Measuring the output from a Transistor Amplifier

[Pushoam]

Homework Statement

I understand that the DC current will not flow through the capacitor once the capacitor gets fully charged.
So, after a long time , only AC component of the net current flow through the capacitor.
How to show that all of the carying voltage appear across the capacitor?
Does capacitor work as conducting wire for an AC component of the net current?

Homework Equations

The Attempt at a Solution

 

[cnh1995]

Does capacitor work as conducting wire for an AC component of the net current?

Not for the net current, but for the output current, which is very small. It provides very low impedance to high frequency ac. Hence, almost all the ac voltage developed between collector and emitter will appear between terminals 1 and 2. Its job is to remove the dc bias in the output.

 

[CWatters]

Does capacitor work as conducting wire for an AC component of the net current?

Short answer is Yes.

Add a load resistor between terminals 1 and 2 and the output stage looks like a high pass filter...

https://en.wikipedia.org/wiki/High-pass_filter

A high pass filter blocks DC and lower frequencies but passes higher frequencies. If you want to pass low frequencies (eg for audio applications) then C and R must be quite large.

 

[rude man]

You should assume a resistive load R_L for the output, otherwise no current ever flows through C so C is never allowed to charge to the dc collector voltage and the dc output voltage = dc collector voltage - V_i where V_i is the (constant) capacitor voltage. Assume R_L >> R_C. (Any good voltmeter will meet this requirement.)

 

[Pushoam]

Thank you all for helping me.
Now what I understood is :
The impedance provided by a capacitor for AC is $\frac 1 {\omega C}$and for DC is infinite.
So, when we put a capacitor in a circuit, it is assumed that the impedance provided by the capacitor for AC( either by having high frequency signal or high capacitance or both) is so small that it can be approximated to wire for AC.
So, -$i_C R_C$ appears at terminal 1 wrt ground.

 

[rude man]

Thank you all for helping me.
So, when we put a capacitor in a circuit, it is assumed that the impedance provided by the capacitor for AC( either by having high frequency signal or high capacitance or both) is so small that it can be approximated to wire for AC.

That statement is too general. Capacitors often serve to establish desired time constants and filter characteristics. So for example in post 3 it may be desired to establish a "corner" or "cutoff" frequency f₀ for which signals below f₀ would be attenuated but above that frequency would be allowed to pass. In that particular circuit the output is only 0.707 of the input at f = f₀ and tails off below f₀ until approaching zero at frequencies << f₀ while approaching 1.0 at f >> f₀.

 

[cnh1995]

Thank you all for helping me.
Now what I understood is :
The impedance provided by a capacitor for AC is $\frac 1 {\omega C}$and for DC is infinite.
So, when we put a capacitor in a circuit, it is assumed that the impedance provided by the capacitor for AC( either by having high frequency signal or high capacitance or both) is so small that it can be approximated to wire for AC.
So, -$i_C R_C$ appears at terminal 1 wrt ground.

Yes. Note that the capacitor doesn't get charged to Vcc (as the underlined part in your OP might suggest). It gets charged to the dc collector-emitter voltage V_CE given by the Q-point.
The collector current is sinusoidal and can be written as
I_c=I_dc +I_csin(ωt).
So the voltage across collector resistance is,
V_R=I_dcR_c+I_cR_csin(ωt).

This means the voltage between terminals 1 and 2 without the capacitor is,
V₁₂=Vcc-V_R

∴V₁₂=Vcc-I_dcR_c-I_cR_csin(ωt),
where the green part is the dc bias in the output (Q-point). In presence of a capacitor (and a highly resistive load as pointed out by rude man), this dc bias is removed and you only get sinusoidally varying voltage between terminals 1 and 2.