Half power point problem help please

C

ChadMan

Guest
Hello All
Trying to figure out the 1/2 power point of the circuit
below. This is my homework, so I just need a nudge please.

When VO is not connected to the load, and points A & B are
connected, I figure the 1/2 power point (where the amplitude
is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.

Now here is where I get stuck. With points A & B disconnected
and VO being taken across RL how do I figure the f response
@ 3dB down? The above formula no longer works because R5 and
C3 are no longer there, f was never stated (unless I'm supposed
to use the previous answer).



VCC
+
|
o----------o
| |
| .-.
| | |
.-. RC| |
| | '-'
RB1 | | |
'-' | + #| C2
| o------#|------o o---------o
| | #| VO |
| | o .-.
|# + | |/ | RL| |
o-----|#----o--------| NPN === | |
|# | |> GND '-'
VI | | |
o | | |
| | | + #| C3 ===
=== | A o B o---#|----o GND
GND | | #| |
| | |
| | |
.-. .-. .-.
| | | | RE | |
RB2 | | | | R5 | |
'-' '-' '-'
| | o
| | |
| | |
o----------o-----------------o
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
 
"ChadMan" <chadhammer@nospamcharter.net> wrote in message
news:vpj87rl1ikhi33@corp.supernews.com...
Hello All
Trying to figure out the 1/2 power point of the circuit
below. This is my homework, so I just need a nudge please.

When VO is not connected to the load, and points A & B are
connected, I figure the 1/2 power point (where the amplitude
is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.

Now here is where I get stuck. With points A & B disconnected
and VO being taken across RL how do I figure the f response
@ 3dB down? The above formula no longer works because R5 and
C3 are no longer there, f was never stated (unless I'm supposed
to use the previous answer).
Click to expand...
Do I Use f = 1 / 2Pi x RL x C2 ?
 
ChadMan wrote:
Hello All
Trying to figure out the 1/2 power point of the circuit
below. This is my homework, so I just need a nudge please.

When VO is not connected to the load, and points A & B are
connected, I figure the 1/2 power point (where the amplitude
is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.
Click to expand...
What about RE?

Now here is where I get stuck. With points A & B disconnected
and VO being taken across RL how do I figure the f response
@ 3dB down? The above formula no longer works because R5 and
C3 are no longer there, f was never stated (unless I'm supposed
to use the previous answer).



VCC
+
|
o----------o
| |
| .-.
| | |
.-. RC| |
| | '-'
RB1 | | |
'-' | + #| C2
| o------#|------o o---------o
| | #| VO |
| | o .-.
|# + | |/ | RL| |
o-----|#----o--------| NPN === | |
|# | |> GND '-'
VI | | |
o | | |
| | | + #| C3 ===
=== | A o B o---#|----o GND
GND | | #| |
| | |
| | |
.-. .-. .-.
| | | | RE | |
RB2 | | | | R5 | |
'-' '-' '-'
| | o
| | |
| | |
o----------o-----------------o
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
Click to expand...
With RL connected to C2:

Draw a voltage source feeding RC, in series with C2 and RL. Now
calculate where the low frequency rolloff point is.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
The answer the were looking for was based on this:
1 / 2Pi x C2 x (RL + RC).

I am not sure why they used RL + RC !?!

ChadMan



"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message news:m%omb.247$R85.231@newsfep3-gui.server.ntli.net...
ChadMan wrote:
Hello All
Trying to figure out the 1/2 power point of the circuit
below. This is my homework, so I just need a nudge please.

When VO is not connected to the load, and points A & B are
connected, I figure the 1/2 power point (where the amplitude
is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.

What about RE?


Now here is where I get stuck. With points A & B disconnected
and VO being taken across RL how do I figure the f response
@ 3dB down? The above formula no longer works because R5 and
C3 are no longer there, f was never stated (unless I'm supposed
to use the previous answer).



VCC
+
|
o----------o
| |
| .-.
| | |
.-. RC| |
| | '-'
RB1 | | |
'-' | + #| C2
| o------#|------o o---------o
| | #| VO |
| | o .-.
|# + | |/ | RL| |
o-----|#----o--------| NPN === | |
|# | |> GND '-'
VI | | |
o | | |
| | | + #| C3 ===
=== | A o B o---#|----o GND
GND | | #| |
| | |
| | |
.-. .-. .-.
| | | | RE | |
RB2 | | | | R5 | |
'-' '-' '-'
| | o
| | |
| | |
o----------o-----------------o
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

With RL connected to C2:

Draw a voltage source feeding RC, in series with C2 and RL. Now
calculate where the low frequency rolloff point is.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
Click to expand...
 
"ChadMan" <chadhammer@nospamcharter.net> wrote in message news:<vpqm7q14k25q87@corp.supernews.com>...
The answer the were looking for was based on this:
1 / 2Pi x C2 x (RL + RC).

I am not sure why they used RL + RC !?!

ChadMan



"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message news:m%omb.247$R85.231@newsfep3-gui.server.ntli.net...
ChadMan wrote:
Hello All
Trying to figure out the 1/2 power point of the circuit
below. This is my homework, so I just need a nudge please.

When VO is not connected to the load, and points A & B are
connected, I figure the 1/2 power point (where the amplitude
is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.

What about RE?


Now here is where I get stuck. With points A & B disconnected
and VO being taken across RL how do I figure the f response
@ 3dB down? The above formula no longer works because R5 and
C3 are no longer there, f was never stated (unless I'm supposed
to use the previous answer).



VCC
+
|
o----------o
| |
| .-.
| | |
.-. RC| |
| | '-'
RB1 | | |
'-' | + #| C2
| o------#|------o o---------o
| | #| VO |
| | o .-.
|# + | |/ | RL| |
o-----|#----o--------| NPN === | |
|# | |> GND '-'
VI | | |
o | | |
| | | + #| C3 ===
=== | A o B o---#|----o GND
GND | | #| |
| | |
| | |
.-. .-. .-.
| | | | RE | |
RB2 | | | | R5 | |
'-' '-' '-'
| | o
| | |
| | |
o----------o-----------------o
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

With RL connected to C2:

Draw a voltage source feeding RC, in series with C2 and RL. Now
calculate where the low frequency rolloff point is.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
Click to expand...
In AC analysis, RC is parallel to RL. I guess, f = 1/2*pi*(Rc//RL)*C2.
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top