Using 0D3's for DC regulation in Aikido

[cbutterworth]

I have been experimenting with PSUDII and with my present PSU, I get to within 2 VDC between PSUDII model and empirical measurement.

Anyway, I think I may regulate the DC on my PSU, which currently is: 275-0-275; 5V4G; 2.5uF; 10H; 60uF; 10H; 80uF = 344 VDC loaded with approx 25.8mA.

I want to add 2 0D3's in series to the PSU as they will regulate to 300V. To estimate resistor value to keep the current seen by the 0D3 to around 10mA, I added a final RC filter to my PSU. The R was initially set at 1K and the cap is set at 0.1uF (recommended as a noise bypass cap for the 0D3s - maybe not really needed?). I then set the current load to my 25.8mA PLUS 10mA for the 0D3's and began to model. I then adjusted the resistor to an NPV that results in a final B+ of around 300VDC. When I allow the 0D3's to draw 10.6mA, I get a PSUDII modelled B+ of 300.6VDC.

From this, I estimate that to use 2 0D3's in series to regulate at 300VDC, with 25.6mA tube current and around 10mA seen by the regulators, THEN I need a 680R resistor between 80uF cap and the 0D3s.

Also when I bypass the 1st 0D3 with a 100K resistor, I allow enough voltage to ensure that the second 0D3 lights-up.

My questions are:

1. Have I calculated the 680R resistor correctly?
2. Will my PSU deliver enough voltage to ensure that the 0D3 light-up properly?
3. Do I need the bypass cap?
4. Do I need the bypass cap for 0D3 #1?

I have attached a schematic.

Thanks,
Charlie

[pchw]

I will give it a try.

1. Have I calculated the 680R resistor correctly?

You want to drop 44V (344 - 300), the current passes the R is
44/680 = 64.7ma. Main circuit draws 25.8ma. The 0D3 will shunt
approx. 39ma which is very close the 40ma limit. There is no room
for a slight voltage rise.

2. Will my PSU deliver enough voltage to ensure that the 0D3 light-up properly?

Marginal. 0D3 need about 180V to start which means approx
330V across the stack. 680 * 25.8ma = 17.5V which just enough to
bump it below 330 mark. Increase R in #1 will further reduce the
V, there may not be enough voltage to start the 0D3's.

3. Do I need the bypass cap?

Yes, just to reduce the 0D3 noise.

4. Do I need the bypass cap for 0D3 #1?

Did you mean the bypass resistor? If so, I will say no.

[cbutterworth]

Fred,

Your numbers seem to fit, although I can't help thinking that you have outlined a circular problem that cannot be solved, ie. if B+ is increased so that the voltage drop caused by drawing 25.8mA (for the tubes) results in more than 330V, then the circuit will draw more current requiring a larger valued resistor, which in turn will result in larger voltage drop.

Do you see what I mean? Of course, my logic may be confounded by my ignorance!

Charlie

[Original Burnedfingers]

Also when I bypass the 1st 0D3 with a 100K resistor,

I hate to ask a dumb question but why do you bypass the first oD3?

[cbutterworth]

The bypass resistor over the forst 0D3 allows for the second to see a sufficiently high voltage to allow it to start-up.

Charlie

[cbutterworth]

OK, I found a datasheet with an explanation of how to calculate the value required for the series resistor.

The calculations are:

R1 minimum = (V1max - V2min) / (Itube max + Iload min)
R1 maximum = (V1min - V2max) / (Itube min + Iload max)
V1 minimum for starting = V1min X R2/(R1+R2)

OK, so what values would I use for:

1) V1max - should I use my present PSU's loaded voltage (346V) OR it's voltage unloaded?
2) V1min - should I use my present PSU's loaded voltage (346V)?
3) Itube min and max are taken from the datasheet (5mA and 40mA respectively).
4) Iload min and max - should I just use the estimated tube current draw (25.6mA)?
5) What about R2? I have no idea what this is, when I model in PSUDII, I use constant current!!!

Please help.

I have attached a redrawn schematic from the Tung-Sol datasheet.

Charlie

[cbutterworth]

Sorry.....

I would also assume that V2 min and V2 max is 300V because I would be regulating to 300VDC!

Charlie

[pchw]

Hi Charlie,

There are 2 variables to play with to use the shunt regulators:
the voltage before the series resistor and the series resistor.

Use your circuit as an example. In here you fix the voltage
before the series resistor to 344V. Therefore, there is only
one variable we can change.
At startup the 0D3's don't conduct and need a minimal of 330V
to start ( 180 + 150, once the first 0D3 fires up, it drops to
150V which passes the extra 30V to the second. hat's why
you don't need the bypass resistor). Before any
OD3's conducts, only the main circuit draws current at 25.8ma.
Hence, PS - 0.0258 * R must be >= 330, where PS is the voltage
before the series resistor. From here, we establish series
R <= (PS-330)/0.0258 ohm (**). Next, the 0D3 must
maintain 5ma to keep the regulation alive. So, the range of
current the 0D3's can shunt is 5ma < I < 40ma. This implies
0.0308 * R < Volt across series resistor < 0.0658 * R (!!). Now use (**)
and plug in PS=344, the series resistor is 542 at max. Now, plug R into the range equation (!!), we have 16.7 and 35.66.
344 - 35.66 > 300 which will force the 0D3's to shunt more current
than 40ma to maintain the 300V. This will overload the 0D3's and kill
it. In conclusion, no R will work.

Suppose you can squeeze another 16V out of the power supply.
Let see if this will work. In order to start the 0D3's, the R <= 30/0.0258 = 1163. Plug this R value into the 0D3's shunting range,
we have 35.8 and 76.5. Dropping 60V across 1163R implies
total current drawn of 51.6ma. Subtract the main circuit load = 51.6 - 25.8 = 25.8ma which is shunted by the 0D3!!!

As you can see, you will have to tweak the 2 variable make this work.

Cheers,

[cbutterworth]

Fred,

OK, if you're correct on your methodology, then I can repeat your calculations. However, I did not understand the part about

Now use (**)
and plug in PS=344, the series resistor is 542 at max.

So I used the value for series resistor calculated in the ** formula which was 542.64R.

When I use 360 for input voltage the circuit works fine.

BUT here's a question:

When I first switch on the PSU, I get very low B+ voltage until the 5V4 has heated up. Then the voltage rises to around 390VDC for a few seconds until the 6SN7's warm-up and begin functioning. At that point B+ drops back down to around 346VDC. If I plug in 390VDC to your equations, then I get 13mA current going to the 0D3s. BUT, the higher voltage seen at B+ initially implies a very low current drawn by the 6SN7's (i.e. more like an unloaded PSU). If I use 0.1mA for the 6SN7s (close to zero load), then I get total current load as 0.15mA total draw with 0.05mA pulled by the 0D3. Does this mean that they will not start?

I expect that as the unloaded B+ voltage drops, then current will rise. This would cause more current to flow across the 0D3's which will allow them to light.

Maybe a simple change in C1 from 2.5uF to 4uF will result in loaded B+ of around 360 or 370VDC.

Charlie

[cbutterworth]

Hmmmm....

I found another way of estimating the value of the series resistor:

On Steve Bench's website, he briefly discusses tube regulators and says that selecting the series resistor is pretty trivial. He suggests selecting a current at which to run the tube, say 15mA. He then explains that if you have 450VDC nominal and want to regulate it to 300VDC at 15mA, it means dropping 150V at 15mA resulting in a 10K resistor.

In my PSU, this would mean dropping 46V (346 - 300), which at 15mA would mean in the region of 3K for the series resistor.

Does this sound reasonable?

Of course, my nominal voltage of 346VDC is at 25.8mA for my tubes, so should I be looking to drop 46V at 40.8mA (25.8mA for tubes and 15mA for regulator)? This would require around a 1K resistor?

Charlie