveform at the emitters of the output stage on another,
unmodified buffer which, unlike my own, still feeds the output the
original way, through C632, 220nF with no 47R in series. It is a bit
worse than the one I have already shown and, yes, it is someone else's:
The signal is produced the same way, by keying when in SSB mode. The
set-on-test resistor 'RX' is still a short-circuit, ready for
calibration. But the buffer is clearly not ready for calibration! The
-ve clip is due to the low value of R604 (470R) and the +ve clip is
Is this a good sinewave? Is it a good replica of the sinewave from the
CODEC? Not by the normal definition, no.
Thanks to Glenn for the photo of this signal on his latest STAR.
Tell me how yours looks
So why is it
clipping if it is not the buffer or DAC?
If you have by chance already read my discussion on biassing levels for
, you may guess what is coming. I wrote the
bus-switch bit some time before finding this first-hand example! Ah
well, at least I already know the cure.
If you look at what I said about biassing bus-switches at mid rail, you
may agree with my conclusion that it is the Wrong Thing To Do. This is
a good demonstration!
The output from the DAC is a signal of 2.83Vpp at full wallop
(data-sheet spec). This, due to ac coupling and mid-rail biassing on
IC604 (FST3125) results in a signal swinging from (2.5V - 1.42V) to
(2.5V + 1.42V), which positive level goes above the turn-off threshold
of the bus switch FET (see the bus switch section; it turns off
completely at 1.5V below the +ve supply, i.e. +3.5V. TI parts have a
worse threshold than Fairchild).
Well, 2.5V + 1.42V is 3.92V
for the signal, but the FET is off with around 3.5V! Also, the
FET on-resistance has been increasing exponentially as the voltage
approches this level, but since
this switch is not in a 50R path the effect is not severe until it is
really and finally 'off'.
The flattening of the waveform above, at the output of the switch,
occurs at 3.7V, a bit more than it might be but clearly it is turning
off and causing clipping, exactly as I said it would! Because this is
not a 50R circuit but several kOhmhs, the clipping shows suddenly as
the FET actually pinches off. If it were a 50R path there would have
been severe non-linearity at somewhat lower levels.
Does it improve after altering the bias point?
Of course it does! Here are two 'after' images. After I
fitted a 5k6 in parallel with R631 (10k) to shift the bias point from
2.5V to 1.3V. As you can readily see, the clipping is gone. No
surprises. The d.c. coupled trace of the bus-switch IC604 pin 8 shows
the new bias point. The waveform neither reaches cut-off nor goes below
0V, as it will not for anything other than a large asymmetrical
waveform (where the capacitive coupling causes much more signal to
appear on one or other side ofthe mean value - this is inevitable in
The first trace is the signal, d.c. coupled, at the bus switch pin 8
showing the corrected bias level after adding 5k6 in parallel with
R631. Once again the origin (0V) is the lowest graticule line. The
flyback line (above the trace) is produced by the 'scope,
The second trace is the resulting waveform at the buffer opamp IC606A
look Mum, no clipping! The crossover point is visible as a small
kink in the waveform near the mid-point - remember that this is the
part-modified buffer with no Class AB crossover bias so far.
Why doesn't this happen on 'my' radio?
It does happen!
In fact, 'you
' have made it as
severe as possible by the setup you
performed! In the Tx calibration, you
fitted a value for 'RX' that
ensures that, on the band with lowest output (from the PA), the DAC
output level is at maximum. On the other bands, where less signal is
needed for full output, you adjust it downwards with the per-band
settings, which scale the digital drive to the DAC. But what you did
was to ensure that the maximum possible signal reaches the bus switch -
giving most likelihood of this clipping mechanism occurring. Had you
set 'RX' to zero and used digital scaling to reduce the DAC signal, the
DAC level would be less and the problem may be invisible. Also, the
negative peak clipping happens. It is a designed-in 'feature'. Take a
OK, now that I have found this first-hand manifestation of a bus-switch
biassing problem that I had already predicted, and successfully applied
the cure (but only to this one position...) I am ready to modify and
try out the proper crossover biassing of the buffer. At last!
Properly implemented crossover biassing should remove that visible
'kink' in the output waveform at the mid-point of the sinusoidal test
signal, and at the same time make the stage safe from thermal runaway.
modification: safe biassing
I am unhappy that the complementary NPN/PNP output pair are biassed
solely from the drop in 2 diodes, with no formal crossover
biassing and no prevention of excessive offsets causing thermal
runaway (the transistors are directly across the 10V supply). We can
only hope that it remains in Class B. If the
transistors warm up enough that their combined Vbe drops below the
forward conduction drop of the two diodes, then current will flow
through the transistors directly from the +10V rail to ground. This
increase the temperature and therefore hFE
transistors and hence further
increase the current whilst reducing re
. This is thermal runaway
recipe for transistor destruction. That it appears so far not to have
a miracle, not a manifestation of good design.
What is needed, if we are not to penny-pinch and take unnecessary
is some padding resistance in the emitter of each transistor, and a
formal method for causing some quiescent current to flow (to prevent
crossover distortion). Ideally this should be possible to implement
reversibly on the present PCB. Also, it would be excellent if no
adjustment pot (common in audio amplifier designs) were needed. Let's
Since the diodes and transistors are all
wire-ended, it is reasonable
to add components into the circuit.
By adding four resistors as shown below (R9, R10, R11, R12), the
criteria are met:
This works as follows (remember we tend to start at the output with
When quiescent, no current flows from the output (labelled Vamp)
through the 2u2 capacitor.
Please bear in mind that the values I have used are, so far, just a
'guess' at what we might end up with. We might need to change them once
the sums are done.
The op-amp has two inputs.
On the '+' input is a mid-rail (+5V) bias due to the two 10k resistors
and the 10V supply.
On the '-' input is the same level as on the point 'Vamp' at the amp
The op-amp will try to alter its output voltage to make these the same;
the nature of feedback circuits.
I have shown an op-amp output resistor R13 of 1R, this was to allow me
experiment with op-amp output resistance tolerances and is set at 1R to
effectively disable it.
Because the Q1, Q2 and associated components circuitry is symmetrical,
be at mid-rail when the op-amp output is also at mid-rail, i.e. no
current is drwan through the op-amp output in order to achieve this
nominal balance (some may flow in order to correct slight imbalance,
but it will be small).
Since no current flows in the Filter (d.c. is blocked by C1 in the
diagram) and an insignificant amount flows in the op-amp output, we can
calculate voltages & currents. I will assume that, at least
nominally, the diode forward drop is equal to the transistor Vbe for
now and that each is 0.6V. Later we can correct this assumption.
The current in the network R1, R9, D1, D2, R10, R2 is therefore easy to
calculate since it flows from +10V to ground. It is (10V - [2 x
0.6V] / (3k3 + 62R +62R +3k3) = (8.8V / 6724R) = 1.31mA
So the surplus voltage on each base is the drop in 62R due to 1.31mA
This causes base current to flow in each
transistor emitter. Because
we have nominally offset the Vbe against the diode drop, the current
that will flow (provided hFE
it) is (81.22mV / 10R), since the excess voltage appears across the
10R emitter resistors, i.e. the quiescent current is (81.22mV / 10R) mA
emitter resistance for now).
Since the transistor hFE
<0.13mA base current is
needed in order to obtain this emitter current, so the 1.31mA in the
base network is not greatly affected. This
If variations (e.g. mismatch or thermal effects) occur, any change in
the conditions will reflect in a change in quiescent transistor bias
8.122mA. Even if the Vbe and diode-drops mismatch by ±50mV, the
quiescent emitter currents will only alter by ±5mA from the
values. In the original design, complete thermal runaway was possible
under such conditions. So we have considered the diode/Vbe match
earlier and found it to be reasonable and to cope with at least
We should consider re
of the transistors. This emitter resistance is made up from the
physical resistance of the bond wires and leads, plus the dynamic
resistance of the device. The latter is the more significant at low
current, and is roughly (26R
/ Ie[mA]). At our quiescent current of 8mA, this becomes (26/8) =
3.5 Ohms. This is lower than the physical resistors (for which I
suggested 10R) which is a good sign. In this calculation I have
deliberately ignored any contribution from the base circuit resistance,
partly because this is not just a simple resistor but is a complex
circuit including the op-amp. Simulation will take care of this later.
Now, if an external input to the op-amp '+' pin appears, the op-amp
try to move the network mid-point in such a way and by such an amount
as will produce the same signal on the '-' input, because of the high
open-loop gain of the op-amp and the nature of feedback circuits. If a
1V positive change occurs on the '+' pin, the op-amp will try to
produce the same +1V change at the output. To do this, it must move
its own output by more than 1V positive because there are two
resistive attenuators: the 62R against the 3k3 (which is very little)
and the 10R against the load of maybe 50R (which is not so little). The
op-amp output will have to swing more than 1V positive by roughly 10R /
50R for a 50R load, or 200mV (note that the load is nominally 100R,
made from the output series resistance (source R) and the nominal load
resistance in series, but for safety we are considering a load of 50R).
Since the op-amp output with +-5V supplies (ours is 'split' 10V) and 2k
load is shown as nominally ±3V under these conditions,
figure 6) we should be OK for an input swing of ± (3V x 50/60)
for a 50R load, i.e. a swing of ±2.5V into 50R. The CODEC
output is 1.42Vpk; this is trimmed down by 'RX' and cannot exceed 1.42V
anyway, so our design is sound even with a 50R load (we have introduced
a series 47R into the load already, making nearer 100R, so this is very
I mentioned "penny-pinching" earlier. In my book, the addition of four
small conventional wire-ended resistors is not an expensive thing to
do. Making the modification to add them is a bit tricky, but why were
they not in the original design (and hence present in the copper
The quiescent power dissipation in each output transistor is around
(8.122mA x 5V), i.e. 45mW, which is well below the transistor maximum
600mW derated at 5mW/°C above 25C.
With thermal resistance of
junction-ambient of 200°C/Watt, the junction rise at 45mW is (200 x
45/1000)°C, only 9°C, which is trivial, so our quiescent bias
not much contribute to device temperature. At peak signal output,
assuming a squarewave (which is the worst possible condition), the
current into the load will be 1.42V / 50R (again assuming a shorted
filter), 28.4mA. Adding this to the quiescent 8mA and assuming that the
transistor still has 5V dropped (it is less for a resistive load), then
the worst-case possible device dissipation is 5V x 28.4mA for half the
time, and maybe the quiescent current for the other half, when the
transistor conducts signal into the load, i.e. a dissipation of 71mW +
This resulting 116mW will raise temperature by 23.2°C above
leaving plenty in hand. Derating at 5mW/°C above 25°C, this
us a local ambient of up to around 125°C, which sounds like no
These 'extra' resistors can be fitted in-line with the existing
component leads, especially if tiny wire-ended resistors are used. This
fulfills my requirement for minimum intrusion.
Oh, the simulated response of the whole
The green line shows the op-amp gain rolloff with frequency due to it's
internal unity-gain compensation; it it the voltage between the op-amp
Light blue is the buffer output, and dark blue is the filter output.
rolloff is due to the coupling capacitors. Note that it is flat
(almost) at 15kHz but cuts the first image at 33kHz by at least 10dB,
as a conventional reconstruction filter should.
Schematic of buffer showing the modifications
Glenn has once again kindly altered his schematic of this area to show
Real measurement of modified buffer
There's nothing like seeing it work, to bear out all that theory!
I (actually Jackie, my XYL, since I have a few bruises at present)
modified my PCB by adding parts as in the schematic. We used different
but similar spec transistors, BC337 (NPN) and BC327 (PNP) because I cut
out the original 2N3904 & 2N3906 to avoid damage to the PCB, and
had no more of them. The pinning on my substitutes is reversed relative
to the case moulding, but that is easily dealt with by mounting them
180deg rotated. So long as your preferred part has no less hFE
it will probably be OK!
Here is a photo:
The 10R's are mounted vertically in the emitter holes in the PCB, taken
up the outside of the TR case and soldered to the bent-up emiter pin. I
used miniature 10R but had none in 62R so used larger parts, as you can
This is the 'scope picture of the op-amp output, no crossover nicks:
Compare this with the first
modification, that didn't use class AB
And similarly at the final output, the emitter side of C632, no
distortion, no surprise:
They look almost identical, because there is now no crossover
distortion; the feedback loop does not need to try
to leap a crossover 'gap' at each mid-point.
The level of quiescent current as designed is 8.122mA, but the actual
value is quite rightly highly
dependent on the precise forward-drops of the diodes, the individual
transistor Vbe values and the eventual value of re
(depending on emitter current).
The acid test is to measure the volt-drop on one of the 10R emitter
resistors, with no signal, to see what quiescent current has been
achieved in my own amplifier. Yours may differ, but should be a few mA,
say 4-12mA, depending on diode-to-transistor variation.
I measured 52mV, corresponding to a quiescent bias of 5.2mA. The design
value, which assumed identical drops in the transistor Vbe and the
diodes, and made no allowance for re
, was 8.122mA. If we
had also allowed 3.5R of re
in series with each 10R emitter
resistor, the expectation would reduce from 8.122mA to 5.9mA - but the re
at 5.9mA would be higher - so achieving 5.2mA is very satisfactory.
of quiescent current is good for a low-power circuit of this type, and
the anti-crossover bias that gives us
class AB operation is clearly all present and correct.
For the addition of 4 resistors and some altered values (and maybe,
like me, a new pair of
transistors after cutting out the originals in order to avoid damage to
the PCB), we have now got rid of any
distortion and, overall, done it like it should have been.
So should we all add these modifications?
The answer is the same as it always is, with a bit more bite this time.
It is entirely
your own choice
I have presented is not, and never will be, an
I have presented all the information, including full design
calculations, in good faith, to help you decide. There are 'scope
traces of various levels of 'before' and 'after' where these are
helpful. There is simulation output. You
. Bear in mind that by modifying your STAR with any
of these changes you will make it 'non-standard'. I can not offer any
I would respectfully suggest that the Tx buffer changes should
definitely be done, since the 'standard' design distorts terribly and
unnecessarily. If you want to minimise the mods, at least replace R604
(470R) with 3k3 to match R615 (3k3) and solder a 5k6 above R631. Not
the full monty, quite easy to do, and at least it will remove from the
Tx Buffer that severe peak clipping that is a 'feature' of the standard
design. But while you are doing that, why not do the full thing!
Now all that is necessary is to check
This done, and happy that the changes have not caused problems,
we can return to the overall description with a reasonable solution