Question about Crack OTL and voltage gain vs current gain

[12AU7-6SN7]

I have a question; I noticed someone say that "a tube amp" (in this case the Crack OTL) is usually better thought of as a current amplifier than a voltage amplifier. Is this so? Are they generalizing tubes incorrectly, especially as we are talking about a high output impedance OTL amp for high impedance cans and not an 8ohm speaker amp, OTC or OTL.

This seems counter-intuitive to me, since the Crack at full bore with typical 300 ohm load pushes around 10v and ~30mA from a much lower voltage source, myself usually having it output less than 1mw at much less than 5mA for safe 80db-83db peak listening levels.

An old school veteran Air Force tube radio technician told me that in audio it's typically a voltage gain situation But I think he said both voltage and current gain are happening. I did some of my own research that lead to some basic circuit designs with tube voltage gain and tube current gain stages explained separately then combined for an OTC power amp, further confusing the issue for me. Another technician online had typed "I was a technician, not an engineer, but I do know that a tube is a voltage amplifier and a transistor is a current amplifier." which I found via google search.

Is the crack, as an OTL headphone amp mainly utilizing voltage gain? or both current and voltage stages?

Can you folks clear this up? I can solder and build, but have no idea what is happening electrically beyond the basics for every type of tube amp, so anything helps.

 

[Paul Birkeland]

I have a question; I noticed someone say that "a tube amp" (in this case the Crack OTL) is usually better thought of as a current amplifier than a voltage amplifier. Is this so?

Which half of the Crack are you referring to? (12AU7 front half or 6080 output half?)  Speedball or loading resistors? (This changes the answer to the question a bit, lol)

Are they generalizing tubes incorrectly

When you lump all tube amps into one generalization, I would say so.

An old school veteran Air Force tube radio technician told me that in audio it's typically a voltage gain situation But I think he said both voltage and current gain are happening.

This will largely depend on the circuit and tubes used.  Some tubes (the 12AX7 for instance) are good at voltage amplification but not so great with current amplification.  Other tubes (the 6080 for instance) are useless as voltage amplifiers but marvelous current amplifiers.  A tube like the 6C45PI does both really well.  On the circuit side, I've seen and listened to 6080's and 12AX7's setup as both current and voltage amplifiers. 

I did some of my own research that lead to some basic circuit designs with tube voltage gain and tube current gain stages explained separately then combined for an OTC power amp, further confusing the issue for me.

Yeah, I would leave the transformer out for now, especially since the discussion is mostly around the Crack.

Is the crack, as an OTL headphone amp mainly utilizing voltage gain? or both current and voltage stages?

One of each.  The 12AU7 stage is far easier to understand, as it's just a common grounded cathode amplifier.  From a DC perspective, the 12AU7 grid is at ground, the cathode voltage is set (or limited if you will) by the 1.57V LED, and the plate is loaded with a 22.1K resistor.  The combination of the 1.5V LED, the 22.1K resistor, the B+ available, and the characteristics of the 12AU7 establish what DC voltage appears at the plate.

On the AC side, signal is applied to the grid of the 12AU7 and modulates the quiescent current flowing through the stage.  If you examine the plate curves for a 12AU7 and put a dot at 165V (our B+) on the voltage axis and a dot on 7.5mA on the current axis (7.5ma is the 165V available divided by 22.1K plate load from Ohm's Law), then you will see that when you move along that line between the 1.5V of bias to the left 1.5V (to the 0V line, though in reality we won't quite get there) and to the right 1.5V (to the imaginary line of 3V of bias), then the plate voltage changes.  The first image posted shows this. For 3V of peak-to-peak voltage at the input, we get about 38V peak-to-peak at the output (so a voltage amplification factor of about 12.5).  You'll notice that the current swings by +/- a ma or two as well.  When you install the Speedball, this load line becomes nearly horizontal, swinging very little current but more voltage (available voltage swing will still be limited by the 1.5V of bias from our LED). So if we put 2V AC into a stock Crack, we expect to see about 25V coming out of the 12AU7 gain stage, but without the current available to really be that useful.


The 6080 is setup as a cathode follower, which has a fixed plate voltage (the plate is tied to the filtered high voltage supply directly) and a load under the cathode.  Back when we looked at the 12AU7, when we applied signal to the grid, we were moving around between the 0V and 3V curve on the charts since the cathode voltage wasn't changing (again, thinking of AC signals, not DC bias).  With the 3K cathode resistor and no bypass capacitor, the cathode's AC voltage isn't tied down.  When the incoming signal swings positive, more current flows through the cathode resistor and when the incoming signal swings negative, less current flows through the cathode resistor.  The reason that this type of stage is called a "Cathode Follower" is because the cathode voltage follows the grid voltage. 

In terms of the charts, we can draw a load line for the 6080 and see everything line up nicely in terms of where the DC voltages will settle when we run the amplifier, but what about the AC voltages?  Since the cathode is following the grid, the AC voltage available at the cathode can only be as much as the AC voltage at the grid (though in reality it is a bit less), so ultimately our AC load line ends up depending a whole lot on the loading impedance, and the cathode follower itself can be treated as a current amplifier.

I've intentionally completely glossed over how feedback/output impedance play into this since you didn't ask and it's a long conversation.

 

[12AU7-6SN7]

Thank you for the very detailed answer. Good to learn something new!

I meant the Crack amp as a whole (with Speedball or without, as I run mine.) I didn't know if the output stage was a cathode follower or not (which the tube basics site I visited pointed out are a current gain stage)

You've pretty much covered everything I was interested in knowing. Wish I could send a beer via the internet for ya!

 

[attmci]

Because you are 12AU7-6SN7. You may want to read this too. LOL

http://www.turneraudio.com.au/basic-tube-1.html

 

[Paul Birkeland]

Is there a difference between a constant current and a quiescent current? The reason for my question is this: If the 6080 acts as a current amplifier, how is it possible for it to have a constant current?

You're asking the right questions.  Quiescent current is what the tube draws when it's sitting there minding its own business and not doing anything.  This is not a limit though!

To say that about half of the B+ is applied to the grid of the 6080 isn't the same as saying that the grid bias is about half of the B+.

Correct.  A 6080 with -75V of bias would have a hard time drawing 30mA!

The reason for this is that the grid voltages shown on datasheets are actually the difference between the cathode voltage and the grid voltage.  Is my understanding correct?

Yes.

3. What would the load line on the 6080 look like for a stock Crack with Speedball and 12AU7s & 6080s at nominal values? Could you please draw it?

Well, if you have nothing plugged into the headphone jack (and we will just ignore the shorting function), the loadline is a horizontal line at 30mA.  As you load the amp, that load line rotates clockwise depending on the loading impedance. 

 

[Paul Joppa]

Short answer - the load line goes through the quiescent point, and if the load is resistive its slope is the load resistance.

If the load impedance has a reactive component, then the load line is an oval, centered on the quiescent point.

If the varies with frequency, then the load line is different at each frequency.

And yes, of course the distortion depends on the path of the load line. It gets complicated.

 

[Paul Birkeland]

Where would the loadline intersect the x & y axes for a 250 Ohm load? What about a 600 Ohm load?

You can put a dot at 75V/30mA on the 6080 plate curves, then draw lines with those slopes to see.  These load lines are going to hit the X axis long, long before they hit the Y axis.

 

[cddc]

Which half of the Crack are you referring to? (12AU7 front half or 6080 output half?)  Speedball or loading resistors? (This changes the answer to the question a bit, lol)

When you lump all tube amps into one generalization, I would say so.

This will largely depend on the circuit and tubes used.  Some tubes (the 12AX7 for instance) are good at voltage amplification but not so great with current amplification.  Other tubes (the 6080 for instance) are useless as voltage amplifiers but marvelous current amplifiers.  A tube like the 6C45PI does both really well.  On the circuit side, I've seen and listened to 6080's and 12AX7's setup as both current and voltage amplifiers. 
Yeah, I would leave the transformer out for now, especially since the discussion is mostly around the Crack.
One of each.  The 12AU7 stage is far easier to understand, as it's just a common grounded cathode amplifier.  From a DC perspective, the 12AU7 grid is at ground, the cathode voltage is set (or limited if you will) by the 1.57V LED, and the plate is loaded with a 22.1K resistor.  The combination of the 1.5V LED, the 22.1K resistor, the B+ available, and the characteristics of the 12AU7 establish what DC voltage appears at the plate.

On the AC side, signal is applied to the grid of the 12AU7 and modulates the quiescent current flowing through the stage.  If you examine the plate curves for a 12AU7 and put a dot at 165V (our B+) on the voltage axis and a dot on 7.5mA on the current axis (7.5ma is the 165V available divided by 22.1K plate load from Ohm's Law), then you will see that when you move along that line between the 1.5V of bias to the left 1.5V (to the 0V line, though in reality we won't quite get there) and to the right 1.5V (to the imaginary line of 3V of bias), then the plate voltage changes.  The first image posted shows this. For 3V of peak-to-peak voltage at the input, we get about 38V peak-to-peak at the output (so a voltage amplification factor of about 12.5).  You'll notice that the current swings by +/- a ma or two as well.  When you install the Speedball, this load line becomes nearly horizontal, swinging very little current but more voltage (available voltage swing will still be limited by the 1.5V of bias from our LED). So if we put 2V AC into a stock Crack, we expect to see about 25V coming out of the 12AU7 gain stage, but without the current available to really be that useful.


The 6080 is setup as a cathode follower, which has a fixed plate voltage (the plate is tied to the filtered high voltage supply directly) and a load under the cathode.  Back when we looked at the 12AU7, when we applied signal to the grid, we were moving around between the 0V and 3V curve on the charts since the cathode voltage wasn't changing (again, thinking of AC signals, not DC bias).  With the 3K cathode resistor and no bypass capacitor, the cathode's AC voltage isn't tied down.  When the incoming signal swings positive, more current flows through the cathode resistor and when the incoming signal swings negative, less current flows through the cathode resistor.  The reason that this type of stage is called a "Cathode Follower" is because the cathode voltage follows the grid voltage. 

In terms of the charts, we can draw a load line for the 6080 and see everything line up nicely in terms of where the DC voltages will settle when we run the amplifier, but what about the AC voltages?  Since the cathode is following the grid, the AC voltage available at the cathode can only be as much as the AC voltage at the grid (though in reality it is a bit less), so ultimately our AC load line ends up depending a whole lot on the loading impedance, and the cathode follower itself can be treated as a current amplifier.

I've intentionally completely glossed over how feedback/output impedance play into this since you didn't ask and it's a long conversation.

Very nice chart on 12AU7!

Would you also draw a chart on the 6080 in Crack, PB? Thanks!

 

[Paul Birkeland]

An AC load line for the 6080 will change a lot depending on the headphone impedance.

 

[cddc]

An AC load line for the 6080 will change a lot depending on the headphone impedance.

Suppose we're using Crack's best mate - Sennheiser HD650/600/6XX headphones of 300 Ohms.

 

[Paul Birkeland]

With or without a Speedball?

 

[cddc]

With or without a Speedball?

My Crack has Speedball, so let's say with Speedball 

 

[Paul Birkeland]

Here's a resistive loadline for 300 ohms.

 

[cddc]

Looks great, thanks very much for drawing the chat, PB!