| Gabevee |
To all,
Ok, I am opening up this subject again because... I found something very intetresting.
On the subject of not needing to connect the grid of a tube in order for it to work, I got some supporters, some who were tactfully saying I was right but wrong, or perhaps just plain old full of it, and one who adamantly said I was an ignoramus, and knew little about anything tubes and electronics.
OK, I recently had the pleasure of building/repairing/modifying two amplifiers from two separate modern kit and amp manufacturers/competitors, Bottlehead and Sun Audio.
Now, I am not promoting or denegrating their products. They are both very good. They are both 2A3 single ended.
They are both direct coupled inputs! That is, the connection from the RCA jack to the tube is a straight wire. The amps worked very well when my CD player was directly connected... pure AC coupled! No DC connection to the cathode except that which anyone who really knows tube technology would tell you, the internal connection.
SO, I guess Doc Bottlehead and the engineers at Sun Audio are also ignoramuses.
Gabe:scratch: :bigeyes: :eek: |
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| SY |
I don't know if those guys are ignorami or not, never having heard of these companies, but if there's no provision for some sort of grid-leak resistance, it's a design which is inherently unreliable (not the first time that latter-day tube "designers" have paid no attention to the lessons of the past). My guess is that, despite the ac coupling of your CD player, there's an output resistor to ground following the output cap- otherwise, the player would have horrible turn-on and turn-off bangs. And that resistor is acting as a remote grid leak.
Direct wire connection to the grid, even with a grid-leak resistor to ground, is poor design practice anyway, an invitation to RFI and oscillation. But I've seen highly regarded (not by me) amps selling for 5 figures that had horrific slewing distortion, blew up output tubes on a regular basis, and had power bandwidths orders of magnitude below spec. I guess that's what their "designers" called "musical." So merely making an amp vulnerable to external noise is not the worst error a "designer" can commit. |
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| Sch3mat1c |
Ok....so....let's go back to the top and tell me exactly WTF we're talking about here? :bawling: :rolleyes: :scratch:
If you mean no DC resistance to the grid, I've done a little experimentation with that. Not much, mind you.
The jist of it is, the tube will self-bias at a few volts below cathode. But, if the tube's been working hard, it's possible that grid emission may take over and it climbs positive! This makes for a runaway situation, of course.
Tim |
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| Gabevee |
SY,
Right, I agree. All my amps have grid resistors.
My contention, though, was that it was not necessary to have it for the tube to work. And if you recall, in my argument I did say the amp would be unstable (I guess that is what is considered "musical", since many also desire the 5Y3 rectifier, a tube that changes the B+ by merely looking at it).
BTW, those amps sound very good, FWIW. Also, there are quite a few other "engineers", if you like, who also design their amps with no input grid resistor or coupling cap.
As for the CD player... I will look in it physically. It does have a mute to disconnect output when not playing. But I am pretty sure there is no resistor after the cap. I mean, why put one there if the next stage is usually a preamp with an input resistor or a stepped attenuator anyway (think "bean counters")?
At any rate, this prompted me to use a coupling cap after the preamp I was using, since its output was designed to mate with a power amp that had a coupling cap in it, so there was none in the output. So it was changing the input tube's bias in the power amp. I isolated it with caps.
Sun Audio is not cheap. They are a Japanese company, if that means anything. They use Tamura iron.
Bottlehead is home of the Audio Asylum forum. They tout Magnequest iron.
Gabe |
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| SY |
| Yeah, I took a look at Bottlehead's website after I read your post. 2.5 watts at 5% distortion for $1200. Barnum was right. |
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| jeff mai |
| quote: | Originally posted by SY
2.5 watts at 5% distortion for $1200. Barnum was right. |
This is all you need to know about an amp to judge its sonic merits? You're good, really good! |
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| Sch3mat1c |
Yeah but at 30mW into your $5,000 107dB horns it only gets .1% THD and it's almost all harmonious 2nd H.! ....
Tim :dead: |
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| fdegrove |
Hi,
| quote: | | My contention, though, was that it was not necessary to have it for the tube to work. |
Yes...it will work.
If I read you correctly the input valve on the amps you worked on had no gridleak resistors and had cathode bias.
The tube will still bias itself through its' internal capacitance form grid to cathode,no problem there.
But in real life as soon as you connect a CDP or preamp to that it will effectively see the bleeder R at the output of these devices.
This may be anything from a few K to a M ohm.
So your gridleak will be that resistor and will always vary anyway you look at it.
Imagine the manufacturers would have used a 100K gridleak R and you hook up a CDP with a 10K bleed R at the output.
Mow you will have two resistors in // and the end value is going to be smaller than the 10 K,right?
Naturally,if you have another fool at the other end of the stick following the same reasoning and he doesn't use any bleeder at the output of his gear you're going to be in for a few surprises too.
The way I see it: it will work alright,another costfactor is reduced,but it sure is not good practice not to terminate the transmission lines and leave it to the inherently nonlinearity of the tube to determine this.
Cheers,
;) |
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| SY |
| quote: | Originally posted by jeff mai
This is all you need to know about an amp to judge its sonic merits? You're good, really good! |
Not all, but enough. I don't buy homeopathic drugs, wear magnetic insoles, or have my auras read, either. |
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| dhaen |
| quote: | | I don't buy homeopathic drugs, wear magnetic insoles, or have my auras read, either. |
He he...me neither..
I don't believe in luck either. There's no such thing! It's a method of diverting blame.
I told this to a colleague who was a bit depressed due to his "bad luck". Unfortunately I never saw him again, because he died sitting in his car in the carpark.:(
Life's odd, but not irrational.
And, UFO's are "unidentified":) Otherwise they're something else!
Cheers, |
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| Joel |
| quote: | Originally posted by SY
there's an output resistor to ground following the output cap- otherwise, the player would have horrible turn-on and turn-off bangs. And that resistor is acting as a remote grid leak.
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Exactly. I think "Doc Bottlehead" (WTF???) is taking a big risk in hoping that all the units you hook this thing up to will have a resistance to ground.
Gabe, I can't believe you're dredging this up again?!!?
You keep confusing two issues - developing signal voltage, and providing a dc path between the grid and cathode. In your example, there is a path in there somewhere, or the tube will not function. |
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| jackinnj |
| GG was very common in amplifiers for radio: |
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| jackinnj |
| Grounded grid 3-400Z amplifier: |
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| Joel |
| Jack, how is that relevant? |
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| fdegrove |
Hi,
Q:| quote: | | Jack, how is that relevant? |
A: It is not.
Cheers,;) |
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| Sch3mat1c |
A rag-chewer eh?
Looks like an ARRL schematic, I'd guess a 60s or 70s Handbook, not 50s since the SS diodes...
Tim |
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| Joel |
Frank, is that really the flag of Belgium? I like it! Very understated.:nod:
I wish there was more blue color in the US flag. |
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| fdegrove |
Hi,
| quote: | | Looks like an ARRL schematic, I'd guess a 60s or 70s Handbook, not 50s since the SS diodes... |
It's your typical grounded grid circuit and as such has no bearing on what Gabe put forward.
Joel,
| quote: | | Frank, is that really the flag of Belgium? I like it! Very understated. |
Actually it's the European Union jack.;)
The Belgian tricolours are like the German one (our first king was German) but the stripes are vertically aligned iso horizontally.
Cheers,;) |
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| Gabevee |
| quote: | Gabe, I can't believe you're dredging this up again?!!?
You keep confusing two issues - developing signal voltage, and providing a dc path between the grid and cathode. In your example, there is a path in there somewhere, or the tube will not function. |
BTW, WTF means "What The F$#%." Oh, BTW means "By The Way."
I only "dredged it up again" because I found a couple of examples to prove my point. Your contention was a path was necessary from the grid to cathode. If you said "internally", I would have agreed and that would have been the end of it. Theories and statements in electronics should require solid proof. So here I have presented this proof.
I am not confusing anything.
Do you know what you get when you put a positive voltage on the anode and a negative voltage on the cathode of a triode with nothing on the grid? A diode. The tube will conduct! The grid is there to control that initial flow of current. All that is really needed is a negative voltage with relation to the cathode in order to control the flow. That control can come from a pure AC source. I repeat myself here, that is not a practical practice, to be sure. But the tube will work. It will flow current. It will transfer the change in grid voltage. It needs a static grid to cathode voltage to bias it at idle. That can come from the cathode alone. There are some tubes whose specs allow them to work still in class A with a grid bias of zero.
I am sorry to sound condescending here, or patronizing, but it seems that with all your "book knowledge" you have missed something.
But, these two amp examples, of many I have come across, show that the tube will work without an external path.
If it is a problem for you to read things that are repeated, just ignore it. That is going to happen alot, since this is a subject with a finite number of subsets.
If you have a problem with me, just click on my profile and choose the "ignore" option below. Then anything I post you won't be informed of or be aware of.
But, you certainly don't dictate here or anywhere else in this country what whomever says. So... unless you like to have your freedoms revoked, get off my case! Because your freedom of speech stops with your trying to supress mine.
And if what I have posted here is wrong in any way, I would hope that the moderator would be the one to tell me so, not you.
Gabe |
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| Joel |
Gabe, first you really need to calm down! :bigeyes:
Second, this isn't some first amendment issue, and I certainly don't have a vendetta against you.
I expressed some.... frustration. I apologize.:nod: But I thought we had already covered this ad nauseum. However, I see there is still some confusion. Your "proof" is actually not proof of anything other than doc bottlehead's lack of a grid resistor!:)
As for your other points:
A triode is not a diode. I do not want, or ask it to behave like a diode. For a triode to "work" (ie. perform the functions of a triode) the grid needs a dc path back to the cathode. And no, this is not "internal" in any tube I've ever seen.
I'll happily repeat again, and I would welcome somebody trying to disprove it, but:
ALL THE ELECTRODES IN ANY TUBE MUST HAVE A DC PATH TO THE CATHODE.
Cheer up!
Joel |
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| Sch3mat1c |
I could argue your point of:
| quote: | | I'll happily repeat again, and I would welcome somebody trying to disprove it, but: ALL THE ELECTRODES IN ANY TUBE MUST HAVE A DC PATH TO THE CATHODE. |
If you cap-coupled an AC signal to the grid, then when the signal goes positive, it will draw current from the G-K diode. Where does this current go? Why, clearly, it constitutes a negative bias voltage due to the rectification of the applied signal.
And don't get ****y with me about leakages...we are ignoring those in this scenario.
So where's the DC g-k path, hmm?
Tim |
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| Joel |
| quote: | Originally posted by Sch3mat1c
it will draw current from the G-K diode. |
The grid and cathode do not form a diode. The grid is negative with respect to the cathode, so how can any diode be formed?
| quote: | | And don't get ****y with me |
Excuse me? |
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| Sch3mat1c |
Ummm....how can they not form a diode....
Besides, a diode doesn't have to be forward-biased to be a diode. |
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| Joel |
Tim, a diode is an anode and a cathode. It functions because we apply a positive potential to the anode. The electrons are attracted to it.
What happens if I apply a negative potential to the anode? Think about a rectifier.
You see? |
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| Sch3mat1c |
I repeat...
It doesn't have to be forward-biased to be a diode.
Or are you saying that, in a rectifier, the tube alternately becomes a diode and a - nothing? - every 60Hz?
Well anyway, still - what about the bias voltage? There's no DC path out of the grid, only in...
Tim |
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| Joel |
Tim, the statement I made about all electrodes needing a dc path to the cathode can be found in the first few pages of Radiotron. It is stated very clearly. I did not make it up, and I'm pretty sure RCA and Mr Langford-Smith knew what they were talking about.
You're going to have to retract later on, but please continue if you feel it's necessary.:dodgy: |
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| fdegrove |
Hi,
Let me get this straight from the start:I don't want to stirr things up any further but:
| quote: | | And no, this is not "internal" in any tube I've ever seen. |
Neither have I, however when you use a cathode bias and don't connect anything to the grid you will still notice a negative voltage developping at the grid.
This take time and by adding the gridleak resistor the biasing process is sped up.
As Gabe pointed out the tube will function.
Sure enough it will function as a diode until a load is added to the grid.
While I toyed around with omitting bleeder resistors from the preamp end I noticed an opening up of the sound,with more air and splashier highs on rimshots.
As if a small veil was lifted.
Quite likely there must have been an impedance mismatch somewhere along the line.
Later on I increased the value of the bleeder resistor from 100K to 1M and I came close to the same result as without the bleeder.
Hope this clarifies some,;) |
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| Gabevee |
Joel wrote... somewhat erroneously:
| quote: |
The grid and cathode do not form a diode. The grid is negative with respect to the cathode, so how can any diode be formed?
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In a triode biased in class B in a detector circuit, the grid and cathode do indeed form a diode detector, with the grid as the "anode". The same function of grid as "anode" or "plate" is true in a pentagrid converter. The second grid is actually the "plate" of the first "triode" grid.
The definition of a diode, functionally (the key word), is to "rectify", or allow current to flow in one direction but not the other. When the grid is biased negative enough to just cut off the tube, or class "B", then anything that brings the grid toward positive turns on the tube. Anything below that voltage level, does not turn the tube on. So... the grid/cathode circuit becomes a rectifier... a diode. Plate current follows the changes and with a load resistor, "amplifies" them.
Don't they have that in the RDH? It is in the seven or eight books, and countless electronics magazines, I have read about it since I was 10.:scratch:
As for the path... when the tube conducts, there is a path already made between the cathode and the grid... simply because the grid is in the way of current flow. The grid can impede the flow of current with AC simply because the negative peaks will cause it.
Re the grid as a plate, again to bring up an icon in the business, Steve Bench has made amplifiers where the output is the grid. He says they sound very good.
Cheers!
Gabe :) |
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| Gabevee |
Joel,
Please refer to the example of DeForest's experiments here:
http://www.tpub.com/neets/book6/20e.htm
It shows that the triode he worked with flowed 5 milliamps with nothing connected to the grid, while it flowed 10 milliamps with a positive voltage to the grid, and 2.5 milliamps with the same amount of negative voltage.
But note the non connected one. Grid not connected to anything... tube works. Tube theory from the one who invented the thing!
A pure AC on the grid will make the tube fluctuate. But, as those pages you posted brings out, one would not want the AC to bring the tube into saturation or cut off, so applying a DC bias voltage allows one to have an AC voltage such that plate current will vary within the cut-off/saturation points.
The RDH was likely written to discourage engineers from designing less stable tube circuits. Same with solid state. I have seen many SS circuits where only a resistor from base to B+ was used, the most unstable circuit imaginable, but the circuits work. In all the textbooks I have read for solid state, this practice is strongly discouraged.
I rest my case.
Gabe |
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| pedroskova |
| quote: | Originally posted by Joel
Exactly. I think "Doc Bottlehead" (WTF???) is taking a big risk in hoping that all the units you hook this thing up to will have a resistance to ground.
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DocB has sold a few thousand of those amp kits, with a very good reputation for reliability. He, along with Paul Joppa and the late John "Buddah" Cammille, have brought some very good products to market. It was DocB and cohorts that reintroduced "Parafeed" output topology and the C4S to the masses - both of which have become well-respected and copied.
The "Barnum" statement posted elsewhere in this thread is just another example of the great unwashed and anonymous opining on the Internet. Take your designs, make them affordable, have people buy them and become faithful customers, then you can make such statements without sounding like an ***. |
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| dhaen |
With no apparent DC path, the grid will most likely drift in one direction or the other, it depends on the valve.
The actual likelyhood is that a DC path does exist through the coupling component or cable leakage.
Cheers, |
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| Joel |
| quote: | Originally posted by Gabevee
But note the non connected one. Grid not connected to anything... tube works. Tube theory from the one who invented the thing! |
Gabe, c'mon. A tube merely showing plate circuit current does not qualify as it "working".
A triode amplifies.
Let's consider the picture below. Where is the discharge path for the capacitor? |
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| SY |
| quote: | | The "Barnum" statement posted elsewhere in this thread is just another example of the great unwashed and anonymous opining on the Internet. |
You're welcome to question my hygiene, but "anonymous" is a lie. A minute spent at my profile and chasing a link of two will demonstrate that I've not done much to hide my identity. What did you think of the mermaid?
(edit: I see that the profile button doesn't exist in the new BB software, but the link to my homepage still does) |
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| dhaen |
Joel,
| quote: | | Where is the discharge path for the capacitor? |
If I may comment....;)
The discharge route is through the leakage of the capacitor.
Which is an unreliable factor.
Cheers, |
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| fdegrove |
Hi,
Is this really worth any more energy?
Leaving the grid to the elements is just not good engineering.Period.
Cheers,;) |
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| Gabevee |
Joel,
As dhaen says the discharge is through the capacitor. When the other side of the cap goes to the opposite polarity, the cap discharges.
Fdegrove,
You are absolutely correct sir. The practice is not good. But the theory is still correct. BTW, 100K grid resistor? Kinda low, IMHO. It does tend to source grid current.
Sy,
While I agree with you, the amplifier that Bottlehead puts out uses a fixed bias (solid state, for that matter) at both the plate and cathode of both the driver tube and the output tube, with DC NFB, so DC bias remains stable. A floating grid will work fine, IMHO.
Gabe |
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| fdegrove |
Hi,
| quote: | | BTW, 100K grid resistor? Kinda low, IMHO. It does tend to source grid current. |
Although it was just an example...the value will depend on the tube used,won't it?
Cheers,;) |
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| Gabevee |
fdegrove,
I suppose. I think it depends on transconductance.
Gabe |
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| Sch3mat1c |
Hummm none of those links say the diode stops being a diode when it's not forward-biased.... :nod:
In fact, the first one even says that a triode's grid will act as a diode plate when positive.
So, is the G-K diode real?
Perhaps I should ask you, is a car a car only when it's running? When it's off, is it no longer a car?
Ok back to the topic. I never said it'd be stable nor useful (except in high-impedance circuits, such as an electrostatic sensor), but it will work with no apparent DC grid bias.
Tim |
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| fdegrove |
Hi,
| quote: | | I never said it'd be stable nor useful (except in high-impedance circuits, such as an electrostatic sensor), but it will work with no apparent DC grid bias. |
And then there are those triodes that are quite happy with a 0 Vg bias.
Then again,I'm sure you knew that already too.:rolleyes:
Cheers,;) |
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| Sch3mat1c |
..And there are those which aren't designed for it, but look beautiful at it...(try 6V6, 150Vp, 150Vs (pentode or triode mode actually), 10 to 15Vp grid drive, load resistance to suit.)
Note also that, with a floating grid, it isn't at 0V. It is, in fact, a few volts below cathode! (This can be proven by alternately connecting the grid to a bias source, then floating it, all the while measuring current.)
Tim |
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| Joel |
Gabe - John said the leakage through the capacitor (if any exists, it should be VERY VERY small) is the path. Remember, a cap has an insulator between it's plates. If you want to build an amp where your AC coupling is dependant on the leakage through an insultor, be my guest!:dead:
In fact, I encourage you to go home, clip all the grid resistors and/or chokes out of every stage in your favorite amp and tell us the result.:bawling:
Frank is right, this is a waste of time. |
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| Gabevee |
Joel,
No Duh.:rolleyes:
I was making two points. When the cap sees the opposite polarity on its plates, what happens? I will let you answer that one.
Gabe |
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| Joel |
| quote: | Originally posted by pedroskova
Take your designs, make them affordable, have people buy them and become faithful customers, then you can make such statements without sounding like an ***. |
$1200 is affordable?
:confused: |
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| pedroskova |
| quote: | Originally posted by Joel
$1200 is affordable?
:confused: |
...with 2 each of Magnequest power trfo's, filter chokes, plate chokes, OPT's...and C4S CCS's, ...yes, they are affordable for what you get.
You could always get the Paramours @ $550/pr. They also employ parafeed and CCS loading, and sound quite nice.
DocB and his wife are very nice and generous people that don't deserve the "Barnum" quip. |
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| fdegrove |
Hi,
| quote: | | DocB and his wife are very nice and generous people that don't deserve the "Barnum" quip. |
Even to a seasoned DIYer this should be a reasonable asking price.
You'll even get unused tubes for the money.:devily:
Cheers,;) |
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| Joel |
| quote: | Originally posted by pedroskova
DocB and his wife are very nice and generous people that don't deserve the "Barnum" quip. |
Hey pedro,
Who said they weren't nice people? Personally, I think they're geniuses if they can get a substantial number of people to pay $1200 for an amp!:nod:
And don't get me started on Magnequest please. $80 chokes?
sheesh.:dodgy: |
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| pedroskova |
| quote: | Originally posted by Joel
Hey pedro,
Who said they weren't nice people? Personally, I think they're geniuses if they can get a substantial number of people to pay $1200 for an amp!:nod:
And don't get me started on Magnequest please. $80 chokes?
sheesh.:dodgy: |
Two amps...sheesh.:dodgy: |
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| SY |
| People tell me that Sylvia Browne is a nice lady, too. So what? A rip off is a rip off. |
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| SY |
| quote: | Originally posted by Gabevee
While I agree with you, the amplifier that Bottlehead puts out uses a fixed bias (solid state, for that matter) at both the plate and cathode of both the driver tube and the output tube, with DC NFB, so DC bias remains stable. A floating grid will work fine, IMHO.
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Wouldn't a 10 cent resistor of 100K or so be a better solution that putting in DC nfb and solid state bias? What happens when the cap leaks a bit, or the humidity changes, or...? Just not good engineering. |
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| jeff mai |
| quote: | Originally posted by Joel
And don't get me started on Magnequest please. $80 chokes?
sheesh.:dodgy: |
Let's see, by the time I sourced the materials to make one choke and wound it myself by hand, I could quite easily buy ten of these for the cost of the time alone. I can't quite see how this or the amplifiers sold by Doc Bottlehead are ripoffs. I'd wager no one here could mass produce these amps for a siginificantly lower price. You aren't going to find enough parts in the dumpster to make 100s of these amps. |
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| Ryder |
Gabe, is it possible you had a really bad day and missed it? I'm told it's on the RCA jack side and not the tube socket.
probably not.
Is it possible the (probably first time) builder screwed up? Wouldn't be the first time.:bulb:
I contacted a guy who has seen them and he says they do have a grid connection. He knows what he's talking about too. ;)
Tempest in a teapot me thinks. Na, that could never happen at an audio forum.:scratch:
Cheers
Craig Ryder |
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| Gabevee |
Ryder,
WHOOPS!!!! You know something!! You are absolutely right!
DOH!:bigeyes:
But I did rewire the Sun amp, bypassing the volume controls and used no grid resistor. Amp still worked fine.
Thanks for pointing that out to me. Yes, I must have been having a bad day at that point. The Bottlehead amps do indeed have a grid resistor to ground. I stand corrected.
When I get back in town I will disconnect it... just to see what happens.
However, As I said above, I did not use one on the Sun amp.
Sy,
And what is wrong with DC NFB and voltage regulated fixed bias? Depending solely on component (tube or transistor)characteristics, quality of capacitors and resistors would be bad engineering... IMHO.
NFB and fixed bias should make an amp more stable, and the output more true to the input... again IMHO.
Gabe |
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| fdegrove |
Hi,
| quote: | | When I get back in town I will disconnect it... just to see what happens. |
Yes,of course it will *work*...as long as you have something connected to the amp.
Others and myself explained all this clearly in previous posts,or so I thought?
Cheers,;) |
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| Gabevee |
fdegrove,
Well... skeptic that I am... and being one who has to be absolutely sure... you could have told me that there was such a thing as server fairy dust, as in that IBM commercial, and I would have had to acquire some for myself and try it in order to see for sure that it works.
IOW, I do not even take the word of the RDH or any of the other books I have in my possesion. I try it for myself to be sure.
And no, some were saying that it wouldn't work.
Gabe |
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| fdegrove |
Hi Gabe,
If you go through the whole thread you will notice I toyed with this too.
There is nothing there that defies neither logic nor physics...bar one thing,you need to consider the termination of your transfer line, AKA transmission line.
Not terminating at either end will yield a frequency response that will be left to the elements, if you know what I mean.
I realise that at least one member said that it won't work,I know,and to some extend I agree.
Looking at things in an isolated manner,naturally this can't possibly work...looking at the interaction between all that is interconnected I say it can work.
Imagine,if you like a preamp with a 100K bleeder R at the output followed by an amp with an input R of 100K...We now have the preamp looking at a 50K input impedance.
What happens when I now clip the the 100K R out at the amp?
Will it still work?
Yes,it will.
Will I now hear a difference in sound?
Quite likely...
Will my amp now change frequency response with whatever I hook up?
Unfortunately,yes!
Try it out for yourself and tell me what you hear.Moreover,I'm sure you're equipped to do some measurements.
That should straighten it out.
Cheers,;) |
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| Sch3mat1c |
| quote: | Originally posted by fdegrove
Hi Gabe,
*snip*
Imagine,if you like a preamp with a 100K bleeder R at the output followed by an amp with an input R of 100K...We now have the preamp looking at a 50K input impedance.
What happens when I now clip the the 100K R out at the amp?
Will it still work?
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Note that, in this case (and I'll use it as a case of 'everything is different'), the reduced load will 1. reduce delivered output power (well there's no load, is there? ;) ), 2. increase output voltage, and 3. reduce distortion.
What I'd like to know is why this thread got so long...
Tim |
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| fdegrove |
Hi,
| quote: | | What I'd like to know is why this thread got so long... |
Beats me too.:rolleyes:
Cheers,;) |
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| Gabevee |
You think THIS is long? You should check out the capacitor one.:D
It is up to six pages now! It's still going on!
I would think that there would be only a small increase in response in the low end, because the capacitor's reactance would see a very high impedance that is the grid.
As for loss of power... how????? If anything, the amp would clip sooner, because the preamp sees an extremely small load (the very high impedance of the grid), the preamps output being a little higher. So... I will just do it for myself, report on it on my site, and wallow in my own ignorance... or edification... whichever one may want to deem it as.
I will end it here since it does seem to be going "ad nauseum".
Gabe |
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| Sch3mat1c |
Well...it depends on what load you're measuring the power in ;) If it happens to be that 100k that you just disconnected, obviously, the amp must be delivering less power! :)
Tim |
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| Joel |
| quote: | Originally posted by Gabevee
WHOOPS!!!! You know something!! You are absolutely right!
DOH!:bigeyes:
...The Bottlehead amps do indeed have a grid resistor to ground. I stand corrected. |
Uh huh... now, can we finally put this to rest?
:dodgy: |
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| Gabevee |
Just for THAT sssssnide remark.... I WONT stop it!
Courtesy of The Great One.
:D
Gabe |
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| Sch3mat1c |
Hmmm.
Circlotron, get yer *** in here and stop this thread!!! :p :p
Tim |
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| SY |
| quote: | Originally posted by Gabevee
And what is wrong with DC NFB and voltage regulated fixed bias? Depending solely on component (tube or transistor)characteristics, quality of capacitors and resistors would be bad engineering... IMHO.
NFB and fixed bias should make an amp more stable, and the output more true to the input... again IMHO.
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To quote the greatest movie ever made ("Brazil"): "My complications have complications." |
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