Better Low Z driver stage design based on our Bootstrapped Gain Stage thread

[SoulFetish]

I made a couple of changes to the driver stage in my amp based on some of the findings we discussed the "Bootstrapped Gain Stage Theory" thread posted here a while back:
https://music-electronics-forum.com/...n-stage-theory

Here is how the stage was previously designed incorporating cathode biased followers. The goal was to use the drivers as a transparent buffer for a DC coupled output stage.



However, one of the more revelatory things to me was the effect bootstrapping has not only on the grid impedance, but also the cathode impedance as well. Therefore, the output impedance of a cathode biased cathode follower (as was the case in my earlier design) would be dramatically increased by the bootstrapping effect of this configuration. Interestingly, I've never seen this mentioned in any of the literature I've read on cathode follower operation or design before. This seems like an important thing to mention, particularly because cathode followers are often used as low impedance buffers.
Did we ever come up with a formula to determine the output impedance of a cathode biased follower? I couldn't find it in the other thread.

In any case, here is the modified circuit incorporating a fixed bias adjustment to the driver stage.


I think this improves it's function as a transparent buffer in a couple of ways. The current biasing more changes the input impedance, dropping it to around 213k (set by the 422k in parallel with 433k). While this might seem to be a drawback from the previous 5M input impedance, 213k more closely approximates the usual 220k (or so) loading at the output of the LTP in a normal AC coupled stage. This also drops the output impedance around 322Ω (≈1/gm) making it a much better buffer.

[Helmholtz]

I understand that you want to avoid blocking distortion by DC coupling. But why do you need a very low driver output impedance? Do you really want to "over-saturate" the ouput tubes by lots of grid current in class AB2 operation? Remember that pentodes like the EL84 have lots of gain and easily saturate in class AB1 without any grid current.

[nickb]

As shown, the output impedance is about 850 ohms, ignoring the effect of source impedance. If you take the output from the cathode, the impedance goes down to 370 ohms as you no longer have the 500 ohm resistor in series with the output. The source impedance does have an effect so for Z src = 50K the cathode output impedance goes from 370 to 400 ohms. All this seems insignificant compared to the grid stopper of likely 1.5K or so.

In addition to what was said above is that a lower drive impedance will improve the HF response. That might not be desirable in a guitar amp.

[Helmholtz]

The source impedance does have an effect so for Z src = 50K the cathode output impedance goes from 370 to 400 ohms.

Do I interprete you correctly that the bootstrapping causes only a very small increase in output impedance probably because the source impedance is relatively small compared to the bootstrapped resistor?
I seem to remember from earlier discussion that high source impedance increases the cathode output impedance with bootstrapping because it increases the positive feedback signal at the grid.

[g1]

nickb that attached pic in post #3 is 'invalid file specified'. You will need to re-size it or zip it.

[nickb]

nickb that attached pic in post #3 is 'invalid file specified'. You will need to re-size it or zip it.

Well that was weird. It would show for me sometimes. A bugette.

[nickb]

Do I interprete you correctly that the bootstrapping causes only a very small increase in output impedance probably because the source impedance is relatively small compared to the bootstrapped resistor?
I seem to remember from earlier discussion that high source impedance increases the cathode output impedance with bootstrapping because it increases the positive feedback signal at the grid.

Change in feedback is fairly small for Zsrc= 50k because of the 470K grid leak resistor. For infinity (quite useless) it's something like 3k for this cct with a 12AU7.

Another thought. The bias voltage is very uncertain as you rely on the tube to set it. It would be more predictable if you were to feed a reference to the grid. That would lower the input impedance and couple power supply noise. This first is not an issue an the second just requires attention to filtering.

[nickb]

Some equations to answer the OP.

Gain = Vcathode / Vgrid = Ra.(R1+R2).Gm/(Ra + R1 + R2 + Ra.(R1 + R2).Gm) ; Ra = plate resistance, Gm = transconductance;

Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for zero source impedance

Output Z = 1/(1/(R1+r2) + Gm.k + 1/Ra) for non zero Zsrc where k = R3/(R3+Zrc). Assumes R1<<R2 and R2<<R3




I don't doubt HH will check my math soon enough

[Helmholtz]

Some equations to answer the OP.

Gain = Vgrid / Vcathode = Ra.(R1+R2).Gm/(Ra + R1 + R2 + Ra.(R1 + R2).Gm) ; Ra = plate resistance, Gm = transconductance;

Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for zero source impedance

Output Z = 1/(1/(R1+r2) + Gm + 1/Ra) for non zero Zsrc where k = R3/(R3+Zrc). Assumes R1<<R2 and R2<<R3

I don't doubt HH will check my math soon enough

Right you are: I am confused.

Gain should be Vcathode/Vgrid. And of course Ra is zero for the cathode follower, so 1/Ra means trouble (singularity).
The second Output Z formula is identical to the one above for zero source impedance and doesn't contain the parameter k.

[nickb]

Right you are: I am confused.

Gain should be Vcathode/Vgrid. And of course Ra is zero for the cathode follower, so 1/Ra means trouble (singularity).
The second Output Z formula is identical to the one above for zero source impedance and doesn't contain the parameter k.

I knew you'd check!

Vcathode /Vgrid is just a typo, fixed (edited).

No, Ra is not zero, it's the internal anode (plate )resistance of the tube, about 5.7k for a 12AU7 as shown.

Yes, the second formua is almost the same, it should be. I did it like that to illustrate the effect of the feedback factor, k.

[R.G.]

As an experienced confuser, I have to add...

You can use power MOSFETs instead of cathode followers and get an even larger decrease in output impedance. There's an illustration of this in the MOSFET Follies on geofex.com.

[Helmholtz]

No, Ra is not zero, it's the internal anode (plate )resistance of the tube, about 5.7k for a 12AU7 as shown.

Sorry, I thought you meant the external plate resistor. I'm used to see ra for internal and Ra for external plate resistance.

Yes, the second formula is almost the same, it should be. I did it like that to illustrate the effect of the feedback factor, k

But the 2 formulae are exactly the same. No k contained in either. But the Zsrc should have some small effect, right ?

[nickb]

Sorry, I thought you meant the external plate resistor. I'm used to see ra for internal and Ra for external plate resistance.



But the 2 formulae are exactly the same. No k contained in either. But the Zsrc should have some small effect, right ?

Ah, another typo.I left out the "k". Fixed. Thx for checking

[Helmholtz]

FWIW, source or output impedance can be determined from the signal reduction caused by adding a load resistor.
Or more easily by using an LCR meter that provides AC resistance measurement (also works for input resistance). To block DC wire a cap in series with the meter.

[nickb]

I'd love to have a decent LCR meter or rather impedance analyzer. The cheap ones have about 1% accuracy which means the real part is wildly ( i.e several hundred percent) off when the phase is close to +/-90 degrees.