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grid resistors first stage of amp

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Posted on June 23, 2012 at 19:38:07
hennfarm
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What value do you use? The Max resistance stated in the tube datasheet? Say, the input of an amp, I've seen commercial buffer stages use 470k grid res. Going lower can help combat
Hum or gnd loops but shunts more of the signal. I know, I know "try it yourself and find out"

 

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RE: grid resistors first stage of amp, posted on June 23, 2012 at 20:07:09
I like 100K to 150K. Going up to 0.5M risks grid current biasing.
At 100K versus 500K, the difference in overall db signal loss is pretty darn small, bordering on unmeasurable. The reason for higher value grid leak resistors is to be able to use lower value coupling caps as well as less high frequency degradation with long and higher capacitance interconnect runs. If you don't have those issues, the 100K is peachy.

 

RE: grid resistors first stage of amp, posted on June 23, 2012 at 22:00:40
Tre'
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I agree generally.

Just remember that grid resistor is in parallel with the plate load of the previous tube stage.

The lower it's value, the more vertical the previous stage's load line will be.


Vertical load lines increase the harmonic distortion and decrease the gain.


Tre'
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RE: grid resistors first stage of amp, posted on June 23, 2012 at 23:14:41
Triode_Kingdom
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I generally use 470K in that application. Many commercial designs use 1Meg or more, but I worry about grid leak bias with values that high.
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RE: grid resistors first stage of amp, posted on June 24, 2012 at 13:16:33
Jim McShane
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Even at 100K the "shunting" is minuscule as Chris wrote.

The value you use is a tradeoff:

a) a higher value is easier to drive for the component upstream

b) a lower value is (all else being equal) quieter.

You need to know the output impedance of the device that is going to be plugged into the amp, and ensure the resistor value you use is large enough to prevent problems upstream. But there is no need to use a larger value than needed to meet condition "a", since that may increase the noise (whether the increase is audible depends on a lot of things).

There is an old rule of thumb that says you need an input impedance that is a minimum of 10X the output impedance of the driving device. It's a pretty good rule of thumb. And of course the value should be in spec for the tube you use.

 

RE: grid resistors first stage of amp, posted on June 24, 2012 at 16:46:55
Eli Duttman
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Jim is spot on about the 1:10 rule. We want waveform fidelity, not power transfer in small signal situations.

Another factor to take into account is Miller capacitance. The 470 KOhm parts previously mentioned are fine with a pentode I/P stage, but can easily cause HF info. to be rolled off, with a 12AX7 section in the I/P stage.

Modern, digital, sources will drive a 10 KOhm load, as that's an IHF "standard". FWIW, I use voltage followers in my designs to do that too. The old standby of a 600 Ω line impedance did not come into being by accident. ;>)

Eli D.

 

RE: grid resistors first stage of amp, posted on June 24, 2012 at 17:28:15
Tre'
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"Another factor to take into account is Miller capacitance. The 470 KOhm parts previously mentioned are fine with a pentode I/P stage, but can easily cause HF info. to be rolled off, with a 12AX7 section in the I/P stage."

How so?

Tre'
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"Still Working the Problem"

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 06:28:51
Naz
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Eli, I asked the same question as Tre did a while back assuming you read the post wrong and never got a reply. How do you figure the grid leak R as opposed to grid stop R reacts with Miller to affect HF?

Naz

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 06:33:42
Naz
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>>as well as less high frequency degradation with long and higher capacitance interconnect runs<<

Chris, I generally agree with your other statements too but how do you figure the grid leak value affects HF?

Naz

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 08:29:28
A grid resistor of 470,000 versus 47,000 will shift down the frequency/phase effects of any tube capacitance by 10x. High pass on interconnects as well.

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 11:05:39
Tre'
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The grid is sitting at the output impedance of the previous stage.

The value of the grid resistor will not cause HF loss.

It is the output impedance of the driving stage vs. the Miller (plus cable capacitance) that causes loss of HF.

Tre'
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"Still Working the Problem"

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 13:14:33
Jim McShane
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Because the screen in a pentode reduces Miller capacitance to near zero. With a triode (12AX7) the Miller effect can be substantial

Once again, Randall Aiken's site provides some excellent info - click the link below.

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 14:34:57
Tre'
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I understand Miller capacitance but it's the drive impedance vs. the Miller capacitance that causes HF loss. (They form a low pass filter.)

"Effect on frequency response

The input capacitance of the tube, in conjunction with the source impedance of the previous stage, forms a simple, single-pole RC lowpass filter with a -6dB/octave (-20dB/decade) slope, and an upper -3dB cutoff frequency equal to: f = 1/(2*pi*R*C)"

It's the source impedance not the value of the grid resistor.

Tre'
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"Still Working the Problem"

 

It dawns on me that the grid resistor does lower the source impedance but..., posted on June 25, 2012 at 14:55:16
Tre'
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only in a terrible design. (or maybe a pentode driver)

If you had a 100k source impedance (without the grid resistor in place) and you used a 100K grid resistor.....the grid resistor being in parallel with the source....your drive impedance would be 50K and your voltage would be -6db.

If you had a 100k source impedance and you used a 10K resistor, your drive impedance would be 100k//10k=9.09k and you would have 1/10th of your drive voltage.

In either case the load line of the driver stage would be vertical (not good for triodes).

To my way of thinking the source impedance should be low enough to drive the Miller capacitance before the grid resistor value is considered. At that point the value of the grid resistor won't matter (except the lower it is, the more vertical the load line for the driver stage).

In any event, it's the source impedance that has to be low enough to fully drive the Miller and the grid resistor value is just a small subset of that.

I hope that is clear.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

A triode driver example, posted on June 25, 2012 at 15:30:26
Tre'
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I use a 5687 as a driver for my 6b4g SET.

The 5687 has a plate resistance of about 2500 ohms.

I use a CCS instead of a plate resistor so the output impedance (at the plate) is 2500 ohms.

I use the mu output of the CCS. It's output impedance is about 500 ohms.

This is without considering the grid resistor of the 6b4g.

If I use a 10k ohm grid resistor that will change the drive impedance since it is in parallel with the output impedance of the CCS.

10k//500ohms=476 ohms

If I use a 470k ohm grid resistor the drive impedance will be 499 ohms.

If I take the output off the plate and use a 10k grid resistor the output impedance would be 2.5k//10k=2k and the load line for the 5687 would be more vertical and I would lose gain and the harmonic distortion would increase.

a 470k grid resistor give us 2486 ohms and the load line would stay pretty horizontal.

With any one of those, the Miller is fully driven and the -3db point is more than a decade out of the audio band.


Tre'


Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 15:41:48
Jim McShane
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I think there is some confusion in this thread between:

a) the "grid-leak" resistor that has the task of providing a path for electrons to escape the tube's grid where they can collect; and is usually connected to the signal input and to ground and is the dominant factor in the input impedance of the stage, and

b) the grid "stopper" resistor which is usually a relatively low value resistor in series between the output of the previous stage and the tube's grid - this resistor would be in series with the input capacitance (Miller and other) of the tube.

I suspect the somewhat confusing terminology didn't serve this thread well.

 

You may be right, posted on June 25, 2012 at 15:56:33
Tre'
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but I never read anyone talking about a grid stop resistor in this thread.

Of course the larger the grid stop resistor the more highs one will lose.

The grid stop resistor is in series with the source impedance and adds to it directly.

The original poster ask "grid resistors first stage of amp
What value do you use? The Max resistance stated in the tube datasheet? Say, the input of an amp, I've seen commercial buffer stages use 470k grid res."

Nobody I know would use a 470k ohm stop resistor.

Thanks

BTW I use the smallest value stop resistor that will do the job.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 18:16:29
Eli Duttman
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CMiller combines with the grid to ground resistance to form a low pass pole. When the resistance is large and CMiller is comparatively large, F3 of the pole will be low enough to have an effect at the top of the AF band.

IMO, it's worth noting that some units with a large I/P grid to ground resistance have a pentode I/P stage. The list includes the H/K Cit. 2, the H/K Cit. 5, and the Dyna ST70. Accident? I think NOT.

Eli D.

 

RE: grid resistors first stage of amp, posted on June 25, 2012 at 21:36:24
Naz
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>>CMiller combines with the grid to ground resistance to form a low pass pole. When the resistance is large and CMiller is comparatively large, F3 of the pole will be low enough to have an effect at the top of the AF band.<<

I thought there may have been confusion between grid to ground (grid leak) and series grid stopper but this confirms we are indeed discussing the grid leak R the as the statement is incorrect.

It is the series resistance including OP impedance, grid stopper and any attenuators in the chain form a low pass filter with C Miller.

The grid to ground (grid leak R) forms a high pass with any coupling cap preceding it, affecting only LF.

Tre has summed it up nicely elsewhere in this thread.

Naz

 

RE: A triode driver example, posted on June 26, 2012 at 16:26:14
Triode_Kingdom
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Note that the OP asked about the first stage of an amplifier. If that stage happens to be a 6SL7 or 12AX7, Miller rears its ugly head in a way that can be difficult to deal with. Many commercial designs from past decades resorted to relatively simple, add-on RC compensation to flatten the response. I see nothing wrong with that so long as the implementation uses good quality parts and doesn't incur a noise penalty.

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RE: grid resistors first stage of amp, posted on June 26, 2012 at 17:17:39
Tre'
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"The input [Miller] capacitance of the tube, in conjunction with the SOURCE IMPEDANCE of the previous stage, forms a simple, single-pole RC lowpass filter with a -6dB/octave (-20dB/decade) slope, and an upper -3dB cutoff frequency equal to: f = 1/(2*pi*R*C)"

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: A triode driver example, posted on June 26, 2012 at 20:01:05
Tre'
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"Note that the OP asked about the first stage of an amplifier. If that stage happens to be a 6SL7 or 12AX7, Miller rears its ugly head in a way that can be difficult to deal with."

Yes. I looked at his system and I see that he uses a 4kn8 choke loaded line stage.

I couldn't find a data sheet for that tube but I did read that the transconductance is higher than a 6dj8.

If I assume that the plate resistance is the same as the 6dj8 or no higher then the output impedance will be the plate resistance with the plate choke in there (at least through the midband and lower everywhere else).

If I assume 3k for the output impedance I don't think I'm too far off.

The Miller of a 12ax7, assuming a in circuit mu of 65, will be about 112pf. Let's call it 120pf.

3k (assuming a very high value grid resistor on the 12ax7) driving 120pf has a -3db point of 442kHz.

If the 12ax7 is actively loaded and the in circuit mu is the mu of the tube then we have 1.7pf (grid to plate) X 101 + 1.8pf (grid to cathode capacitance) for a total of 173.5pf

The -3db point would now be 305kHz.

I don't see any reason to load down the preamp tube with a low value grid resistor on the first stage of the power amp (even if it's a 12ax7) when you have a 3k drive impedance to start with.

Unless there is a lot of cable capacitance HF roll off shouldn't be a problem.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: A triode driver example, posted on June 27, 2012 at 21:14:01
hennfarm
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Thank you all for the replies, I caught the confusion and was not confused by it. Tre, its a 4kn8 loaded with that cheapo Hammond 150h choke, 3 foot interconnect driving a 12hg7,, 100k grid resistor. The 4kn8 draws more current in the same socket as 6dj8 so it's rp is lower eh? JH

 

RE: A triode driver example, posted on June 28, 2012 at 09:13:50
Tre'
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Location: So. Cal.
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That should be fine. The 100K grid resistor on the 12hg7 defines the load line on the 4kn8.

100k on a 6dj8 is close to horizontal.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

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