# Thread: mosfet follies - cathode follower

1. ## mosfet follies - cathode follower

Hello, can anybode gude me on this:
- http://www.geofex.com/Article_Folder...osfetfolly.htm
The part about mosfet cathode followers.

1. The idea is that the mosfet follower amplifies the output current, providing lower output impedance. So it is much better than taking the output signal from the anode with a voltage divider, right?

2. How to calculate the source resistor. It will set the maximum current consumption of the mosfet, right?

3. The output voltage swing at the follower is going to be the same as the anode voltage swing at the gate, right?
- I guess, the source resitor could be two resistors in series (voltage divider) and I will attenuate to the voltage I need.

4. How to calculate the output impedance of the mosfet cathode follower? I guess it depends on the current set by the source resistor and the voltage swing.

Regards

2. Originally Posted by emosms
1. The idea is that the mosfet follower amplifies the output current, providing lower output impedance. So it is much better than taking the output signal from the anode with a voltage divider, right?

2. How to calculate the source resistor. It will set the maximum current consumption of the mosfet, right?

3. The output voltage swing at the follower is going to be the same as the anode voltage swing at the gate, right?
- I guess, the source resitor could be two resistors in series (voltage divider) and I will attenuate to the voltage I need.

4. How to calculate the output impedance of the mosfet cathode follower? I guess it depends on the current set by the source resistor and the voltage swing.
1. Yes, it will have lower output impedance but it it won't attenuate the signal like a voltage divider.

2. I'd use the same value as would be used for a similar cathode follower circuit.

3. Yes, the output voltage swing will be the same as the anode voltage.
- The source resistor could be two resistors but then you have lost the low output impedance of the source follower. To attenuate with low output impedance the anode resistor could be split and then buffered by the source follower.

4. Output impedance is ~1/gm

3. Originally Posted by emosms
1. The idea is that the mosfet follower amplifies the output current, providing lower output impedance. So it is much better than taking the output signal from the anode with a voltage divider, right?
That's one way to say it. I'd modify that a little by saying that MOSFETs don't have a current gain; or rather, their current gain is nearly infinite. They have a transconductance, where a change of voltage on the gate-source, causes a current to flow in the drain/source. The gate of the MOSFET is so close to an open circuit that it takes very, very fancy equipment to tell them apart, so the tube anode is unloaded.

The MOSFET source is a low impedance output, and is loaded by both its source resistor and whatever load is connected in parallel to the source. A signal on the source is loaded by both the source resistor and the output load, so the source signal voltage sags a tiny bit.

The gate, being held almost perfectly to the preceding anode, does not sag, and so the voltage from gate to source sags a little - and in the perfect direction to make more current flow into the source from the drain. So tiny, tiny changes in the source voltages can make bit current changes through the MOSFET.

And that gets us back to your statement: very high impedance to the anode? Check. Amplifies the current and provides low output impedance? Check, but in a slightly subtler way. Better than taking signal from the anode? If you are driving a load lower than 10x the anode source resistance, check.

2. How to calculate the source resistor. It will set the maximum current consumption of the mosfet, right?
Yes, it sets the idle current flow through the MOSFET. The gate is held at some relatively fixed DC voltage, the source will be about 4-6V lower, and the current in the MOSFET at idle will be the source voltage divided by the source resistor. That means the source resistor dissipates as heat the value of the resistance times the voltage across it squared. You can burn up a lot of power in one of these.

3. The output voltage swing at the follower is going to be the same as the anode voltage swing at the gate, right?
With only tiny differences for transconductance losses, yes. If you're puzzling over these issues, yes, think of it as being identical.
- I guess, the source resitor could be two resistors in series (voltage divider) and I will attenuate to the voltage I need.
Yes. But there are many ways to get less signal voltage.

4. How to calculate the output impedance of the mosfet cathode follower? I guess it depends on the current set by the source resistor and the voltage swing.
This is actually calculated the same way that a CF output impedance is calculate for a triode, which is an equation I can't remember. The trick is this: the output impedance of a CF tube or SF MOSFET depend on the inverse of the transconductance. A 12AX7 tube has a trsconductance of something like 1.6mMho, or 1.6 ma/volt of change on the grid. A MOSFET worthy of the name will have a transconductance of 1000 - 2000 mMho, or a thousand times more, so the output impedance of a MOSFET SF is a thousand times lower, typically.

This is so much lower than a tube CF that it makes the source impedance relatively unimportant unless you're trying to do stuff like drive low impedance reverb tanks or speakers from a single SF stage. You aren't, are you?

4. I almost understand .

1. The source resistor sets the idle current for the mosfet. How much would be the max current?
- In the case of the triode stage, the max possible current is set by the anode resistor.

2. The output impedance is... not dependant on the idle current of the mosfet, nor the output voltage swing..

---
Lets put an example why I am asking these:

At the moment I have a tiny transformer to supply a double triode and probably two mosfet followers. Transformer is rated 250v AC, 10mA.
Say I want to set 100k anode resistor (played a bit with the load line), 300v DC rectified.
This is 2 x 3mA current for the double triode. This should be the max possible consumption, regardles of voltage swing.
(as far as I get it, the anode resistor limits the max possible current)

There are 2mA per a mosfet left. This is the total allowed curent. Wether it is the max current or iddle current, I don't know .
The first mosfet follower is supposed to drive 100k Zin stage.
The second is supposed to drive 4 stages with 100k Zin in parallel. I guess, the total load impedance should be 25k. Say 20k .

Would that 2mA suffice...
Just using ohm's law: 1,227352 v RMS (3,472v peak to peak) / 20k (worst case) = .. it will suffice . If not missing something.
NOTE: the figure I am getting (0,0613676 mA) ^ is added on top of the 2mA idle current ??

Voltage attenuation is next thing..

5. Since we're on the subject here's a practical question from few days ago. On the pic below is an LND150 based FX loop also know as Metro Zero Loss FX loop. There's no CF there but the first stage is wired as inverting feedback amplifier which has a very low output impedance assuming the role of a CF.
So after firing this thing up I measured 2 Volts at the send stage drain which is not normal. At the other end of the 100k resistor voltage was 254V which indicated ~2.5mA of current draw which is strange. After replacing components and scratching my head for couple of hours without any result I added a 1M resistor from gate to ground and very slowly the voltage at the LND drain in question went to 220V while at the other resistor's end to 305V. The only thing missing from my build is that R4 220k resistor.
So I was wondering what happened and why? Shouldn't the loop work without that resistor to ground as per schematic?

6. Originally Posted by GainFreak
The only thing missing from my build is that R4 220k resistor.
I think that's the answer. Without R4 (or your 1M) the gate of VT1 has no DC reference to ground.

7. Originally Posted by emosms
1. The source resistor sets the idle current for the mosfet. How much would be the max current? - In the case of the triode stage, the max possible current is set by the anode resistor.
Note that what you say is true if the active device, tube or MOSFET, is shorted. Then the current is just the power supply divided by the resistor. It does limit the max possible current, but you'll probably never see that current.

Most 12AX7s can't really conduct more than 1-2ma no matter what you do to the grid, and they can't really "saturate" like solid state devices. As a practical matter, a triode with a 100K anode resistor from 300V will see on the order of 1 to 1.5ma.

In the case of the MOSFET, it can sure act more like a short ciruit than a triode can, but you can also predict things better. The triodes will have their anodes at a voltage between 90 and 250V, set by the DC value of the cathode resistor for the triode. Let's say you set that to 150V by careful tinkering.

The MOSFET gate will be at that same voltage; I recommend using a 100R to 1K "grid stopper" right on the MOSFET's gate for other reasons, but that won't affect the gate voltage. The MOSFET source will stabilize one Vt down from the gate. Vt (threshold voltage) is what it takes to get an enhancement mode MOSFET to start conducting. This varies from device to device, but is trivially small compared to the plate voltage on the gate. So say the source voltage will be 150-5 =145V.

Now you get to pick the MOSFET's current. You do that by saying "I want this MOSFET to conduct Xma at idle". X will be in the range of 0.5 to 2ma, depending on what you decide X is. Let's say 1ma. So the source resistor has 145V across it and 1ma through it. Georg Ohm says the resistance must be 145/0.001 = 145K. That's clumsy, if you don't have a full bag of 1% resistors, so pick either 140K or 150K.
2. The output impedance is... not dependant on the idle current of the mosfet, nor the output voltage swing..
What I'm telling you now is heavily simplified, but I think that's what you need right now. Really, the output impedace does depend a little on the idle current and swing IIRC, but that's way down on the list of your worries. And it doesn't depend much.

At the moment I have a tiny transformer to supply a double triode and probably two mosfet followers. Transformer is rated 250v AC, 10mA.
And this is the point where I started understanding your design needs. You're power supply liimited.

Is that transformer's 10ma "RMS" or "rectified DC", and what kind of rectifiers and filters are you using. This can make for a biggish difference.

On the other hand, you're probably only headed for 5-6ma, so you may be fine either way.

Say I want to set 100k anode resistor (played a bit with the load line), 300v DC rectified.
This is 2 x 3mA current for the double triode. This should be the max possible consumption, regardles of voltage swing.
(as far as I get it, the anode resistor limits the max possible current)
See above. You're likely to only see 1-1.5ma per plate.

There are 2mA per a mosfet left. This is the total allowed curent. Wether it is the max current or iddle current, I don't know .
You get to pick whether it's 1ma 1.5ma, or 2ma with your choice of the source resistor, and it's DC idle current.

[QUOTE]
The first mosfet follower is supposed to drive 100k Zin stage.
The second is supposed to drive 4 stages with 100k Zin in parallel. I guess, the total load impedance should be 25k. Say 20k .
Those are AC current loads. I'd give it a try with 150K source resistors, and see if the loads don't pulll down the signal level for you.

You **might** get into a situation where the SF tries to pull up the output AC load but can't get enough current from the power supply. This will happen at bass frequencies first. And you could run the SF out of headroom with big enough signals., I'd say to try it before buying a bigger power supply.

8. Originally Posted by R.G.
Those are AC current loads. I'd give it a try with 150K source resistors, and see if the loads don't pulll down the signal level for you.

You **might** get into a situation where the SF tries to pull up the output AC load but can't get enough current from the power supply. This will happen at bass frequencies first. And you could run the SF out of headroom with big enough signals., I'd say to try it before buying a bigger power supply.
That might be bad. Don't know what to expect if it clips. (bass guitar)
It would be nice if the amplitude is just limited and sinusoide is rounded, so that I get a compression .

On the other hand, how to calculate these AC currents.. I simplified as just adding up to the idle DC current.
Both cases (7v Rms /100k Zin, 1,2v RMS/20k), if I calculate currents Zin, it is < 0,1mA.

The transformer is rated 250 v AC, 10mA.... + 6v, 600mA. I guess should be RMS.
I calc it as 350v DC (250x1,414) bridge rectified +cap. Haven't tried it yet, but the calc is pretty much correct. For lower voltage transformers..
Then I will drop it to 300v, with 2 resistors and 2 smoothing caps. (f.ex. 22uF)

The tube I consider is 6n1p-ev (more like 12AY7, military grade "ev", NOS). So it mby can go closer to 3mA. Setting it with 100k anode resistor would yiled a bit more gain and a possiblitiy to swap with 6n2p-ev (12ax7 like)

---

Btw, abt. the topology itself. Here we have valve CF without any grid reference or biasing whatsoever:

But Blencowe suggests grid biasing (much like the GainFreak circuit):

I will be happy to have it the simple way - just gate resistor added.

The idea to have voltage divider at the anode resitor to attenuate signal..
Keeping AC swing and current low.. Sounds good to me now .

9. Originally Posted by emosms
On the other hand, how to calculate these AC currents.. I simplified as just adding up to the idle DC current.
I'm not sure what current you are trying to calculate here. If it's the total current the power supply has to provide then you could just add up the idle currents. These are class A amplifier stages. The power supply current won't change much between idle and full output.

10. Originally Posted by Dave H
I'm not sure what current you are trying to calculate here. If it's the total current the power supply has to provide then you could just add up the idle currents. These are class A amplifier stages. The power supply current won't change much between idle and full output.
Sounds reasonable, thx . A lot more need to consider and understand than ohms law. I ve been reading abt class A amps, but...

11. I think that's the answer. Without R4 (or your 1M) the gate of VT1 has no DC reference to ground.
I was left with the impression that in theory the zero potential at the output cap should provide the "virtual earth" to the grid but in practice there was ~0.8V at the the grid which made the LND draw more current.

12. Originally Posted by GainFreak
I was left with the impression that in theory the zero potential at the output cap should provide the "virtual earth" to the grid.
There's no resistive path to ground at the output cap so it won't necessarily be at zero potential. It will charge up to some other potential depending on leakage currents which is probably different from the 0.8V you measured with the meter connected.

13. Hi again. I tested my psu with 22k load on the HT and the heater of actual 6n1p tube.
The AC HT drops to 233v, which is ok. I can also test with rectivier/RC network and check the DC, then recalc the loadline and biasing.

But, the heater voltage drops to 5.7v AC. This is the lower allowed heater voltage.
According to the tube specs, it has steepnes of 4.35 ma/v (I guess for nominal heater voltage) and 3.2 steepnes for heater @ 5.7v.
This makes me reconsider the psu transformer.. If I want to really put 6n1p in proper operation.
(For 6n2p it seems ok - HT under 22k load is 251v AC, heater under filament load is 6 v.)

So far so good. Moreover, I found that Rod Eliot has solution for this (as well..) .

Voltage Followers

He claims the circuit has very low output impedance (330 ohm, R4) and very good driving capabilities.
But...

"The current limit is around 25mA with a 330 ohm resistor. "

How does he calculate this ?????

"Increase the value of R5 if you don't need the drive capacity provided by the 22k source resistor. Output impedance is not affected if you change the value of R5, but the ability to provide a high level signal into low impedances is reduced. The circuit above can provide well over 5V RMS into a 2.2k load impedance."

If R5 (source resistor) limits the max current (disregarding R4), 250v/22k = 11,36mA....
How is that calculated?

My reasons are again - power supply limit - I can order 15VA transformer, giving 40mA for HT and 700mA for heater.
Is that correct, that the circtuit above draws 25mA ??? How it is possible?
If it is correct, how to halve the max current? Changing R5?
"Increase the value of R5 if you don't need the drive capacity provided by the 22k source resistor."

Since I don't know how these 25mA are calculated (HT - 250v), I don't know how to set the circuit to draw half the current .
Except - building a prototype and test..

14. Originally Posted by emosms
But...

"The current limit is around 25mA with a 330 ohm resistor. "

How does he calculate this ?????
I guess the FET needs about 4V Vgs to turn it on. Subtracting 4V from the 12V zener leaves 8V across the 330R resistor so the current limit is 8/0.33 or around 25mA. This is only when sourcing current. It can't sink 25mA. I think it can only sink a peak current equal to the standing current i.e. 119/22 or about 5mA. 5mA peak limits the output voltage for a 2.2k load to 11V peak but it's limited to 10V peak anyway by the 10V zeners. 10V peak is about 7V rms which is 'well over' 5V rms

To set it to draw half the current double the value of the 22k source resistor. This will halve the voltage drive to a 2k2 load to 2.5V rms (I think)

15. This is correct but, as emosms says, even if the transistor is shorted, you cannot get higher current in this circuit than 11 mA. So the information about 25 mA limit is "general" and it applies to other values of R5 resistor and/or other power supply voltages.

Mark

16. Originally Posted by GainFreak
Since we're on the subject here's a practical question from few days ago. On the pic below is an LND150 based FX loop also know as Metro Zero Loss FX loop. There's no CF there but the first stage is wired as inverting feedback amplifier which has a very low output impedance assuming the role of a CF.
So after firing this thing up I measured 2 Volts at the send stage drain which is not normal. At the other end of the 100k resistor voltage was 254V which indicated ~2.5mA of current draw which is strange. After replacing components and scratching my head for couple of hours without any result I added a 1M resistor from gate to ground and very slowly the voltage at the LND drain in question went to 220V while at the other resistor's end to 305V. The only thing missing from my build is that R4 220k resistor.
So I was wondering what happened and why? Shouldn't the loop work without that resistor to ground as per schematic?

I have a question regarding this schematic. What is the purpose of the R2 pot? It shorts the input of the first stage of the loop with the output. Is it correct, or it's a mistake on the schematic?

Mark

17. Originally Posted by MarkusBass
but, as emosms says, even if the transistor is shorted, you cannot get higher current in this circuit than 11 mA. So the information about 25 mA limit is "general" and it applies to other values of R5 resistor and/or other power supply voltages.
Isn't 11mA is just the max current you can get in the 22k source to ground resistor (R5). The FET can source more current to a 2k2 load than that.

18. Originally Posted by MarkusBass
I have a question regarding this schematic. What is the purpose of the R2 pot? It shorts the input of the first stage of the loop with the output. Is it correct, or it's a mistake on the schematic?
It's adjustable negative feedback to set the send level.

19. 5vRms/2k2 is 2.27mA.

So, how does the value of R5 (22k) affects the driving capability ??

For the gate threshold voltage Vgs, values are 2 to 4 v indeed.

So, 12 (zenner)-4 for the Vgs is 8/330, or abt. 25mA.
Eventhough, I do not understand - is that 4v dropped in the gate-source, so we get only the difference of 8 volts?

Further on, if R4 dissipates that current, should'nt we add the output current to that 25mA? So, consumption is over 25mA?

Finally, If R4 eats that current, what if I substitute it with 620 ohms. Then we have 8/620 = 0,13mA.
The "nominal" impedance is 620 ohms, like standard, and the output driving capability is not impaired.
(Whatever the reason that it depends on R5 is...)

p.s. Also, I hope, that the G-S zenner of 12v does not clip the tube output swing to 12v.....

20. Originally Posted by emosms
So, how does the value of R5 (22k) affects the driving capability ??
It sets the negative clipping point. The current through R5 is 119/22 = 5.4mA. When the FET shuts off (negative clipping point) there's no current through the FET so 5.5mA peak is pulled from the load. If the load is 2k2 that's 2.2 x 5.4 ~ 11V peak.

Originally Posted by emosms
So, 12 (zenner)-4 for the Vgs is 8/330, or abt. 25mA.
Eventhough, I do not understand - is that 4v dropped in the gate-source, so we get only the difference of 8 volts?

Further on, if R4 dissipates that current, should'nt we add the output current to that 25mA? So, consumption is over 25mA?
You've got it in the first line The voltage between the gate and bottom of R4 is limited to 12V by the zener and the FET needs 4V to turn on so there's 8V max across R4 or 25mA.

25mA is the fault current if you like. It's the current you'd measure if you shorted across R5 with an ammeter. The current consumed from the power supply is the no signal (quiescent) current through R5 i.e. 5.4mA. The current through R5 (and the load) will vary with signal but the average from the power supply will be 5.4mA.

Originally Posted by emosms
Finally, If R4 eats that current, what if I substitute it with 620 ohms. Then we have 8/620 = 13mA.
The "nominal" impedance is 620 ohms, like standard, and the output driving capability is not impaired.
(Whatever the reason that it depends on R5 is...)
Yes you could do that but all of that 13mA isn't available to drive the load some of it is the current through R5. I'll try and work out where the positive clipping point is with R4 at 620R to see if it's OK.

21. Hi, I think I am getting it, thanks .
The fault consumption of 25 ma is 25 * 8volts drop = 0,2 VA of the PSU. This is the case when we have cold cathode and almost HT at mosfet's gain. Lets leave R4 as it is.
This is not to be considered uner normal operation. There would be no voltage exceeding 12v across gate-source, because the source "follows"

---
Trying to calc some worst case scenarious. If the valve is 6n2p, 100k anode load, 300v DC - the load line shows that anode voltage could go over 240v DC...
It is not that much for 6n1p, so taking the value of 220v as worst case.
Disregarding the drop in R4 and the voltage needed for mosfet-on.

220v between R4 and R5, 220/2k2 - we have 10mA consumption - "worst case scenario"
If the voltage is higher, I can rise R5 - 25k, 27k, 33k... (more than worst case scenario)

- Here a side question. When setting the load line, all the grid voltage curves - is that RMS voltage ???? Needed to more precisely do the math.
F.ex. if I measure the guitar output with multimeter, I can get the RMS AC voltage (I guess it is not peak to peak).
Then, I can use that to get the max grid volgate and the max possible anode voltage at given HT/anode load.

---
Further on.
Suppose I order transformer 250v AC 40mA. This is 250 * 0,04 = 10VA. Bridge rectifier and caps resutls in 250*1.414 = 353.5v DC.
10VA/353.5 = ~28mA expected possible current draw without sag.
28mA is 11(mosfet CF)+11(mosfet CF)+3 (tube load)+3(tube load). Provided the tube will operate lower than 3mA, I will have > 11mA per mosfet.

More considerations - In another forum, people want me to get even bigger transformer. Suggesting ratio of 2/2.2 to 1 - AC to DC current.
I did the math above and the result is 40mA/ 250v AC vs 28mA/ 353v DC. This is a ratio of 1.42 to 1.

The only thing I am missing - how much extra current/power would eat 2 electrolytic capacitors in the CRC chain ??? Is it that much? Is it something to consider at all ? .

22. Originally Posted by emosms
Hi, I think I am getting it, thanks .
The fault consumption of 25 ma is 25 * 8volts drop = 0,2 VA of the PSU. This is the case when we have cold cathode and almost HT at mosfet's gain. Lets leave R4 as it is.
This is not to be considered uner normal operation. There would be no voltage exceeding 12v across gate-source, because the source "follows"

---
Trying to calc some worst case scenarious. If the valve is 6n2p, 100k anode load, 300v DC - the load line shows that anode voltage could go over 240v DC...
It is not that much for 6n1p, so taking the value of 220v as worst case.
Disregarding the drop in R4 and the voltage needed for mosfet-on.

220v between R4 and R5, 220/2k2 - we have 10mA consumption - "worst case scenario"
If the voltage is higher, I can rise R5 - 25k, 27k, 33k... (more than worst case scenario)
Thoughtfully considered. Well done!

Originally Posted by emosms
Here a side question. When setting the load line, all the grid voltage curves - is that RMS voltage ???? Needed to more precisely do the math.
F.ex. if I measure the guitar output with multimeter, I can get the RMS AC voltage (I guess it is not peak to peak).
Then, I can use that to get the max grid volgate and the max possible anode voltage at given HT/anode load.
Data points on the load line plot are instantaneous voltage. So if you impose a signal on the bias, use peak-to-peak values to see how the anode voltage will track, positive and negative from the bias 'point'.

Originally Posted by emosms
how much extra current/power would eat 2 electrolytic capacitors in the CRC chain ??? Is it that much? Is it something to consider at all ? .
The only time the caps are 'eating' current is when they are charging. Initial inrush figures into things like sizing caps to prevent destroying tube rectifiers; normal half-cycle charging under load figures into ripple. Under a steady-state condition, the caps do not consume any power, so do not 'eat' any current.

23. ^ seems almost ok (still I don't know the max peak to peak input levels)

But I am still confused. The signal after the CF is going to feed opamp EQ, +-15v supply.
The signal needs to be attenuated. I attached all tube circuit (matchless hotbox) with CF. The attenuation is achieved with 100k resistor.
- How to calc the attenuation resistor?
- How does that resistor affect the ouptut impedance??

Next, the clipping diodes (2 x 10v) set the max output swing (peak to peak??). I guess, for a +-15v circuit is better to set 2 x ~6v diodes.
The output clipping diodes must be after the attenuator resistor ^.
All possible signal clipping is not going to be tube clipping, but zenner clipping.. .

24. That 100k 'distortion output' control looks funny. Do you have another drawing, or an actual unit, to compare?
The attenuation will be = (wiper resistance)/(wiper resistance + 100k). The part of the volume knob that goes to the output might actually be connected (or jumpered to) the wiper.
If you want to sub a mosfet for the CF with the 56k on the source, just 'plug it in' following Merlin's guidance on gate resistor and zener Vgs protection (and use an 'enhancement mode' device!). It really is that simple. A tube CF or Mosfet SF work almost identically in terms of gain and impedance buffering.

edit: I see your other comments. The input impedance for an OP amp should be high enough that your interstage load doesn't matter for any tube type circuit. A couple dozen kOhms to a couple hundred kOhms won't matter. You are feeding the interstage divider into an OP amp, right? select your diodes to clip before you hit the +/-15vdc power rails. However, (train of thought thinking here) depending on the feedback in an active Op amp-based EQ, you may still clip the rails even with a lowish input signal.

25. Originally Posted by eschertron
edit: I see your other comments. The input impedance for an OP amp should be high enough that your interstage load doesn't matter for any tube type circuit. A couple dozen kOhms to a couple hundred kOhms won't matter. You are feeding the interstage divider into an OP amp, right? select your diodes to clip before you hit the +/-15vdc power rails. However, (train of thought thinking here) depending on the feedback in an active Op amp-based EQ, you may still clip the rails even with a lowish input signal.
Leave the matchless hotbox ..

Does the above mean, I can feed the opamp EQ without CF whatsoever? No loss of clarity (especially the highs)?
Anyway, I think I know what transformer I need, with or without mosfet CF's.
CF should be solved... I hope

What I am digging now is another story (full train of thoughts).

The EQ is suggested to have input voltage of ... 0.775 volts (I guess RMS).
If I calc 0,775*2*SQRT(2) - we have 2,19v peak to peak. 2,19v * 15db band boost could result in ... ~12v peak to peak. Near the power rails.
There might be something more, I have to analyze the circuit. I am not much better with opamps, than with tubes

My idea is to put the eq in between two triode stages. I just imagine, a heavily boosted EQ band would yiled plenty of nice harmonics and clip softly, if necessary. The whole amp is going to be hybrid stuff, solid state power amp.
What do you think?

26. Look around at some designs. One of the designer's first decisions is whether they want a tube in a SS amp, or some SS components in a tube amp.

Just thinking out loud here, but if I were to want a SS EQ in a 'tube' amp, I'd choose between on the one hand putting the SS after all the tube drive components (including a SS power stage) and on the other hand use the SS for getting a clean and clear signal through the (SS) preamp and EQ to feed the tube distortion stages. I see two advantages: 1) I've chosen a topology that lends itself well to either a high-gain 'sculpted' sound, or a classic clean/light crunch. 2) Minimize transitions from SS to tube and you've minimized the headache or trying to match level and impedance without adding tons of extraneous components.

If you want versatility, make everything rackmount, matching standard line level and impedance, and patch it together as desired for the occasion.

edit: .775vrms is specified to be 0dBu

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