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Question from Sedra and Smith (7th Edition)
Question: The given bias arrangement is to be used for a common-base amplifier. Design the circuit to establish a dc emitter current of 1 mA and provide the highest possible voltage gain while allowing for a maximum signal swing at the collector of ±2 V. Use +10 V and -5 V for power supplies.
So far, I have calculated Rb = 0 and Re = 4.3k ohms. But I am unable to proceed further, for Rc. I tried the following approach,
$$ I_{E} = I_{C} = 1 mA $$
$$ V_{E} = – 0.7 V $$
$$ V_{C} = V_{CC} – I_{C}*R_{C} $$
$$ R_{C} = \frac{V_{CC} – V_{C}}{I_{C}} $$
$$ R_{C} = \frac{10\pm2}{1 \times 10^{-3}} $$
Answer: $$ R_{C} = 8.4k \Omega $$
What am I doing wrong here?
Edit:
The text before the question is as follows: Note that if the transistor is used with the base grounded (i.e. the common base configuration), then Rb can be eliminated altogether. On the other hand, if the input signal is to be coupled to the base, then Rb is needed.
Also, my textbook considers that \$ V_{BE} = 0.7 V \$, and as \$ V_{B} = 0 \$, hence \$ V_{E} = -0.7 V \$.
Best Answer
So far so good.
Next you need to consider the midpoint voltage on the collector that would allow maximum p-p swing for an output signal without clipping top or bottom. Clearly the output could swing as high as +Vcc so that's 50% done but how low can Vc swing?
The simple and straightforward answer (assuming that I didn’t make a stupid error, which of course I did lol) is that Vc can swing no lower than Vb - 0.7 volts else the CB region is forward biased hence the midpoint is half way between +10 volts and +4.3 volts. However, I'd give some margin at the low end to ensure that the base-collector region doesn't saturate (you lose beta when that happens) so I'd say the optimum midpoint of Vc is +7.5 volts.
That means that Rc drops 2.5 volts at 1 mA or conversely, has a resistance of 2.5 kohm.
This calculation assumes that the base resistance Rb is zero ohms (as is usually the case in common-base to prevent miller effects).
Edited section due to mistake in the above calculation.
I made a mistake above by assuming Vb is at 5 volts but of course it’s at 0 volts hence, the lowest collector voltage that can be comfortably reached is 0 volts (or maybe a tad less) but I’m comfortable with 0 volts. The implication is that the Vc midpoint is 5 volts and therefore Rc drops 5 volts and is a value of 5 kohm for 1 mA collector current.