- #1
ii = ie + αreie/R4rude man said:ii = ie + reie/R4.
finally giving
Ai = ic/ii = α/(1 + αre/R4) ~ α.
NascentOxygen said:ii = ie + αreie/R4
...
Negligible difference, but the algebra looks neater.
hisotaso said:I have had a lot of problems with this text so far, approximations, incorrect solutions...However, I was using this other text on google books as a reference:
http://books.google.com/books?id=bXsWq0sq-WEC&pg=PR3&lpg=PR3&dq=biased+common+base+circuit+analysis&source=bl&ots=An5Z7MIMDV&sig=-qRuW6p_znsC9qMFrja7bXWAgkU&hl=en&sa=X&ei=kfyjUYn0BejXiAKfnIAY&ved=0CFEQ6AEwBA#v=onepage&q=biased%20common%20base%20circuit%20analysis&f=true
If you scroll down to page 1-22 you see this author is using the same equation, I don't understand it :/. Thank you for your input.
rude man said:The reason you don't understand it is that it's wrong. Read my post #5 again.
hisotaso said:As far as confusing symbols, yes I am aware that the symbols are not the same, so let me be more specific. In the attachment that shows my actual problem statement, my small signal drawing, and the given solutions, how does one arrive at the divider term R4/(R4 + Ri)? Ri in that problem is the resistance from the source, not input resistance, which is labeled Rin.
hisotaso said:Thank you for your input. The use/lack of use of ro was a source of confusion for me, and I asked about that somewhere else. I was told that not only would it make a very messy equation, but in fact it had little effect on the results. (I wish the author had explicitly stated this in the text, but that is another issue)
As far as confusing symbols, yes I am aware that the symbols are not the same, so let me be more specific. In the attachment that shows my actual problem statement, my small signal drawing, and the given solutions, how does one arrive at the divider term R4/(R4 + Ri)? Ri in that problem is the resistance from the source, not input resistance, which is labeled Rin.
Once again thank you for taking the time to reply.
hisotaso said:I see now that my error was attempting to explicitly solve for ie and ii and calculate the ratio, rather than summing currents at the emitter. I often stare so hard at a problem that I get lost in it and get stuck looking at it one way, not seeing the trees for the forest as it were. You have been an immense help, and I thank you for having the patience and taking the time to reply. Also thank you to everyone else who has contributed.
Regards,
JC
hisotaso said:Thank you for your input. The use/lack of use of ro was a source of confusion for me, and I asked about that somewhere else. I was told that not only would it make a very messy equation, but in fact it had little effect on the results.
A common-base amplifier is a type of electronic circuit that amplifies the current of a signal while maintaining a constant input voltage. It is one of the three basic configurations for bipolar junction transistors (BJTs), along with common-emitter and common-collector amplifiers.
The current gain of a common-base amplifier is defined as the ratio of output current to input current, and is denoted by the symbol α (alpha). It is typically less than 1, meaning that the output current is smaller than the input current. The current gain is affected by the characteristics of the BJT and the external components in the amplifier circuit.
The current gain for a common-base amplifier can be calculated using the formula α = IC/IE, where IC is the collector current and IE is the emitter current. It can also be expressed as α = β/(1+β), where β is the current gain of the BJT in the common-emitter configuration.
The current gain of a common-base amplifier can be affected by various factors, including the resistances and capacitances in the circuit, the operating frequency, and the temperature. The characteristics of the BJT, such as the base width and doping levels, also play a role in determining the current gain.
The current gain of a common-base amplifier is typically lower than that of a common-emitter amplifier. This is because the common-base configuration provides less voltage gain, but offers a wider bandwidth and better high-frequency response. Common-base amplifiers are often used in high-frequency applications, while common-emitter amplifiers are more commonly used for low-frequency applications.