[luvdunhill]

Quote:

Originally Posted by regal /img/forum/go_quote.gif Guys the voltage drop is determined by the tube. You are setting current fixed via the IXYS and cathode bias via your cathode resistor, so the tube pushes its plate voltage to match. Based on the plate curve. Its backwards of how we were originally taught biasing tubes but this is how it works.

ah, that's what I thought, but I was confused by the terminology used, i.e. "this means dropping ~100V across the CCS" all is good and the world is still round...

 

[regal]

It is dropping across the CCS, the tube just determines how much.



Dsavitsk, just noticed that new DAC on your site, can't wait for the details.

 

[Pars]

So in terms of calculating what I would want to set the CCS up for (mA), what I posted would be correct (~18mA)? Or that is the correct way to do it?

 

[regal]

Pars if I read your post correctly you currently have 18 mA bias on your tube (pre CCS.) R3 in the ciruit I posted is just there to ensure that if you mistakenly turn the pot to zero the chip won't fry.

Rtest is just used when you hook of the CCS to a 9V battery and dial in the current you want (18 mA). Use a 100 ohm resistor and measure the Vdrop across Rtest. So dial the pot to 1.8V. Then remove or jumper Rtest.

You are now ready to sub for the existing plate resistor. You will need a big heatsink if you stay with 18mA (*180V) 3W. You can add a big 3W 5K resistor at the end of the CCS to share the load and reduce the heatsinking requirements.

 

[dsavitsk]

Quote:

Originally Posted by Pars /img/forum/go_quote.gif So in terms of calculating what I would want to set the CCS up for (mA), what I posted would be correct (~18mA)? Or that is the correct way to do it?

I think you are thinking about this wrong. You are not dropping voltage due to Ohm's law. The CCS does not have "resistance" per se. Rather, it has a changeable resistance determined by need.

The tube has an operating point. Usually you set plate voltage and bias, and the current is determined by the tube. With a CCS, you set current and bias, and the plate voltage is set by the tube. The CCS will then drop the correct amount of volts -- this is what it means to be a CCS: the current is constant regardless of outside influences. The CCS circuit will adjust what voltage needs to be dropped to met the constant current requirements. Really, go read the MEHA page. The notes on the Bottlehead CCS page might help a little, too.

 

[Pars]

OK. I read the MEHA page and started in on the Bottlehead page as well as the boozehound labs page, and will spend some more time going over this. About time to learn tubes I guess

A couple of questions:
1) Initially, with a plate resistor(s), current flowing thru the tube can be calculated by the voltage drop across the resistor(s) divided by the resistance, or 250 - 70 = 180 / 10,051 ohms = ~18mA... correct? So this would be the approximate current the design is looking for on the plate?

2) Looking at your MEHA page, the second load graph, you drew a load line thru the point. I don't quite understand how you determined the slope of this line. You say:
In the figure below, this line passes through our operating point, and contacts the X axis at ~23V (which is the plate voltage at 0mA of current) and the Y axis at 6mA (which is the plate current at 0 ohms).
Huh? Where did you get the 6mA and the 23V from? It takes two points to determine slope and you have 1 point defined. Or is this what is really going on:

The first is that the load line contacts the X axis at ~23V. I am assuming that we are using a 24V power supply. The 23V is to allow for the fact that that the cathode is biased up about a volt.

I'll do some more reading... thanks for the hints and help!

Chris

 

[dsavitsk]

I picked an operating point at 2mA/16V. Then, I knew that I had a 24V supply. When the tube is not conducting, when current flow is 0, the voltage on the plate will be the same as the B+ since there is no current flowing across the loading resistor and thus no voltage drop. So, 0mA/24V is the second point. Draw a line that connects them and extend it to the Y axis. It will contact 0V at some point -- 6mA in this case. So, this is when the tube is conducting 6mA all of the voltage will be dropped across the load leading to 0V on the plate. Now, it can't ever really reach these two extremes, but this is sufficient for estimation purposes. Then, you can use V = IR to determine the size of this resistor. The V is the change in voltage (24 - 0 = 24) and the I is the change in current (.006 - 0 = .006). Thus, 24/.006 = 4000.

 

[Pars]

Gotcha. The way the text read it didn't really explain where you came up with the points from. And reading the boozehound text, it appears the key is the change in voltage over the change in current to determine slope.

 

[TimmyMac]

I went to the dump today to take some junk, and I couldn't help noticing air conditioners people had thrown out with HUGE caps in them... looked like oil caps. Checked the specs on one - 45mfd 370VAC - would this be good to use in an amplifier somewhere? Film or oil caps that large are tough to find normally.

 

[Pars]

Quote:

Originally Posted by Pars /img/forum/go_quote.gif <snip> At any rate, I attached the schematic and a couple of pics. Schematic: Another modified pre (user):

I spent the weekend changing out the line stage coupling caps (the center 4 in the picture). Running RMAA on this, there is more rolloff below 50-30 Hz than there used to be (down 1.5dB at 20 Hz) but the graph looks much more rolled off than before. I did not change R17 form the 100K yet. A couple of questions:

• Looking at R17, it is attached to the wiper that is used for biasing. If I increase R17 to say ~500K, will this have an effect on biasing (i.e., will I be able to adjust it)? This is adjusted for lowest THD.
• As in the pictured modified unit, I have seen another one done like this that uses what I presume to be 8uf Dynamicaps (the big pair) which matches the stock value. The position these are in however is C10... not C8? C10 is followed by a 1M resistor to gnd. C8 by the 100K to gnd. Any reason you can think of for doing this, as it would seem that even with a 1uf / 1M on C10, the -3dB point is 0.16Hz?

Any other comments welcome.

 

[regal]

1. R17 is a grid leak resistor , it doesn't affect the tube bias. The pot doesn't either, it just sets gain. Increasing R17 isn't going to lower your -3dB point either. There is no reason to increase it.

2. C8 is an input coupling cap.

C10 may be followed by a 1M resistor but it will be in parralell with your amp's input impedance (meaning the 1M becomes negligible.) So just ignore that the 1M is there and calculate your -3dB for C10 using your amp's impedance.

 

[Pars]

Quote:

Originally Posted by dsavitsk /img/forum/go_quote.gif Lots of people have good things to say about the Dynamicaps. I don't think they'll be a bad choice. 1uF is probably fine on the output, and you should shrink C8 to that size as well. In fact, make R17 larger (like 500K) and you can shrink C8 even more. Does this really run with 42V on the plate of a 6DJ8? Remind me not to buy a Counterpoint.

Quote:

Originally Posted by regal /img/forum/go_quote.gif 1. R17 is a grid leak resistor , it doesn't affect the tube bias. The pot doesn't either, it just sets gain. Increasing R17 isn't going to lower your -3dB point either. There is no reason to increase it. 2. C8 is an input coupling cap. C10 may be followed by a 1M resistor but it will be in parralell with your amp's input impedance (meaning the 1M becomes negligible.) So just ignore that the 1M is there and calculate your -3dB for C10 using your amp's impedance.

Seems like R17 should still function as part of a high pass filter... at any rate the pot is used in adjusting for lowest THD.

The actual plate voltage is 52V in one channel and 57V in the other, from a B+ of 237V. I should probably raise the B+ a bit (schematic shows 253V). I'm not sure if raising the B+ would also raise the plate voltage or force R10 to dissipate more voltage drop across it (has 180V across it now).

 

[Pars]

Put the preamp on the bench today because the L channel went out for some reason... haven't found a problem as it is working fine

At any rate, Regal, you were absolutely right. I patched one of the 8uf caps in parallel with C8 and then C10 and ran RMAA on it. C8 had no effect. C10 took it out flat to 10Hz or less, down 0.2dB IIRC at 20Hz (vs 2.6dB previously). I think I will order bigger caps for C10 / C110. I had failed to remember that the output impedence with the 1M resistor was in parallel with the next device input impedence.

 

[regal]

You didn't believe me?

 

[Logistics]

I vote keeping the stock value in microfarads, but upgrading to a non-polarized (bipolar) capacitor from a known quality manufacturer. I don't mean something like a BlackGate, I mean something like a Panasonic SU.