samy555
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I know that the design fig.2 safes battery more than fig1, and that have greater input impedance?
Are there differences that are more important?
Thank you
My question is what is the difference between the two designs, one of them has (1mA, 1.5V) operating point and the other has (0.1mA, 1.5V) operating point?
On any base or on what basis is operating current selected at a certain value?
It may be 100mA, 10mA, 1mA or even 0.1mA
I understand that the value of VCE must be equal to half the value of the battery in order to get the symmetric swing of a resulting signal?
This was the best answer I've ever so far, so thank youYou generally use as small a collector current as will give satisfactory operation in you circuit. But small currents mean higher collector load resistors and the resulting high output impedance has to drive the load impedance (including any stray capacitances) at the highest signal frequency of interest. That is usually the limiting factor for how low in operating current you can go.
Another factor may be circuit noise. Higher value resistors have higher thermal (voltage) noise and you want to keep that below any signal noise.
Good pointBandwidth will be affected by the current in the transistor.
Keith
I have read all what is stated in the PDF fileSearch for ' bandwidth of common emitter amplifier'. Something like https://www.learnabout-electronics.org/Amplifiers/amplifiers14.php might be helpful.
There are plenty of explanations of common emitter amplifiers on the internet.
While you original question is valid, it makes some assumptions about the design such as the operating point, the basing method and lack of emitter resistor. Designing an amplifier will normally be based on requirements such as bandwidth, current, gain, voltage swing, gain stability, input and output impedance etc. From those requirements the appropriate design is chosen and the operating point and number of gain stages selected.
Keith
What you said is true in the case of a particular design and we change ICQSamy555,
I am a bit surprised that - up to now - the influence of the collector current Ic on the gain of the stage was not mentioned.
The transconductance gm that gives - together with the collector resistor - the gain of the stage (without emitter feedback) is proportional to Ic.
We have: gm=Ic/26mV
and the gain G=-gm*Rc.
In case of a emitter degeneration (Re feedback) - as proposed by Keith the gain expression is: G=-Gm*Rc/(1+gm*Re)
...... so the second is better.
I think that's because both circuits have almost identical gain, Rc is scaled with Ic respectively (inverse proportional).I am a bit surprised that - up to now - the influence of the collector current Ic on the gain of the stage was not mentioned.
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