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LW1DSE > TUBES 18.03.12 15:27l 322 Lines 13251 Bytes #999 (0) @ WW
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Subj: The White Cathode Follwer
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The White Cathode Follower
--------------------------
Mr. White's improvement on the Cathode Follower was to create a
buffer that boasted a much lower output impedance and the ability to sink as
well as source current, i.e. push-pull operation. The lower output impedance
results from the use of a feedback loop from the plate resistor to the bottom
tube and the use of two tubes allows the buffer actively to draw current in
both directions.
ÚÄÄÄÄÄÄÄÂÄı±±±ÄÄÄÄÄÄo +B
³ ³
³ ³ Ra
³ ³
³ V2 ³
Vi ßßß ³
oÄÄÄÄÄÄÄ----- ³
ÚÄÄÄ¿ ³
³ ÄÁÄ C1
³ ÄÂÄ
³ ³
³ ³ C3 ³³
ÃÄÄÄÄÄÄÄÄÄ(ÄÄÄÄÄÄÄ´ÃÄÄÄo Vo
³ ³ ³³
³ ³
ßßß ³
-----ÄÄÄÄÄÄÄ´
ÚÄÄÄ¿ V1 ³
³ ³
ÃÄÄÄ¿ ³
³ ³ ³
± ÄÁÄ ±
Rk ± ÄÂÄ C2 ± Rg
± ³ ±
³ ³ ³
ÄÁÄ ÄÁÄ ÄÁÄ
/// /// ///
Figure 1:ÿSchematic of White Cathode Follower.
Because of the increased complexity of the circuit, the math is much
more complicated than that of a simple Cathode Follower.
As an example, given a setup that consists of a 6DJ8 with a bypassed
cathode resistor of 200ê and a 10Kê plate resistor, the results are * :
Gain = 0.97
Zo = 3.44 ê
PSRR = -65 dB.
æý + ærp/Ra
Gain = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
(æý + æ + 1) + (æ+2)rp/Ra
1
Zo = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
æ(æ + 1)
1 + ÄÄÄÄÄÄÄÄÄÄ
æ + 1 1 + rp/Ra
ÄÄÄÄÄÄÄÄÄ + ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
rp + Ra rp
2rp + (2æ + 2)Rk
PSRR = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
[2rp+(2æ+2)Rk+Ra+ærp+(æý+æ)Rk] [(Ra+rp)/(æ+1)+rp+(æ+1)Rk)/(æRa)]
If this seems too good to be true, that's because it is too good to
be true. Yes, the gain is almost unity and the Zo is amazingly low, yet the
circuit can't deliver very much current into a low impedance load, such as a
Grado headphone, 32 ê. Imagine a car with 340 horsepower, yet which could
only do 10 miles per hour. Surprisingly, if we try to output more than a few
millivolts into the 32 ohm load, we will overdrive the circuit, as we will
break out of Class A operation.
Here is what happens in detail. Any variation in the current flowing
though the top triode will produce a variation in the voltage developed
across the plate resistor. In turn, this voltage will be transmitted to the
bottom triode's grid, which can only see a few positive volts before it is
driven into positive grid voltage or, if the voltage swings negatively, it is
completely turned off. The greater the value of the plate resistor, the
easier it is to overdrive the bottom triode, as a smaller amount of current
is needed to develop a large voltage change across the plate resistor.
On the other hand, if we make the plate resistor smaller in value,
we gain dynamic headroom, but lose the stellar specifications. In fact, if we
set the plate resistor 0ê, we end up with a classic Cathode Follower with an
active load, i.e. the bottom triode. Of course, if the load we wish to drive
isn't a punishingly low 32 ê, the headroom issue is much less of an issue.
But if the load is a high impedance one, such as a 100Kê potentiometer, then
we must ask: Why do we need to use a super low output impedance buffer?
Optimal White Cathode Follower
------------------------------
We found that too large a value plate resistor limits the potential
output current from this buffer and that too low a value reduces the buffer's
specifications. So the question is what would be the optimal value for a
given load and a given desired out voltage swing?
This was the question I had asked myself, when I was disappointed by
the results of an experiment wherein I had built a White Cathode Follower
with the aforesaid tube and resistor values for driving a much more reasonable
load: the Sennheiser headphones, which have an impedance of 300 ê. After
only a few millivolts, clipping occurred. My expectation was that the circuit
should be able to deliver the idle current of at least 10 mA into this load,
if not almost 20 mA, which would conform to classic Class A, push-pull
amplifier standards. I then replaced the plate resistor with a 10Kê potentio-
meter with its center tab connected to one of its outside tabs, which allowed
for easy adjustment of the plate resistor value.
After adjusting the pote, I found the optimal value according to the
trace on the oscilloscope to be 100ê. The lowness of the value surprised me.
I then wondered what the optimal value would be for the 32ê load represented
by the Grado headphones. Even more surprising was that the same 100ê plate
resistor value yielded the best performance into the 32ê, in spite of this
load being 10 times lower in value than the previous load. Moving to the other
extreme, I replaced the 32 ohm resistor with a 3Kê resistor and retested. The
100ê plate resistor value once again made for the biggest and most symmetrical
voltage swings. After some mathematical introspection, everything made perfect
sense to me.
For any push-pull tube amplifier to work well, there most be an
almost identical signal presented to each tube. (The signals must differ in
phase.) In this circuit, if the top triode sees an increase in its
grid-to-cathode voltage, then the bottom triode must see an equal decrease in
its grid-to-cathode voltage. How do we ensure equal drive voltages for top
and bottom triodes?
Let us start our analysis with the severest load possible, not Grado
headphone, but 0ê, in other words, a dead short to ground via a large valued
capacitor.
White Cathode Follower with a shorted output.
--------------------------------------------
The top triode now functions as a Grounded Cathode amplifier and
doesn't see the bottom triode at all. The amount of current flowing from
ground into the capacitor then into the cathode of the top triode is given by
the formula:
Ip = VgGm',
where
æ + 1
Gm' = ÄÄÄÄÄÄÄÄÄ
Ra + rp
Now as the bottom triode current flow is governed by the top triode's
current flow into the the plate resistor, the amount of current flowing from
the bottom triode's plate into the capacitor is given by the formula:
Ip = VgGm
where Gm is the transconductance and
Gm = æ / rp.
By rearranging the formulas for current we get Vg = Ip / Gm' for the
top triode and Vg = Ip / Gm for the bottom triode. Obviously, the only way
that the two grid voltages can match is if Gm' = Gm. Expanding this formula
out yields:
æ + 1 æ
ÄÄÄÄÄÄÄÄÄ = ÄÄÄÄ
Ra + rp rp
which when we solve for Ra becomes
Ra + rp æ + 1
ÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄ
rp æ
rp (æ + 1)
Ra = ÄÄÄÄÄÄÄÄÄÄÄ - rp
æ
rp æ rp
Ra = ÄÄÄÄÄÄ + ÄÄÄÄ - rp
æ æ
Ra = rp / æ
and as rp / æ = 1 / Gm
Ra = 1 / Gm.
Thus, the only way the Vg of the top triode can equal Vg of the
bottom triode is if the plate resistor equals the inverse of the transconduc-
tance of the triodes being used. (The test to put any tube circuit equation
through is to try the equation with a 6AS7 and then with a 12AX7 to check the
equation for absurdities.)
What happens if we chose to start with infinite load instead of 0ê?
The answer is the same, the optimal value for the plate resistor is the
reciprocal of the Gm of the triodes used, or what is the same quantity, rp/æ.
The math can become quite thick here, but if we think abstractly
enough, it won't be too difficult to follow. We know that if the top triode
sees a +1 volt pulse at its grid, its cathode will follow to some degree less
than +1 volt. Whatever this outcome may be, we will refer to it as "Vg." Now
Vg/rp equals the increase current (Ip) flow through the entire circuit, as
all components are in current series with each other. Ip times the plate
resistor (Ra) equals the voltage pulse that the bottom triode sees, which
times the Gm of the bottom triode will equal Ip, if the right value of Ra has
been chosen. Thus,
Ra æ Vg
Vg ÄÄÄÄ ÄÄÄÄ = ÄÄÄÄ
rp rp rp
which when we solve for Ra equals:
Vg Ra Vg
æ ÄÄÄÄÄÄÄ = ÄÄÄÄ
rpý rp
Ra
æ ÄÄÄÄ = 1
rp
æ Ra = rp
rp
Ra = ÄÄÄÄ
æ
Okay, what if we choose a load impedance somewhere between zero and
infinity, say, 10Kê. Same result, Ra = rp/Gm. In this case, the load impedance
is in parallel with the rp of the bottom triode. So Vg/(rp//RL) equals the
increase current (Ip) flow through the top triode and IpRa equals the pulse
voltage to the bottom triode. In this case, like the one with a shorted
output, we have true Class A output current swing capability, so as the
bottom tube approaches cutoff, the top tube's current conduction will near
twice its idle value. And, of course, vice versa for negative input voltage
swings. Thus,
Ra æ Vg
Vg ÄÄÄÄÄÄÄÄ ÄÄÄÄ = ÄÄÄÄÄÄÄÄ
rp//RL rp rp//RL
which when we solve for Ra equals:
Ra æ Vg
Vg ÄÄÄÄÄÄÄÄ ÄÄÄÄ = ÄÄÄÄÄÄÄÄ
rp//RL rp rp//RL
Ra
Vg æ ÄÄÄÄ = Vg
rp
Ra
æ ÄÄÄÄ = 1
rp
æRa = rp
rp
Ra = ÄÄÄÄ
æ
Optimization and Zo
-------------------
We can use the stock, long, complex equation for output impedance
for the White Cathode Follower or we can realize that we have.
Now let us step back and look at what is happening with this circuit
in broad terms. Without an external load the rp of the bottom triode will
define the sole load impedance for the top triode, remember we had defined an
infinitely high impedance load. Since the gain of this circuit is less than
unity, the cathode voltage will slightly lag the grid's and this gap is the
change in the grid-to-cathode voltage that will prompt a change in current
flow through both the top triode and plate resistor, which in turn will give
rise to a change in voltage across that resistor, which will then be relayed
to the bottom triode's grid. We need to ensure that that bottom tube receives
an identical grid-to-cathode voltage signal as the top tube.
Stipulated that Gm' = Gm as a condition of satisfaction in the quest
for the optimally valued Ra, and we found that Gm' = (æ + 1)/(Ra + rp). In
effect, what we have actually done by specifying the correct value for Ra is
to balance the push-pull aspect of the circuit, which includes each triode
offering the same output impedance to the load. Consequently,
Zo = 1 / 2Gm,
or
Zo = rp / 2æ.
Conclusion
----------
We find once again that we can't get something for nothing: spectacularly
low output impedance came at the price of a disappointingly low input
overload voltage and a miniscule output current ability. But what we did get,
when we gave the White Cathode Follower the optimal plate resistor value to
work with, was a buffer circuit twice as good as a textbook Cathode Follower:
half the output impedance and a symmetrical output current swing with twice
the output current swing than a single triode Cathode Follower.
* This data has been confirmated from the Seely's book pages 120/121.
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º Osvaldo F. Zappacosta. Barrio Garay, Almte. Brown, Buenos Aires, Argentina.º
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