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Grid Resistors
I often see volume control pots with their wiper connected directly to
the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian |
Grid Resistors
On Wed, 14 Nov 2007 11:05:10 +0000, Ian Thompson-Bell
wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. Disconnected grids are bad news. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes. From 0V or bias voltage to the grid. AND better you replace the pot after cleaning it if that doesn't stop the sratchiness. Patrick Turner. |
Grid Resistors
"Patrick Turner" wrote in message
... Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes. From 0V or bias voltage to the grid. AND better you replace the pot after cleaning it if that doesn't stop the sratchiness. And I would also suggest ensuring that there is no DC across the pot (such as that due to a leaky coupling cap from a previous amplifier stage). Even a pot in pristine condition will create noise when adjusted if there is DC across it. David. |
Grid Resistors
"Don Pearce" wrote in message ... On Wed, 14 Nov 2007 11:05:10 +0000, Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. Yes. This was discussed some time ago on RAT. I think it was Sander who raised the subject. He too recommended R to be 10 x the pot value. Iain |
Grid Resistors
On Nov 14, 11:05 am, Ian Thompson-Bell
wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Buy good Swiss pots and the problem should not arise. I like the DACT pots which are Swiss SMD stepped ladders made for medical instruments and repurposed for audio. Shunt a high value resistor from the grid to 0V as an additional grid leak, if it still worries you. The late Bill May used to say, "Prayer also helps," on occasions when I ran out of ideas. Andre Jute Visit Jute on Amps at http://members.lycos.co.uk/fiultra/ "wonderfully well written and reasoned information for the tube audio constructor" John Broskie TubeCAD & GlassWare "an unbelievably comprehensive web site containing vital gems of wisdom" Stuart Perry Hi-Fi News & Record Review |
Grid Resistors
Don Pearce said
Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian |
Grid Resistors
I've never had that problem, even using cheaper Alpha pots...
Sealed military grade pots (PEC) or ladder attenuators (as mentioned) should help. Andrew Ian Thompson-Bell wrote in message ... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian |
Grid Resistors
David Looser wrote:
"Patrick Turner" wrote in message ... Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes. From 0V or bias voltage to the grid. AND better you replace the pot after cleaning it if that doesn't stop the sratchiness. And I would also suggest ensuring that there is no DC across the pot (such as that due to a leaky coupling cap from a previous amplifier stage). Even a pot in pristine condition will create noise when adjusted if there is DC across it. David. Yes I have come across that on old broadcast receivers. Leaky coupling cap buggers up grid bias on output tube - result distortion and hot tube. often need to replace cap and tube to repair. Ian |
Grid Resistors
On Wed, 14 Nov 2007 18:55:28 GMT, "Ian Iveson"
wrote: Don Pearce said Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I was considering 9.72, but on balance... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian Because the anode current will rise until the valve hits saturation. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
"Don Pearce" wrote in message ... On Wed, 14 Nov 2007 18:55:28 GMT, "Ian Iveson" wrote: Don Pearce said Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I was considering 9.72, but on balance... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian Because the anode current will rise until the valve hits saturation. d Most input stages are cathode biased. Iain |
Grid Resistors
On Thu, 15 Nov 2007 08:02:14 +0200, "Iain Churches"
wrote: "Don Pearce" wrote in message ... On Wed, 14 Nov 2007 18:55:28 GMT, "Ian Iveson" wrote: Don Pearce said Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I was considering 9.72, but on balance... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian Because the anode current will rise until the valve hits saturation. d Most input stages are cathode biased. Iain Cathode biasing needs the grid to be held down. What happens if the grid becomes disconnected is that the valve (I am assuming a triode) becomes effectively a diode, wired across HT to ground, with current limiting provided only by the anode and cathode resistors in series. Generally the anode resistor is much bigger than the cathode, so it does most of the work in this regard. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
"Patrick Turner" wrote in message ... Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes. From 0V or bias voltage to the grid. AND better you replace the pot after cleaning it if that doesn't stop the sratchiness. Patrick Turner. ....AND better no pot at all, but if you must, use a step attenuator with quality resistors. west |
Grid Resistors
"West" wrote in message news:TAS_i.21505$763.20213@trnddc07... "Patrick Turner" wrote in message ... Ian Thompson-Bell wrote: I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Cheers Ian Yes. From 0V or bias voltage to the grid. AND better you replace the pot after cleaning it if that doesn't stop the sratchiness. Patrick Turner. ...AND better no pot at all, but if you must, use a step attenuator with quality resistors. Hi West. Better still, no preamp at all:-) With the addition of a stepped attenuator, one can feed the CD player straight to the power amp. Morgan Jones states: "No preamplifier is better than any preamplifier!" I agree with that sentiment. Many take this a step further and add an input selector switch as well as the SA to the power amp. This makes sense too. Cordially, Iain |
Grid Resistors
"Iain Churches" wrote in message i.fi... "West" wrote ...AND better no pot at all, but if you must, use a step attenuator with quality resistors. Hi West. Better still, no preamp at all:-) With the addition of a stepped attenuator, one can feed the CD player straight to the power amp. Morgan Jones states: "No preamplifier is better than any preamplifier!" I agree with that sentiment. S'what I did with my 2A3 SET which is effectively a power amp with a volume control - my thinking was that the 'pre' section could be a nightmare for a beginner to get right (quiet) and that there are/were any number of 'ready-made' pre's that could be used if necessary, until such time I built my own! It has worked so well without a pre I never got round to building one! Many take this a step further and add an input selector switch as well as the SA to the power amp. This makes sense too. Except that it eliminates the option of a 'single source direct' and condemns the amp to possibly 'inescapable' crosstalk with the extra internal wiring.... |
Grid Resistors
"Andre Jute" wrote in message
ups.com... On Nov 14, 11:05 am, Ian Thompson-Bell wrote: The late Bill May used to say, "Prayer also helps," on occasions when I ran out of ideas. Andre Jute Is this where the phrase "Plug and pray" originated ? |
Grid Resistors
On Thu, 15 Nov 2007 09:06:56 -0600, John Byrns
wrote: In article , (Don Pearce) wrote: On Thu, 15 Nov 2007 08:02:14 +0200, "Iain Churches" wrote: "Don Pearce" wrote in message ... On Wed, 14 Nov 2007 18:55:28 GMT, "Ian Iveson" wrote: Don Pearce said Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I was considering 9.72, but on balance... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian Because the anode current will rise until the valve hits saturation. d Most input stages are cathode biased. Iain Cathode biasing needs the grid to be held down. What happens if the grid becomes disconnected is that the valve (I am assuming a triode) becomes effectively a diode, wired across HT to ground, with current limiting provided only by the anode and cathode resistors in series. Generally the anode resistor is much bigger than the cathode, so it does most of the work in this regard. But what happens if some of those electrons flying past the grid on their way from the cathode to the anode get stuck on the grid making it negative? Couldn't that keep a high mu input stage from saturating? No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
"Don Pearce" wrote in message
... No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. If the electrons hit the grid with enough energy to cause secondary emission then the grid potential would be driven positive until it reached the same potential as the anode. Whether secondary emission would occur would depend on the anode potential and the design and material of the grid. David. |
Grid Resistors
On Wed, 14 Nov 2007 11:05:10 +0000, Ian Thompson-Bell wrote:
I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? I was taught that you should *never* allow DC on a volume or tone control. The input to an amp should follow the sequence of resistor to ground, isolating cap, pot, isolating cap, grid resistor to ground, grid (or grid stopper). In light of this I've also wondered the same thing! -- Mick (Working in a M$-free zone!) Web: http://www.nascom.info http://mixpix.batcave.net |
Grid Resistors
In article ,
(Don Pearce) wrote: On Thu, 15 Nov 2007 09:06:56 -0600, John Byrns wrote: In article , (Don Pearce) wrote: On Thu, 15 Nov 2007 08:02:14 +0200, "Iain Churches" wrote: "Don Pearce" wrote in message ... On Wed, 14 Nov 2007 18:55:28 GMT, "Ian Iveson" wrote: Don Pearce said Yes, it is a good idea. Make the resistor about ten times the value of the pot, or you will be changing the way the volume control reacts. 10-1 rule again :-) I was considering 9.72, but on balance... I often see volume control pots with their wiper connected directly to the grid of the following tube. Trouble is as they wear they get a bit scratchy and often lead to the grid becoming disconnected. Would you agree it is good practice to add a resistor (say 1Meg) directly across the grid? Disconnected grids are bad news. Why, in this case? cheers, Ian Because the anode current will rise until the valve hits saturation. d Most input stages are cathode biased. Iain Cathode biasing needs the grid to be held down. What happens if the grid becomes disconnected is that the valve (I am assuming a triode) becomes effectively a diode, wired across HT to ground, with current limiting provided only by the anode and cathode resistors in series. Generally the anode resistor is much bigger than the cathode, so it does most of the work in this regard. But what happens if some of those electrons flying past the grid on their way from the cathode to the anode get stuck on the grid making it negative? Couldn't that keep a high mu input stage from saturating? No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? Regards, John Byrns -- Surf my web pages at, http://fmamradios.com/ |
Grid Resistors
On Thu, 15 Nov 2007 15:59:47 GMT, John Byrns
wrote: No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? I can't be absolutely definitive on that, because I don't have a valve to try it with. I'm dimly aware of it being caused by a space charge rather than actual travelling electrons headed for the anode, but when you have a few hundred volts sucking hard a couple of mm away, how much space charge can you build up? It is a phenomenon I only really associate with the spaces between electrodes when they tend towards the same potential. How close to saturation would you consider close enough? d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
John Byrns wrote:
That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? Yep, its the same as grid bias, you sometimes see it in use, cathode to 0v, 1M or so resistor between the grid and 0v, it will bias itself up as you suggest. TL did it on the d3a in the second version of his LCR phono stage. But in this case, the anode load won't have been selected for this sort of bias, so you will still get more current than you expect. -- Nick |
Grid Resistors
On Thu, 15 Nov 2007 16:25:19 +0000, Nick Gorham
wrote: John Byrns wrote: That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? Yep, its the same as grid bias, you sometimes see it in use, cathode to 0v, 1M or so resistor between the grid and 0v, it will bias itself up as you suggest. Like grid leak biasing - where the coupling cap from the previous stage gets charged up and provides negative grid bias? TL did it on the d3a in the second version of his LCR phono stage. But in this case, the anode load won't have been selected for this sort of bias, so you will still get more current than you expect. Isn't this something you'd expect to find only on low voltage tubes. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
"Don Pearce" wrote in message
... On Thu, 15 Nov 2007 15:59:47 GMT, John Byrns wrote: No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? I can't be absolutely definitive on that, because I don't have a valve to try it with. I'm dimly aware of it being caused by a space charge rather than actual travelling electrons headed for the anode, but when you have a few hundred volts sucking hard a couple of mm away, how much space charge can you build up? It is a phenomenon I only really associate with the spaces between electrodes when they tend towards the same potential. How close to saturation would you consider close enough? As a quick experiment I've just tried disconnecting the grid of the second half of the ECC83 in a Quad QC2 pre-amp that happens to be on my bench at the moment. With 280V HT the anode voltage is normally 235V (50K anode load, 2.2K cathode bias resistor), with the grid disconnected it drops to 160V. Trying the same trick with the first half (470K anode load, 4.7K cathode resistor) only results in the anode voltage dropping from 114V to 105V. David. |
Grid Resistors
Don Pearce wrote:
On Thu, 15 Nov 2007 16:25:19 +0000, Nick Gorham wrote: John Byrns wrote: That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? Yep, its the same as grid bias, you sometimes see it in use, cathode to 0v, 1M or so resistor between the grid and 0v, it will bias itself up as you suggest. Like grid leak biasing - where the coupling cap from the previous stage gets charged up and provides negative grid bias? Yes, except, you don't actually need a cap, as long as grid current generated across the grid resistor generates the required voltage. TL did it on the d3a in the second version of his LCR phono stage. But in this case, the anode load won't have been selected for this sort of bias, so you will still get more current than you expect. Isn't this something you'd expect to find only on low voltage tubes. Yes, but again thats the sort of tube you would expect to have a pot connected to the grid. -- Nick |
Grid Resistors
On Thu, 15 Nov 2007 16:48:13 -0000, "David Looser"
wrote: "Don Pearce" wrote in message ... On Thu, 15 Nov 2007 15:59:47 GMT, John Byrns wrote: No. Firstly it would only be electrons that actually chanced to hit the grid that would stick, and secondly once even quite a small charge had built up, the rest of the electrons would just get deflected slightly away from the negative grid wires and no further charge would build up. That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? I can't be absolutely definitive on that, because I don't have a valve to try it with. I'm dimly aware of it being caused by a space charge rather than actual travelling electrons headed for the anode, but when you have a few hundred volts sucking hard a couple of mm away, how much space charge can you build up? It is a phenomenon I only really associate with the spaces between electrodes when they tend towards the same potential. How close to saturation would you consider close enough? As a quick experiment I've just tried disconnecting the grid of the second half of the ECC83 in a Quad QC2 pre-amp that happens to be on my bench at the moment. With 280V HT the anode voltage is normally 235V (50K anode load, 2.2K cathode bias resistor), with the grid disconnected it drops to 160V. Trying the same trick with the first half (470K anode load, 4.7K cathode resistor) only results in the anode voltage dropping from 114V to 105V. David. OK - that is about what I would expect. The ECC83 has an internal resistance of 80kohm at 100V, so that is about as close to saturated as you could expect it to get. I suspect that if you could actually remove the grid wire, the voltages would not have come out far different. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
On Thu, 15 Nov 2007 16:54:41 +0000, Nick Gorham
wrote: Don Pearce wrote: On Thu, 15 Nov 2007 16:25:19 +0000, Nick Gorham wrote: John Byrns wrote: That's fine, but the question is would enough charge buildup to prevent the tube from saturating? I would think the situation would be similar to the so called "contact bias" scheme used in the first audio stage of old US built tube radios? Yep, its the same as grid bias, you sometimes see it in use, cathode to 0v, 1M or so resistor between the grid and 0v, it will bias itself up as you suggest. Like grid leak biasing - where the coupling cap from the previous stage gets charged up and provides negative grid bias? Yes, except, you don't actually need a cap, as long as grid current generated across the grid resistor generates the required voltage. Except that in the case we are looking at, there is no grid resistor; it has been disconnected. TL did it on the d3a in the second version of his LCR phono stage. But in this case, the anode load won't have been selected for this sort of bias, so you will still get more current than you expect. Isn't this something you'd expect to find only on low voltage tubes. Yes, but again thats the sort of tube you would expect to have a pot connected to the grid. Always expect the unexpected! d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
Don Pearce wrote:
As a quick experiment I've just tried disconnecting the grid of the second half of the ECC83 in a Quad QC2 pre-amp that happens to be on my bench at the moment. With 280V HT the anode voltage is normally 235V (50K anode load, 2.2K cathode bias resistor), with the grid disconnected it drops to 160V. Trying the same trick with the first half (470K anode load, 4.7K cathode resistor) only results in the anode voltage dropping from 114V to 105V. David. OK - that is about what I would expect. The ECC83 has an internal resistance of 80kohm at 100V, so that is about as close to saturated as you could expect it to get. I suspect that if you could actually remove the grid wire, the voltages would not have come out far different. d I think you are confusing the rp with Rp, looking at the diode line http://www.mif.pg.gda.pl/homepages/f...37/1/12AX7.pdf I think its more like 46k, but the numbers still make sense, 160v across the diode line would expect about 3.4ma, you are getting a bit less (280-160)/50000 = 3.2ma, so again near enough, maybe its biasing itself off a little. -- Nick |
Grid Resistors
Don Pearce wrote:
Except that in the case we are looking at, there is no grid resistor; it has been disconnected. == very high value resistor. -- Nick |
Grid Resistors
"Keith G" wrote in message ... "Iain Churches" wrote in message i.fi... "West" wrote ...AND better no pot at all, but if you must, use a step attenuator with quality resistors. Hi West. Better still, no preamp at all:-) With the addition of a stepped attenuator, one can feed the CD player straight to the power amp. Morgan Jones states: "No preamplifier is better than any preamplifier!" I agree with that sentiment. S'what I did with my 2A3 SET which is effectively a power amp with a volume control - my thinking was that the 'pre' section could be a nightmare for a beginner to get right (quiet) and that there are/were any number of 'ready-made' pre's that could be used if necessary, until such time I built my own! It has worked so well without a pre I never got round to building one! Yes. That's a very good solution. What is ther sense of a preamp with excess gain which you need to attenuate? I like to have the gain control at about 2 oclock for monitoring at a "healthy" level Many take this a step further and add an input selector switch as well as the SA to the power amp. This makes sense too. Except that it eliminates the option of a 'single source direct' and condemns the amp to possibly 'inescapable' crosstalk with the extra internal wiring.... Never found that to be a problem, unless of course you switch to other unused and open inputs. Once they are terminated, one can get a fig of 85dB. If you need, you can use relay input switching, with the unused inputs just looking into a dummy load. Cheers Iain |
Grid Resistors
On Thu, 15 Nov 2007 17:35:27 +0000, Nick Gorham
wrote: Don Pearce wrote: As a quick experiment I've just tried disconnecting the grid of the second half of the ECC83 in a Quad QC2 pre-amp that happens to be on my bench at the moment. With 280V HT the anode voltage is normally 235V (50K anode load, 2.2K cathode bias resistor), with the grid disconnected it drops to 160V. Trying the same trick with the first half (470K anode load, 4.7K cathode resistor) only results in the anode voltage dropping from 114V to 105V. David. OK - that is about what I would expect. The ECC83 has an internal resistance of 80kohm at 100V, so that is about as close to saturated as you could expect it to get. I suspect that if you could actually remove the grid wire, the voltages would not have come out far different. d I think you are confusing the rp with Rp, looking at the diode line http://www.mif.pg.gda.pl/homepages/f...37/1/12AX7.pdf I think its more like 46k, but the numbers still make sense, Quite so. Something about 40k looks like the absolute lower limit. 160v across the diode line would expect about 3.4ma, you are getting a bit less (280-160)/50000 = 3.2ma, so again near enough, maybe its biasing itself off a little. So any charge collected by the grid is doing next to nothing by way of bias. d -- Pearce Consulting http://www.pearce.uk.com |
Grid Resistors
"Iain Churches" wrote Yes. That's a very good solution. What is ther sense of a preamp with excess gain which you need to attenuate? I like to have the gain control at about 2 oclock for monitoring at a "healthy" level 12 o' clock is my psychological best for 'starting to get loud'!! :-) Many take this a step further and add an input selector switch as well as the SA to the power amp. This makes sense too. Except that it eliminates the option of a 'single source direct' and condemns the amp to possibly 'inescapable' crosstalk with the extra internal wiring.... Never found that to be a problem, unless of course you switch to other unused and open inputs. More of a problem when more than one source is running, of course. Once they are terminated, one can get a fig of 85dB. If you need, you can use relay input switching, with the unused inputs just looking into a dummy load. Or simple toggle switches, in the case of valve amps - very satisfying to use! |
Grid Resistors
Keith G wrote:
"Iain Churches" wrote in message i.fi... "West" wrote ...AND better no pot at all, but if you must, use a step attenuator with quality resistors. Hi West. Better still, no preamp at all:-) With the addition of a stepped attenuator, one can feed the CD player straight to the power amp. Morgan Jones states: "No preamplifier is better than any preamplifier!" I agree with that sentiment. S'what I did with my 2A3 SET which is effectively a power amp with a volume control - my thinking was that the 'pre' section could be a nightmare for a beginner to get right (quiet) and that there are/were any number of 'ready-made' pre's that could be used if necessary, until such time I built my own! It has worked so well without a pre I never got round to building one! Keith, Exactly what I do as well. Many take this a step further and add an input selector switch as well as the SA to the power amp. This makes sense too. Except that it eliminates the option of a 'single source direct' and condemns the amp to possibly 'inescapable' crosstalk with the extra internal wiring.... I do this with a separate shielded cable from each input jack to the selector switch terminal, all grounds connected at the input jack end only. I have had no crosstalk issues. Raymond |
Grid Resistors
"Nick Gorham" wrote in message
... Yes, except, you don't actually need a cap, as long as grid current generated across the grid resistor generates the required voltage. Sorry no, the cap is an important part of the circuit. Firstly it charges to the signal peak, meaning that grid current only flows on signal peaks, not all the time. Secondly it connects the signal to the grid without providing a DC shunt path for the high-value grid resistor. David. |
Grid Resistors
David Looser wrote:
"Nick Gorham" wrote in message ... Yes, except, you don't actually need a cap, as long as grid current generated across the grid resistor generates the required voltage. Sorry no, the cap is an important part of the circuit. Firstly it charges to the signal peak, meaning that grid current only flows on signal peaks, not all the time. Secondly it connects the signal to the grid without providing a DC shunt path for the high-value grid resistor. David. Ok, fair enough, but what do you think would happen without an input signal or any other connection other than the high resistance connection between the grid and cathode circuit. I agree that practically it needs to be there to prevent the grid being pulled to ground, but I think you will find the idea of charging it up to signal "peaks" is just a bit strange. Whats going to charge up, the grid side of the cap is referenced to ground, and will it all go wrong after a period of silence? We could play, hunt the hidden cap, can you say miller? -- Nick |
Grid Resistors
Flipper wrote:
Oops. My mistake. I meant "least negative," not most negative. In particular, it'll bias to the zero grid current point, since there's no grid current. Right, seems a good point. I would expect it to be determined by the thermionic relationship between grid and cathode, more or less regardless of the anode current. The reason I asked was that, considering the amp will stop working, it seems unlikely that any damage would be done as a result of the pot failing. Unless it's a power stage, in most cases the valve would tolerate the extra current for a long time, and it should be unlikely that resistors will get too hot. The main thing to worry about if the pot fails is the silence. cheers, Ian |
Grid Resistors
Ian Iveson wrote:
Flipper wrote: Oops. My mistake. I meant "least negative," not most negative. In particular, it'll bias to the zero grid current point, since there's no grid current. Right, seems a good point. I would expect it to be determined by the thermionic relationship between grid and cathode, more or less regardless of the anode current. The reason I asked was that, considering the amp will stop working, it seems unlikely that any damage would be done as a result of the pot failing. Unless it's a power stage, in most cases the valve would tolerate the extra current for a long time, and it should be unlikely that resistors will get too hot. The main thing to worry about if the pot fails is the silence. cheers, Ian Just FYI, I made a similar mistake breadboarding a 2a3 output stage, I forgot to reference the grid to the cathode circuit, the current was double what I expected, so that would have left me with a glowing anode if I had not noticed. There isn't the anode load in that case to help out with the DC current. I agree though, in normal use, the pot will go to a small signal valve, and the sky won't fall. -- Nick |
Grid Resistors
"Nick Gorham" wrote in message
... I agree that practically it needs to be there to prevent the grid being pulled to ground, but I think you will find the idea of charging it up to signal "peaks" is just a bit strange. I don't find it remotely strange. Whats going to charge up, the capacitor. the grid side of the cap is referenced to ground, and will it all go wrong after a period of silence? During silence the charge will leak away via the grid resistor. With no-signal the grid is at chassis potential and the valve is operating under zero-bias conditions. The capacitor is charged to the whatever voltage is on the other end of it which maybe quite high if it comes from the anode of a previous valve, or it might be zero if direct from a volume control; let's assume the later. Now apply a signal, as the signal swings positive the grid-cathode diode is forward biased thus dramatically reducing the input resistance of the valve, the capacitor now starts to charge up, aiming for the signal voltage and with a time constant set by the value of the capacitor and the total circuit resistance (source resistance plus valve grid-cathode resistance). Once the signal has passed it's peak value the grid voltage starts to fall, turning off the grid-cathode diode, thus significantly increasing the time constant. The charge on the capacitor that it acquired during the positive half cycle is thus retained creating a small amount of negative bias for the valve. This charge will leak away slowly via the high value grid leak so that the charge on the capacitor, and thus the negative bias, roughly follows the envelope of the signal The thing is analogous to the valve TV sync separator where positive-going sync pulses drive the valve into grid-current forcing the coupling capacitor to charge to the instantaneous anode voltage of the preceding video amplifier. The difference here is that the signal amplitude is high enough to cause cut-off of the sync separator valve during the subsequent picture portion of the video waveform. David. |
Grid Resistors
On Fri, 16 Nov 2007 10:18:13 -0000, "David Looser"
wrote: I agree that practically it needs to be there to prevent the grid being pulled to ground, but I think you will find the idea of charging it up to signal "peaks" is just a bit strange. I don't find it remotely strange. Whats going to charge up, the capacitor. It is actually a very standard technique called clamping. It was used in televisions to keep black levels constant. It isn't the best way to do things, because the level will drift around a bit with signal amplitude, but when component count is important and volts are scarce it is certainly good enough. d -- Pearce Consulting http://www.pearce.uk.com |
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