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1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency .

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Supplement: Time constant (high-pass cut-off) of the output RC combination shown circuit

From the circuit diagram it is evident (as mentioned already) that the time constant - as derived from the discharging procedd - is T=C(RC+RL). In particular, this is true because the BJT is treated as a (ideal) current source (no current into the collector during discharging of C).

Question: Will we arrive at the same result for the charging process? For this purpose, we have to generate the differential equation for the case that a current source I (input step) is charging the capacitor C.

• Curent through RC is ic(t)=Vc/RC (Vc=collector voltage)

• Current through RC is io(t)=Vo/RL (Vo as shown in the drawing).

• Current through C is the same: io(t)=C[d(Vc-Vo)/dt]

• With Vc and Vo from the first two equations and with I=ic(t)+io(t) we arrive at the equation (dropping the brackets (t) for clarity):

• io=C[(d(icRC - ioRL)/dt]=[(d(I-io)RC - ioRL)/dt]

• Beause d(I)/dt=0 we can write (after some minor manipulations)

• io=-C(RC+RL)d(io)/dt.

• Setting (Ansatz) io(t)=îo * exp(t/T) the solution of the diff. equation is

• io(t)=îo * exp(t/T)=-C(RC+RL)(1/T)îo * [exp(t/T)]

• From this: T=-C(RC*RL)

Result (Summary): : The time constant T for the output circuitry can be calculated using (a) the discharging or (b) the charging process of the capacitor C.

1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency .

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Supplement: Time constant (high-pass cut-off) of the output RC combination shown circuit

From the circuit diagram it is evident (as mentioned already) that the time constant - as derived from the discharging process - is T=C(RC+RL). In particular, this is true because the BJT is treated as a (ideal) current source (no current into the collector during discharging of C).

Question: Will we arrive at the same result for the charging process? For this purpose, we have to generate the differential equation for the case that a current source I (input step) is charging the capacitor C.

• Curent through RC is ic(t)=Vc/RC (Vc=collector voltage)

• Current through C is io(t)=Vo/RL (Vo as shown in the drawing).

• Current through C is the same: io(t)=C[d(Vc-Vo)/dt]

• With Vc and Vo from the first two equations and with I=ic(t)+io(t) we arrive at the equation (dropping the brackets (t) for clarity):

• io=C[(d(icRC - ioRL)/dt]=C[(d(I-io)RC - ioRL)/dt]

• Beause d(I)/dt=0 we can write (after some minor manipulations)

• io=-C(RC+RL)d(io)/dt.

• Setting (Ansatz) io(t)=îo * exp(t/T) the solution of the diff. equation is

• io(t)=îo * exp(t/T)=-C(RC+RL)(1/T) * îo * [exp(t/T)]

• From this: T=-C(RC*RL)

Result (Summary): : The time constant T for the output circuitry can be calculated using (a) the discharging or (b) the charging process of the capacitor C.

1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency .

Supplement: Time constant (high-pass cut-off) of the output RC combination shown circuit

From the circuit diagram it is evident (as mentioned already) that the time constant - as derived from the discharging procedd - is T=C(RC+RL). In particular, this is true because the BJT is treated as a (ideal) current source (no current into the collector during discharging of C).

Question: Will we arrive at the same result for the charging process? For this purpose, we have to generate the differential equation for the case that a current source I (input step) is charging the capacitor C.

• Curent through RC is ic(t)=Vc/RC (Vc=collector voltage

• Current through RC is io(t)=Vo/RL

• Current through C is the same: io(t)=C[d(Vc-Vo)/dt]

• With Vc and Vo from the first two equations and with I=ic(t)+io(t) we arrive at the equation (dropping the brackets (t) for clarity):

• io=C[(d(icRC - ioRL)/dt]=[(d(I-io)RC - ioRL)/dt]

• Beause d(I)/dt=0 we can write (after some minor manipulations)

• io=-C(RC+RL)d(io)/dt.

• Setting (Ansatz) io(t)=Iexp(t/T) the solution of the diff. equation is

• io(t)=Iexp(t/T)=-C(RC+RL)(1/T)I[exp(t/T)]

• From this: T=C(RC*RL)

Result (Summary): : The time constant T for the output circuitry can be calculated using (a) the discharging or (b) the charging process of the capacitor C.

1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency .

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Supplement: Time constant (high-pass cut-off) of the output RC combination shown circuit

From the circuit diagram it is evident (as mentioned already) that the time constant - as derived from the discharging procedd - is T=C(RC+RL). In particular, this is true because the BJT is treated as a (ideal) current source (no current into the collector during discharging of C).

Question: Will we arrive at the same result for the charging process? For this purpose, we have to generate the differential equation for the case that a current source I (input step) is charging the capacitor C.

• Curent through RC is ic(t)=Vc/RC (Vc=collector voltage)

• Current through RC is io(t)=Vo/RL (Vo as shown in the drawing).

• Current through C is the same: io(t)=C[d(Vc-Vo)/dt]

• With Vc and Vo from the first two equations and with I=ic(t)+io(t) we arrive at the equation (dropping the brackets (t) for clarity):

• io=C[(d(icRC - ioRL)/dt]=[(d(I-io)RC - ioRL)/dt]

• Beause d(I)/dt=0 we can write (after some minor manipulations)

• io=-C(RC+RL)d(io)/dt.

• Setting (Ansatz) io(t)=îo * exp(t/T) the solution of the diff. equation is

• io(t)=îo * exp(t/T)=-C(RC+RL)(1/T)îo * [exp(t/T)]

• From this: T=-C(RC*RL)

Result (Summary): : The time constant T for the output circuitry can be calculated using (a) the discharging or (b) the charging process of the capacitor C.

1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency

1. In AC analysis, shouldn't both the resistors (RC and RL) be in parallel?

Answer: Yes, correct. This is because - as seen from the output of the current source (BJTs collector) - the ouput current Ic is split between Rc and RL. That means: For gain calculations both resistors are in parallel.

For the cut-off frequency we have to analyze the circuit from the capacitor side (and NOT from the collector side). Simplest method for finding the time constant of an RC combination: Find the current through the connected resistors - when the (charged) capacitor discharges. As we can see - the capacitor discharges through the resistors (left and right from the capacitor).

Thus, we must consider the series combination: Time constant T=C(RL+RC).

2. We don't we consider the RB (input resistance) for cutoff frequency?

Answer: When there is is no (external) coupling capacitance there is no input time constant and no corresponding lower cut-off frequency .

Supplement: Time constant (high-pass cut-off) of the output RC combination shown circuit

From the circuit diagram it is evident (as mentioned already) that the time constant - as derived from the discharging procedd - is T=C(RC+RL). In particular, this is true because the BJT is treated as a (ideal) current source (no current into the collector during discharging of C).

Question: Will we arrive at the same result for the charging process? For this purpose, we have to generate the differential equation for the case that a current source I (input step) is charging the capacitor C.

• Curent through RC is ic(t)=Vc/RC (Vc=collector voltage

• Current through RC is io(t)=Vo/RL

• Current through C is the same: io(t)=C[d(Vc-Vo)/dt]

• With Vc and Vo from the first two equations and with I=ic(t)+io(t) we arrive at the equation (dropping the brackets (t) for clarity):

• io=C[(d(icRC - ioRL)/dt]=[(d(I-io)RC - ioRL)/dt]

• Beause d(I)/dt=0 we can write (after some minor manipulations)

• io=-C(RC+RL)d(io)/dt.

• Setting (Ansatz) io(t)=Iexp(t/T) the solution of the diff. equation is

• io(t)=Iexp(t/T)=-C(RC+RL)(1/T)I[exp(t/T)]

• From this: T=C(RC*RL)

Result (Summary): : The time constant T for the output circuitry can be calculated using (a) the discharging or (b) the charging process of the capacitor C.