5.9 Frequency Response of the Common-Emitter Ampjstiles/412/handouts/5.9 Frequency Response of...

4/27/2011 section 5_9 Frequency Response of the CE Amp 1/1

Jim Stiles The Univ. of Kansas Dept. of EECS

5.9 Frequency Response of

the Common-Emitter Amp

Reading Assignment: 491-503

Amplifiers made with BJTs are similar to amplifiers made with op-amps—the both exhibit finite bandwidth.

HO: AMPLIFIER BANDWIDTH The gain within the bandwidth is usually constant with respect to frequency—we call this value the mid-band gain.

HO: MID-BAND GAIN Large capacitors (e.g., COUS) determine the low-frequency limit of amplifier bandwidth. We can explicitly determine this value be analyzing the low-frequency small-signal circuit.

HO: THE LOW-FREQUENCY RESPONSE Parasitic capacitors (e.g., Cπ) determine the high-frequency limit of amplifier bandwidth. We can explicitly determine this value be analyzing the high-frequency small-signal circuit.

HO: THE HIGH-FREQUENCY RESPONSE

4/20/2011 Amplifier Bandwidth 1/16


Amplifier Bandwidth BJT amplifiers are band-limited devices—in other words, they exhibit a finite bandwidth. Q: ??? A: Say the input to a BJT small-signal amplifier is the eigen function of linear, time-invariant system:

{ }cos Re jωtin inV ωt V e−=

Since the small-signal BJT amp is (approximately) a linear system, the output will likewise be the eigen function—an undistorted sinusoidal function of precisely the same frequency ω as the input! Q: Of course that’s true! We know that:

cosinV tω+

−

I

Amplifier

cosoutV tω+

−



( ) ( )out vo inv A vt t=

Therefore the magnitudes of the input and output sinusoids are related as:

out vo inV A V= Right? A: Not necessarily! The small-signal, open-circuit voltage gain of a BJT amplifier depends on the frequency ω of the input signal! Q: Huh!?! We determined earlier that the small-signal voltage gain of this amplifier:

.3 7 K

.2 3K

1K

1K

15 V 15 V

COUS

( )iv t

( )Ov t

100β =

+ -

COUS



was: 200o

voi

vAv

= = −

So then if the small-signal input is:

( ) cosi inv V ωtt =

isn't the small-signal output simply:

( ) 200 coso inv V ωtt = − ?????????

A: Maybe—or maybe not! Again, the gain of the amplifier is frequency dependent. We find that if ω is too high (i.e., large) or too low (i.e., small), then the output might be much less than the 200 times larger than the input (e.g., only 127.63 times larger than the input—Doh!). Now, the signal frequencies ω for which

( ) 200 coso inv V ωtt = − is an accurate statement, are frequencies that are said to lie within the bandwidth of this amplifier (ω is just right!). Conversely, frequencies ω for which:

( ) 200 coso inv V ωtt −≠



are frequencies ω that lie outside this amplifier’s bandwidth. Fortunately, the frequencies that compose an amplifier’s bandwidth typically form a continuum, such that the frequencies outside this bandwidth are either higher or lower than all frequencies within the bandwidth. Perhaps a plot would help. The frequencies between Lω and Hω thus lie within the bandwidth of the amplifier. The gain within the bandwidth is sometimes referred to as the midband gain. For signals with frequencies less than ( )L Lω f , the amplifier gain will be less than the midband gain—likewise for frequencies greater than ( )H Hω f .

ω

( )voA ω

Lω Hω

200



Q: So what then is the value:

200ovo

i

vAv

= = −

determined for the example amplifier? It doesn’t seem to be a function of frequency! A: The value -200 calculated for this amplifier is the midband gain—it’s the gain exhibited for all signals that lie within the amplifier bandwidth. Your book at times uses the variable AM to denote this value: Q: So it’s actually the midband gain that we’ve been determining from our small-signal circuit analysis (e.g.

200MA = − )? A: That’s exactly correct!

ω

( )voA ω

Lf Hf

MA



Q: So how do we determine the frequency dependent gain ( )voA ω ? More specifically, how do we determine midband gain

MA , along with Lf and Hf ? A: The function ( )voA ω is simply the eigen value of the linear operator relating the small-signal input and the small signal output:

( ) ( ){ } ( ) ( ) ( )o o vo iiv V A Vvt ω ω ωt= ⇒ =L Q: Yikes! How do we determine the eigen value of this linear operator? A: We simply analyze the small-signal circuit, determining ( )oV ω in terms of ( )iV ω .

Specifically, we must explicitly consider the capacitance in the small-signal amplifier—no longer can we make approximations! So, instead of vaguely labeling large capacitors as Capacitors Of Unusual Size, let’s explicitly consider the exact values of these large capacitors:



Likewise, we must consider the parasitic capacitances of the BJT—specifically μC and πC . The small-signal circuit—when we explicitly consider these capacitances—is thus:

.3 7 K

.2 3K

1K

1K

15 V 15 V

EC

( )iv t

( )Ov t

100β =

+ -

iC



We analyze this circuit to determine ( )oV ω , and then the eigen value—the small-signal gain—is determined as:

( ) ( )( )

ovo

i

V ωA ωV ω

=

Q: So what again is the meaning of ( )iV ω and ( )oV ω ? A: It’s the Fourier transform of ( )

iv t and ( )ov t !

( ) ( ) jωti iV ω v t e dt

∞

−∞

= ∫ and ( ) ( ) jωto oV ω v t e dt

∞

−∞

= ∫

Q: So—I can’t recall—what’s the relationship between ( )

iv t and ( )

ov t ?

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K

1 K

+ -

0 5 K.

1ijωC

1πjωC

1EjωC

1μjωC



A: If: ( ) ( ) ( )o vo iV A Vω ω ω=

Then in the time domain, we find that the input and output are related by the always enjoyable convolution integral!!!

( ) ( ) ( )o iv t g t t v t dt

∞

−∞

′ ′ ′= −∫

where the impulse response of the amplifier is of course:

( ) ( ) jωtvog t A ω e dω

∞−

−∞

= ∫

Q: What the heck? What happened to solutions like:

( ) ( )200o iv t v t= − ??

A: This result implies that the impulse response of the amplifier is:

( ) ( )200g t δ t= −

Such that:

( ) ( ) ( )

( ) ( )

( )

200

200

o i

i

i

v t g t t v t dt

δ v t dtt t

v t

∞

−∞∞

−∞

′ ′ ′= −

′ ′= − ′−

= −

∫

∫



Q: You say that the result:

( ) ( )200o iv t v t= −

“implies” that the impulse response of the amplifier is:

( ) ( )200g t δ t= −

Are you saying the impulse response of the common-emitter example is not this function? A: It is definitely not that function. The impulse response

( ) ( )200g t δ t= −

is ideal—the impulse response of an amplifier with an infinite bandwidth! Q: So all our small-signal analysis up to this point has been incorrect and useless??? A: Not at all! The small-signal gain we have been evaluating up to this point (e.g., -200) is the amplifier midband gain AM . As long as the small-signal input ( )iv t resides completely within the amplifier bandwidth, then the output will be:

( ) ( )200o iv t v t= −



The problem occurs when the input signal lies—at least partially—outside the amplifiers bandwidth. In that case, we find that:

( ) ( )200o iv t v t−≠ And instead:

( ) ( ) ( )o iv t g t t v t dt

∞

−∞

′ ′ ′= −∫

where:


∞−

−∞

= ∫

and the eigen value ( ) ( ) ( )

vo o iA ω V ω V ω= is determined by evaluating this small-signal circuit:

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K

1 K

+ -

0 5 K.

1ijωC

1πjωC

1EjωC

1μjωC



Q: What do you mean when you say that a signal lies “within the amplifier bandwidth” or “outside the amplifier bandwidth? How can we tell? A: Use the Fourier Transform! If we plot the magnitude of the Fourier Transform ( )iV ω of the input signal ( )iv t , we can see the spectrum of the input signal: For example, if you are attempting to amplify a signal representing the audio of symphonic music, the spectrum ( )iV ω will include low-frequency signals (e.g., from the tubas),

mid-range frequency signals (e.g., from the trumpets), and high-frequency signals (e.g., from the flutes).

ω

( )iV ω



Now, if it is your desire to reproduce exactly this music at the output of your amplifier, then the amplifier bandwidth must be wide enough to include all these spectral components ! For the case above, the input signal resides completely within the bandwidth of the amplifier (i.e., between Lf and Hf ), and so we find (for 200MA = − ) that:

( ) ( ) ( )200 io ω ωV V=

and: ( ) ( )200o iv t v t−=

However, if the input spectrum resides (at least partially) outside the amplifier bandwidth, e.g.:

f Lf Hf

( ) ( ) ( )voo i ωAV ω ω V=



then we find that:

( ) ( ) ( )200 io ω ωV V≠

and: ( ) ( )200o iv t v t−≠ !!!!

Instead, we find the more general (and more difficult!) expressions:

( ) ( ) ( )o vo iV A Vω ω ω=

and:

( ) ( ) ( )o iv t g t t v t dt

∞

−∞

′ ′ ′= −∫

f Lf Hf

( ) ( ) ( )voo i ωAV ω ω V=



where the impulse response of the amplifier is:


∞−

−∞

= ∫

Q: So just what causes this amplifier to have a finite bandwith? A: For mid-band frequencies mf (i.e., between L m Hf f f< < ), we will find that the Capacitors Of Unusual Size exhibit an impedance that is pretty small—approximately an AC short circuit:

( ) 1 0COUS mm ous

Z ωω C

= ≅

Likewise, the tiny parasitic capacitances μC and πC exhibit an impedance that is very large for mid-band frequencies—approximately an open circuit:

( ) 1πC m

m π

Z ωω C

= ≅ ∞

However, when the signal frequency ω drops too low, the COUS will no longer be a small-signal short. The result is that the amplifier gain is reduced—the values of the COUS determine the low-end amplifier bandwidth Lf .



Likewise, when the signal frequency ω is too high, the parasitic caps will no longer be a small-signal open. The result is that the amplifier gain is reduced—the values of the parasitic capacitors determine the high-end amplifier bandwidth Hf .

4/22/2011 Midband Gain 1/4


Mid-band Gain Q: So, to find the mid-band gain of this amplifier: we must find the analyze this small signal circuit:

.3 7 K

.2 3K

1K

1K

15 V 15 V

COUS

( )iv t

( )Ov t

100β =

+ -

COUS

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K

1 K

+ -

0 5 K.

1ijωC

1πjωC

1EjωC

1μjωC



to determine:

( ) ( )( )

ovo

i

V ωA ωV ω

=

and then plotting the magnitude: we determine mid-band gain MA , right? A: You could do all that, but there is an easier way. Recall the midband gain is the value af ( )voA ω for frequencies within the amplifier bandwidth. For those frequencies, the AC coupling capacitors (i.e., COUS) are approximate AC short-circuits (i.e., very low impedance) .

ω

( )voA ω

Lω Hω

MA



Likewise, for the signal frequencies within the amplifier bandwidth, the parasitic BJT capacitances are approximate AC open-circuits (i.e., very high impedance). Thus, we can apply these approximations to the capacitors in our small-signal circuit: Now simplifying this circuit (look, no capacitors!):

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K

1 K

+ -

0 5 K.

( )iV ω RC =1 K

( )oV ω + vbe -

0 37 K. 200 bev

+ -



Q: Hey wait! Isn’t this the same small-signal circuit that we analyzed earlier, where we found that:

( ) ( )200o iv vt t= − ??

A: It is exactly! All of the small-signal analysis that we performed previously (i.e., the circuits with no capacitors!) actually provided us with the mid-band amplifier gain. Taking the Fourier transform of the equation above:

( ) ( )

( )

200200

o ijπ

i

V Vω ωe V ω

= −

=

Thus, the midband gain of this amplifier is:

200 200jπMA e= − =

4/22/2011 Low-Frequency Response 1/11


Low-Frequency Response Q: OK, I see how to determine mid-band gain, but what about determining amplifier bandwidth? It seems like I have no alternative but to analyze the exact small-signal circuit (explicitly considering all capacitances):

And then determine:

( ) ( )( )

ovo

i

V ωA ωV ω

=

and then plot the magnitude:

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K

1 K

+ -

0 5 K.

1ijωC

1πjωC

1EjωC

1μjωC

( )voA ω

Lf Hf

MA

f



And then from the plot determine the amplifier bandwidth (i.e., determine Lf and Hf )? A: You could do all that, but there is an easier way. An amplifier frequency response ( )voA ω (i.e., its eigen value!) can generally be expressed as the product of three distinct terms:

( ) ( ) ( )vo L M HA ω F ω A F ω=

The middle term is the of course the mid-band gain—a number that is not frequency dependent. The function ( )LF ω describes the low-frequency response of the amplifier—from it we can determine the lower cutoff frequency Lf . Conversely, the function ( )HF ω describes the high-frequency response of the amplifier—from it we can determine the upper cutoff frequency Hf . Q: So just how do we determine these functions ( )LF ω and

( )HF ω ?? A: The low-frequency response ( )LF ω is dependent only on the large capacitors (COUS) in the amplifier circuit. In other



words the parasitic capacitances have no affect on the low-frequency response. Thus, we simply “ignore” the parasitic capacitances when determining ( )LF ω ! For example, say we include the COUS in our common-emitter example, but ignore μC and πC . The resulting small-signal circuit is: To simplify this analysis, we first determine the Thevenin’s equivalent circuit of the portion of the circuit connected to the base. We start by finding the open-circuit voltage:

( )iV ω 1 K

( )oV ω

( )be ωV

+

−

( )200 beV ω

B C

E

0.37 K

1 K

+ -

0 5 K.

1ijωC

1EjωC



And the short-circuit output current is: And thus the Thevenin’s equivalent source is:

( ) ( )

( )( )

( )0.37

1 0.37

octh o

ii

i

V Vω ωjω CV ω jω C

=⎛ ⎞⎟⎜= ⎟⎜ ⎟⎟⎜ +⎝ ⎠

( )iV ω 0.37 K +

-

1ijωC

( )ocoV ω

( ) ( )

( )( )

( )

10.37

0.37

0.371 0.37

i

oco i

jωC

ii

i

V Vω ω

jω CV ω jω C

⎛ ⎞⎟⎜ ⎟⎜= ⎟⎜ + ⎟⎟⎜⎝ ⎠⎛ ⎞⎟⎜= ⎟⎜ ⎟⎟⎜ +⎝ ⎠

( )iV ω

0.37 K + -

1ijωC

( )scoI ω

( )1

( )( ) ( )i

sc io i i

jωC

V ωI ω V ω jωC= =

( )thV ω + -

( )thZ ω



( )( )

( )

( )

( )

( )( )

10.371 0.37

0.371 0.37

oco

th sco

i

ii

i

V ωZ ω I ωjω C

jωCjω C

jω C

=

⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜= ⎟ ⎟⎜ ⎜⎟ ⎟⎟⎟⎜ ⎜+ ⎝ ⎠⎝ ⎠

=+

Likewise, the two parallel elements on the emitter terminal can be combined:

( )1

1

11 1

E

E

jωCE

jωC E

Z ω jωC= =

+ +

Thus, the small-signal circuit is now: From KVL:

1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

0 5 K.

( )EZ ω

( )thV ω + -

( )thZ ω



( ) ( )0 00.5 1

0.5 101

th b b Eth

thb

th E

V I I ZZ β

VIZ Z

+ − − =+ +

⇒ =+ +

From Ohm’s Law:

0.5be bV I= Therefore:

( ) ( )

( )( )

( )

200 1

200 0.5 0.5 101100

0.5 101

o be

th

th E

thth E

V ω VV ω

Z Z

V ωZ Z

= −

= −+ +

⎛ ⎞− ⎟⎜= ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠

Inserting the expressions for the Thevenin’s equivalent source, as well as ZE .

( ) ( )

( )( )

( )( )( )

1000.5 101

2000.37 2 2020.37 11 0.37 1 10.37

o thth E

ii

ii E

V ω V ωZ Z

jω CV ω jω C jω C jωC

⎛ ⎞− ⎟⎜= ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠⎛ ⎞− ⎟⎜⎛ ⎞ ⎟⎜⎟ ⎟⎜ ⎜= ⎟ ⎟⎜ ⎜⎟ ⎟⎟⎜ + +⎜+⎝ ⎠ ⎟⎜ ⎟⎜ + +⎝ ⎠



Now, it can be shown that:

( )( )

( )( )

20312 202 1 2030.37 1

1 10.37

E

E

i E

jω Cjω C

jω C jωC

≅+

+ ++ +

Therefore:

( ) ( )( )

( )

( )( )

( )2030.37 2001 1 2030.37

Eio i

i E

jω Cjω CV ω V ω jω C jω C⎛ ⎞⎛ ⎞ ⎟⎟⎜⎜ ⎟= ⎟⎜⎜ −⎟⎟⎜⎟⎜ ⎟⎜+ +⎝ ⎠⎝ ⎠

And so:

( ) ( )

( )

( )( )

( )2030.37 2001 1 2030.37

Eivo

i E

jω Cjω CA ωjω C jω C

⎛ ⎞⎛ ⎞ ⎟⎟⎜⎜ ⎟= ⎟⎜⎜ −⎟⎟⎜⎟⎜ ⎟⎜+ +⎝ ⎠⎝ ⎠

Now, since we are ignoring the parasitic capacitances, the function ( )HF ω that describes the high frequency response is:

( ) 1HF ω =

And so: ( ) ( ) ( ) ( )vo L M H L MA ω F ω A F ω F ω A= =

By inspection, we see for this example:

200MA = − We knew this already!



And:

( ) ( )

( )

( )( )

2030.371 1 2030.37

EiL

i E

jω Cjω CF ωjω C jω C

⎛ ⎞⎛ ⎞ ⎟⎟⎜⎜ ⎟= ⎟⎜⎜ ⎟⎟⎜⎟⎜ ⎟⎜+ +⎝ ⎠⎝ ⎠

Now, let’s define:

11 2.7

0.37Pi i

ωC C

= = and 2203

PE

ωC

=

Thus,

( ) ( )( )

( )( )

1 2

1 21 1P P

LP P

j jω ω ω ωF ωj jω ω ω ω

⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟= ⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜+ +⎝ ⎠⎝ ⎠

Now, functions of the type:

( )( )1

P

P

j ω ωj ω ω

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ +⎝ ⎠

are high-pass functions:

( )( )

( )

( )

2 2

21 1P P

P P

j ω ω ω ωj ω ω ω ω

=+ +

with a 3dB break frequency of Pω .



Thus:

( )

( )

2

2

1.0

0.51

0 0

P

PP

P

for ω ω

ω ω for ω ωω ω

for ω

⎧≅ >⎪⎪⎪⎪⎪⎪⎪= =⎨⎪+ ⎪⎪⎪⎪⎪ =⎪⎩

As a result, we find that the transfer function:

( ) ( )( )( )

( )( )

( )1 2

1 2200

1 1

vo L M

P P

P P

A ω F ω Aj jω ω ω ω

j jω ω ω ω

=⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟= −⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜+ +⎝ ⎠⎝ ⎠

will be approximately equal to the midband gain 200MA = − for all frequencies ω that are greater than both 1Pω and 2Pω .

( )

( )[ ]

2

21P

P

ω ωω ω

dB+

logωlog Pω03−



I.E.,:

( ) 1 2200 if vo M P PA ω A ω ω ω ωand≅ = − > >

Hopefully, it is now apparent (please tell me it is!) that the lower end of the amplifier bandwidth—specified by frequency

Lω —is the determined by the larger of the two frequencies

1Pω and 2Pω !

The larger of the two frequencies is called the dominant pole of the transfer function ( )LF ω . For our example—comparing the two frequencies 1Pω and 2Pω :

11 2.7

0.37Pi i

ωC C

= = and 2203

PE

ωC

=

it is apparent that the larger of the two (the dominant pole!) is likely 2Pω —that darn emitter capacitor is the key! Say we want the common-emitter amplifier in this circuit to have a bandwidth that extends down to 100Lf Hz= The emitter capacitor must therefore be:



( )

22032

203 203 32,300 !!!!2 2 1000

L PE

EL

πf ωC

C μFπf π

> =

⇒ > = =

This certainly is a Capacitor Of Unusual Size !

4/22/2011 High Frequency Response 1/4


High-Frequency Response To determine the high-frequency response of our example common-emitter amp, we simply consider explicitly the parasitic capacitances in the small-signal model, while approximating the COUS as small-signal short-circuits: Now, since we are ignoring the COUS, the function ( )LF ω that describes the low- frequency response is:

( ) 1LF ω =

And so:

( ) ( ) ( ) ( )vo L M H M HA ω F ω A F ω A F ω= =

( )iV ω

3.7 K 1 K

( )oV ω

( )bev ω

+

−

( )200 bev ω

B C

E

2.3 K + -

0 5 K.

1πjωC

1μjωC



We will find that the high-frequency response will (approximately) have the form.

( ) ( ) ( )3 4

1 11 1H

P PF ω j jω ω ω ω

⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜= ⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

Now, functions of the type:

( )1

1 Pj ω ω⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠

are low-pass functions:

( ) ( )

2

21 1

1 1P Pj ω ω ω ω

=+ +

with a 3dB break frequency of Pω .

( )[ ]2

11 Pω ω

dB+

logωlog Pω0

3−



Thus:

( )2

1.0

1 0.51

0

P

PP

for ω ω

for ω ωω ω

for ω

⎧≅ <⎪⎪⎪⎪⎪⎪⎪= =⎨⎪+ ⎪⎪⎪⎪⎪ → ∞⎪⎩

As a result, we find that the transfer function:

( ) ( )

( )( ) ( )3 3

1 1200 1 1

vo M H

P P

A ω A F ω

j jω ω ω ω

=⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜= ⎟ ⎟⎜ ⎜− ⎟ ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

will be approximately equal to the midband gain 200MA = − for all frequencies ω that are less than both 3Pω and 4Pω . I.E.,:

( ) 3 4200 if vo M P PA ω A ω ω ω ωand≅ = − < >

Hopefully, it is now apparent (please tell me it is!) that the higher end of the amplifier bandwidth—specified by frequency Hω —is the determined by the smaller of the two frequencies 3Pω and 3Pω !

The smaller of the two frequencies is called the dominant pole of the transfer function ( )HF ω .



Generally speaking, to increase the value Hω , we must increase the DC bias currents of our amplifier design!

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