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), 

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

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

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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)"

“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

 

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), 

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

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

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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