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-   -   6922 as a CCS, a question ? (http://www.diyaudio.com/forums/tubes-valves/31734-6922-ccs-question.html)

Shoog 6th April 2004 07:02 PM

6922 as a CCS, a question ?
 
I am trying to design a simple preamp. I am going to use one half of a 6922 as a CCS. I will be running the tube at B+ of 50V (anode at 0V and cathode at -50V) delivering a constant current of about 7mA. I think I need to hold the grid at about -35V by using a voltage divider. Anyone like confirm that i'am on the right track ???

SY 6th April 2004 09:01 PM

The grid has to be negative with respect to the cathode by a couple of volts. What I'd do is connect the cathode to the -50 rail via a resistor (one of the sim guys can probably suggest a value; if you want to use my method of solder-and-try, start with 390-470 ohms), then connect the grid directly to the -50 rail. The resistor will bias up the cathode by a few volts positive, which in a relative sense accomplishes the same thing. The cathode resistor's degeneration will also stiffen the CCS, a good thing.

cogsncogs 7th April 2004 07:07 AM

7mA
 
Quote:

I will be running the tube at B+ of 50V (anode at 0V and cathode at -50V) delivering a constant current of about 7mA.
Hi

I don't think you can pass 7mA thru a 6DJ8 with only 50V anode to cathode. I ran a few sims and according to the two models I have you need about 165V plate to cahtode with a cathode resistor of about 470 ohms. One model with 50V a-k was about 4mA with a 30 ohm cathode resistor.

Wayne

cogsncogs 7th April 2004 07:17 AM

Hi again

If you paralleled the two sections you can get 7mA barely. 30 ohms with one model, 82 ohms with another.

Wayne :smash:

Shoog 7th April 2004 12:26 PM

As you can probably quess my knowledge of tubes is scant so you will have to bare with me.
I was originally going to use a 7K resistor to apply the 7mA to the cathode. I have used a 6922 with 100V B+ and generated a 7mA current through it and things sound fine. In this configeration the 7ma was generated with a 14K resistor. I am trying to improve the whole circuit by replacing that resistor with a constant current source, unfortunately in doing that I think I am halving my availble voltage over the two halves of the triode, so instead of having 100v availble to the valve, I only have 50V each half. Will I have to accept that if I go down this rought that I will only have the much lower current to play with. My question really is, if I attempt to get this working at either 3ma or 7ma, what potential will I have to hold the grid at, and what resistor wiill generate that potential, assuming that I use a single biasing resistor, referenced to the negative rail.

If I get this right I think a lot of people may be interested in the result.

Please forgive my ignorance on this, but despite all my internet research I have a very slight grasp on the mechanics of valves.

Shoog

SY 7th April 2004 12:36 PM

I had assumed that you were going to parallel sections. Given Wayne's sim, you'd do well to increase the available voltage, elsewise you're trying to swim upstream. More voltage = higher cathode resistor = better CCS performance.

EC8010 7th April 2004 12:59 PM

The Mullard curves suggest that you can pass 7mA at Va = 50V if you set Vgk to -0.9V. As SY says, insert a cathode resistor of 0.9V/7mA = 129 Ohms. Nearest standard value is 130R.

cogsncogs 7th April 2004 09:04 PM

Hi

Yes, I have to add that the models I have aren't very accurate in the high current region. So experiment with the resistor values.
As Sy said the higher the V a-k and resistor value the better the regulation. There may be other valves that will give you what you need.
You could use a FET (cough). :xeye:

Wayne :)

Shoog 7th April 2004 09:34 PM

I omitted to mention the possably vital information that the CCS is to be at the bottom of a Cascoded pair, and that the line out is at the junction of the two. Hence it functions as a less than unity gain line amp.
I am certain that this will effect the implementation, as none of the example of this use just a cathode resistor. I have seen this done with both a single resistor from the neg rail to the grid and a voltage divider from with one leg to earth. There is then a cathode resistor. I am not certain of the mechanism that this uses, but it is a varient of the grounded cathode amplifier. I am begining to think that in this situation the B+ voltage will not be at 0V.

Again forgive my total ignorance- but I am really trying hard to understand a new field.

Shoog

EC8010 7th April 2004 11:25 PM

Quote:

Originally posted by Shoog
I am begining to think that in this situation the B+ voltage will not be at 0V.
It probably won't be, it will be a few volts positive, and depending on valve choice, even +10V. It won't matter much.


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