Tube Loads

[StarTrucker]

Finally reading the RCA receiving tube manual from the 50s and came across the statement about the benefit of tetrodes being that there is little relationship between plate voltage and plate current. I want to make sure I am understanding plate loads and the creation of signal correctly.

A tube amplifier stage has a set DC voltage on the plate, coming through a load resistor off the B plus line. With no current, both sides of the resistor should show the same voltage. When constant current flows through the tube, a constant voltage drop appears across the resistor because IR=V. By changing plate current, the resistive load of the plate resistor (say 100k) produces a voltage swing to match, which becomes our signal output for the stage.

Now take for instance two differential amplifier stages in a row. If the plate resistor of V1 produces a voltage signal, we can tap that through a capacitor and apply it to a second stage as an input signal. Doesn't the grid leak resistor of the second stage also present a load to the first stage in addition to Rpk? How do the two load resistors interact since they're both loads but not in parallel?

If we connect a capacitor from V1 plate to ground (parallel to a V2 grid leak) I believe that would result in zero voltage swing on V1 plate, but does it also load the tube and drag down B+ as the tube tries to supply more current into an AC short, or does it return us back to the case where there is only Rpk with no current flowing because there is no longer a voltage swing and therefore no plate current?

I've had volume controls change the signal coming off of the plate as you turn them past a certain point, is this dual load why?

 

[merlin]

The anode resis

Doesn't the grid leak resistor of the second stage also present a load to the first stage in addition to Rpk? How do the two load resistors interact since they're both loads but not in parallel?

There is a DC load line created by the main anode resistor, and an AC load line formed by the two effectively in parallel. You can think of it as the load line rotating around the bias point.

If we connect a capacitor from V1 plate to ground (parallel to a V2 grid leak) I believe that would result in zero voltage swing on V1 plate, but does it also load the tube and drag down B+

It loads the tube for AC but not DC, so it doesn't affect the B+. Basically the AC load line rotates so it becomes vertical; current swing but no voltage swing.

 

[StarTrucker]

It loads the tube for AC but not DC, so it doesn't affect the B+. Basically the AC load line rotates so it becomes vertical; current swing but no voltage swing.

So the ac load line is created by the V2 grid leak resistor. Input signal affects DC?

How do you load one but not the other? When a signal is applied to the grid the plate current swings as AC across the plate resistor and pulls B+ down.

Could you look at is as the plate resistor creates the AC signal out of DC supply, but only DC is loaded (through the cathode) until connection from the plate to signal ground (gird leak or capacitor) is made to load AC. If the current flowing through the tube affects DC, wouldn't a larger AC load mean more AC current and more DC current being drawn?

If you demand more AC by adding a cap to ground does it not load DC, it just distorts and clips because the tube's current output is fixed and the only way to supply more current is by giving it more input (loading DC line), not by increasing the load?

 

[merlin]

When a signal is applied to the grid the plate current swings as AC across the plate resistor and pulls B+ down.

It pulls the anode voltage down, it doesn't pull the B+ down.
The average DC can't be affected by anything after the coupling cap. Caps block DC!

 

[StarTrucker]

You mean the anode voltage ie signal AC?

 

[merlin]

The voltage at the tube anode is AC riding on top of some DC (unless it is so heavily loaded with a cap to ground that it cannot swing the AC part)

 

[CJ]

Here are a few experiments,

Post in thread 'Meta-Electronics 101-Part 2: A C Theory' https://groupdiy.com/threads/meta-electronics-101-part-2-a-c-theory.8506/post-585427

 

[StarTrucker]

Awesome, thank you for your time on this guys

 

[CJ]

I went to school with Steve Jobs and Steve Wozniac, they took the same electronic classes, we had the same teacher, Mr. McCollum, so they ran those same experiments listed above, so that means if you do those experiments, you will become the CEO of a multi million electronics company.

 

[StarTrucker]

The future is bright.

My compressor is on fire but my software company is having a great quarter.

 

[Mr. Moscode]

You'll need a CFO who went to Trump University for ethics.

Back to the load line: The information is correct if the grid leak Rs go to ground. But if the 2nd stage has bootstrapped Grid leakers then the grid leak value can become inconsequential.

I attached a sketch using common R values. 100K plate R, 1meg grid Rs. In small signal tubes, you could go higher than 1 meg grid R. I've used 2 meg grid Rs with no problem with 6DJ8s, 12AX7s, AU7s.
The larger the grid R allows for smaller value interstage coupling cap values.

The 1st stage of the sketch has no bootstrap and could be used as it is referred to above as the 2nd stage, too. The 2nd stage is boot strapped which raises the input impedance. I don't know the formula for figuring that value out but I would be grateful if someone could post it.

That said:
You need to consider the value of the grid leak R. With 100k stage 1 plate R and a 1meg 2nd stage grid R the load imposed by the grid leakers is almost insignificant.

If the 2nd stage uses boot strapped grid R's then the impedance of the 2nd stage becomes very high. By bootstrapped I mean the there would be 2 cathode resistors in series with each cathode which then connect together at the bottom and the grid leakers come off the center tap. The voltage created at the the center tap raises the input impedance of the 2nd stage because the cathode is in phase with the grid, allowing even smaller interstage coupling caps, handy if size is an issue. It will alter the gain of the 2nd stage.

I post this in the interest of furthering the conversation. If I'm wrong point it out to all. The last time I was wrong was 1972.

 

[StarTrucker]

The last time I was wrong was 1972.

Don't tell his wife he said this

 

[StarTrucker]

Since tubes already have a high input impedance, why would you bootstrap them? I've seen negative feedback applied like this but I don't think I've seen it as a grid reference.

 

[Mr. Moscode]

Hi impedance is a relative term. Compared to (most) bipolar inputs for sure.
There are plenty of single ended tube and fet circuits that are bootstrapped especially in test equipment and microphones where it's hard or expensive to get above a gigohm input z for the capsule which is only a few pf and acts as the pickup and the input cap.

Also done in cap coupled cathode followers to minimize/isolate loading on the amplifier ahead closer to its ideal transfer curve. Sometimes used if the cathode follower has to work with different supply voltages or with a servo.

Like I said the higher the input z in the smaller the coupling cap in front.
Also the greater the isolation.

Also used with driven shields on scope input probes.