Bottlehead Forum
General Category => Technical topics => Topic started by: Mikey on December 20, 2010, 07:55:43 AM
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Hi Gang,
I'd like to try to use the Bottlehead C4S as a differential amplifier cathode load.
According to the C4S manual, this is possible by configuring the kit with MJE340's and 2N2222's.
The C4S would be used in place of an existing triode CCS (see attached file).
Is the replacement as simple as cutting out the triode and inserting the C4S in it's place?
Mike
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Yes it is that simple. Be sure to check the operating limitations, 300 volts and 0.67 watts dissipation for the MJE340. (You can add an Aavid 5775 heat sink and get up to 1.67 watts dissipation.) Run at least 10% of the output current in the bias string; 20% is better but not always practical.
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Hi PJ, thanks for the reply!
So....just to document the retrofit for the forum archives:
1. Remove the existing triode and it's bias components from the circuit
2. attach C4S terminal A to the B- supply
3. attach C4S terminal B to the cathodes of the vacuum tube
4. attach C4S terminal C to GND
Is that correct?
If so, now I need to attach some values to resistors R1 and R2.
Measurements of the existing circuit show the following:
- B+ supply measures +250 VDC
- B- supply measures -250 VDC
- voltage drop across plate resistor R1 (49.9K) -> 125V drop = 2.5mA
- voltage drop across plate resistor R6 (49.9K) -> 125V drop = 2.5mA
My rudimentary math skills indicate that the existing CCS is set up to pass 5.0mA, no?
If so, how do I calculate the proper values for R1 and R2? Do the same formulas apply
when using the C4S as a plate load?
If my logic is correct, let me know and I will try to calculate the proper R1 and R2 values
in a follow-up post...
Mike
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R1 = 0.855/current, approximately. So 5mA = 0.005A, and R1 = 171 ohms
R2 passes 20% of 5mA = 1mA, and drops 250v, so R2 = 250K ohms.
R2 dissipation is 250v * 1mA, or 0.25 watts. I always downrate resistors by at least a factor of 3, so use a 1-watt resistor. Make sure it's rated for well over 250v!
The transistor drops 252v (assuming the differential pair biases at 2v) so the dissipation is 252 * 0.005, or 1.26 watts. It needs a heat sink.
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Hi Paul,
Let's assume for a moment that I didn't read your post above, and decided to calculate R1 and R2 from the procedure outlined in the C4S manual.
Let's also assume that I would make the calculations based upon the 'high current' application, and using 2mA (preferred?) LED bias.
Going strictly 'by-the-C4S-book', here is what I would have come up with:
R1 (from the chart on page 11) = 205 ohms
R2 (from the formula on page 15) = E/I = B-/.002A = 250V/0.002A = 125K ohms
Now that I've read your post, it seems you're OK with 1mA (20%) LED bias in this case. That certainly explains the different R2 value that you calculated.
So, is the 'R1 = 0.855/current' formula a better way (rather than the chart on page 11) of determining the value of R1?
Mike
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One more quick question:
If two constant current sinks were laid out on a small circuit board,
could they share the bias components (LED's, R2) between them?
The CC sinks may or may not be running at the same current limit...
MIke
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...
So, is the 'R1 = 0.855/current' formula a better way (rather than the chart on page 11) of determining the value of R1?
...
One more quick question: If two constant current sinks were laid out on a small circuit board,
could they share the bias components (LED's, R2) between them?...
The old manual was based on the old LEDs, which have not been in production for years. The current HLMP-6000 LEDs fit the 0.855/current estimate pretty well. The thing is, there are several different red LED formulations, each of which had a different voltage drop and different deviations from perfect linearity - if you change the LED, you have to re-calculate the resistances. If you have a stash of the old LEDs that Buddha originally specified (they had a clear package and radial round wire leads) then use his chart.
Yes, you can share the bias components - remember to double the bias current if you are feeding double the current sources. I don't think it buys you anything in performance unless it keeps the bias current above 1mA. Saves a 10-cent resistor and a couple 10-cent LEDs though.
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Thanks PJ,
I've got the HLMP-6000 LED's, so I'll stick with your formula.
I'll scrap the idea of a shared bias circuit...not worth the hassle.
Mike
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Yes it is that simple. Be sure to check the operating limitations, 300 volts and 0.67 watts dissipation for the MJE340. (You can add an Aavid 5775 heat sink and get up to 1.67 watts dissipation.) Run at least 10% of the output current in the bias string; 20% is better but not always practical.
Hi Paul,
Looking at the datasheet for the MJE340, it seems as thought this part can handle up to 20 watts dissipation.
Is this simply a matter of bolting on an even bigger heatsink?
Mike
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Yes, you are correct. However, the full 20 watts requires a huge heatsink - think 4" by 5" by 2" deep - and good circulation of room-temperature air, not in the summer without air conditioning. I try to limit the internal chip temperature to 100 degrees C even though they are rated for 150C - this is based on various real-world failures. With the clip-on heat sinks we use, you are limited to the dissipation I mentioned. You can use TO-220 transistors - we use the MJE5731A in PNP alpplications - and get 1 watt bare, 2.5 watts with the clip-on we use for the Paramount soft-start upgrade, and 5.5 watts for the big extruded one used in Eros and the Speedball.
As usual, transistors are much touchier than tubes. A 20-watt tube will run, by itself, at 20 watts for its full life; practical transistors can be counted on for at most 1/20 of their rating alone, and 1/4 of the rating with generous heatsinks. Tubes can take transient plate voltages of at least twice their rated maximum, some of them 5 times for short intervals; transistors are damaged ijmmediately with the tiniest excess.
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Hi Paul,
Here's another thought:
A vacuum tube acts like a NPN transistor, correct?
So, could a rugged triode vacuum tube take the place of the
MJE340? I'm guessing that it would have to bias a bit differently...
Would the calculations to derive R1 and R2 remain the same,
or would they need to be re-formulated?
Mike
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Ought to work; make sure the tube bias voltage and current are within the limitations of the remaining small transistor.
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Is the replacement as simple as cutting out the triode and inserting the C4S in it's place?
Hi Paul,
I think I should have originally written:
Is the replacement as simple as cutting out the triode and it's cathode resistor, and inserting the C4S in it's place?
Would that be more accurate?
Mike
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Sorry, I didn't see the attachment on the first post. Now I see it but I can't open it. I assumed there was more than just a resistor to operate the triode as a current source, so now I'm just confused!
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Hi Paul,
I've attached an earlier version of the Word document, hopefully you can open this one...
Mike
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Ah! I get it! Thanks, Mikey.
You can remove the current source triode, the 401K resistor, the two Zeners, the 499K resistor, the cap (is it 47uF or .47uF?) - and the triode. Replace them all with a C4S. You can keep the triode and replace the MJE340 with it if you prefer, grid is base, cathode is emitter, and plate is collector.
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Hi Paul,
Thank you for the help!
I'm glad you were able to open the second attachment...
It seems as though I've got one last decision to make:
- Should I use the MJE340, or use the existing vacuum tube triode (in this case, a 6SN7).
The rework to the existing circuit board will be about the same in either case so that's not an issue.
However, if at a later date I decided to really 'ramp up' the current through this circuit it seems the vacuum
tube might hold an advantage over the transistor in terms of dissipation. From the 6SN7GTA datasheet,
each plate can dissipate 3.75 watts (when both are operating). That seems like it might be better than
a scathing hot transisitor with a HUGE heatsink!
So my questions are:
- In your opinion, does one option hold a technical advantage over the other?
- Would you expect one option to sound better than the other?
Mike
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In my opinion there is not going to be any noticeable sonic difference; the device is largely controlled by the small transistor. But the tube will be more robust - less chance of blowing something if anything goes wrong - and it will get a heat source up on top and out of the chassis insides, which is always a good thing. Downsides are of course heater power and real estate.
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Hi Paul,
I will try the 'hybrid' C4S first, using the vacuum tube in place of the MJE340.
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07452.jpg&hash=5eab5b2852ffb3a44c66c1f51d56d448432c9bd1)
You can see the existing 6SN7 constant current sink in the above photo.
I'll splice in nearby, and leave the C4S board supported by the wires for
the time being....photos to follow!
Mike
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Hi Paul,
Well, at the last minute, I decided to install the C4S 'as is',
without utilizing the existing triode in place of the MJE340.
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07508.jpg&hash=5defbfef5241ccceaad3fbfbe8df8c6b8e479b4d)
One channel is up and running with the C4S sink, I'll convert the
second channel tonight....
Mike
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Mike, is that battery on a cathode?
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Mike, is that battery on a cathode?
Hi Grainger,
The input stage of this phono preamp is a cascode, and I replaced the bottom
tube (2X 12AT7's in parallel) with a MAT12 transistor. With the transistor in
the circuit, I can use the battery to bias the top tube in the cascode...
Mike
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I retrofitted the second channel last night, so I will
be able to give the new setup a listen this weekend.
Stay tuned...
Mike
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Hi Grainger,
The input stage of this phono preamp is a cascode, and I replaced the bottom
tube (2X 12AT7's in parallel) with a MAT12 transistor. With the transistor in
the circuit, I can use the battery to bias the top tube in the cascode...
Mike
Got'cha! There was a mod of the Paramour where you biased the driver, changed to a 76, with a 9V rechargeable. Once it was charged the circuit kept it charged.
Should this be quieter?
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Should this be quieter?
Grainger,
I honestly don't know, it was the lazy way out....
This way, I didn't have to construct a voltage divider.
Mike
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In my opinion there is not going to be any noticeable sonic difference; the device is largely controlled by the small transistor. But the tube will be more robust - less chance of blowing something if anything goes wrong - and it will get a heat source up on top and out of the chassis insides, which is always a good thing. Downsides are of course heater power and real estate.
Hey PJ,
One additional thought:
Could an IRF820 (see attachment) be substituted for the MJE340 without any issues?
It looks like a pretty beefy transistor, and I happen to have a few of them on hand...
Mike
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...
Could an IRF820 (see attachment) be substituted for the MJE340 without any issues?
It looks like a pretty beefy transistor,...
It's not a transistor, it's FET - MOSFET to be specific. It's enhancement more so it needs several volts of positive bias, more than can be supplied by an LED. I don't know to what extend the cascode arrangement will reduce the potential for distortion from the high and nonlinear capacitances. We never experimented with FETs, though I know Tucker and the late John "Buddha" Camille tried several circuits a decade or more ago and rejected them as inferior sounding.
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Hi Paul,
I got some time on the newly C4S'ed preamp, and I liked what I heard! In fact, I liked it enough
to tackle the next iteration...
In a previous post, I mentioned the desire to try a 'hybrid' C4S, using the existing 6SN7 triodes in
place of the MJE340s. Well, I perfomed the modification this evening, and it works like a charm!
Here's a photo:
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07513.jpg&hash=95650386a4a39d467580b1df9641d9bb67ab47de)
The swap required a bunch of 'cut-and-splice' to the circuit board, but it was worth it. Now the
vacuum tube can dissipate the heat when I bump up the current thru the circuit. Although the
C4S board is dangling from the wires at the moment, I'll be able to mount it on the underside of
the existing circuit board...without the MJE340's, it is a very low profile assembly!
Thanks again for all of your 'hand holding' throughout this project, I couldn't have done it without
your assistance!
Mike
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Hi PJ,
I wanted to wrap up this thread with a few photos of the completed installation,
and also ask one additional question...
I mounted the C4S boards to the underside of the existing circuit boards. To bolt
them in place, I drilled a couple of holes in the existing board, and used 1/4" long
nylon spacers with some #4-40 hardware. Jumper wires were made to splice the
C4S into the existing curcuit:
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07528.jpg&hash=6ea878da969b27c818b8b5da93b020a9228c576e)
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07525.jpg&hash=0083f5d03c58117fa579f4765a14b841c2da4f2f)
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi3.photobucket.com%2Falbums%2Fy80%2Fmpaschetto%2FAtma-Sphere%2520MP-1%2FDSC07529.jpg&hash=f1750050a11ae5518c6cc8b4cd5cade031164eea)
The C4S certainly helped reject any last vestige of power supply noise from this
circuit....al that remains is a bit of tube rush.
With the 6SN7 in place of the MJE340, I think I'm going to be limited in the amount
of current I can draw through this arrangement. With both triodes (of the 6SN7) in
use, the RCA data sheet specifies the maximum dissipation of the tube is 7.5 watts,
or 3.75 watts per triode. By my reckoning, that equates to a current of 15mA
through the C4S.
Does that sound about right? Is 15mA the upper limit of this scheme?
Mike
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With a 250v drop, then yes, 15mA is your maximum without over-heating the 6SN7. You can draw up to 20mA, the 6SN7 maximum rating, if the low voltage is reduced to maintain the maximum heating.
To be careful, check the 6SN7 bias voltage and compute the dissipation in the small transistor. There are clip-on heat sinks for those tiny metal cans, and I recommend them if you exceed 1/2 of the free-air dissipation limit.
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Paul, you wrote:
"To be careful, check the 6SN7 bias voltage"
What voltage am I measuring?
In other words, where do I place the voltmeter probes?
"and compute the dissipation in the small transistor"
Earlier, you showed me how to calculate the dissipation in Q2, but I didn't see
anything in the C4S manual about the dissipation in Q1....how is it calculated?
Mike
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In normal use, the bias on Q2 is +0.7v, leaving about 2.5 drop across Q1 (2*1.6v across the LEDs, minus 0.7v base to emitter on Q2). If the 6SN7 bias is (for example) -8v then the drop across Q1 is 11.2v. Measure ground to the 6SN7 cathode in you case. Then if the current is 15mA, the dissipation in Q1 is (0.015A * 11.2v) or 168mW. The 2N2222 is rated fro 500mW without a heat sink, so I would not run it at more than 250mW without a heat sink. The above calculation says that's OK but the margin is not large.
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"Measure ground to the 6SN7 cathode in your case"
PJ, did you mean to write:
"Measure B- to the 6SN7 cathode in your case"
If I measure from ground to the cathodes, I get some funky numbers.
Assuming you meant B-, I measure 10.0v between B- and the 6SN7 cathode.
Plugging into your equation, the drop across Q1 is 13.2v. At the moment, this
C4S is set for 8mA, so the 2N2222 is dissipating 105mW the way things sit today.
Of course, that assumes that I measured the 6SN7 bias correctly.....
Can't I measure the voltage drop across Q1 directly? Sure, it may be somewhat
cumbersome because the 2N2222's three terminals are out of reach, but I can
follow the circuit board traces to points that are more accessible. It would be
nice to compare the results both ways to see if they match!
Mike
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The 2N2222 emitter is connected to B- (yes, that's what I meant) through R1 which drops about 0.855 volts. Its collector is connected to the 6SN7 cathode. So B- to cathode is a slightly conservative estimate of the voltage drop, and much easier to measure.
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I know this is an old thread, but I didn't want to hi-jack someone else's thread off-topic, so this seemed like the right place to start.
I am designing a pre-amp using a balanced differential pair of tubes. The two kathodes will be tied together and will feed a CCS current sink to ground. I'm curious to know whether the C4S will work in my circuit. The voltage drop between the kathodes and ground will only be about 9 or 10 volts, so there will be very little across the CCS. Would this be enough for the C4S to operate? There shouldn't be much voltage swing in this part of the circuit. Its the nature of a differential pair to hold the kathodes at a nearly constant voltage.
Thanks for any replies or comments.
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Yes, if the voltage swing is small that should be enough for the C4S. Try whenever possible to allow at least 5V, or whatever the rms signal swing is, plus a couple of volts extra headroom to keep the stage out of the space charge region.
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The other strategy that I've used when I don't have enough compliance from cathode to ground is to put a voltage doubler on the heater winding, then ground the positive side to make a negative DC rail for the CCS.
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Thanks to both of you for the quick reply. I'll likely buy a C4S kit to experiment with. I bought one 10 or 15 years ago, but I know I lost the booklet when we moved and I suspect the parts are somewhat changed since then. Do the current kits include the appropriate parts for a current-sink application?
I'm not sure I understood the solution offered by Caucasian Blackplate. Is this sort of like "filament bias" where the extra voltage is consumed in a resistor between filament and ground? (but in this case, consumed by the C4S).
If it helps any, I'm going to use a Coleman Regulator on each filament.
Thanks again!
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Thanks to both of you for the quick reply. I'll likely buy a C4S kit to experiment with. I bought one 10 or 15 years ago, but I know I lost the booklet when we moved and I suspect the parts are somewhat changed since then. Do the current kits include the appropriate parts for a current-sink application?
Not really. If you bought something like the SEX C4S kit, you could cut the board in half and then substitute an MJE-340 in for the MJE5731A/MJE350, as well as a 2N2222A for the PN2907. The LED's also fit in the other way, and for a current sink, the ground terminal goes to B+, then the "I" pad is the input (ground or negative rail) and the "O" pad goes out to your cathodes.
I'm not sure I understood the solution offered by Caucasian Blackplate. Is this sort of like "filament bias" where the extra voltage is consumed in a resistor between filament and ground? (but in this case, consumed by the C4S).
If it helps any, I'm going to use a Coleman Regulator on each filament.
I didn't consider the possibility that you're using a differential DHT. In that case, my alternative will not work. There's also a third alternative that I credit to PJ (not sure if he came up with it or ran across it elsewhere) where you can add a resistor between ground and the negative output of your power supply. Given that you're drawing constant B+ current, you can use Ohm's Law to drop whatever voltage you like across that resistor. One end of the resistor will be at ground potential, and the other end will provide a negative bias voltage.
In your case, I'd just try running it with the bias voltage you're intending to use.
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I'm sorry, I should have described what I was doing better. I try to use DHTs whenever possible.
Is this what you are describing for the C4S? (See attached concept) Each tube will pass 6 mA from the plates (total of 12 mA), so there won't be much heat.
I could probably push the tube bias up so there is 10 volts across the C4S if I increased the Vp a bit, but I think the #26 sounds best near where I've specified it.
Thanks again for your advice. You guys are the best!!
Brian
(http://)
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A couple of things:
1. You shouldn't need/use two separate filament regulators unless heat is an issue.
2. R3 isn't drawn in the right spot.
3. The resistor above R3 appears to serve no purpose.
4. R2 could be a 5K resistor and R1 would be about 142 Ohms.
5. I would tie the two filaments together, then use a pair of 10 Ohm resistors to make a virtual center tap, then connect the output of the C4S there. Do note that you may end up needing a hum pot.
6. The AC balance provided by the C4S will be exceptional! The DC balance will depend a bit on the tubes themselves, and you may find that transformer performance will improve with matched pairs of tubes.
(and of course, I'd throw in that a single ended circuit with a C4S plate load, parallel feed capacitor, and parallel feed output transformer would be worth building to compare against the differential circuit)
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Thanks for reviewing my concept circuit. I've not used a C4S as a current-sink before, so I wasn't sure about the connections and where the resistors go. Let me respond to your comments as best I can, and perhaps I'll feel more confident about how the circuit works.
1. You shouldn't need/use two separate filament regulators unless heat is an issue.
Response: Rod Coleman (inventor of Coleman Regulators) advised against running more than one tube per regulator. He says this results in poor sound from mixing of the plate currents.
2. R3 isn't drawn in the right spot.
Response: I wasn't sure how to set up a C4S as a current sink. Where should it go?
3. The resistor above R3 appears to serve no purpose.
Response: This is a 10-turn, 20 Ohm potentiometer. Its purpose is to allow for minor balancing of the tubes by adjusting their bias. The tubes should be matched, but the pot allows for minor tweaking. If both tubes are perfectly matched this pot would be centered. If one tube is stronger and passing more current I can adjust the bias to favor the other tube. Balancing the currents should improve the performance of the transformer.
4. R2 could be a 5K resistor and R1 would be about 142 Ohms.]
Response: My schematic has no resistors labeled R1 or R2. Where should these go? (and what is the formula for calculating them?)
I would tie the two filaments together, then use a pair of 10 Ohm resistors to make a virtual center tap, then connect the output of the C4S there. Do note that you may end up needing a hum pot.
Response: A hum-bucking pot would probably require the two filaments to be tied together and be fed by one regulator. This is advised against. Anyway, with this regulator there really shouldn't be any hum. These regulators operate with hum measured in micro-volts.
6. The AC balance provided by the C4S will be exceptional! The DC balance will depend a bit on the tubes themselves, and you may find that transformer performance will improve with matched pairs of tubes.
(and of course, I'd throw in that a single ended circuit with a C4S plate load, parallel feed capacitor, and parallel feed output transformer would be worth building to compare against the differential circuit)
Response: I've got quite a stash of #26 tubes and plan to acquire more. :) I'm hoping to find a matched quad to within a half-mA or better. I'm sure the tranny will appreciate the effort. And for parafeed? Actually, I've got another design I'm working on using chokes and parafeed out. I still haven't decided which direction I will go. However, both options have a CCS in the tail, so I need to work out this part of the circuit first.
Question: If the Rp of each tube is 7400 Ohms, what is the impedance as seen by the transformer in a balanced/push-pull configuration with a CCS in the tail? Is it just doubled? (i.e. 14,800 Ohms), or are other factors involved? There is no kathode resistor (and no bypass cap). The CCS acts like an almost infinite resistor. However, since each tube is operating in opposite phase there is no voltage change at their joined kathodes. With no voltage change, there is no degeneration, and there should be no multiplication by Mu to affect the plate resistance. At least that is my (crude) understanding.
This is all very tentative in my mind and I'm still reading my books to learn how this works. I'll need to know the impedance in order to narrow my search for an output transformer. If you have a clue on how to determine this, I'm all ears.
Thanks again for all the assistance.
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Response: Rod Coleman (inventor of Coleman Regulators) advised against running more than one tube per regulator. He says this results in poor sound from mixing of the plate currents.
That explanation isn't super satisfactory. In your circuit, the CCS forces balanced AC plate current, the regulators do not control that aspect of the circuit.
Response: I wasn't sure how to set up a C4S as a current sink. Where should it go?
The LED's start at ground and then the LED bias resistor is between the second LED and B+. R4 also goes between the emitter of the 2N2222A and ground. (your transistors are also out of order)
Response: This is a 10-turn, 20 Ohm potentiometer. Its purpose is to allow for minor balancing of the tubes by adjusting their bias. The tubes should be matched, but the pot allows for minor tweaking. If both tubes are perfectly matched this pot would be centered. If one tube is stronger and passing more current I can adjust the bias to favor the other tube. Balancing the currents should improve the performance of the transformer.
With such low plate current, you'll probably want more adjustment than this.
Response: My schematic has no resistors labeled R1 or R2. Where should these go? (and what is the formula for calculating them?)
R1 is generally labeled as the current set resistor on our boards, R2 is the LED biasing resistor. R1 is approximately 0.855/current. R2 should bias the LED's with at least 10% of the C4S current, and in your case 1-2mA is OK.
Response: I've got quite a stash of #26 tubes and plan to acquire more. :) I'm hoping to find a matched quad to within a half-mA or better. I'm sure the tranny will appreciate the effort. And for parafeed? Actually, I've got another design I'm working on using chokes and parafeed out. I still haven't decided which direction I will go. However, both options have a CCS in the tail, so I need to work out this part of the circuit first.
Try loading the tube with a CCS. Having a CCS under the filaments in a single ended circuit will require a large capacitor across the CCS to function. The C4S will provide higher impedance than the choke and it won't be susceptible to magnetic noise.
Question: If the Rp of each tube is 7400 Ohms, what is the impedance as seen by the transformer in a balanced/push-pull configuration with a CCS in the tail? Is it just doubled? (i.e. 14,800 Ohms), or are other factors involved? There is no kathode resistor (and no bypass cap). The CCS acts like an almost infinite resistor. However, since each tube is operating in opposite phase there is no voltage change at their joined kathodes. With no voltage change, there is no degeneration, and there should be no multiplication by Mu to affect the plate resistance. At least that is my (crude) understanding.
This is a differential circuit, not exactly push-pull. When you setup a gainstage like this, you get half the mu out of each tube and each half of the transformer primary will see a 7.4K source.
This is all very tentative in my mind and I'm still reading my books to learn how this works. I'll need to know the impedance in order to narrow my search for an output transformer. If you have a clue on how to determine this, I'm all ears.
I'd be happy to talk about design potentials with you for something like this. I've build a few differential preamps that don't look that different than what you have on paper, and all of them have been pulled apart and repurposed for other projects. Doing a single ended parallel feed preamp with the 26 won't be too tough to design and won't be so tough to build either.
-PB
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Thanks again for the help. I think the design is coming together. I've attached a revised schematic that incorporates my understanding of your comments. I've reversed the transistors and moved the resistors. This is starting to make more sense to me now.
I'm assuming the LEDs will drop a total of 3 volts so their bias resistor will need to drop 6 volts at 1.5mA (a 4k resistor). This leaves 10.5mA for the transistors, so the current-set resistor would be 0.855/0.0105 = 81.5 Ohms. Is my reasoning correct on this?
I added a cap across the whole CCS. Why is this needed? What is it's purpose?
This is a differential circuit, not exactly push-pull. When you setup a gainstage like this, you get half the mu out of each tube and each half of the transformer primary will see a 7.4K source.
From my experience running a #26 single-ended, these tubes like to see a load of at least 150 Henrys. With a 4.5:1 step-down, the reflected impedance from the secondary will be over half a megOhm, so the inductance of the transformer will dominate (as it should).
For a balanced differential circuit, will each plate see only half the Henrys of the entire transformer, or is there some form of inductive coupling between halves of the primary? As a waveform reaches a crest on one tube it will be reaching a trough for the other tube. I can almost visualize current passing between each half of the tube pair (like a tennis match) through the coil from one end to the other. If this visualization is correct, all the Henrys in the primary would be available to load each plate. If this visualization is incorrect, and all the current comes from the center-tapped B+ connection, then only half the Henries of the entire primary are available to each plate.
I'd be happy to talk about design potentials with you for something like this. I've build a few differential preamps that don't look that different than what you have on paper, and all of them have been pulled apart and repurposed for other projects. Doing a single ended parallel feed preamp with the 26 won't be too tough to design and won't be so tough to build either.
I gather that (since you pulled them apart) you were not too happy with the sound?
For the moment, I'm committed to a balanced design. All my sources are balanced and it seems like a waste to throw away a perfectly good split-phase signal. One of the signals will be on a 40-foot cable which could easily pick up common-mode noise. A differential preamp will likely kill most of this.
I intend to tackle this design in two steps. The first iteration will use 6SN7 tubes. These are quite linear, cheap and have nearly identical plate resistance to the #26. They take a bit less current so there would be some resistors to adjust. The big advantage is a simpler heater supply. Once I get this working, a conversion to the #26 shouldn't be too difficult.
My C4S kit was purchased more than 10 years ago and I've moved twice since then. Not much remains from those early days. I'm quite sure the original LEDs are no longer available. What LEDs are currently recommended for the C4S circuit?
Thanks for all your thoughtful replies. This forum is an amazing resource.
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I made a bit of a typo earlier. R2 needs to go to B+, otherwise the resistor and LED string negate the high impedance of the C4S (then you end up with a push-pull circuit after all!). Take your B+ voltage and divide by 0.002A and you'll get an approximate resistor value, then recalculate your current set resistor for 6mA.
Balanced refers to the cable interface, and when you use an input transformer (which you have drawn into the schematic), then you satisfy the conditions for noise rejection. There isn't anything about a balanced cable interface that both the hot and cold legs are actually driven by signal. This is a huge assumption that a lot of modern HiFi manufacturers would like their customers to make, but it simply isn't true.
When you consider the primary of the transformer, it's important to note that the center tap is at AC ground, so one tube will not have the whole primary available to it.
Is your old C4S kit with the off-white PC boards? The HLMP-6000 is the current LED that meets the requirements for the C4S.
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The LED drops about 1.55v each, and the smaller transistor shows about 0.7v base to emitter, leaving 0.855v for the current-set resistor R1.
On the schemo, you have reversed the transistor orientation but have not reversed the rest of the circuit. The base of the 2N2222 goes to the junction of the two LEDs, and the MJE340 base goes to the junction of R2 and the top LED.
There should be no cap across the C4S (from ground to cathodes), and as PB said, R2 should go to the high voltage (I get 70K ohms). Both of these changes are to maintain a high AC impedance at audio frequencies, which is the reason for using a C4S in the first place.
In class A, where both triodes are conducting all the time, the most meaningful inductance at signal frequencies is that of the whole winding as driven by the two triodes in series. The two halves of the primary winding are very tightly coupled and the diff-amp cannot be analyzed as two independent halves.
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Ok, I think I've got it. Please verify the attached schematic. Here are my responses:
In class A, where both triodes are conducting all the time, the most meaningful inductance at signal frequencies is that of the whole winding as driven by the two triodes in series. The two halves of the primary winding are very tightly coupled and the diff-amp cannot be analyzed as two independent halves.
The combined plate resistance of the two triodes in series is about 14.8k. Based on my understanding of your comment, I would want an output transformer with a total primary inductance of about 300 Henrys. This would be just over 2.5 times the plate resistance at 20 Hz. Does that sound about right?
There isn't anything about a balanced cable interface that both the hot and cold legs are actually driven by signal. This is a huge assumption that a lot of modern HiFi manufacturers would like their customers to make, but it simply isn't true.
In some cases, I would agree; it depends on the circuit. A lot of pro audio equipment is balanced and does have signal on both legs. The signal coming off a phono cartridge is balanced and has signal on both legs. If my phono preamp uses fully balanced gain stages throughout the circuit, the output (into this preamp) will have signal on both legs. On the other hand, I've seen many balanced connections that are just cosmetic.
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The combined plate resistance of the two triodes in series is about 14.8k. Based on my understanding of your comment, I would want an output transformer with a total primary inductance of about 300 Henrys. This would be just over 2.5 times the plate resistance at 20 Hz. Does that sound about right?
That's a good target to shoot for.
In some cases, I would agree; it depends on the circuit. A lot of pro audio equipment is balanced and does have signal on both legs.
A lot of pro audio equipment (especially a lot of the high dollar vintage gear and copies) have transformer coupled inputs and outputs to provide balanced inputs and outputs.
The signal coming off a phono cartridge is balanced and has signal on both legs. If my phono preamp uses fully balanced gain stages throughout the circuit, the output (into this preamp) will have signal on both legs. On the other hand, I've seen many balanced connections that are just cosmetic.
This is only true if the + and - connections to the cartridge for each channel see equal impedances to ground. If you connect -L and -R to hard ground at the input of your phono preamp, then you no longer have a balanced cable interface between your cartridge and phono preamp. If you put a step-up transformer between your MC cartridge and your phono preamp, you can actually add a few switches and resistors to listen to both a balanced and unbalanced cable interface. I would suggest trying and evaluating this setup.
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You have understood my erratic comments on the circuit correctly. :^)
I'll just add that the notion of primary inductance is pretty fuzzy when you have no airgap in the core. The effective inductance depends on signal level and frequency, and can vary over a very wide range. Usually the manufacturer's spec will indicate a suitable impedance level, i.e. the inductance will be acceptable at that impedance.