Wouldn't the current at output be lesser than collector current in a Common Emitter amplifier circuit

Thread Starter

Doubtician

Joined Mar 19, 2022
18
I am having trouble understanding the CE amplifier.When the output current is measured,wouldn't that be less than the collector current since current will divide at a junction(KCL application)?

Document 93_1_2.jpg
 

ericgibbs

Joined Jan 29, 2010
16,819
hi D,
The Collector Current refers to the actual current flowing into the Transistors Collector path to Emitter and 0V.
The current flowing in the Collector resistor will be the same as the Collector current if there is no output loading.
If the transistor drives an external as shown in your diagram, the Collector does not source the current into the external load, the current in the collector resistor is the sum of the actual Collector current and the external load current.

Do you follow, OK.?
E
 

Jerry-Hat-Trick

Joined Aug 31, 2022
209
I think the main point about this arrangement is that with a change in base voltage, the emitter "follows" that voltage so that current is drawn through Rc. So if Rc is bigger than Re the voltage output changes by Rc/Re times the base voltage change. Most of the current flows through the Rc, the collector to emitter and Re. With a resistor (load) across the output current will flow through Rc and that load but the transistor will still try to satisfy the above. Capacitive coulpling on the output will allow the static circuit to sit in its happy place with only dynamic changes being seen by the output load
 

Thread Starter

Doubtician

Joined Mar 19, 2022
18
hi D,
The Collector Current refers to the actual current flowing into the Transistors Collector path to Emitter and 0V.
The current flowing in the Collector resistor will be the same as the Collector current if there is no output loading.
If the transistor drives an external as shown in your diagram, the Collector does not source the current into the external load, the current in the collector resistor is the sum of the actual Collector current and the external load current.

Do you follow, OK.?
E
So what is the point of current amplification(hfe)?
 

ericgibbs

Joined Jan 29, 2010
16,819
hi,
Consider the effective impedance of the bypass capacitor in parallel with the emitter resistor, at different frequencies
Their combined impedance determines the Negative feedback of the amplifier.
E
 

LvW

Joined Jun 13, 2013
1,586
The circuit diagram shown below is your classic common emitter amplifier.
Without C2 the AC gain is -RL/RE independent of frequency.
RE introduces negative feedback. When the impedance of RE//C2 decreases, negative feedback decreases also.
Since the impedance of C2 is a function of frequency, the AC gain becomes a function of AC frequency.
The sentence "When the impedance of RE||C2 decreases....." could be misinterpreted having the given expression (see post#17) for the gain (-RL/RE) in mind.
This could lead to the impression that the gain would be larger and larger (when the effective feedback impedance RE||(1/wC2) goes down with rising frequencies.
This, of course, not the case.
Therefore, it must be added that the expression RL/RE is a rough approximation only.
The real gain expression (with negative feedback) is:
Av=-gm*RL/(1+gm*RE) which reduces to
Av=-RL/RE for RE>>1/gm , and to
Av=-gm*RL for RE=0 (with transconductance gm=d(Ic)/d(Vbe) )
 
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