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Jan 25 2015

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Shared Cathode Drain Feedback

T-Junction

A “T”-junction of THREE resistors.

Two cathode resistors R1, R2, of two triodes, with their respective currents routed to ground, but via a common, shared resistor R3.

Boring?

DC Conditions?

Things get interesting, actually, as there are also AC signals present here.
In-phase, or out-of-phase.

This is a story of a shared cathode drain feedback mechanism.
Positive or Negative.

You choose.

The same actually applies per analogy to a related story of a shared transistor emitter drain feedback mechanism.

All voltages in the following are referenced to ground:

U1 = I1 R1  +  I1 R3  +  I2 R3    {1}

U2 = I2 R2  +  I2 R3  +  I1 R3    {2}

R3 = ( U1  -  I1 R1 ) /  I     {3}

R3 = ( U2 – I2 R2 ) / I        {4}

Whereby {3} or {4} can be easily derived from {1} or {2}, respectively.

Whereby  I = I1 + I2,   or simply the joint current flowing through R3.

U2 = I2 R2 + ( I1 + I2) ( U1 – I1 R1 ) /  ( I1 + I2 )    {5}, by putting {3] into the {2}.

U2 = I2 R2 + U1 – I1 R1        {6}, simplified {5}

This is logical, as after some regrouping we see that:

U1 – I1 R1 = U2 – I2 R2 =  U3     {7}, whereby U3 represents the voltage on R3

Let us now extract R2 from {7}:

I2 R2 = ( U2 – U1 ) + I1 R1

R2 = ( U2 – U1 ) / I2  +  R1  ( I1 / I2 )

OK.

So now we have some interesting formulas to play with.

But in order to play, we need to fill the sandbox with sand.

So, for examples sake, let us assume that:

V1 is an input stage triode.   U1 is 2 Volts, a cathode bias that results in an I1 cathode current of 2 mA.

V2 is an output stage triode. U2 is 50 Volts, a cathode bias that results in an I2 cathode current of 100 mA.

If it were not for a joint R3 resistor, … two “totally separate” triode stages would result, each connected directly to ground, and then the calculation of R1 and R2 would be trivial.  With R3 = 0, the whole task is trivialized down to:

R1 = U1 /  I1 = 2 V / 2 mA = 1 k = 1000 R.

R2 = U2 / I2 = 50 V / 100 mA = 500 R.

These are the obvious border conditions for the case that there is no common R3 path for the joint currents I1 and I2 of the tubes, as there is no such thing as a common connection between them.  In other words, R3 does not exist, it is equal zero.

Now, let us assume, that since V1 has a small, trickling current of I1=2mA, which is a small fraction of the I2 = 100 mA current of V2, we can simply say that R1 is equal to zero and that we route ALL of the V1 cathode current I1 so as to directly joint the “mainstream” current I2=100mA.

In other words, let us assume that:   R1 = 0.   Having said that, Let us now find R2 and R3.

R2 = ( U2 – U1 ) / I2  +  R1  ( I1 / I2 )

R2 = ( 50 V – 2 V ) / 100 mA + 0 ( 2 mA / 100 mA ) = 480 R.

R1 is zero, remember? So the whole second section falls out.

R3 = ( U1 – I1 R1 ) / I        

R3 = ( 2 V – 2 mA 0 R ) / 52 mA = 38.462 R.

Interesting to observe, that:

U3 = I3 * R3 = 52 mA * 38.462 R = 2 V

Which is exactly the level of bias voltage that we need on the V1 tube.

So, now we basically have nailed down the “border” conditions for the values of each of the resistors:

R1 may accept values from the following range:
1000 R   — equivalent to a total separation of the triode currents (U3 =0).
0 R  – equivalent to a full integration of the triode currents (U3 = U1).

R2 may accept values from the following range:
500 R   — equivalent to a total separation of the triode currents  (U3 =0).
480 R  – equivalent to a full integration of the triode currents (U3 = U1) .

R3 may accept values from the following range:
0 R   — equivalent to a total separation of the triode currents (U3 =0).
38.462 R  — equivalent to a full integration of the triode currents (U3 = U1) .

Between these “border” cases, there is, for the given voltage and current conditions, that I accepted for the examples sake, a whole myriad of possible scenarios that can be adopted.

But why Is this significant and what is the possible benefit of such a scenario ?

To answer that question, it suffice that you look at U3 and think about the implication of the U3 upon each of the triodes, but specifically, on its implications for the triode V1.

Remember, that apart from the DC voltage, as represented by U3, there is also a “u3″ signal component, or AC voltage, which is “fed back” to the cathode of V1.

An AC signal voltage, stemming from the “output tube” which is fed back to the cathode of the input tube.

How about an example right “in the middle” ?

Suppose that out of that 2V of bias for V1, we would like to have 1V on the “shared” resistor, i.e. so that exactly one half of the cathode autobias voltage “sees” the AC signal from the output tube.

Therefore, R1 = 500 R  (which is 1/2  of the maximum possible value of 1000 R):

R2 = ( U2 – U1 ) / I2  +  R1  ( I1 / I2 )

R2 = ( 50 V – 2 V ) / 100 mA + 500 R ( 2 mA / 100 mA ) = 490 R.

R3 = ( U1 – I1 R1 ) / I        

R3 = ( 2 V – 2 mA 500 R ) / 52 mA = 19,23 R.

Let us make a cursory check of the value of the U3 voltage:

U3 = U1 – I1 R1

U3 = 2V – 2 mA 500R = 1 V,

or better still:

U3 = I R3

U3 = 19,23 R  52 mA = 1 V

If the AC signals on the cathodes of V1 and of V2 happen to be in “in-phase”, that would mean that for any attempted signal swing on the grid, hence cathode of V1, say of 100mV in value, about half of that signal swing will be countered, or bucked, by a counter signal swing  coming back from the output tubes cathode current, resulting in a counter voltage swing of 50 mV on the shared resistor R3, which is then conveyed as a counter reaction to the cathode of V1.

A very simple and interesting measure of establishing a required level of overall loop gain of the circuit, and / or providing it with a level of added linearity and stability.

Mind you, this value of 50% in terms of U3 to U1 does not directly or easily reflect upon the value of the overall closed loop gain of the circuit. The formulas are slightly more complicated. But in essence, now you have a general feel for the concept and enough plain and basic maths knowledge to get a feel of what you are playing with.

As funny as it seems, it turns out that the current of the input tube may be “beneficially” drained to ground by a resistor which is “shared” with some other tube, which is further down the signal path.

Mind you, if you happen to join, in such a manner, two tubes, the cathodes of which are in “anti-phase”, a positive feedback scenario shall result, providing for a decrease in stability, and most probably oscillations.

 

Cheers,

Ziggy.

 

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