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Common Emitter Config with Negative FB

W

Warren

Jan 1, 1970
0
I see the common emitter configuration like the following
frequently:

http://imageshack.us/photo/my-images/828/groundedemitter.png

This is not a configuration I would normally use, but I would
like to understand it better. It seems to appear frequently in
60s/70s era designs.

To bias this, you take the ratio of R3 to R2, to arrive at a
collector voltage (target usually about half of Vcc). This
part is easy- in this example the ratio is 10:1. So the
resulting collector voltage is approx (10 + 1) times Vbe,
giving about 7 volts.

The link above is a LTspice simulation using 2N3904, with
a DC Beta of 300.

QUESTIONS:

(I only care about audio frequencies here)

Since there is negative feedback involved:

1. How do you compute the small signal gain?
2. Input impedance?
3. Output impedance?

LTspice output of a small AC wave form showed a gain
of about 219 (9.9mV amplitude in, 2.165V amplitude out).

But with Beta at 300, I would have thought the gain to
be near 300 minus the signal feedback (1/10th),
which would be about 270. But the simulation showed about
219 instead.

I'm guessing the input impedance to be approx R2||R3+R1
(nearly just R2). But with the negative fb, I suspect that the
input impedance is raised considerably.

I'm also guessing the output impedance to be approx R1||R3+R2
(nearly just R1).

Thanks, Warren
 
W

Warren

Jan 1, 1970
0
Tim Wescott expounded in
...
Jacob Millman, "Microelectronics", McGraw-Hill, 1979. Very
good book. I haven't read a better one for transistor
circuits. (But then, I haven't read _any_ other book for
transistor circuits -- I'm not a great buyer of books).

That's one I don't own yet- I may acquire it used. My wife
will be thrilled with another addition. :)
1:

You compute the small signal gain by observing that the
input impedance is very low, and thus that the small signal
gain will be set by the transistor's g_m and collector
load. Kv = g_m * R_c.
Ok.

Note that g_m is set by the emitter
current -- at room temperature it's around I_c / 26mV,
although that 26mV changes proportionally to absolute
temperature, and the B-E voltage (which contributes
significantly to emitter current) changes with temperature,
too.

The sim shows I_e = 0.841 mA. So if I_c ~= I_e, then according
to this g_m = 0.841mA/26mV => 0.032. Thus Kv ~= 265.

Given the sim gave approx 219, working backwards, we could
achieve that result if the 26mV were 31mV instead. That seems
a bit high (just my gut feeling).
3:

If the source impedance is truly zero then the input will
short out any feedback effect from R3, so the output
impedance will be pretty close to R1. If the source
impedance is higher (and as the source impedance varies
with frequency) the output impedance will stiffen up with
increasing feedback.

That's my bad (I hadn't thought about the source resistance).
Let's assume about R_s=600 ohms. I'll try this tonight when I
get home (where I have the file saved). This might explain why
LTspice had some wild numbers for the dc transfer, when I last
tried it.

Warren
 
J

John S

Jan 1, 1970
0
Il Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha scritto:
Tim Wescott expounded in
[..]
1:

You compute the small signal gain by observing that the
input impedance is very low, and thus that the small signal
gain will be set by the transistor's g_m and collector
load. Kv = g_m * R_c.

Ok.
Yes, but now you have to calculate R_c, which is *not* R1 in your circuit.

The gm*vbe current splits among R1 and R3. To simplify, if we assume an
infinite gain then the input will not move from ground, that is you have
R1//R3 as R_c. Numerically this gives 7.3k which multiplied by gm gives
around 230 (again, as in my back-of-the-envelope calculation).
How high is VA - the Early voltage - in the model you are using?

M

I examined a 2N3904 in LTSpice and found the Early voltage to be -99.

John S
 
W

Warren

Jan 1, 1970
0
Michele Ancis expounded in
Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:
Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha scritto:
Tim Wescott expounded in
[..]
You compute the small signal gain by observing that the
input impedance is very low, and thus that the small
signal gain will be set by the transistor's g_m and
collector load. Kv = g_m * R_c.

Ok.

Yes, but now you have to calculate R_c, which is *not* R1
in your circuit.

The gm*vbe current splits among R1 and R3. To simplify,
if we assume an infinite gain then the input will not
move from ground, that is you have R1//R3 as R_c.
Numerically this gives 7.3k which multiplied by gm gives
around 230 (again, as in my back-of-the-envelope
calculation). How high is VA - the Early voltage - in the
model you are using?

M

I examined a 2N3904 in LTSpice and found the Early voltage
to be -99.

John S

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is
118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 = 220

I'd say it's pretty close :-D

M

Indeed.

What is the effect of R3 on input impedance in this circuit?
(Given the negative FB) Can I glibly approximate as:

R2 || R3 + ( R1 || ro )

?

Warren
 
W

Warren

Jan 1, 1970
0
Michele Ancis expounded in
Il Fri, 29 Jul 2011 12:25:32 +0000 (UTC), Warren ha
scritto:
Michele Ancis expounded in
Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:
On 7/28/2011 12:41 PM, Michele Ancis wrote:
Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha
scritto:
Tim Wescott expounded in
[..]
You compute the small signal gain by observing that
the input impedance is very low, and thus that the
small signal gain will be set by the transistor's g_m
and collector load. Kv = g_m * R_c.

Ok.

Yes, but now you have to calculate R_c, which is *not*
R1 in your circuit.

[..]. How high is VA - the Early voltage - in the
model you are using?

M

I examined a 2N3904 in LTSpice and found the Early
voltage to be -99.

John S

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is
118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 =
220

I'd say it's pretty close :-D

M

Indeed.

What is the effect of R3 on input impedance in this
circuit? (Given the negative FB) Can I glibly approximate
as:

R2 || R3 + ( R1 || ro )

?

Warren

Uhm...I would put some extra parentheses in there,
otherwise I cannot order the operations..

|| has higher precence than '+' :)
Anyways, I think
that your question "What is the effect of R3 on input
impedance in this circuit?" can be answered in three ways:

I'll have to look at this at home (I'm at work). And see
if I can grok this..
..
Rin = R2//rb//(R3/220)

What does the simulator say?

M

LTspice is off the planet when I try to answer this. It is
probably my fault:

..tf V(Out) V2

Small Signal AC Analysis:
AC Amplitude: 0.01
AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer
v2#Input_impedance: 2.1252e+016 impedance
output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

http://imageshack.us/photo/my-images/535/commonemitter2.png

but I had similar results, with the former sim as well.

Warren
 
W

Warren

Jan 1, 1970
0
Michele Ancis expounded in
...
Awrighty then :)
So you don't need the parentheses in your expression :-DD

Yes- for clarity only (albeit inconsistently). :)
So...Last time I used Spice's netlist interface was too
many years ago so I cannot check syntax here.

LTspice doesn't require all that arcane data entry - you can
enter a schematic graphically.
I'd say -
however - you could put your AC amplitude to 1V, which
gives you "normalized" results ...
Also, if you define V2 to output 1V AC, you don't need to
use .tf but just plot V(out) and you're done :-D

M

Surely you don't mean 1V for generator V2? With a stage gain
of approx 220, this leads to some trouble at the collector.
This is precisely why there is truly a "small signal"
supplied. :)

The LTspice (*.asc) file is provided below, for anyone wanting
to try it:

Version 4
SHEET 1 968 680
WIRE 464 -112 304 -112
WIRE 304 -80 304 -112
WIRE 304 48 304 0
WIRE 304 48 128 48
WIRE 336 48 304 48
WIRE 128 80 128 48
WIRE 304 144 304 48
WIRE 464 144 464 -112
WIRE -160 192 -208 192
WIRE -16 192 -80 192
WIRE 128 192 128 160
WIRE 128 192 48 192
WIRE 240 192 128 192
WIRE -208 272 -208 192
WIRE 128 272 128 192
WIRE -208 400 -208 352
WIRE 128 400 128 352
WIRE 128 400 -208 400
WIRE 304 400 304 240
WIRE 304 400 128 400
WIRE 464 400 464 224
WIRE 464 400 304 400
WIRE 304 448 304 400
FLAG 304 448 0
FLAG 336 48 Out
IOPIN 336 48 Out
SYMBOL npn 240 144 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL res 288 -96 R0
SYMATTR InstName R1
SYMATTR Value 8.2k
SYMBOL res 112 256 R0
SYMATTR InstName R2
SYMATTR Value 6.8k
SYMBOL res 112 64 R0
SYMATTR InstName R3
SYMATTR Value 68k
SYMBOL Misc\\battery 464 128 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 15V
SYMBOL voltage -208 256 R0
WINDOW 3 6 116 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 0.01V 60Hz)
SYMATTR InstName V2
SYMATTR Value2 AC .01 0
SYMBOL cap 48 176 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 47µF
SYMBOL res -176 208 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName Rs
SYMATTR Value 600
TEXT -88 536 Left 0 !;tran 0 1 0 .0001
TEXT -112 -152 Left 0 ;From Art of Electronics, Fig 2.40.
TEXT -88 576 Left 0 !.tf V(Out) V2
TEXT 176 24 Left 0 ;DC 7.318V
TEXT 192 -88 Left 0 ;0.936 mA
TEXT 320 88 Left 0 ;0.71V\nAmplitude
TEXT 136 224 Left 0 ;3.1 mV\nAmplitude
TEXT 320 248 Left 0 ;AC Gain \n229
TEXT 40 80 Left 0 ;98.1uA
TEXT 320 352 Left 0 ;0.841mA

Warren
 
W

Warren

Jan 1, 1970
0
Rich Grise expounded in
That's almost the same circuit I used when I won the blue
ribbon in the Science Fair, in 5th grade, ca. 1959. ;-D

Cheers!
Rich

Humph!

I didn't get any ribbon for my high frequency Tesla coil. The
spark would just tickle the skin and the ozone generated would
eliminate onion smells, even.

My classmate stole the show with his mockup of an open heart
surgery thingy (his mother was a nurse).

I'm not bitter after all these years!
Humph!

Warren
 
T

tm

Jan 1, 1970
0
Warren said:
Rich Grise expounded in

Humph!

I didn't get any ribbon for my high frequency Tesla coil. The
spark would just tickle the skin and the ozone generated would
eliminate onion smells, even.

My classmate stole the show with his mockup of an open heart
surgery thingy (his mother was a nurse).

I'm not bitter after all these years!
Humph!

Warren

Hey, maybe the judges were trying to listen to the ball game :).


tm
 
W

Warren

Jan 1, 1970
0
tm expounded in
Hey, maybe the judges were trying to listen to the ball
game :).
tm

:)

Just a trifle crackle on the band.

It wasn't like the harmless little voltage doubler with big
caps that I made for shocking classmates with..

Warren
 
J

John S

Jan 1, 1970
0
Michele Ancis expounded in
Il Fri, 29 Jul 2011 12:25:32 +0000 (UTC), Warren ha
scritto:
Michele Ancis expounded in

Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:
On 7/28/2011 12:41 PM, Michele Ancis wrote:
Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha
scritto:
Tim Wescott expounded in
[..]
You compute the small signal gain by observing that
the input impedance is very low, and thus that the
small signal gain will be set by the transistor's g_m
and collector load. Kv = g_m * R_c.

Ok.

Yes, but now you have to calculate R_c, which is *not*
R1 in your circuit.

[..]. How high is VA - the Early voltage - in the
model you are using?

M

I examined a 2N3904 in LTSpice and found the Early
voltage to be -99.

John S

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is
118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 =
220

I'd say it's pretty close :-D

M

Indeed.

What is the effect of R3 on input impedance in this
circuit? (Given the negative FB) Can I glibly approximate
as:

R2 || R3 + ( R1 || ro )

?

Warren

Uhm...I would put some extra parentheses in there,
otherwise I cannot order the operations..

|| has higher precence than '+' :)
Anyways, I think
that your question "What is the effect of R3 on input
impedance in this circuit?" can be answered in three ways:

I'll have to look at this at home (I'm at work). And see
if I can grok this..
..
Rin = R2//rb//(R3/220)

What does the simulator say?

M

LTspice is off the planet when I try to answer this. It is
probably my fault:

.tf V(Out) V2

Small Signal AC Analysis:
AC Amplitude: 0.01
AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer
v2#Input_impedance: 2.1252e+016 impedance
output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

http://imageshack.us/photo/my-images/535/commonemitter2.png

but I had similar results, with the former sim as well.

Warren

Using AC signal analysis:

V(n003)/I(Rs) = 288.316 < -11.2908 degrees @ 60Hz
or
V(n003)/I(Rs)= 282.736 - 56.449j

V(n003) is between the 600 ohm source resistor and the input capacitor.

Hint: This is easy to do using waveform arithmetic.

Cheers,
John S
 
J

John S

Jan 1, 1970
0
Il Fri, 29 Jul 2011 14:01:02 +0000 (UTC), in sci.electronics.design hai
scritto:
Michele Ancis expounded in
Il Fri, 29 Jul 2011 12:25:32 +0000 (UTC), Warren ha
scritto:

Michele Ancis expounded in

Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:
[..]

I examined a 2N3904 in LTSpice and found the Early
voltage to be -99.

John S

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is
118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 =
220

I'd say it's pretty close :-D

M

Indeed.

What is the effect of R3 on input impedance in this
circuit? (Given the negative FB) Can I glibly approximate
as:

R2 || R3 + ( R1 || ro )

?

Warren

Uhm...I would put some extra parentheses in there,
otherwise I cannot order the operations..

|| has higher precence than '+' :)

Awrighty then :)
So you don't need the parentheses in your expression :-DD
I'll have to look at this at home (I'm at work). And see
if I can grok this..


LTspice is off the planet when I try to answer this. It is
probably my fault:

.tf V(Out) V2

Small Signal AC Analysis:
AC Amplitude: 0.01
AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer
v2#Input_impedance: 2.1252e+016 impedance
output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

http://imageshack.us/photo/my-images/535/commonemitter2.png

but I had similar results, with the former sim as well.

Warren

So...Last time I used Spice's netlist interface was too many years ago so I
cannot check syntax here. I'd say - however - you could put your AC
amplitude to 1V, which gives you "normalized" results and doesn't affect
anything else, being AC a linear simulation (I mean, with linear components
in the network). Then, to calculate the Rin, find a way to get the current
flowing through the input source (normally it is accessible somewhere, but
you may need to place a probe in series, dunno), and calculate "one over
that".
It shouldn't pose any problem at all.
Also, if you define V2 to output 1V AC, you don't need to use .tf but just
plot V(out) and you're done :-D

M

The current through C1 is available. I like to rotate components such
that the current probe shows current going from source to load
(usually). Warrens is the opposite from that but can still be handled,
of course.

John S
 
J

John S

Jan 1, 1970
0
Il Fri, 29 Jul 2011 16:46:44 +0000 (UTC), Warren ha scritto:
[..]
LTspice is off the planet when I try to answer this. It is
probably my fault:

.tf V(Out) V2

Small Signal AC Analysis:
AC Amplitude: 0.01
AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer
v2#Input_impedance: 2.1252e+016 impedance
output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

http://imageshack.us/photo/my-images/535/commonemitter2.png
So...Last time I used Spice's netlist interface was too
many years ago so I cannot check syntax here.

LTspice doesn't require all that arcane data entry - you can
enter a schematic graphically.
I'd say -
however - you could put your AC amplitude to 1V, which
gives you "normalized" results ..
Also, if you define V2 to output 1V AC, you don't need to
use .tf but just plot V(out) and you're done :-D

M

Surely you don't mean 1V for generator V2?

Oh yes, indeed.
With a stage gain
of approx 220, this leads to some trouble at the collector.
This is precisely why there is truly a "small signal"
supplied. :)

Ohi ohi, I guess we need to understand each other a little better here :)
Are you simulating:

1 - AC small signal
2 - Transient with sinusoidal excitation

If 1 holds, then, *per definition* you don't care about the absolute
amplitude of the signals, since everything in your network is simulated
with linear components, which (unlike REAL stuff) don't bother about the
amplitude of their drive, they always respond with a fixed proportionality
constant. Like if you put 1V across base-emitter, at the collector you read
230V, but that is fine provided you understand what you read (That is,
gain).
If 2 holds, then why are you doing it? Things like "input impedance" and
"gain" and "output impedance" are indeed meaningful and correctly defined
*only* under linear assumptions..
Anyways, to plot the input impedance you still have to plot

V2/I2

Where V2 is the voltage of your input source and I2 is the current through
it. This is the definition of input impedance seen by your source.
The LTspice (*.asc) file is provided below, for anyone wanting
to try it:

Version 4
[..]

Warren

I will download LTspice sooner or later..;-D

M

IMHO, the input impedance to the circuit is between the source
resistance and the input terminal of the circuit, that is, at the
junction of the 600 ohm resistor and the 47uF capacitor.

John S
 
J

John S

Jan 1, 1970
0
Surely you don't mean 1V for generator V2?

Oh yes, indeed.
With a stage gain
of approx 220, this leads to some trouble at the collector.
This is precisely why there is truly a "small signal"
supplied. :)

Ohi ohi, I guess we need to understand each other a little better here :)
Are you simulating:

1 - AC small signal
2 - Transient with sinusoidal excitation

If 1 holds, then, *per definition* you don't care about the absolute
amplitude of the signals, since everything in your network is simulated
with linear components, which (unlike REAL stuff) don't bother about the
amplitude of their drive, they always respond with a fixed proportionality
constant. Like if you put 1V across base-emitter, at the collector you read
230V, but that is fine provided you understand what you read (That is,
gain).
If 2 holds, then why are you doing it? Things like "input impedance" and
"gain" and "output impedance" are indeed meaningful and correctly defined
*only* under linear assumptions..
Anyways, to plot the input impedance you still have to plot

V2/I2

Where V2 is the voltage of your input source and I2 is the current through
it. This is the definition of input impedance seen by your source.
The LTspice (*.asc) file is provided below, for anyone wanting
to try it:

Version 4
[..]

Warren

I will download LTspice sooner or later..;-D

M

If you use waveform arithmetic V(n003)/I(Rs) you don't need to change
the source to 1V. Of course, you may if you wish.

John S
 
J

John S

Jan 1, 1970
0
Il Fri, 29 Jul 2011 12:31:23 -0500, John S ha scritto:

[..]
IMHO, the input impedance to the circuit is between the source
resistance and the input terminal of the circuit, that is, at the
junction of the 600 ohm resistor and the 47uF capacitor.

John S

I was referring to the first circuit:

http://imageshack.us/photo/my-images/828/groundedemitter.png

which don't have the source resistance yet, so that we are taking the same
point :-D
Yes.

Anyways, I would tend to "ignore" the bypass cap as its function is to
allow for proper DC decoupling but otherwise should be "transparent".
That is, if I were to talk about the input impedance of such circuit, I
would do it mid-band, where the cap is a short (at least, in theory).

Well, he emphasizes 60Hz on the schematic. Otherwise, it can be done any
way one wishes.
Still better ;-D
I would *first* evaluate the input impedance *after* the cap and consider
it the somewhat "intrinsic" impedance. Now, this will be resistive at the
beginning and - hopefully - greater than the impedance of the source we
intend to drive the circuit with. From this value we *then* decide which
bypass cap is suitable to obtain the lower corner frequency of the circuit
:-D

M

Of course. It is up to the designer to decide what data is most useful
to him/her.

John S
 
J

John S

Jan 1, 1970
0
Still better ;-D
I would *first* evaluate the input impedance *after* the cap and consider
it the somewhat "intrinsic" impedance. Now, this will be resistive at the
beginning and - hopefully - greater than the impedance of the source we
intend to drive the circuit with. From this value we *then* decide which
bypass cap is suitable to obtain the lower corner frequency of the circuit
:-D

M


That impedance is 282.7 ohms. Phase shift is insignificant.

John S
 
C

Charmed Snark

Jan 1, 1970
0
Michele Ancis expounded in
Il Fri, 29 Jul 2011 13:45:41 -0500, in
sci.electronics.design hai scritto:


So if I got you right, you say that the input impedance is
dominated by what I called "intrinsic" impedance, and if I
remember correctly we're stimulating @60KHz, which means

Actually 60 Hertz, just to be clear. ;-)

Warren
 
W

Warren

Jan 1, 1970
0
Tim Wescott expounded in
When SPICE does AC analysis it linearizes the circuit at
the quiescent operating point, then doesn't check that it's
linearizing assumptions are valid. You could use 10^12V
for your generator, and SPICE will give you the same
answers for gain, phase, impedance, etc.

So you want to use 1V or 1A, etc.

Ok but when I tried it again, LTspice produced the same screwy
results.

Warren
 
W

Warren

Jan 1, 1970
0
John S expounded in
Michele Ancis expounded in
Il Fri, 29 Jul 2011 12:25:32 +0000 (UTC), Warren ha
scritto:

Michele Ancis expounded in

Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:
On 7/28/2011 12:41 PM, Michele Ancis wrote:
Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha
scritto:
Tim Wescott expounded in
[..]
You compute the small signal gain by observing that
the input impedance is very low, and thus that the
small signal gain will be set by the transistor's
g_m and collector load. Kv = g_m * R_c.

Ok.

Yes, but now you have to calculate R_c, which is
*not* R1 in your circuit.

[..]. How high is VA - the Early voltage - in the
model you are using?

M

I examined a 2N3904 in LTSpice and found the Early
voltage to be -99.

John S

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is
118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 =
220

I'd say it's pretty close :-D

M

Indeed.

What is the effect of R3 on input impedance in this
circuit? (Given the negative FB) Can I glibly
approximate as:

R2 || R3 + ( R1 || ro )

?

Warren

Uhm...I would put some extra parentheses in there,
otherwise I cannot order the operations..

|| has higher precence than '+' :)
Anyways, I think
that your question "What is the effect of R3 on input
impedance in this circuit?" can be answered in three
ways:

I'll have to look at this at home (I'm at work). And see
if I can grok this..
..
Rin = R2//rb//(R3/220)

What does the simulator say?

M

LTspice is off the planet when I try to answer this. It is
probably my fault:

.tf V(Out) V2

Small Signal AC Analysis:
AC Amplitude: 0.01
AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer
v2#Input_impedance: 2.1252e+016 impedance
output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

http://imageshack.us/photo/my-images/535/commonemitter2.
png

but I had similar results, with the former sim as well.

Warren

Using AC signal analysis:

V(n003)/I(Rs) = 288.316 < -11.2908 degrees @ 60Hz
or
V(n003)/I(Rs)= 282.736 - 56.449j

V(n003) is between the 600 ohm source resistor and the
input capacitor.
John S

I'll have to chew on this over the long weekend.

Thanks, Warren
 
J

John S

Jan 1, 1970
0
Il Fri, 29 Jul 2011 13:45:41 -0500, in sci.electronics.design hai scritto:


So if I got you right, you say that the input impedance is dominated by
what I called "intrinsic" impedance, and if I remember correctly we're
stimulating @60KHz, which means the 47uF looks like a 1/(2*Pi*6e4*47e-6)
which is on the neiborhood of 56mOhm...Yes, I'd say is negligible ;-)

Now, my Buglar way to calculate the input resistance gave me

Rin = R2//rb//(R3/K_v) with rb = Vt/Ib = 26e-3*300/0.841e-3 = 9.2k

Rin = 6.8k//9.2k//309 = 286 Ohm

Not bad ;-D

M

Actually, the stimulus is 60Hz as Warren pointed out elsewhere. Also,
the 282.7 ohms with negligible phase shift is measured at the junction
of the capacitor and the base. So, the 47uF is out of the picture
completely. There is actually a phase shift in the millidegrees which is
probably attributable to the transistor's intrinsic capacitance and can
be ignored.

But, you're right; not bad at all.

John S
 
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