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Old 29th May 2010, 05:10 PM   #1
alexage is offline alexage  United States
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Default Concertina Splitter question - Need help

Attached is a schematic I made of the driver and concertina splitter from an old Lectrolab amp that was given to me after the last tech gave up. There is no available schematic of the amp, model S 950. I checked this a few times, the circuit is as I describe it.

The 47K resistor at the bottom right of the attached schematic was unconnected at one end, as shown. Can any one tell me:

- What function this resistor serves, where it might connect? The circuit would look like a standard self-biased concertina splitter without it (i.e., if there was a connection to ground in place of the 47K resistor).

- Why the driver cathode is connected to the cathode-end of the concertina?

Thank You!
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Old 29th May 2010, 05:36 PM   #2
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You should be able to trace the 47 KOhm resistor to the O/P trafo secondary. It's part of the NFB loop. The 47 KOhm and 680 Ω parts form the gain set voltage divider.
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Old 29th May 2010, 05:44 PM   #3
alexage is offline alexage  United States
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Thank you, Eli.

So current for the bottom half of the splitter flows through the driver-stage cathode resistor (680 ohm)?
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Old 29th May 2010, 06:06 PM   #4
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The current flowing in the NFB loop is tiny. Remember, the NFB loop is in parallel with the loudspeaker. For practical purposes, only a voltage is present. The values you provided feed back only 1.4% of the voltage applied to the loudspeaker. I suspect you have misread a resistance value. For instance, 4.7 KOhms combined with 680 Ω feed back 12.6% of the net O/P voltage, which strikes me as being a more realistic number.
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Old 29th May 2010, 08:28 PM   #5
alexage is offline alexage  United States
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I understand the NFB, and yes, we usually see a ratio of 1/20 or higher, so what you say makes perfect sense.

My last question doesn't relate to NFB, though. I'm sure this is garbled, but let me try again:

I'm wondering about the current source/path for the CATHODE of the tube on the right in the schematic, which is the cathodyne/concertina splitter. If there was no NFB, you would see it connect to ground (where the 100K and 47K meet in the schematic). This is a very common configuration.

Since there is no ground there in this case, where does cathode connect to ground? I see the path as going through the 1k8 resistor (generating the bias for the grid), then through the 100K, generating the current for the cathode, then over to the 680 ohm resistor which serves as the cathode resistor for the OTHER tube, the driver. Does it make sense for the cathodes of BOTH tubes to share that last resistor, and why?
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Old 29th May 2010, 08:31 PM   #6
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I wonder if anybody else has tried the concertina strap to the cathode of the first stage and the results ? The 100K top and bottom resistors seem pretty high.
I'm assuming the concertina operates with no distortion but what benefit is there other than making the first stage operate more linearly ? I wouldn't have thought it was worth it.

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Old 29th May 2010, 08:53 PM   #7
SY is offline SY  United States
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I suspect that's a mistake, since it would be detrimental to performance- the OP suggested that someone had been screwing around with the amp, fruitlessly. Perhaps that's where the trouble began?
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Old 29th May 2010, 11:51 PM   #8
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Quote:
Originally Posted by alexage View Post
I understand the NFB, and yes, we usually see a ratio of 1/20 or higher, so what you say makes perfect sense.

My last question doesn't relate to NFB, though. I'm sure this is garbled, but let me try again:

I'm wondering about the current source/path for the CATHODE of the tube on the right in the schematic, which is the cathodyne/concertina splitter. If there was no NFB, you would see it connect to ground (where the 100K and 47K meet in the schematic). This is a very common configuration.

Since there is no ground there in this case, where does cathode connect to ground? I see the path as going through the 1k8 resistor (generating the bias for the grid), then through the 100K, generating the current for the cathode, then over to the 680 ohm resistor which serves as the cathode resistor for the OTHER tube, the driver. Does it make sense for the cathodes of BOTH tubes to share that last resistor, and why?
Normally, unbypassed 680R at V1's cathode would be a negative feedback.
Reducing voltage gain and linearity of Mu, in return for an inreased linearity
of Gm. Assuming this is a voltage gainstage driving a very high impedance,
maybe a NFB here that only helps linearize Gm isn't a worthwhile tradeoff.

What you see is a positive feedback. Cancelling some or maybe all the NFB.
Result in fixed voltage at V1's cathode. Just like a bypass, but without the
extra capacitor. Allowing V1 triode's Mu to be in control, rather than Gm.

It is an attempt to be no cathode feedback, by virtue of PFB cancellation.
Thats why it looks screwy, you thinking all feedbacks must be negative...

----------

V2 wired as concertina, has entirely sacrificed Mu linearity for Gm linearity.
This type of linearity is fine too, but you could replace V2 with a transistor.
Function of Mu is made irrelevant by huge unbypassed cathode resistance.

Contrast V2's linear Gm against linear Mu function of V1!

Last edited by kenpeter; 30th May 2010 at 12:15 AM.
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Old 30th May 2010, 01:54 AM   #9
Enzo is offline Enzo  United States
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In another forum, it was suggested that the circuit made more sense if the wire along the lower left connected to the bottom of the 680 ohm resistor instead of the top, ie. ground.
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Old 30th May 2010, 03:04 AM   #10
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I was right to be curious on this: My 1960 dusty files came up with the circuit below: Well, it's lacking the o/p tranny global nfb stuff, but close enough to think about the combination.. My reckoning, in this circuit, oscillation and instability would have occured if the anode load of the first stage obtained it's volts from the conventional B+ way; But as the anode volts derived from the second stage cathode, the cathode voltage is in phase with the first stage triode anode (neg fbk) so stability results. The two cathodes joined is the positive feedback connection. I think I've got this right. So we have positive and negative feedback circuit. This circuit is a high gainer, except Miller effect makes effective gain slightly more than a sigle pentode stage. x100 x200.
Probably the ideal arrangement for an ECC83 but no further. Might try doing a 10W amp with this circuit and see how it turns out.The trick is to find the neutralised values of both pos and neg feedback to keep circuit stable. More math necessary.

Yes SY, someone probably botched alexages original.

richy
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