LTP input impedance and clipping

[Chuck H]

I'm vaguely recalling reading here and elsewhere that the input impedance of the typical LTP inverter has a relatively high input impedance, BUT that the input impedance tanks at the onset of clipping. I can't find any relative historic threads and I'd like to have a better understanding of this. It seems like it's a somewhat unaddressed subject that dramatically affects the choice if input cap value. I did some simulation with a spice type program and found that changing the input load from 3Megs (as I've read might be typical for the input of a LPT) to 470k dramatically affect the source LF to the LTP. On the order of about 5dB! In the end my ears will be the judge of what cap value to use but it would be helpful to know more about this so I can apply it to whatever sort of rough design I'm drawing up. Does the input impedance of the LTP tank on clipping? How low does it go? Is the NFB loop to be considered?

I'm formula impaired so a rough explanation will do.

TIA

[jazbo8]

"that changing the input load from 3Megs (as I've read might be typical for the input of a LPT) to 470k dramatically affect the source LF to the LTP."

1M is typical, have not seen many above that...

The input impedance of any tube would drop like a rock once you clip its input and the grid begins to conduct. You can use large grid stopper to limit the grid current but at the expense of HF rolloff and noise. The formula for determing the coupling capacitor is C = 1/6.28*2*Rg*F, where Rg is the grid input resistance and F is the -3dB corner frequency of interest. As usual, Merlin has a page on this.

[Alan0354]

OK, I am quite new with tubes, so I just join in just as a discussion, so this is not the answers, just how I think. To me tubes and transistor has more common than people think.

1) For small signal where the grid to cathode is still negative. The input tube of the long tail pair is nothing more than common cathode type except the resistance the cathode of the input tube sees is 1/gm of the second tube in parallel with the cathode tail resistance. Say gm of 12AX7 is 5000umho=0.005, so 1/gm =200ohm. So the cathode of the input tube is seeing less than 200 ohm. Therefore you can count on the input impedance is higher than a normal common cathode stage with cathode bypass cap. At least it should be way in Mohms region.

2) For clipping where gate to cathode is 0 or positive, the grid conducts. This is where tubes are different and I don't dare to say anything definite. My guess is the input should be just like regular common cathode stage when grid starting to conduct.

If you change the input grid bias resistance to 470K, your low frequency cut off should be governed by the 470K resistor when the gate cathode voltage is negative. The -3dB cutoff frequency should be f = 1/(2 X \pi X R X C). if the coupling cap is 0.022uF, f =1/(6.28 X 470K X 0.022uF)=15.4Hz. This can be confirmed by a typical Fender that uses 0.022uF coupling cap for Bass amps.

I don't know how to calculate when the tube have 0 or positive bias. But I think you will see the same charge up of the coupling cap and cause the LTP tube cut off just like other common cathode stage.

Sorry.

[jazbo8]

Alan,

I know you got the differential amplifier/pair operation down cold, so why are you looking at just 1/2 of the circuit? You also read RDH4 Chapter 2 and know about the grid/input impedance for both positive and negative bias. So you must be putting us on... good one!

[Alan0354]

Alan,

I know you got the differential amplifier/pair operation down cold, so why are you looking at just 1/2 of the circuit? You also read RDH4 Chapter 2 and know about the grid/input impedance for both positive and negative bias. So you must be putting us on... good one!

No, this is how differential pair is in BJT, They analyse using half circuit!!! the emitter sees the re of the other transistor. The input impedance of the input transistor is beta time twice of the re or 2/gm.

This is one article talk about analysis with half circuit:http://faculty.mu.edu.sa/public/uplo...6lecture11.pdf. Just look at the diagram of the first 2 pages and you'll see.

I have to read up the RDH4 again tonight. I am not that good with tubes!!!! Honestly. There are a lot of electronic theory that are totally device independent. The theory of amplification is like 80% or more common between tubes, BJT, JFET and MOSFET. You'll be surprised how much you can jump in knowing just one device in detail. Like you tube guys, you can learn transistor in no time. Just concentrate on the 20% difference and you'll get it. Circuit are pretty much the same...... In fact, tubes has more unique characteristics than BJT and FETs, it is easier for you tube guys to learn transistor than me learning tubes.

(1) Common cathode/emitter/source(typical stage like Fender first stage with cathode bypass),

(2) common grid/base/gate(use in RF application to get rid of Miller capacitance),

(3) common plate/collector/drain ( cathode follower configuration).

They are all the same characteristics. A differential pair can be looked at as a common cathode(input tube) couple with a common grid stage( the other tube). The differential pair with the power tube and OT is just a variation of a discrete opamp with negative feedback. You drive into the +ve input, the NFB goes back to the -ve input of the differential amp. It's just implemented in tubes!!!

Then the input capacitance, Miller capacitance, the output capacitance work exactly the same way in all devices. Look at the drain curve of FETs vs plate curve of Pentodes, they are almost the same. BJT just have much more horizontal collector curves but same different. I have to learn the plate curve of triodes that is totally different from Pentodes. You guys don't even have to learn any new curves!!!!

As I said before, I have been treating tubes like transistors on the first pass and learn the particulars of the tubes like the +ve grid input current. I am just interested in this thread and follow up with this.

[teemuk]

There is a "bootstrap" in typical LTP: Grid resistors do not shunt to ground, they hook up to low impedance voltage source that follows the input signal, that is the cathode circuit. This artificially increases the impedance of those grid resistors, the circuit may "see" them as several times higher impedance than what their generic resistive value is.

But what happens when the stage clips and the cathode signal no longer follows the input signal?

Note that I'm not talking about grid conduction. During LTP clipping the grid signal can still be "intact" while the signal at the other end of the bootstrap isn't.

To tell the truth, the impedance aspect of this is a phenomenon I have never studied on, but this is one theory explainin sudden change in impedance during clipping.

[Alan0354]

There is a "bootstrap" in typical LTP: Grid resistors do not shunt to ground, they hook up to low impedance voltage source that follows the input signal, that is the cathode circuit. This artificially increases the impedance of those grid resistors, the circuit may "see" them as several times higher impedance than what their generic resistive value is.

But what happens when the stage clips and the cathode signal no longer follows the input signal?

Note that I'm not talking about grid conduction. During LTP clipping the grid signal can still be "intact" while the signal at the other end of the bootstrap isn't.

To tell the truth, the impedance aspect of this is a phenomenon I have never studied on, but this is one theory explainin sudden change in impedance during clipping.

1) Can you explain Bootstrap?

2) What is "they hook up to low impedance voltage source that follows the input signal".

I don't understand all the terminologies here.

Thanks

[teemuk]

Signal at the cathode follows the grid, right? Look how the grid resistors in a typical LTP circuit do not connect to ground but actually almost to cathodes (there's usually a "smallish", 470-ohm - few kilo-ohm resistor in between in practice but you get the basic idea).

Now what's the difference in between current flow through a resistor connecting from a voltage source to ground, or through a resistor connecting from one voltage source (the grid) to practically identical voltage source (the cathode)? If R=U/I, and U (which is our signal voltage at grid) stays constant while current flow, I, decreases what happens to R in latter case?

That's "bootstrapping".

[Dave H]

Below clipping the input impedance of the LTP is also affected by negative feedback around the output stage. Without feedback the LTP input impedance could be 2M but with feedback it could be more than twice that.

[Alan0354]

Signal at the cathode follows the grid, right? Look how the grid resistors in a typical LTP circuit do not connect to ground but actually almost to cathodes (there's usually a "smallish", 470-ohm - few kilo-ohm resistor in between in practice but you get the basic idea).

Now what's the difference in between current flow through a resistor connecting from a voltage source to ground, or through a resistor connecting from one voltage source (the grid) to practically identical voltage source (the cathode)? If R=U/I, and U (which is our signal voltage at grid) stays constant while current flow, I, decreases what happens to R in latter case?

That's "bootstrapping".

Can you explain again using this drawing? I still don't get it. R1 and R2 are grid bias resistors. R3 is 470 ohm bias setting resistor that develop about 1.7V for 12AX7 to set up the plate current of V1 and V2.



1) What is the voltage source? Do you mean the output of the preamp driving into the grid of V1?

2) Identical voltage source ( the cathode): Are you talking about point A labeled in the drawing?

3) What is R? Are you referring to R1?


From analyzing the circuit, the reason they do this is to use This configuration is to use R4 to develope a lot of voltage across it to set up the current. With so much voltage, small changes at the grid of V1 does not affect the voltage across R4 and not affecting the current. This together using negative feedback to the bottom of the R4 will create an almost constant current for the differential pair. this is done before the days of constant current circuit was invented.

R3 is used to create the negative bias voltage for the grid through R1 and R2. R3 is the main resistor that set up the current. R4 is used to drop a lot of voltage ( 100V) to simulate constant current source.

But I still don't quite get the Bootstrap thing.

Thanks

[Alan0354]

Below clipping the input impedance of the LTP is also affected by negative feedback around the output stage. Without feedback the LTP input impedance will be between 1M and 2M but with feedback it could be more than twice that. (4M?)

Chuck did not say whether he has NFB, but yes, NFB increase the input impedance before clipping.

[Dave H]

Chuck did not say whether he has NFB, but yes, NFB increase the input impedance before clipping.

And if Chuck has negative feedback and drives the amp to clipping the input impedance of the LTP will suddenly drop to half because the feedback loop is ineffective.

[Alan0354]

And if Chuck has negative feedback and drives the amp to clipping the input impedance of the LTP will suddenly drop to half because the feedback loop is ineffective.

Yes. If the PI clipped, the loop opens and no more feedback.

[GainFreak]

LTP explained:

The Long-Tail Pair

Designing Long-Tail Pairs - The Load Line Approach