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Remove the four 680 ohm resistors at R 308 and R310.Replace the R 308 resistors with 1.2k ohm resistors and the R310 gets 820 ohm resistors.
(1) the weaknesses of the SRPP topology, (2) I plan on running my Clarinet into two devices simultaneously: a power amp (100K ohm input impedance) and an active subwoofer (12K ohm input impedance).
(1) The "weakness" of SRPP in general is it is NOT all cure though the topology does attracts tons of DIYers out there to try building it. It is designed to drive certain range of load. Otherwise it will not be better than any basic cathode followers.
If the gain of yr driving preamp is high enough, why NEED to go thru the hazzle of building a SRPP. Wouldn't a simple 12AU7 cathode follower do a good joB.
Don't forget to have 2 separate heater circuits for the upper half tube & for low half tube as the cathode voltages of both upper & low halves are different by a huge margin, risking blowing the tubes if no such precation is taken.
(2) Taking the rules-of-thumb, the ideal I/P impedance should be min 10X, so O/P Z of the SEPP should be 12/10=1.2KR. But the SRPP get 3KR O/P Z. A bit too high for yr sub.
Don't forget to have 2 separate heater circuits for the upper half tube & for low half tube as the cathode voltages of both upper & low halves are different by a huge margin, risking blowing the tubes if no such precaution is taken.
In the next formula, we can determine the low frequency shoulder (-3dB cutoff) of the Clarinet. Where C is the output capacitor's value in microFarads (uF) and R is the output impedance (Zo) of the Clarinet:Hz-3dB=159155/C/RUsing the stock values (1uF coupling cap and 3K output impedance):159155/1/3000=53Hz [NOTE: The specifications for the Clarinet state a bandwidth of 10-100K, so I must be missing something here.]
The following formula is for determining the optimal impedance (Zopt)for the SRPP circuit. In other words, the optimal input impedance that the Clarinet should see "downstream."
It's again a question: what the formula optimizes? At first glance it looks like the formula from Blencowe's article, so it is for optimal power, not voltage!
In effect, the external load is mitigating the tube's transconductance, so the value of resistor Rak (R2) must be increased by 2Rload/mu. In other words, a different load impedance, a different Rak value. If the triode offered infinite mu, one Rak resistor value would work with all load impedances. This means that setting up an SRPP circuit takes much more thought than many solder slingers can marshal, but armed with a calculator and the formula finding the optimal resistor value is not difficult. The actual difficulty occurs when we try to use our correct value in an actual circuit. For example, a 6H30-based impedance-multiplier circuit working into a 32 ohm load requires an Rak value of 73 ohms, which when used with a 100V differential from cathode to plate, will result in an idle current draw of over 40mA, which comes dangerously close to exceeding the tube's dissipation limit. In addition, we may only want to allow 20mA of idle current. The solution to this problem is the SRPP+, as its additional resistor allows us to set the idle current to 20mA and to retain the impedance-multiplier ratio of 2. Resistor Rak has been replaced by two resistors, R1 and R2, in the SRPP+ circuit. The first step is to set value of resistor R1 and R2 combined, as it must match that of the bottom triode's cathode resistor. For example, the 6H30 requires a cathode resistor of 220 to establish an idle current of 20mA with cathode-to-plate voltage of 100V. The next step is to use this value as Rk in the following formulas: R2 = rp/2mu + Rload/mu + Rk/2 R1 = Rk - R2
Also I doubt that we should drop the lower and higher cathode resistors difference in case of modified values and more than that - the presence of the anode resistors comparable in the value with the cathode resistors.
Okay, I gathered a bit of that from your earlier statement in regards to Blencowe. Also, Broskie hints at that when discussing current swing effects due to raising Rk. Perhaps these facts are of more importance when the goal is some sort of output transformerless application - such as a headphone amplifier.
I guess what I am most curious about is how Jim H went about the change from the stock Rk values to the modified Rk values. In the stock build, both cathode resistors are 680 ohm, which seems to match the description of the circuit better. It can be assumed that tests of this build proved favorable in Jim's setup. So what puzzles me most is not how raising the values of the cathode resistors led to an improvement (though I think that aspect is worth discussing), but how imbalancing them led to an improvement (i.e. 680 & 680 vs 1200 & 820).
Here he refers to Rk2 as Rak. He is discussing breaking up Rk2 into a voltage divider consisting of R1 and R2 with the output being taken from the join of R1 and R2:Still trying to wrap my head around all of this.
Poty, would you please restate this? I'm not sure I follow.
For example, the following formula tells us what the optimal load impedance is for a given triode and Rak value: Rload = (mu*Rak – rp) / 2Thus, if we use a 6H30 with 250V B+ power supply and 300-ohm cathode and Rak resistors (mu = 15.4 and rp = 1380), the optimal load impedance will be 1620 ohms, which is great if that is the load we wish to drive. But what if we planned on driving 300-ohm headphones instead? Then we would use the following formula which gives the optimal Rak value for a specified triode and load impedance: Rak = (2Rload + rp) / mu
So I modified the Zo formula to account for the 220 ohm plate load resistors (Rp) in the Clarinet. Basically, I can't see any reason they shouldn't be added in series with the dynamic value rp:Zo=[(rp2+Rp2)(Rk+rp1+Rp1)]/[rp1+Rp1+rp2+Rp2+Rk(mu2+1)]
Using circuit values;Zo=[(8900+220)(680+8900+220)]/[8900+220+8900+220+680(16.6+1)]=2959Adding the series output resistance (R311);Zo=2959+220=3179
... the circuit design for his SRPP+ ... differs from Jim's Clarinet design ... there are no plate load resistors and no grid stopper on the upper triode. A series output resistor is not specified.The "+" in SRPP+ is meant to indicate an SRPP circuit substituting a voltage divider (R1 & R2 above) for Rk2. The voltage divider maintains the desired amount of resistance for a certain bias point of an SRPP circuit, while also tapping the circuit's output from a point where the current contribution between the tubes is better equalized, lowering distortion.
So my main concern is that while I'm fine running the Clarinet into the 100K ohm load my tube amp would present, running the Clarinet into that same amp and an active subwoofer simultaneously (12K ohm input impedance; combined load of about 10K) would have some drawbacks. Exhausting any sort of tweak to fine-tune the Clarinet circuit for this sort of duty, I'm probably best off making only one change; increasing the output capacitor to 2.0uF: Hz-3dB=708/[(1+1834*C*(Zo + Zload)] =708/[1+1834*.000002*(1890+10000)] = 16[an output capacitor of only 1.0uF with a 10K load yields a -3dB of only 32Hz, which is too high]
Did you use the formula for bypassed Rk1? The Clarinet is unbypassed.
I forgot to mention earlier, but adding the series resistance - why do not you add the impedance of the output C303?
You have described the differences right, but you should read the purpose of the "+" before sinking into it. As I understand the "+" - we should go this way ONLY if the classic SRPP can't do the duty for low-resistance load. You mentioned much more prominent load that even "optimum" for classic SRPP - why you should be bothered by SRPP+ at all? "+" here doesn't mean "better sound", "advanced", "better quality" at all! It means - suitable for certain conditions outside the normal SRPP usage! That is why (while we do not have such conditions in the preamplifier) the SRPP+ is not suitable for Clarinet.
What is more: Depending on the sensitivity of your amplifier and subwoofer I'd add another R311, C303, R312 to the circuit for the second output and tweak R311 resistances (make them higher) to lower demands for C303s.
Not sure I understand the theory behind such a change. Would it be to tailor each output to the input sensitivity of each amplifier? Might not be a bad idea.
As far as the Clarinet's concerned, I have a couple NOS Amperex Bugle Boy 12BH7s that I intend on running in it. As far as I know, the 12BH7 is a drop-in replacement for the 12AU7, providing the H+ circuit can supply the extra current. As I learn more, I may find the circuit can be better tailored to the 12BH7 and may have a few changes to make along those lines.
Beyond all that, I have a soft-start delay module that will fit inside that I plan on using, as well as a solid-state rectifier substitute for the 5Y3GT. Additionally, I also have a pair of dual Voltage Regulators/Constant Current Sources that I plan on using between the power supply and R306.
If we use a 12k resistor in series with the subwoofer input we effectively lower the sensitivity of the subwoofer to 0.5-1V and raise the input impedance to 24k!
Well... first addition is more "service" than sound. Constant Current Sources? Where do you intent to put them?
Anyhow, most of my current reading on the circuit is in regards to finding the purpose of the 220 ohm plate load resistors in the Clarinet schematic. Curious that most SRPP designs omit these resistors.
The high voltage shunt regulators / constant current sources are the second group of items seen here. One for each channel will go between C301 and R306 in the Clarinet schematic. Coincidentally, if you visit the link and click on "Power Amplifier" on the left menu, you will see the same tube amplifier I am now using and plan to use in conjunction with the Clarinet.
BTW, your list of Russian tubes was very interesting. I was particularly curious about employing a pair of 6Н6П tubes in a circuit like the Clarinet...
I don't think they have some real purpose, but can't say for sure.
I'd better use 6Н1П.