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Old 20th April 2003, 03:52 PM   #1
Morse is offline Morse  United States
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Default screen grid decoupling cap question....

Hi everyone and thanks in advance for your help!

I've searched here and in my (limited) electronics library and have not come up with an answer to this, so here goes:

When do I need to use a decoupling cap between the screen grid series resistor in a pentode and ground? And, what is the general guideline for selecting an appropriate value?

Some of the schematics I've seen for both PP and SET's use 'em and some don't, so I'm a little lost in seeing a pattern to this.

Sorry it's a dumb question, but my design/building experience has been all on triodes - until now. FWIW the amp I'm building now is a 6BM8 (ECL82).

Thanks again and all the best,
Morse
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Old 20th April 2003, 04:13 PM   #2
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You want a low impedance for the lowest audio frequency. Why? Screen current increases when the plate swings lower. The voltage drop at the screen will distort the negative half of the output waveform (both halves in push-pull). So you want to maintain a steady screen voltage.

How low is low? 10-22 uF should be plenty for a push-pull pair, or even single ended. Best to share a bypass for push-pull, as there is some cancellation (screen currents are out of phase).
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Old 20th April 2003, 04:45 PM   #3
EC8010 is offline EC8010  United Kingdom
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Default It's a bit fiddly, but...

The screen grid can be considered to be both a grid and an anode.

If you think of it as a grid, then you must hold its voltage constant unless you want to apply feedback at this point (for example, ultra-linear).

If you think of it as an anode, then it must have an anode resistance, which will affect the value of the decoupling capacitor required:

r(g2) =[ra(triode) x (Ia + Ig2)] / Ig2

Most pentodes give the value for ra(triode), and the current ratio is generally about 5:1, so the "anode" resistance seen at the screen grid is generally about five times that seen at the anode.

The capacitor is connected from the screen grid to ground, so to caclulate its value, we need to know the Thevenin resistance it sees to ground from its other terminal (R'). This will be r(g2)//R to HT//R to ground. Once you know this value, you can then use:

C = 1/(2 x pi x f x R')

You will generally want to set f to 10Hz or less.

For push-pull, you can be even more cunning and connect the capacitor between the screen grids. You now need only one capacitor of half the original value, with a much reduced voltage rating.
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Old 20th April 2003, 07:56 PM   #4
Morse is offline Morse  United States
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Thanks EC8010!

That's exactly the sort of info I was looking for!! Time to break out the RCA Manual, pencil, paper and calculator.....

Thanks Tom!

Okay, the values you've given me are a starting point, and with the info that EC8010 has given me I should be able to get a good handle on this.

Thanks again everyone!!
Morse
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Old 20th April 2003, 09:15 PM   #5
Morse is offline Morse  United States
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Default Oops! Question for EC8010....

Hi EC8010;

Sorry, but I posted too soon; what did you mean by:

r(g2)//R to HT//R to ground

Okay, r(g2) is the calculated resistance of the screen grid, treating it like an anode, but is r(g2)//R "the resistance of the screen grid in parallel to the grid resistor R"? And, what is HT//R?

Here is what I have so far:

B+ = 247V relative to ground (drops 19V by OPT's and cathode is floated 16V above ground so there's about 212V on the plate relative to the cathode).

Rp = 20k, Ip = 35mA, Ig2 = 7mA (from RCA Manual)
R(g2) = (20,000)(.035 + .007)/(.007) = 120k

The grid resistor I'd like to use is 2,700 ohms so does r(g2)//R = 2640 ohms? Is this right, and where do I go from here?

Sorry for being so dense....
Thanks again,
Morse
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Old 20th April 2003, 10:28 PM   #6
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Default Nobody minds being asked questions by someone who genuinely wants to learn...

Yes, you are on the right lines. Sometimes people use a potential divider to feed g2, hence the R to ground, and the R to HT. Your calculation is correct. Now you can apply the capacitor equation:

C= 1 / (2 x pi x 10Hz x 2640) = 6uF

If you use the 22uF suggested by Tom, the screen grid will be held solid down to 2.7Hz. I would use 22uF, or perhaps even more if I had it in stock, perhaps 47uF.
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Old 20th April 2003, 11:24 PM   #7
Morse is offline Morse  United States
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Thanks very much EC8010!!

Okay, it looks like 22uF to 47uF it is (I've got to order some more parts tomorrow anyway...).

Interesting point you raised about the use of a voltage divider here. The RCA Manual stated that I was fine with a series resistor for a pentode but that I should use a voltage divider for a tetrode or a beam power valve. If I do use a voltage divider, would I compute the values of the screen grid resistor as I have already done, then use R(to ground) = R(g2) * (Vg2)/{(B+) - (Vg2)} ?

Thanks again and all the best,
Morse
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Old 20th April 2003, 11:36 PM   #8
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It's a real pig trying to understand equations not written by an equation editor, isn't it?

Unfortunately, no. Remember that g2 draws DC current. The purpose in using a potential divider to feed g2 rather than a single resistor is to hold g2 more firmly at the required voltage. I'm afraid that you really need to learn about Thevenin and Norton here. Without invoking T&N, if you had 250V on g2, and it drew 5mA, you could consider it to be like a 50k resistor in parallel with the lower resistor of your potential divider, and design your potential divider from that starting point.

You really are reaching the point where Thevenin and Norton will make your life easier...
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Old 21st April 2003, 12:05 AM   #9
Morse is offline Morse  United States
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Hi EC8010;

Thanks again for all your help; I just looked up Thevenin's Theorem and it looks like I've got some studying to do.....

All the best,
Morse
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Old 21st April 2003, 12:21 AM   #10
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Default You're welcome.

One of my best electronics teachers used to say, "Be wise, Thevenise," and he was right, it's a very powerful tool.
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