Hi all,
Since we've been on the topic of cathode feedback and the like in the past couple of days, I was wondering if someone would take the time to explain something to me that I am having my problems in understanding fully.
I've been working my way through trying to understand the various nuances of Claus Byrith's EL34 SE amp, and I'm confused with the global NFB loop.
The schematic can be found here.
Please forgive what's likely to be a very basic electronics question - but I've been working hard on trying to understand this and I just don't get it.
The design uses cathode feedback over the output transformer secondary as well as global parallel injected voltage NFB in which voltage feedback is mixed back into the grid of the driver stage. The amount of global NFB is variable from 0 - 10 dB over a 100kOhm potentiometer. I'm having problems coming up with the calculation as to how this circuit delivers 10dB of NFB.
The way I understand it....
10dB of NFB means that the feedback factor in this case shoud be around 0.041.
The stage has an open loop gain of 53. So, 53 / (1 + 0.041 x 53) = a closed loop gain of 16.3.
20log (16.3 / 53) = -10dB.
So far, so good.
The feedback factor is calculated using the voltage divider rule R1 / (R1 + R2). Calculating the case for the full 10dB of NFB, R1 is the 100k pot in parallel with the 22k resistor which gives about 18k.
18k / (150k + 18k) = 0.107, not 0.041 - we're out by a factor of 2.5. So in this case this the feedback factor of 0.107 would be approx 16.5dB NFB and it's supposed to be 10dB so obviously something else plays a role here.
But what?
I'm "guessing" that this has something to do with the fact that cathode feedback is taken off the secondary as well as is global NFB. So the signal is being divided before reaching the global NFB circuit, and only 1 / 2.5 of the feedback signal is further divided by the global NFB circuit and mixed back at the grid of the driver?
But what resistances exactly play a role in this division and how?
The resistance on the lower global NFB limb seen by this mysterious divider could be 22k || 100k || 150k ~ 16k. Which would mean that the other limb of this mysterious divider shoud have an R of around 24k. Right?
How do I calculate this? The cathode resistor has a value of 330R. The EL34 has an internal resistance. I can see the primary of the OPT as well.
Help?
Hopelessly yours,
Griz
Since we've been on the topic of cathode feedback and the like in the past couple of days, I was wondering if someone would take the time to explain something to me that I am having my problems in understanding fully.
I've been working my way through trying to understand the various nuances of Claus Byrith's EL34 SE amp, and I'm confused with the global NFB loop.
The schematic can be found here.
Please forgive what's likely to be a very basic electronics question - but I've been working hard on trying to understand this and I just don't get it.
The design uses cathode feedback over the output transformer secondary as well as global parallel injected voltage NFB in which voltage feedback is mixed back into the grid of the driver stage. The amount of global NFB is variable from 0 - 10 dB over a 100kOhm potentiometer. I'm having problems coming up with the calculation as to how this circuit delivers 10dB of NFB.
The way I understand it....
10dB of NFB means that the feedback factor in this case shoud be around 0.041.
The stage has an open loop gain of 53. So, 53 / (1 + 0.041 x 53) = a closed loop gain of 16.3.
20log (16.3 / 53) = -10dB.
So far, so good.
The feedback factor is calculated using the voltage divider rule R1 / (R1 + R2). Calculating the case for the full 10dB of NFB, R1 is the 100k pot in parallel with the 22k resistor which gives about 18k.
18k / (150k + 18k) = 0.107, not 0.041 - we're out by a factor of 2.5. So in this case this the feedback factor of 0.107 would be approx 16.5dB NFB and it's supposed to be 10dB so obviously something else plays a role here.
But what?
I'm "guessing" that this has something to do with the fact that cathode feedback is taken off the secondary as well as is global NFB. So the signal is being divided before reaching the global NFB circuit, and only 1 / 2.5 of the feedback signal is further divided by the global NFB circuit and mixed back at the grid of the driver?
But what resistances exactly play a role in this division and how?
The resistance on the lower global NFB limb seen by this mysterious divider could be 22k || 100k || 150k ~ 16k. Which would mean that the other limb of this mysterious divider shoud have an R of around 24k. Right?
How do I calculate this? The cathode resistor has a value of 330R. The EL34 has an internal resistance. I can see the primary of the OPT as well.
Help?
Hopelessly yours,
Griz
Hi Griz,
Basic electronics are very important, and sometimes sadly neglected. I will try to help here.
Firstly, we can treat the 2 feedback loops (secondary of transformer to EL34 and global) as entirely separate. The signal is not "divided" in that sense. The EL34 feedback determines the gain for that stage irrespective of global feedback (in fact, the G2 connection to a tap on the output transformer constitutes further local feedback). So you treat that as a block with whatever resultant gain it has, and forget about it when looking at global feedback.
What we are given is that for no feedback, an input of 650mV gives an output of 8W into 8 ohm, i.e. an output voltage of 8V. That presents a gain from voice coil to input of 8/0.65 = 12.31. I do not find the open loop gain of 53 you mention, but it seems realistic for the circuit - not to mind, we will not be using it. Also, your R1 (the 100K pot) is designated R13 on the scematic, and R2 the ECC83 anode load resistor. So I am going to go through the argument using the given schematic annotations, trusting that it would make matters clear. (Explanation will be basic; if I underestimate your knowledge I apologise).
In a parallel feedback circuit of this kind and where the loop gain is high, the signal at G (ECC83) is usually very low and the impedance there as a result of the feedback is also low. This leads to the simplification that one can see that point as a virtual earth or common. In that sense the value of the 100K pot will have negligible effect. The fed-back signal simply occurs over R16 (150K) and the input over R1 (22K), in that ratio. For full feedback this will give 8V over 150K, or [8/150] x 22 = 1.17V input with feedback. But this does not correspond with 2V input because we do not have a high feedback factor. G is not a virtual earth for low feedback factors.
Instead we go the following route. For 8V output we find that 532mV is required at G1(ECC83) - that is the given 650mV divided down by 22K in serie with 100K. Since the 8V and 532mV is out of phase when feedback is applied, there will be 8.53V over the 150K (R16). Since the current in the circuit is the same (we neglect the 100K pot again for the moment), it would require [8.53/150K] x 22K or 1.25V over R1 (22K) to maintain balance. The total input with respect to common will then be 1.25V + 0.53V = 1.78V. If you now bring the 100K pot into the picture, which will also draw current as there is 532mV over it, you will find the actual signal over the 22K to be about 1.38V (I did not do the exact calculation which will involve all the elements; I am just trying to indicate the way). This brings us to an input with respect to common of 1.91V, close enough to the quoted 2V - I presume values were rounded off somewhat.
The 10dB of feedback you find in the ratio of (input with feedback)/(input without feedback), or 2V/0.65V = 3.08, which gives 9.8 dB; again presumed somewhat rounded off to 10dB.
Regarding your logic, note that the 100K pot never acts in parallel with the 22K as the signals across them are not the same. The "something else" that plays a role is the impedance at G1 of the ECC83 under feedback conditions. One can of course draw up a full set of circuit node equations for accurate analysis; I am mathematically lazy and believe the above steps bring one within an engineering solution. There are tolerances anyway.
Lastly, if you also desire an analysis of what is happening in the EL34 stage itself, we do not have enough information for that. It will need knowledge of the % screen tap, valve characteristics under those conditions, etc. There is currently a discussion going under the thread "Tubes: UL to triode" as well as elsewhere should you be interested.
Hope this helped.
Basic electronics are very important, and sometimes sadly neglected. I will try to help here.
Firstly, we can treat the 2 feedback loops (secondary of transformer to EL34 and global) as entirely separate. The signal is not "divided" in that sense. The EL34 feedback determines the gain for that stage irrespective of global feedback (in fact, the G2 connection to a tap on the output transformer constitutes further local feedback). So you treat that as a block with whatever resultant gain it has, and forget about it when looking at global feedback.
What we are given is that for no feedback, an input of 650mV gives an output of 8W into 8 ohm, i.e. an output voltage of 8V. That presents a gain from voice coil to input of 8/0.65 = 12.31. I do not find the open loop gain of 53 you mention, but it seems realistic for the circuit - not to mind, we will not be using it. Also, your R1 (the 100K pot) is designated R13 on the scematic, and R2 the ECC83 anode load resistor. So I am going to go through the argument using the given schematic annotations, trusting that it would make matters clear. (Explanation will be basic; if I underestimate your knowledge I apologise).
In a parallel feedback circuit of this kind and where the loop gain is high, the signal at G (ECC83) is usually very low and the impedance there as a result of the feedback is also low. This leads to the simplification that one can see that point as a virtual earth or common. In that sense the value of the 100K pot will have negligible effect. The fed-back signal simply occurs over R16 (150K) and the input over R1 (22K), in that ratio. For full feedback this will give 8V over 150K, or [8/150] x 22 = 1.17V input with feedback. But this does not correspond with 2V input because we do not have a high feedback factor. G is not a virtual earth for low feedback factors.
Instead we go the following route. For 8V output we find that 532mV is required at G1(ECC83) - that is the given 650mV divided down by 22K in serie with 100K. Since the 8V and 532mV is out of phase when feedback is applied, there will be 8.53V over the 150K (R16). Since the current in the circuit is the same (we neglect the 100K pot again for the moment), it would require [8.53/150K] x 22K or 1.25V over R1 (22K) to maintain balance. The total input with respect to common will then be 1.25V + 0.53V = 1.78V. If you now bring the 100K pot into the picture, which will also draw current as there is 532mV over it, you will find the actual signal over the 22K to be about 1.38V (I did not do the exact calculation which will involve all the elements; I am just trying to indicate the way). This brings us to an input with respect to common of 1.91V, close enough to the quoted 2V - I presume values were rounded off somewhat.
The 10dB of feedback you find in the ratio of (input with feedback)/(input without feedback), or 2V/0.65V = 3.08, which gives 9.8 dB; again presumed somewhat rounded off to 10dB.
Regarding your logic, note that the 100K pot never acts in parallel with the 22K as the signals across them are not the same. The "something else" that plays a role is the impedance at G1 of the ECC83 under feedback conditions. One can of course draw up a full set of circuit node equations for accurate analysis; I am mathematically lazy and believe the above steps bring one within an engineering solution. There are tolerances anyway.
Lastly, if you also desire an analysis of what is happening in the EL34 stage itself, we do not have enough information for that. It will need knowledge of the % screen tap, valve characteristics under those conditions, etc. There is currently a discussion going under the thread "Tubes: UL to triode" as well as elsewhere should you be interested.
Hope this helped.
Johan Potgieter said:
OK. Got it.
Johan Potgieter said:
Exactly what I was looking for. This makes sense to me, now.
Johan Potgieter said:
Yes. This is the next thing I need to look at. I'll have a read through the discussion that you mention.
Johan Potgieter said:
It did indeed. Thanks very, very much for your time, Johan. I appreciate it greatly.
Regards,
Griz
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