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Dave the bass
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#46

Post by Dave the bass »

S'true, the data sheet even states its primarily a 'transmitter triode' but that doesn't stop you fella's :)

OK, so i've just done a quick search for '1626 operating points' and the boozehound lab site shows a proposed point at 200V Anode -18V bias and current through valve at 25mA. http://boozhoundlabs.com/howto-dc/

I'll mark my point and mark a tangent.

That OK so far?

Image

DTB
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#47

Post by Nick »

Yep, thats good, though the chart would indicate that that point woulr need -24v on the grid, not the -18v mentioned before. But other than that, yep fine.

Just found this on xkcd

"Warning: this comic occasionally contains strong language (which may be unsuitable for children), unusual humor (which may be unsuitable for adults), and advanced mathematics (which may be unsuitable for liberal-arts majors)."
Little known fact, coherent thought can destructively interfere with itself leaving no thought at all, that’s why I prefer incoherent thought.
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#48

Post by Dave the bass »

Ah, I was wondering about that, the bias lines don't tie up with the figures the website say. Errr...

1st thing that struck me about the new blue line was that its parallel to your red line.
2nd thing was your comment about it not being that linear for a Class A amp, as you probably already know a fair few of those 'Darling' amps have been built using these valves and word on the street is corrrrrrrrrrrr mostly. Low power natch but deffo corrr. Even though the op is on the curve? Maybe I haven't picked a good example to do this on. I thought the text book and general wisdom pointed to picking a point on the straight linear part of the curve... but with limits at 250V and 25mA its a bit limiting isn't it or am I missing summat?

I was gonna do some maths to find ra along the new blue load line just to see what I find out and to make sure last nights brain bashing has gone in. I've got a new battery in my calculator btw :)

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#49

Post by Nick »

Ok, first off, it may be that the curves look less than linear but it sounds great. I have heard 6080 used as a SET and it seemed to sound great, so don;t jump to conclusions.

Second. Don't be fooled into thinking that the bit where they are straight is the linear bit. What that means is the ra and hence the gm is not varying at that point. However the signal that comes from the valve is controlled by a load line that runs across those lines, and how linear that makes the results is more a matter of how the gaps vary between the grid lines, not the shape of the grid lines themself.

However what the straight lines do show is that with the actual elliptical load line that a reactive load provides the top and bottom of that curve will be passing through a area of roughly the same gm and ra.
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#50

Post by Dave the bass »

Oooo. I've just being working out the ra on that new blue line and the figure comes out the same pretty much everytime.

It would wouldn't it... as ra is the dynamic resistance of the valve.

Back-tracking, remember I wanted to work out the gm of a 1626 but didn't have the ra?... well we have now.

So from the data sheet, mu (amp factor) is 5 and ra is (as discovered how many times over!) approx 2580R.

Formula is u/ra = 5/2580 = 0.001937S = 1.9mS or 1937uS.

Is that right?

I need to know before going any further, my brain is proper frazzled through lack of sleep + number work.

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#51

Post by Nick »

Well, I don't know for certain if thats the correct number, but it certainly should be by the maths.

It does depend on if the gain is correctly reported. You can estimate that from the curves as well, MJ goes into how to find all three values from the curves, but I will leave looking that up as an exercise for the reader :-)

But, yep, I think you are doing things correctly there Dave.

I would have expected the ra to come out much the same with your blue line as my red one, because, as you noticed the two are parallel, so they have the same slope, and the slope represents the resistance, so the ra's wil be about the same.

There are considerable errors in using the curves to work these out, but as the curves may not represent the valve you have exactly anyway, its more than close enough. The only time knowing the ra of a valve with less than 10% erro mattersr I can think off is when designing a phono stage.
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#52

Post by Dave the bass »

Phew! Thanks Nick, rollercoaster ride this innit.

Now as educationally beneficial as it has been for me I'm going to bring the thread back on topic (:shock:) for a moment... remember the 100H chokes resulting in lack of bass in 6EM7? Well, that sterling fella MrI has kindly lent and posted me 2 more at >200H chokes which are identical to the 2 I was hoping to use but found they were O/C.

Well 2 in series in each channel has restored all the LF but its still go that detailed mid (zingy/alive, you know the words i mean). I think ultimately the HF is suffering but I'm tired and its been a very long day so I'm not at my best so I'll not jump to conclusions.

Attached is the bodge to try and cram 4 chokes onto the bread board :-) Classy eh.

Right, thats enough on-topic, lets get back to educational off topic stuff.

Where shall i go next? What can I do with these newly honed powers of maths and 2 triangles full of SI units?

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#53

Post by Andrew »

Dave the bass wrote: Where shall i go next? What can I do with these newly honed powers of maths and 2 triangles full of SI units?

DTB
Phono Stage :?:
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#54

Post by Nick »

Andrew wrote:
Dave the bass wrote: Where shall i go next? What can I do with these newly honed powers of maths and 2 triangles full of SI units?

DTB
Phono Stage :?:
Or maybe a common cathode amplifier stage. Start with valve and B+, pick an op point, decide on a anode resistor, then pick the cathode resistor, calculate the size of bipass cap to use, and finally use the curves to predict the gain, and max rms output from the stage. And finally predict the output impedance of the stage.
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#55

Post by simon »

Dave,

I reckon you should have a go at a phono stage cos then I can learn how to do it too ;).

I agree with Nick that a simple common cathode stage is a good place to start. Things dropped in to place a lot more once I worked through an example. It could make an interesting project too - a pre to try with your amps. Something like a 45 or a 26, or a 31, or a 2A3, or...

Being direct heated there'd be issues with hum on the filaments but nothing that isn't easily sorted with a simple current reg, but they'd have the benefit of low gain so could work well. Doesn't have to be a keeper but the learning from designing and building a common cathode stage would be great :D.
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#56

Post by Dave the bass »

Sorry for delay, I had a sudden career change yesterday, I went Labouring on a building site. Phase 2 of the Skatepark is being built right now and I offered a days labour FOC.

Anyway, back to schoolin'.

Phono stage! :shock: *falls off chair*

:lol: Bunchajokers :lol:

OK, Common Cathode pre-amp stage it is then. Its gotta be as simple as it can get, I'd like to nominate for myself an IDHT, just to make things even easier, a 76 or a 27 both at the low end of medium mu? I've also got both to hand.

http://www.nj7p.org/Tube4.php?tube=27

http://www.nj7p.org/Tube4.php?tube=76

Lets get busy.

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#57

Post by Nick »

Ok, so first step, pick a valve.

Then you need to pick a operating point. Both of the valves you have chosen have suggested op points. But the first thing I would do is take the valve of choice and draw the max power curve on the graph. However the 27 doesn't seem to tell us that.

So, it does state some op points. First step pick one.

Second stage, decide what b+ you have on hand, or plan to use.
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#58

Post by Dave the bass »

I'll make it easy then, it'll be a 76. I'm spoilt for HT rails.... pick a point between 0 and 350V up to 200mA. I've got my own variable PSU to play with.

Lets say HT of 350V. Thats real-world innit?

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#59

Post by simon »

76 is a good choice, and easier than a DHT so probably much better than my selections :). And you've got some too ;).

Agree with Nick again, start with the curves and plot things like max power, max voltage etc. If you've got a variable PSU then it becomes a lot easier :). As a quick ready-reckoner for HT, if you were to choose the 250V 5mA OP then you'd need 250V + 13.5V grid volts + voltage drop across the load resistor (yer old V=IR again) say 50V. But the beauty of a variable PSU is you can play around till your heart's content. And then you can try your choke loaded anode too :D.
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#60

Post by Nick »

Ok, so the 76 spec gives two op points. Lets start with those.

First 250v @ 5 ma -13.5v on grid
Second 100v @ 2.5ma -5v on grid

Start with the first one. Ok, if we have 350v B+ and 250v on the anode we need to be dropping 100v across the anode resistor when we are passing 5ma. So from ohms law, that makes the anode resistor 100 / 0.005 or 20k. Ok, lets plot the load line for a 20k anode resistor with a b+ of 350v.

So lets consider two potential conditions. First one, the valve is not conducting at all, its entirly shut off. In that condition there will be 0 current. And if there is 0 current, there will be no voltage dropped across the resistor, so the voltage on the anode will be the same as b+. so lets put a point on the plate curve, at 0ma current 350v voltage.

Now, the second imaginary (in that it can't actually happen) condition, what would happen if the valve was perfectly conducting? There would be no voltage across the valve, so we are on 0v on the voltage axis. If the valve was a perfect conductor all the b+ would then be across the load resistor, so what current will flow with 350v across the 20k resistor. Ohms law again gives us the answer, 350 / 20000 = 17.5 ma. So you can put the other point at the 0v 17.5 ma point on the graph,

Now every other thing the valve can do with that resistor in its anode and 350v across the pair is shown by drawing a line between the two points we have. Draw that, and it gives your load line.

Notice it will pass through the 250v 5ma point we started with.

So now, if you do that Dave, then we can go onto the next step. If it makes sense, follow the same process to draw the load for the second op point.
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