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Old 21st May 2012, 08:39 PM   #81
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Quote:
Originally Posted by yurgs View Post
Hello,
@Chris
there is LF Oscillation and the squeeging with the feedback loop open,
there is LF Oscillation as soon as i feed a signal <40Hz from the pc to the amp. it's visible at the bias balance leds.
Initially I was thinking of accusing the LF sub oscillation at an overvalued interstage coupling cap (C46? 0.47uF) but you having mentioned LF oscillation with the global feedback loop open, that prettywell rules that out. Now I'm becoming suspicious of a dirty signal source from a pc.
It may have alot of RF i.e badly filtered but maybe well above the 1st stage zobel R/C pole.

richy
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Old 21st May 2012, 10:08 PM   #82
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Tom, Pardon me, I don't mean to be offensive.

But I am still perturbed to find my suspicions are founded - some portion of a DataSheet 101 is really required. I can see that the Simulator has lulled you into thinking that the data sheet values for Ia+Ig2 are peak values, when in fact they are always dc or average values. This is true for old UK data sheets (The Symbol I is defined as dc current in a data-sheet British Standard of the time), and U.S. originated ones as well. It is true for an 813 transmitting tube, just as it is an Audio Beam tube.

There's no need for me to prove this by dragging out the measurements for the my old KT88 amp - it can be demonstrated in minutes using traditional graphical analysis. It is very easy in this case because our KT88 amp is run in text book operating conditions.

As per KT88 data sheet, we use:

Va(0) = 453V
Ia+g2(0) = 2x 50mA
Ia+g2(max sig)= 2x 140mA
RL A-A = 4K
bias = -59V
Vg1 (ac)(crest)=114V (ie peak to peak)
Pout = 70W

Textbook UL conditions are used, so a load line can be imposed on the UL curves given in the data sheet.

Construct the load line in the usual way. For PP, we analyse a single valve's curve. Each PP half drives half the voltage, so the impedance presented to a single valve is 4K/(N^2) = 4K/4 = 1K. We draw the load line from the bias point (453V, 50mA), and take a 1K gradient to the point where Vg1 = -2V: i.e.: (-59 + 114/2)V.

Read off the peak currents: Ia = 360mA peak, Ig2 = 36mA peak.


Therefore,

Ia+Ig2 peak = 396mA
I(av) = I peak / (2^0.5) = 280mA

The B+ capacitor carries out the peak-to-average conversion: that is exactly what it is there for. Therefore, the dc supply current becomes the data-sheet value of 280mA dc.


I would never be parted from my SPICE simulator, and use it every day for professional purposes. For quantification, and experimentation with values and choices, it save me so much time. But I never use it to tell me how something works... I will always work out a circuit by hand first. The day one stops doing this, one risks losing control over one's designs....
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Last edited by Rod Coleman; 21st May 2012 at 10:18 PM.
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Old 21st May 2012, 10:12 PM   #83
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Alternatively, we can look at the data sheet of the General Electric 6550, a close relation of the KT88, and see the current labelled as dc, explicitly.
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Old 22nd May 2012, 02:12 AM   #84
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Quote:
Originally Posted by Rod Coleman View Post
Read off the peak currents: Ia = 360mA peak, Ig2 = 36mA peak.
I agree.

So why does my simulation line up so well with the datasheet values? Happy coincidence? Did I somehow get off by exactly sqrt(2)?

Quote:
Originally Posted by Rod Coleman View Post
The B+ capacitor carries out the peak-to-average conversion: that is exactly what it is there for. Therefore, the dc supply current becomes the data-sheet value of 280mA dc.
How does a capacitor average current? Given that the reservoir cap is constantly recharged, I can see how it would average the B+ voltage, but not the current...
The inductance of the OPT primary winding, on the other hand would do a fine job of averaging the current. Naturally, this would lead to a change in the waveform shape from the full-wave rectified waveform you described earlier to something with a lower di/dt.

Quote:
Originally Posted by Rod Coleman View Post
But I never use it to tell me how something works... I will always work out a circuit by hand first. The day one stops doing this, one risks losing control over one's designs....
I use simulation tools to assist me in the learning process. I use that along with graphical methods like load line analysis, etc. for my designs. And I do appreciate you taking the time to draw up that load line. I hadn't taken the time to do that. Thanks.
There are many ways to accomplish one's goals. I hope the Internet is big enough to allow us to coexist despite any differences in our design or learning approaches.

~Tom
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Old 22nd May 2012, 03:53 AM   #85
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Rod,

I took the parasitic resistance out of the OPT model. Now, I'm able to reproduce the datasheet values as you describe. I guess the sqrt(2) difference was just pure coincidence.

I think I misunderstood your comment about the B+ capacitor averaging the current. I took "averaging" to mean "smoothing". My bad. The current isn't changed by the capacitor, but the voltage across the capacitor is proportional to the integral (or "average") of the current into (or out of) it.

I think we agree now. Thanks for the discussion. Yurgs - sorry we polluted your thread.

~Tom
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Old 22nd May 2012, 04:50 PM   #86
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Tom, Great. Agree, plenty of room for many approaches.

While we have the UL load line to hand, may as well examine the effect of the apparent mismatch. As you say, connecting 8R to the secondary of an OT intended to go from 5K to 3.85R - will present over 10K to the Anodes.

This load reduces the peak anode current to 200mA or so, and the dc Ia+g2 drops to about 155mA per channel.

This could well explain why Yurgs is seeing only a small drop in the dc supply, when clipping takes place.

4R load (eg 2x parallel 8R) certainly recommended, Yurgs!
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Old 22nd May 2012, 05:01 PM   #87
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The other risk with that high load - the load-line hits Vg1=0 below the knee in the curves. This means that the peak Ig2 can suddenly increase as Va drops below about 60V, and reach damaging levels.

The G2 stoppers are there to keep this effect from getting out of control - but still, operating with such a light load is a risk. Overheated screens are the top killer for End-tubes.
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Old 22nd May 2012, 06:01 PM   #88
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Quote:
Originally Posted by richwalters View Post
Initially I was thinking of accusing the LF sub oscillation at an overvalued interstage coupling cap (C46? 0.47uF) but you having mentioned LF oscillation with the global feedback loop open, that prettywell rules that out. Now I'm becoming suspicious of a dirty signal source from a pc.
It may have alot of RF i.e badly filtered but maybe well above the 1st stage zobel R/C pole.
Yeah... I'm not convinced the problem is in the amp, actually.

- We've beaten the supply concerns to death. Yeah, it's not an ideal design, but it doesn't seem to be significantly worse than most other stuff out there.
- The amp shows the same signs of misbehaving when the global negative feedback loop is open. So it's not a loop stability issue.
- While the CCS'es do eat up a good 50 V of headroom (at least the one in the input stage does) and could be improved, there is really nothing downright wrong with them.

Yurgs stated that the problem started at f < 40 Hz when the input was turned up. How high does the volume need to be turned up in order for the problem to be present? I.e. what's the output power into a 4R load when the problem is present?

~Tom
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Old 22nd May 2012, 09:43 PM   #89
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Quote:
Originally Posted by tomchr View Post
The schematic in Post #2 says 5k:3.85ohm for the OPT. I interpret this as a 5k primary when loaded by 3.85 ohm on the secondary. If this is correct, then I suggest using a 3.9 ohm or 4 ohm dummy load. If an 8 ohm load is used, the primary impedance will be closer to 10 kOhm, resulting in lower output power.
~Tom
tom , this is correct, the opt offers a variety of load matching options, i initialy set it up that way thinking that a 10k load is easier to drive than a 8k load, and i could run the amp on this setting for 4 and 8r speakers. i also asumed that most speakers average on something like 6r anyway.
i am away for work reasons for the whole week, and therefore not able to change or measure anything
i can try with a 4r load on the weekend if everything here at work runs smoothly.

@rob & tom, seeing that it helps me understanding some basics of psu design,
no, i dont mind these things beeing discussed here, as long as it sort of helps the idea of the original thread.

i also wanted to add that i use KT90 in the amp now.

Chris mentioned earlier that he would focus on getting rid of parasitic oscillation first( Morgan Jones as well), the only thing i tried in this direction is replacing the gridstoppers on the output tubes with 10k, but with no change.
the noise i called squeeging might in fact be something else, i believe its produced by the right output tube of the chanel under test. ( because it sounds different when i touch the envelope of the tube with my fingernail)
so maybe this is connected to the load - matching of the opt as well.
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Old 22nd May 2012, 11:29 PM   #90
tomchr is offline tomchr  United States
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Quote:
Originally Posted by yurgs View Post
i initialy set it up that way thinking that a 10k load is easier to drive than a 8k load, and i could run the amp on this setting for 4 and 8r speakers. i also asumed that most speakers average on something like 6r anyway.
I would recommend running the amp at its specified load impedance. Though, you're right. Most speakers' impedance vary widely across frequency.

~Tom
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