I am trying to understand the bias circuit of the Tubelab SE and am not getting it. Maybe just blocking, but I could use some help.
The Bias voltage, B- is approximately -150V and sets the bias for the output tube through R14 and the circuit of R11-R13 somehow interacting with R15 and Q1.
How do I figure out what the actual grid bias voltage is on the output tube the way he has this circuit set?
I could just take George's circuit at face value, but I would like to understand it a little better. I think the circuit calculation is just a little to complicated for my tired mind, or maybe I am making it too complicated.
Thanks for any help!
The Bias voltage, B- is approximately -150V and sets the bias for the output tube through R14 and the circuit of R11-R13 somehow interacting with R15 and Q1.
How do I figure out what the actual grid bias voltage is on the output tube the way he has this circuit set?
I could just take George's circuit at face value, but I would like to understand it a little better. I think the circuit calculation is just a little to complicated for my tired mind, or maybe I am making it too complicated.
Thanks for any help!
Read this, which may answer some of your questions.
TubelabSE
Here is the discription from the site you are asking about.
"MOSFET Follower:
In many amplifiers the coupling capacitor is simply connected to the grid of the output tube, and a high value (100K up) resistor goes from the grid to the bias network. The evils of this arrangement were documented by Norman Crowhurst over 50 years ago, yet this continues to be a popular design. Just look up the schematic of your favorite 6SN7-300B amp. Several engineers have tried to overcome this problem with novel circuits. The two most common are the interstage transformer and the cathode follower. I have experimented with these circuits, but have found that a MOSFET follower simply works better. The purpose of any of these circuits is to present a high impedance load on the driver and a low impedance source to the grid of the output tube. I called the combination of the CCS and the MOSFET, PowerDrive, for lack of a better name, and in reference to its function. More details are on the PowerDrive page."
TubelabSE
Here is the discription from the site you are asking about.
"MOSFET Follower:
In many amplifiers the coupling capacitor is simply connected to the grid of the output tube, and a high value (100K up) resistor goes from the grid to the bias network. The evils of this arrangement were documented by Norman Crowhurst over 50 years ago, yet this continues to be a popular design. Just look up the schematic of your favorite 6SN7-300B amp. Several engineers have tried to overcome this problem with novel circuits. The two most common are the interstage transformer and the cathode follower. I have experimented with these circuits, but have found that a MOSFET follower simply works better. The purpose of any of these circuits is to present a high impedance load on the driver and a low impedance source to the grid of the output tube. I called the combination of the CCS and the MOSFET, PowerDrive, for lack of a better name, and in reference to its function. More details are on the PowerDrive page."
Thanks for the reply, I have read pretty much everything on George's website and all the manual several times.
I do understand what the circuit does, what I can't seem to figure out is how this circuit arrives at an appropriate bias voltage on the grid of the output tube.
How does the B- of -150 get to a grid voltage of around -65 or so appropriate for a 300b for example. Understanding this would help me understand why the values of the components are what they are.
I presume it is a matter of circuit analysis and determining the voltages at various points in the sub-circuit involved, but apparently the FET throws me off and I just can't seem to get it.
Again, any ideas for me?
I do understand what the circuit does, what I can't seem to figure out is how this circuit arrives at an appropriate bias voltage on the grid of the output tube.
How does the B- of -150 get to a grid voltage of around -65 or so appropriate for a 300b for example. Understanding this would help me understand why the values of the components are what they are.
I presume it is a matter of circuit analysis and determining the voltages at various points in the sub-circuit involved, but apparently the FET throws me off and I just can't seem to get it.
Again, any ideas for me?
The bias pot is really setting the bias point of the FET, which in turn is setting the bias point of the output tube. The FET behaves much like a tube. So if the FET is biased to around 5mA or so, it will create a voltage across R15, which then becomes the bias voltage for the tube.
Last edited:
It's simple voltage divider. Take the two extremes. With the pot all the way to the right the bias voltage to the FET gate is 150V*122K/252K, or about 72V. All the way to the left, it's 150V*22K/252K, or about 13V. So you can adjust between those values. Now accounting for the MOSFET: A typical MOSFET will need about 3-5V positive bias (Gate to Source) to conduct. So a bias voltage of, say 45V on the FET gate will result in a bias voltage on the output grid of about 40-42V. If you use a depletion mode device, ala DN2540, it requires about 3V or so negative bias, so the grid would see about 48V at the same pot setting.
Sheldon
The Circuit is set as a follower.
The Voltage at the Source will be the same as the voltage at the gate, minus a constant value that depends on the load line of the fet .The pot and 2 resistors R13 and R11 set a DC voltage at the gate between ground and -150V. I am having trouble reading the resistor values in the schematic, but I suspect the range is about -10v to -70 volts or so at the Gate of the FET.
Sheldon posted at the same time, and he may have a more coherent answer.
The Voltage at the Source will be the same as the voltage at the gate, minus a constant value that depends on the load line of the fet .The pot and 2 resistors R13 and R11 set a DC voltage at the gate between ground and -150V. I am having trouble reading the resistor values in the schematic, but I suspect the range is about -10v to -70 volts or so at the Gate of the FET.
Sheldon posted at the same time, and he may have a more coherent answer.
Thank you Sheldon and Doug! That is exactly what I was trying to figure out! Now I really understand What this is doing. Excellent. I guess I should read up on FETs and such.
It is hard to learn 80 years of electronics theory in a short time and via tube amps. I guess I could just build other peoples designs and not ask so many questions, but where is the fun in that? Just less drain bamage...
So, this allows a lower grid leak resistor and input impedance. Why? And how is the 20k R14determined and what exactly is it doing? Is it only an appropriate load for the FET now?
It is hard to learn 80 years of electronics theory in a short time and via tube amps. I guess I could just build other peoples designs and not ask so many questions, but where is the fun in that? Just less drain bamage...
So, this allows a lower grid leak resistor and input impedance. Why? And how is the 20k R14determined and what exactly is it doing? Is it only an appropriate load for the FET now?
As Doug explained, the source voltage follows the gate voltage, so it works by sensing the voltage on the source. In other words, as the current increases, the voltage at the source increases in direct proportion. Consider a follower connected to ground. The voltage would be 0V at the source, no matter what the voltage at the gate. So it couldn't very well follow the Gate. In principle, the source follower is most linear with a very large source resistor (R14), or a current source in that position. However, a large resistor needs a much higher negative supply, and a current source requires at least a couple extra components. But a reasonably high source resistor works just about as well. You want it small enough to pass a reasonable amount of current, but large enough to make the follower linear, all with a reasonable negative supply value. George has picked his best engineering compromise.
A grid leak resistor is used to prevent electrons from accumulating on the grid and changing it's potential. It's provides a DC path to ground to drain away any charge that accumulates. No grid leak resistor is required, nor is there one used here, as the follower has a very low output impedance (a few ohms), which means for DC purposes that the grid has a very low impedance path to ground. The resistor you see on the grid is a grid stopper. It's there to prevent oscillations, which can occur when the inductance of the circuit connected to the grid, and the capacitance of the grid form a circuit which can oscillate at a characteristic frequency. The stopper resistor damps the oscillation.
Sheldon
Last edited:
Ah! R14 is a source resistor for the FET rather than a grid leak (which would go to ground). I was wondering why R14 was tied to the B-.
Just for kicks, could one put a CCS in the R14 position? Using a simple one like the one used as a plate load for the 5842, assuming you had the parts laying about?
Thanks again for your explanations. Very helpful and kind. I feel I understand this better now.
Just for kicks, could one put a CCS in the R14 position? Using a simple one like the one used as a plate load for the 5842, assuming you had the parts laying about?
Thanks again for your explanations. Very helpful and kind. I feel I understand this better now.
Don't worry about the theory part. I don't think Doug was questioning the validity. It will work as long as your current source can handle the voltage. From a precision driven engineering perspective, it's the preferred approach. Will it sound better to you? Not necessarily. You'll have to try. No need to measure the current. You can estimate it from the expected voltage over R14 (grid voltage - bias supply voltage). For -40V on the grid, you'd see about 5mA. That's a good starting point.
Sheldon
I was questioning if the current source would be an improvement in that application.
Since the 20K resistor is much larger than the 2 ohm output Z, I would guess that the difference would be very small.
That is in stark contrast to the CSS plate load, which helps a great deal.
I would trust that George tried it, and he judged it was not worth the extra complexity.
Since the 20K resistor is much larger than the 2 ohm output Z, I would guess that the difference would be very small.
That is in stark contrast to the CSS plate load, which helps a great deal.
I would trust that George tried it, and he judged it was not worth the extra complexity.
I agree, and don't think it would be an audible improvement. But part of the fun of DIY is to try stuff as a learning experience.
There is one possible modest advantage of the current source in this application, and that is that the quiescent current through the FET remains the same, regardless of the output tube. With a resistor, the current can vary significantly between tube types. Does it matter? Unlikely, but for those obsessive types...
Sheldon
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.