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schematic

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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 \$.

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  • \$\begingroup\$ OK. I have edited the question. Why does Re need to be zero for maximum gain? \$\endgroup\$ – Mohammed Arshaan Aug 23 '18 at 17:24
  • \$\begingroup\$ Actually, now that I re-read the edited question it seems that I misread it. I failed to notice "common-base." My mistake. I'm removing my comment. \$\endgroup\$ – jonk Aug 23 '18 at 17:29
  • \$\begingroup\$ @MohammedArshaan What is the minimum collector voltage you can get at the collector before you enter the saturation region? \$\endgroup\$ – G36 Aug 23 '18 at 17:52
  • \$\begingroup\$ Vce(sat) = 0.2 V, so Vc = -0.5 V. \$\endgroup\$ – Mohammed Arshaan Aug 23 '18 at 17:53
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    \$\begingroup\$ For example, if you pick Vc = 5V your Vc voltage can swing from 5V to 7V and swing back from 5V to 3V. So, as you can see we have swing +/-2V electronics.stackexchange.com/questions/301617/… And your task is to find the Vc that will give you the max Rx and +/-2V swing. \$\endgroup\$ – G36 Aug 23 '18 at 18:41
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So far, I have calculated Rb = 0 and Re = 4.3k ohms. But I am unable to proceed further, for Rc

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.

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  • \$\begingroup\$ By using Vce(sat) to find out the minimum Vc, I get Vc = -0.5 V. What is the reason for not using this method to find Vc? \$\endgroup\$ – Mohammed Arshaan Aug 23 '18 at 20:41
  • \$\begingroup\$ @MohammedArshaan nothing wrong with that. I’m thinking I made an error. Rewrite coming up. \$\endgroup\$ – Andy aka Aug 23 '18 at 21:51
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The task is:

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 you need to find the largest \$R_C\$ as possible.

The minimum collector voltage is \$V_{Cmin} = V_E+0.2V = -0.5V\$

Hence the \$R_{Cmax} = \frac{10V - (-0.5V + 2V)}{1mA} = 8.5k\Omega\$

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