Bottlehead C4S as a differential pair constant current sink?

[Mikey]

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

[Paul Joppa]

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.

[Mikey]

"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

[Paul Joppa]

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.

[Horns Forever]

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.

[Doc B.]

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.

[Paul Birkeland]

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.

[Horns Forever]

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!

[Paul Birkeland]

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. 

[Horns Forever]

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://)

[Paul Birkeland]

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)

[Horns Forever]

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.

[Paul Birkeland]

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

[Horns Forever]

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.

[Paul Birkeland]

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.