Microelectronic Circuits - 1

REFERENCE BOOK

I decided to crack open my Microelectronic Circuits book (Sedra/Smith) to start running through it. The goal will be to keep at it from cover to cover or until I get bored or distracted and want to move on to the next thing.

Sedra AS, Smith KC. Microelectronic Circuits 5th Ed. Oxford University Press 2004.

ISBN 0-19-514252-7

The knowledge from this set of exercises is useful in anything for which we need a working knowledge of circuit theory like robotics, microprocessor chip design, electronics modules in automobiles/planes/spaceships, and any DIY gadgets you may want to engage with (check out this guy who has been tinkering with spectral imaging of wifi).

 

PROBLEM

To start things off, I jumped into Appendix B of Sedra/Smith for a quick refresher of 2-port linear networks. In particular, the application of 2-port networks on a small signal equivalent model of the bipolar junction transistor (BJT). Let's have a look:

<LEARNING OPPORTUNITY>

First thing to observe is how this pertains to a BJT:

r_x is the input resistance, seen by current driving to the Base -- very small
r_μ is the Base-to-Collector resistance -- very large
r_π is the Base-to-Emitter resistance -- significantly larger than r_x and significantly smaller than r_μ
g_m is the transconductance from Collector to Emitter, and it is in parallel with
r_O, the output resistance -- large, not significant smaller than r_μ


We can think of r_O as the resistance seen by any downstream node that reads the output of this transistor, thus we can model it as a 2-port network.

SOLUTION

We are to find the H parameters.

Here's how I solved this problem:

COMMENTS / NOTES

• The answer given for h₂₂ in Sedra/Smith has incorrect units. Current / Voltage should have units of inverse Ohms, or Siemens.
• I was able to find the other three H parameters easily, but h₂₁ made me struggle, ngl. In particular, I wasn't sure how to deal with a transconductance in parallel with a short circuit. 
• Remember that for a short circuit, all parallel resistors are eliminated. 
• Contrary to the popular saying, current does not always take the "path of least resistance". However, current does always take the path of no resistance.
• It turns out that you can't eliminate transconductances in parallel with short circuits. Instead, you just apply them to KCL.
• Observe how the short circuit affects the potentials in this circuit. 
• In particular, a transconductance is not a resistance and in this small signal equivalent, it is shown as connected between two nodes of 0 V. 
• That means that r_μ and r_π are in parallel, both with one node at v_π and the other at common.
• Therefore, we can think of it as there being a potential -v_π across resistor r_μ (sorry, looks like my highlight didn't show up in the image).
• Now KCL works perfectly by showing that the current flowing into the node, I₂, is balanced by the two currents flowing out, the transconductance and the potential drop across resistor r_μ (which is negative, so the current ends up flowing into the node from this IR as well).