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Trans Linear Principle

In this blog, I want to introduce an interesting but very handy topic, known as "Trans Linear Principle (TLP)". After giving an overview of this, slowly we shall take up some problems, and in the process of solving that problem, we shall try to understand how TLP actually works.

Trans Linear Principle (TLP):

“In a closed loop containing an even number of ideal junctions, arranged so that there are an equal number of clockwise-facing and counter-clockwise-facing polarities, with no further voltage generators inside the loop, the product of the current densities in the clockwise direction is equal to the product of the current densities in the counter-clockwise direction.”

- Barrie Gilbert

Now, before applying any fancy principle, we should be aware of the conditions and assumptions behind that principle, otherwise there is a possibility of blunder.

So there are some conditions to apply this principle:

There must be an even number of base-emitter junctions in the loop.
Half of the junctions must be oriented in one direction and half in the other.
There should not be further voltage generators inside the loop.

Also there are some assumptions before using the results:

All emitter areas are equal.
All transistors are operating in forward active with large beta.
All transistors have infinite output resistance i.e. ignoring early effect

Proof:

By KVL,

$V_{BE1}+V_{BE2}=V_{BE3}+V_{BE4}$

Suppose the transistors follow the ideal equation

$V_{BE}=V_{T}ln\left ( \frac{I_C}{I_S} \right )$

Substituting,

$V_{T}ln\left ( \frac{I_1}{I_S} \right )+V_{T}ln\left ( \frac{I_2}{I_S} \right )=V_{T}ln\left ( \frac{I_C}{I_3} \right )+V_{T}ln\left ( \frac{I_4}{I_S} \right )$

Therefore,

$I_1 I_2=I_3 I_4$

Mathematically, if a loop of base-emitter voltages satisfies,

$\sum_{CW}^{}V_{BE}=\sum_{CCW}^{}V_{BE}$

Then according to TLP, it is also true that,

$\prod_{CW}^{}J_{C}=\prod_{CCW}^{}J_{C}$

Bonus:

If you want to dig further down into this, if you wonder what will happen if there is finite beta or if the emitter areas are not equal or there is finite output resistance, then you can watch this famous video where the master himself explaining Trans Linear circuits! The video was recorded in February, 1991.

If you have understood this, try to solve Problem-1, Problem-2 by observation using TLP. You will understand, how easily you can solve big circuits without writing and solving so many equations.

Trans Linear Circuits-2

In previous blog, I have introduced an interesting but very handy topic, known as "Trans Linear Principle (TLP)". Now, we shall take up more problems, and in the process of solving those problems, we shall try to understand how TLP works.

Q) Find out the expression of $I_Z$ in terms of $I_X$ & $I_Y$

Assume all emitter areas are equal.
Assume all transistors are operating in forward active with large beta.
Assume all transistors have infinite output resistance i.e. ignore early effect

Solution:

Current $I_Z$ will split between Q5 and Q6. Let's assume $aI_Z$ will go through Q5 and the rest $(1-a)I_Z$ will go through Q6.

First, we shall try to solve this problem with the help of simple KVL and basic BJT equation.

By KVL in Loop Q2-Q1-Q5-Q7,

$V_{BE1}+V_{BE2}=V_{BE5}+V_{BE7}$

Suppose the transistors follow the ideal equation

$V_{BE}=V_{T}ln\left ( \frac{I_C}{I_S} \right )$

Substituting,

$V_{T}ln\left ( \frac{I_X}{I_S} \right )+V_{T}ln\left ( \frac{I_X}{I_S} \right )=V_{T}ln\left ( \frac{aI_Z}{I_S} \right )+V_{T}ln\left ( \frac{I_Z}{I_S} \right )$

Therefore,

$I_X^{2}=aI_Z^{2}$ -------- (1)

By KVL in Loop Q4-Q3-Q6-Q7,

$V_{BE4}+V_{BE3}=V_{BE6}+V_{BE7}$

Substituting,

$V_{T}ln\left ( \frac{I_Y}{I_S} \right )+V_{T}ln\left ( \frac{I_Y}{I_S} \right )=V_{T}ln\left ( \frac{(1-a)I_Z}{I_S} \right )+V_{T}ln\left ( \frac{I_Z}{I_S} \right )$

Therefore,

$I_Y^{2}=(1-a)I_Z^{2}$ -------- (2)

Combining (1) and (2),

$I_Z^{2}=I_X^{2}+I_Y^{2}$

$I_Z=\sqrt{I_X^{2}+I_Y^{2}}$

Intuition:

The question is how to solve this circuit by observation only?

So, there is an interesting and useful technique called "Trans Linear Principle (TLP)"

- Barrie Gilbert

Mathematically, if a loop of base-emitter voltages satisfies,

$\sum_{CW}^{}V_{BE}=\sum_{CCW}^{}V_{BE}$

Then according to TLP, it is also true that,

$\prod_{CW}^{}J_{C}=\prod_{CCW}^{}J_{C}$

Check out this blog to know in details about TLP.

Now, in the above diagram, we can easily find 2 closed loops (2-1-5-7 & 4-3-6-7) of ideal junctions.

But let's check whether the conditions are satisfied or not:

There are 4 base-emitter junctions in each loop which is an even number.
For loop 2-1-5-7, half of the junctions (1 & 2) are oriented in one direction and half (5 & 7) in the other. Similarly, for loop 4-3-6-7, half of the junctions (3 & 4) are oriented in one direction and half (6 & 7) in the other.
There are no further voltage generators inside the loops.

So, by TLP, we can easily tell that,

Loop 2-1-5-7 => $I_X^{2}=aI_Z^{2}$ -------- (1)

Loop 4-3-6-7 => $I_Y^{2}=(1-a)I_Z^{2}$ -------- (2)

Combining (1) and (2),

$I_Z^{2}=I_X^{2}+I_Y^{2}$

$I_Z=\sqrt{I_X^{2}+I_Y^{2}}$

In future, I will take up more examples of such circuits and we shall try to solve them only by observations. You can also try Problem-1 . For more content click here INDEX

Trans Linear Circuits-1

In previous blog, I have introduced an interesting but very handy topic, known as "Trans Linear Principle (TLP)". Now, we shall take up a problem, and in the process of solving that problem, we shall try to understand how TLP works.

Q) Find out the expression of $I_D$ in terms of $I_A$,$I_B$ & $I_C$

Assume all emitter areas are equal.
Assume all transistors are operating in forward active with large beta.
Assume all transistors have infinite output resistance i.e. ignore early effect

Solution:

First, we shall try to solve this problem with the help of simple KVL and basic BJT equation.

By KVL,

$V_{BE1}+V_{BE2}=V_{BE3}+V_{BE4}$

Suppose the transistors follow the ideal equation

$V_{BE}=V_{T}ln\left ( \frac{I_C}{I_S} \right )$

Substituting,

$V_{T}ln\left ( \frac{I_A}{I_S} \right )+V_{T}ln\left ( \frac{I_B}{I_S} \right )=V_{T}ln\left ( \frac{I_C}{I_S} \right )+V_{T}ln\left ( \frac{I_D}{I_S} \right )$

Therefore,

$I_C I_D=I_A I_B$

$I_D=\frac{I_AI_B}{I_C}$

Intuition:

The question is how to solve this circuit by observation only?

So, there is an interesting and useful technique called "Trans Linear Principle (TLP)"

- Barrie Gilbert

Mathematically, if a loop of base-emitter voltages satisfies,

$\sum_{CW}^{}V_{BE}=\sum_{CCW}^{}V_{BE}$

Then according to TLP, it is also true that,

$\prod_{CW}^{}J_{C}=\prod_{CCW}^{}J_{C}$

Check out this blog to know in details about TLP.

Now, in the above diagram, we can easily find a closed loop of ideal junctions.

But let's check whether the conditions are satisfied or not:

There are 4 base-emitter junctions in the loop which is an even number.
Half of the junctions (1 & 2) are oriented in one direction and half (3 & 4) in the other.
There are no further voltage generators inside the loop.

So, by TLP, we can easily tell that,

$I_C I_D=I_A I_B$

$I_D=\frac{I_AI_B}{I_C}$

In future, I will take up more examples of such circuits and we shall try to solve them only by observations. You can also try Problem-2 . For more content click here INDEX

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