BJT/Tube?
- Thread starter analag
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I could not tell you if it is going to work or not, but it is certainly interesting. I am having alot of fun trying to sort this circuit out. I just started playing around with using the cathode in alternate ways the otherday. Confused by a couple things, but some time with the schematic and some paper should sort it all out.
adam
adam
Larrchild
Well-known member
Might drift. Put a zero pot on the first stage, too. 6DJ8's are less consistent than bjt's.
Output impedance seems high. Was that gonna drive a transformer?
Output impedance seems high. Was that gonna drive a transformer?
pstamler
Well-known member
What is D2? If it's just a regular silicon diode, dropping 0.65V, you won't develop any appreciable output from the current source. (If it's an LED, then you're rollin'.)
Also, would the two pots, unless they track super-duper well, not mess up the common-mode rejection of the circuit?
Peace,
Paul
Also, would the two pots, unless they track super-duper well, not mess up the common-mode rejection of the circuit?
Peace,
Paul
[quote author="pstamler"]What is D2? If it's just a regular silicon diode, dropping 0.65V, you won't develop any appreciable output from the current source. (If it's an LED, then you're rollin'.)
Also, would the two pots, unless they track super-duper well, not mess up the common-mode rejection of the circuit?
Peace,
Paul[/quote]
Oops I forgot to mention it's an red LED, my bad. I always use a pot to describe gain control, but we know for superior performance a rotory swith is the way to go. Thanks for pointing that out.
analag
Also, would the two pots, unless they track super-duper well, not mess up the common-mode rejection of the circuit?
Peace,
Paul[/quote]
Oops I forgot to mention it's an red LED, my bad. I always use a pot to describe gain control, but we know for superior performance a rotory swith is the way to go. Thanks for pointing that out.
analag
:thumb:
Are you planning to use a transistor array for this one?
Matching discrete devices is a pain and temp. drift can be a real problem.
Othervise i like the topology. Reminds me of some MM class-B guitar amps where the cathodes of the output tubes are driven in the same manner.
Are you planning to use a transistor array for this one?
Matching discrete devices is a pain and temp. drift can be a real problem.
Othervise i like the topology. Reminds me of some MM class-B guitar amps where the cathodes of the output tubes are driven in the same manner.
This is the sort of circuit that probably gives good results in a simulation but is difficult to translate to a design that works well in practice.
So, several points:
1) The current through the first stage is set by the small difference between the forward volts of an LED and the base-emitter junction of a silicon transistor, both of which have a large tolerance and both change with temperature. You've given yourself 50V at the current source to play with (but where are you going to get a -50V supply in practice?), so why not improve the reliability of the design by using a 6.8V zener diode (best temperature coeff voltage) instead of the LED and increasing the emitter resistor to get the required current?
2) The BC550s are high-voltage transistors, but otherwise unremarkable. The maximum voltage they need to work with is the full common-mode range of the input plus the bias volts of the 6DJ8s, a total of (say) 25V. Most standard transistors could handle that, and you could choose ones that have lower noise figures than the BC550. You only need to go sqrt(2) times better to need only one transistor per phase and then you avoid...
3) If you stick with using two transistors in parallel, you need a way of ensuring both the d.c. and the signal current is equally shared between them. Separate emitter resistors are all very well, but they don't take account of things like component tolerance and Vbe differences. Maybe a 20R pot between the low ends of the resistors?
4) In a similar vein, you need a way of trimming the input stage for d.c. balance between the two current paths The wipers of the 20R pots from point (3) could themselves go to another 20R pot whose wiper goes to the current source.
5) CMRR. The only concession the design makes to common-mode rejection is the compliance at the collectors of the input transistors and the current source. From there on, common-mode has the same gain as signal. But that is a consequence of ....
6) Balance between the phases. The circuitry after the anodes of the first stage comprises two completely independent single-ended gain of 3 amplifiers, with no attempt to match any of the parameters between the phases. I think you should be considering designs that have more cross-coupling between the phases.
7) The open-loop gain of the output stages is a bit less than 30, so there's not much spare for the negative feedback to work with for distortion reduction and to keep the output impedance low.
Apologies if this sounds like a severe critique, but they were the thoughts that came into my head when I glanced at the schematic. Interesting post, though!
So, several points:
1) The current through the first stage is set by the small difference between the forward volts of an LED and the base-emitter junction of a silicon transistor, both of which have a large tolerance and both change with temperature. You've given yourself 50V at the current source to play with (but where are you going to get a -50V supply in practice?), so why not improve the reliability of the design by using a 6.8V zener diode (best temperature coeff voltage) instead of the LED and increasing the emitter resistor to get the required current?
2) The BC550s are high-voltage transistors, but otherwise unremarkable. The maximum voltage they need to work with is the full common-mode range of the input plus the bias volts of the 6DJ8s, a total of (say) 25V. Most standard transistors could handle that, and you could choose ones that have lower noise figures than the BC550. You only need to go sqrt(2) times better to need only one transistor per phase and then you avoid...
3) If you stick with using two transistors in parallel, you need a way of ensuring both the d.c. and the signal current is equally shared between them. Separate emitter resistors are all very well, but they don't take account of things like component tolerance and Vbe differences. Maybe a 20R pot between the low ends of the resistors?
4) In a similar vein, you need a way of trimming the input stage for d.c. balance between the two current paths The wipers of the 20R pots from point (3) could themselves go to another 20R pot whose wiper goes to the current source.
5) CMRR. The only concession the design makes to common-mode rejection is the compliance at the collectors of the input transistors and the current source. From there on, common-mode has the same gain as signal. But that is a consequence of ....
6) Balance between the phases. The circuitry after the anodes of the first stage comprises two completely independent single-ended gain of 3 amplifiers, with no attempt to match any of the parameters between the phases. I think you should be considering designs that have more cross-coupling between the phases.
7) The open-loop gain of the output stages is a bit less than 30, so there's not much spare for the negative feedback to work with for distortion reduction and to keep the output impedance low.
Apologies if this sounds like a severe critique, but they were the thoughts that came into my head when I glanced at the schematic. Interesting post, though!
[quote author="Larrchild"]Might drift. Put a zero pot on the first stage, too. 6DJ8's are less consistent than bjt's.
Output impedance seems high. Was that gonna drive a transformer?[/quote]
I think it's something like 90 ohms per CF.
Output impedance seems high. Was that gonna drive a transformer?[/quote]
I think it's something like 90 ohms per CF.
Larrchild
Well-known member
If that 90 ohms is predicated on lots of feedback, it will rise with gain.
And output will be load sensitive. Edcor, dude.
If people could get away with 2 CF's instead of output iron, designers in the last 60 years would be right on that. Hell, I'd be on it!
With negative feedback, the output impedance is about 10 Ohms at minimum or maximum gain settings, and less inbetween.
That's estimated as follows:
The gm of an ECC88 (6DJ8) is 12.5mS, giving an output resistance as a cathode follower of 80 Ohms (1/gm). If you lift the output by 1 volt, that 1 volt goes round the feedback circuit (150K/47K) and becomes 0.25V at the input grid. The gain of the circuit (anode resistor/cathode resistor of first stage) is -30 (less in practice because of the anode impedance), so the 0.25V becomes -7.5V at the cathode end of the output resistance. That gives a total voltage change of 8.5V across the virtual 80 Ohm resistor, causing 106 mA to flow. So our 1V lift results in about 100mA, giving an effective output resistance of 10 Ohms.
However, as Larrchild pointed out, another shortcoming of the circuit design is that the amount of negative feedback varies with the setting of the gain control, so not only will the output amplitude not be what is expected, but the output impedance, bandwidth and circuit stability will change with gain setting.
I have a JFet variation of this, with higher B+ and White CF the output. Input caps are 5.1uF. Low frequency response with various mics is outstanding, as is detail and everything else we expect in a good mic pre.
This is my favorite pre.
analag
This is my favorite pre.
analag
fucanay
Well-known member
I would love to see analag put out some boards too. And his handle is a great name for a company. It's a play on three different words. Anal (slang for "to pay excessive attention to detail"), analog, and lag (slang for being slow, which I am with DIY :razz
.
Umm sorry, this is way off topic huh. damnit.
Matt
.Umm sorry, this is way off topic huh. damnit.
Matt
[quote author="Boswell"]However, as Larrchild pointed out, another shortcoming of the circuit design is that the amount of negative feedback varies with the setting of the gain control, so not only will the output amplitude not be what is expected, but the output impedance, bandwidth and circuit stability will change with gain setting.[/quote]
No the feedback will always hover around the same amount, because we will always be using the first gain stage to drive up the second stage to pretty much the same output swing, if we record at consistent levels. If the source is weak we turn up the gain, if the source is loud we turn down the gain, in essence we are always sending a certain signal level to the second gain stage. That is why I don't care if I have a control there, because it allow signal consistency between the first and second stage.
analag[/quote]
No the feedback will always hover around the same amount, because we will always be using the first gain stage to drive up the second stage to pretty much the same output swing, if we record at consistent levels. If the source is weak we turn up the gain, if the source is loud we turn down the gain, in essence we are always sending a certain signal level to the second gain stage. That is why I don't care if I have a control there, because it allow signal consistency between the first and second stage.
analag[/quote]
Larrchild
Well-known member
He meant this. As you lose feedback headroom, the output z becomes dynamic. In a transformer output, thats not fatal. A CF to the outside world is, tho.
You miss the point. The 500K gain control that you show in the schematic, whether it be implemented as a pot or as switched resistors, has a wiper that drives a summing junction (virtual earth) through 47K. At its ends, the wiper looks like a zero ohm source and in the middle it looks like 250K. This has two effects:
1) The fraction of the output that is fed back to the first grid by the negative feedback loop is 47K/(150K+47K) = .238 at the two ends of the gain pot but (47K+250K)/(47K+250K+150K) = .664 at the middle. The closed-loop gain of the output stage will therefore vary from 3.19 down to 0.505 and back to 3.19 over the range of the gain pot.
2) Because of the loading effect of the 47K to virtual earth, the law of the gain pot will not be linear, logarithmic or whatever is expected. With switched resistors, it would be possible to calculate the values that would give you (say) 3dB steps overall, taking into account the change in closed-loop gain of the output stage, but it would take a complicated spreadsheet to get that right. Also, it would have to take into account that the virtual earth is not zero signal volts, because the open-loop gain is only about 30.
From what you say, in practice, it sounds as though you wind the knob up and down until you get the output signal level that you want. Fair enough, and the linearity or calibration of the gain control has little importance in that method of operation. But I would be seriously worried by the effect that the (more than) 16dB change of negative feedback over the operating range of the control would have on signal quality, output impedance and damping factor.
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