This is a common-emitter amplifier, which amplifies the input voltage about 10 times.
The capacitor and the 110k and 10k resistors bias the transistor's base at about 1.7 V, so that the average value of the input is moved up to that level. The base-emitter junction acts like a diode, so that the emitter will be a diode drop lower than the base. Since the transistor stays in forward-active mode, the collector current will be 100 times the base current.
The emitter voltage fluctuates with the input voltage, and so the current across the 1k resistor fluctuates proportionately. Since the collector resistor has the same current across it but has 10 times the resistance, the collector voltage swings 10 times as much (and with phase opposite to the input).
Note that the peak value of the output is not 5 V as this analysis would predict. The actual gain is more like 9.5 times for various reasons. For example, the base-emitter drop is not constant, but varies with the base current. -- Credits: Mr. Paul Falstad.
|Ahmad said: (Sep 13, 2011)|
|I need the circuit operation.|
|Praburaj said: (Feb 13, 2013)|
|Is this only for AC? Can it be used for DC?|
|K.Ravi said: (May 17, 2013)|
|How to phase shift in ce configuration explain?|
|Athar Zahoor said: (Apr 5, 2014)|
|Why is the phase shift of 180 in CE amplifier?|
|Taha said: (Apr 9, 2014)|
|Create a MATLAB GUI for designing Common Emitter Display of circuit diagram.|
|Kenez Morotz said: (Dec 11, 2014)|
|"Why is the phase shift"?
Lets imagine the increasing first quarter of a sine wave on the input.
Base current is then increased, which ends up in the increasing of the collector current.
The output voltage of the amplifier before the decoupling cap equals to Vcc-Ic*Rc.
(Vcc is the supply voltage of the collector;Ic is the collector current, Rc is the collector resistor).
So to summarise, Vcc, Rc is constant,
So if Ic is increasing, then the output will be decreasing.
(Ic*Rc will decrease the Vcc more than before.)
And to take into account the capacitor, only the AC part of the signal forms the output. But this is irrelevant from the inverting mechanism of the common emitter circuit.
|Kenez Morotz said: (Dec 12, 2014)|
|"Is this only for AC? Can it be used for DC?".
It is only for AC. DC amplifying is an interesting thing. Of course, people want to amplify DC voltage, but firstly it is not important, what can be realized, what restrictions have to be taken into account. With an operational amplifier, DC voltage can be amplified, but only up to the supply voltage voltage-some value. But "true" amplifiers exists, called DC-DC converters. But because there isn't exists ideal amplifiers, which consumes zero "energy", or "power", and produce some nonzero power, a higher (amplified) DC-voltage will reduce the maximum amount of current, the output can give. And that can be an important factor, when a low-resistance load has to be driven from the output.
|Mehwish Awan said: (Jun 10, 2015)|
|How phase shift of 180 produced?|
|Kishan Subudhi said: (Jul 20, 2016)|
|180* phase shift produced whenever an input AC is applied which is to be amplified ... it gets amplification at output and whenever that output amplified signal further used as input of another amplifier circuit it get amplified at that stage we have got a strong wth higher amplifying output.
Hence, in first amplification it get a 90* phase shift and when further it get amplified then another 90* phase shift occur. So finally we got 180* phase shift.
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