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LW1DSE > TUBES 06.01.18 22:45l 152 Lines 7007 Bytes #999 (0) @ WW
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Subj: The SRPP
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Sent: 180106/1956Z @:LU7DQP.#LAN.BA.ARG.SOAM #:26644 [Lanus Oeste] FBB7.00i
From: LW1DSE@LU7DQP.#LAN.BA.ARG.SOAM
To : TUBES@WW
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The Shunt Regulated Push-Pull (SRPP)
====================================
The SRPP is an elegantly simple circuit, and is really a small output
transformerless (OTL) amplifier. It was widely used in TV circuits for
delivering significant current into heavy capacitative loads. It was first
patented in 1940 by Henry Clough of Marconi, and has returned in may guises
since. Interestingly though, it was not referred to as the SRPP until quite
recently and instead had a variety of other names like 'bootstrap follower'
and 'shunt regulated amplifier'; no one seems to know when the SRPP name
first took hold, but it looks like some time in the 1980s.
ÚÄÄÄÄı±±ÄÄÄÄÄÄo +B
³
³ Ra
³
³ V2
ßßß
ÚÄÄÄ-----
³ ÕÍÍ͸
³ ³ C3 ³³
³ ÃÄÄÄÄÄÄÄÄÄ´ÃÄÄÄÂÄo Vo
³ ³ ³³ ³
³ ³ ³
³ ± ³
³ ± Rk2 ³
³ ± ±
³ ³ ± Rl
³ ³ ±
ÀÄÄÄ´ ±
C1 ³ ³
Vi ³ ³
³³ ßßß ³
oÄÄÄ´ÃÄÂÄÄÄÄÄÄÄ----- ³
³³ ³ ÕÍÍ͸ V1 ³
³ ³ ³
³ ÃÄÄÄ¿ ³
³ ³ ³ ³
± ± ÄÁÄ ³
± Rg Rk1 ± ÄÂÄ C2 ³
± ± ³ ³
³ ³ ³ ³
ÄÁÄ ÄÁÄ ÄÁÄ ÄÁÄ
/// /// /// ///
Figure 1:ÿSchematic of SRPP.
Sadly, the SRPP isn't actually much use for guitar, as it is really
only capable of driving relatively fixed loads (not tone stacks!) and it does
not clip pleasantly at all. One useful place to use it might be as a reverb
driver.
The circuit uses two triodes, usually in the same envelope. Each triode
is biased the same, so half the HT is dropped across each one. The lower
triode acts as a common-cathode gain stage with an active load, and the upper
triode acts as common-anode gain stage with an identical active load. This is
about as close to a complementary transistor pair as valves get!
The reason the circuit is push-pull and not single ended is that the
signal reaching the bottom triode (V1) causes the signal on the grid of the
top triode (V2) to be in anti-phase with it. When the top triode conducts
more, driving current into the output coupling capacitor, the other conducts
less. When the top triode conducts less, charge stored in the capacitor is
returned and flows down into the lower triode. Unlike the æ-follower, the
output should be taken only from the cathode of the upper triode.
Since the circuit is really a small power amplifier, high current
valves are preferred, but the following example uses an ECC83 (12AX7) with an
HT of 300V.
This circuit is often used as an output transformerless (OTL) power
output stage, so high current valves are preferred, but let's see how well
(or badly) and ECC83 does. For perfect balance, V2 ought to have an anode
resistor Ra equal to Rk if the cathode is bypassed, or 2Rk if the cathode is
unbypassed. However, the unbalance is very small for high-æ valves.
The two cathode resistors should equal:
Rk1 = Rk2 = (2Rl + ra) / æ
Where Rl is the following load resistance. If we were driving a 10Kê
load, say;
Rk1 = Rk2 = (20Kê + 65Kê) / 100
= 850 ohms. (820 ohms is the nearest standard).
It is usual to add a cathode bypass capacitor to the lower cathode.
Leaving it out will hardly affect the gain in this case, but it would
increase the anode impedance which makes the stage more susceptible to noise
and increases output impedance. Since the frequency response will be hardly
afftected, there is little point calculating the bypass capacitor's value
carefully, any value greater than 1uF should do.
The quiescent current is given by:
Iq = HT / (2ra + 2æRk)
Iq = 300 / (130Kê + 2 * 100 * 820ê)
= 1 mA
And since the SRPP can only operate in Class A, the peak current
delivered into the load is 2Iq per triode, making 4mAp-p in total, or about
1.4mArms- which isn't as spectacular as we might have hoped. The maximum
undistorted voltage across the load must therefore be 4mA * 10Kê = 40Vp-p, or
20mW. The maximum input signal before clipping is simply 2Iq * Rk, which is
about 1.6Vp-p (so the circuit must therefore have a gain of 25).
Normal rules apply for the grid-leak resistor Rg, and 1Meg is usual.
If both triodes are identical and biased the same, the output impedance
will be:
(ra + 2Rk)[ra+Rk(æ+2)]
Zout = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
[2ra + 2Rk (æ + 2)]
[65000ê + (2 * 820ê)] [65000ê + (820ê * 102)]
Zout = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
[2 * 65000ê + 2 * 820 * 102]
Zout = 32.7Kê
This is about half the normal value for regular unbypassed ECC83 gain
stage. But of course, this figure is of limited use, since the circuit is
still only capable of driving maximum current into the optimum load impedance
it was designed for.
Heater considerations: Because the cathode of the upper triode will be
at roughly half HT, the heater supply will probably need to be elevated to
avoid exceeding the valve's maximum heater -cathode potential- always check
the data sheet.
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º Osvaldo F. Zappacosta. Barrio Garay, Almte. Brown, Buenos Aires, Argentina.º
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