1 Frequency Response

Untill now the discussion for amplifiers had been easy and straightforward as wehad assumed that the capacitors are infinite in value. Though far from true, thisis theoretically convenient notion and it enabled us to make the estimate of gain,input and output impedances. It is now time to face the hard fact that capacitorsare indeed finite. It is more disturbing to note that to realize a moderate capacitorof 1F you need 500 football ground areas separated by 1 meter. The capacitors thatare typically used in microelectronics have values ranging from few 1pF to hundredsof μF. The goal of this lesson is to demonstrate that with practical capacitors we willstill be able to design useful amplifiers.

Rsi

Vcc

vi

R1

R2

Cc

Rc

RE

vo

Figure 1: A typical amplifier circuit

Fig.(1) shows a typical amplifier with capacitor Cc appearing in series with theinput signal and base. Let us consider the effect of this capacitor (assuming somefinite value, say 1μF) at frequencies approaching 0Hz. The impedance offered by theCc is

limω→0

Xc =1

ωC= ∞. (1)

This clearly indicates that the capacitor behaves as open circuit at small frequen-cies and the signal is not coupled to the amplifier any more. Thus at low frequenciesthe signal would not even reach the amplifier in the first place. We sense some problemin this frequency range.

It should be noted that BJT is made of junctions between terminals such asemitter, base and collector. Whenever a junction is made a depletion region wouldresult causing a capacitance associated with it. For different reasons there is diffusion

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capacitance associated with junctions. To summarize, any junction will introduce acapacitor. It is simply impossible to make an active device without capacitors. Thesejunction capacitances act as short circuits at extremely high frequencies harming thetransistor action. Such capacitors are called parasitic and stray capacitors. Thiscauses rapid fall of gain at very high frequencies. Similar capacitors are found inMOSFETs too.

To summarize every transistor amplifier will have gain which is falling down at lowfrequencies because of capacitors that are included in the circuit. At high frequenciesthe gain rolls off due to capacitors that are inside the transistors. The gain willassume constant value between these two extremes as shown in Fig.(2) below. We

Figure 2: Frequency Response of a Typical Amplifier

notice from figure that the amplifier’s performance is satisfactory between a lowercut-off frequency, fl and an upper cut-off frequency fu. The difference is termed asbandwidth of the amplifier,

Bandwidth = BW = fu − fl. (2)

That the amplifier’s bandwidth is finite presents itself as a new challenge to thecircuit designer. Trust me, circuits can still be designed. For example, if you wantto build an amplifier for audio signals (20 to 20kHz), it is necessary that we havefl < 20Hz and fu > 20kHz. Using approprite combination of resistors and capacitorswill get the job done. The region in which the gain is constant (and also highest) iscalled as the mid-band region of the amplifier.

If you want to build a video amplifier the requirements are more stringent. Videoamplifiers need a bandwidth range covering from 5Hz to 5MHz. It is to be noted thatas generations pass by transistor sizes keep on diminishing. Downscaling transistor

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size results in smaller junction areas. We know that C = εAdand hence the capacitor

values drop in proportion. This will push the upper cut-off frequencies to muchhigher range. As technology advances, we will be able to make smaller transistorswhich translates to higher speed.

A typical example that I would quote is that Cell phones in the consumer mar-ket 10 years ago provided voice transmission and texting features to the maximum.Nowadays at the same price you could find a phone that could handle video down-loading and internet live-game playing features. This is all enabled by reduction intransistor size. The impact it has on our lives is tremendous.

1.1 First Order Systems

With this motivation, we will delve into the topic of frequency response in a morerigorous manner. In general a transistor amplifier will have multiple capacitors andas such dealing all of them simultaneously is nearly impossible. Rather we begin ouranalysis with first order systems which provide a simplified approach.

1.1.1 The series capacitor

Fig.(3) shows the circuit with a capacitor connected in series between the input andoutput.

Rs Cs

RP

vi vo

Figure 3: Series Capacitor

If we work in Laplace domain, we have,

vo(s)

vi(s)=

Rp

Rp +Rs +1

sCs

(3)

=sRpCs

1 + s(Rp +Rs)Cs(4)

which can be rearranged and written as,

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vo(s)

vi(s)=

Rp

Rp +Rs

s(Rp +Rs)Cs

1 + s(Rp +Rs)Cs(5)

= Ksτs

1 + sτs(6)

where K = Rp

Rp+Rsis the gain of the circuit (when the capcitor is a short circuit)

and τs = (Rs +Rp)Cs is the time constant of the circuit.This equation is a first order equation as the largest power of s in the denominator

is unity. Before we look quantitatively, we look at certain qualitative aspects of thecircuit. First, at very low frequencies the capacitor impedance becomes large andapproaches ∞. Hence the capacitor at low frequency tends to be open circuit and viis no longer connected to vo. At high frequencies the capacitor tends to be a shortcircuit and vo gets coupled to vi. It is instructive to study this behavior with propermathematical support.

We shall call the ratio vo(s)vi(s)

as the transfer function or frequency response of the

circuit, denoted by T (s).

T (s) =vo(s)

vi(s)=

Rp

Rs +Rp

sτs1 + sτs

(7)

T (jω) =Rp

Rs +Rp

jωτs1 + jωτs

(8)

1.1.2 Magnitude Response

The above equation suggests that the gain is a complex number with real and imagi-nary components. Rather, we prefer to express gain in terms of magnitude and phase.The magnitude of the above equation is,

|T (jω)| =Rp

Rs +Rp

(ωτs√

1 + ω2τ 2s

)(9)

⇒ |T (jf)| =Rp

Rs +Rp

(2πfτs√

1 + (2πfτs)2

). (10)

Considering the above expression in log or decibel scale,we get:

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|T (jf)|dB = 20 log10 |T (jf)| (11)

= 20 log10

[Rp

Rs +Rp

(2πfτs√

1 + (2πfτs)2

)](12)

= 20 log10Rp

Rs +Rp+ 20 log10 (2πfτs)

− 20 log10 (√1 + (2πfτs)2). (13)

We will try to plot the equation given above as shown in Fig.(4). In section (a) ofthe figure, the first term 20 log10

Rp

Rs+Rpis plotted. Since Rp +Rs > Rp, the argument

is always less than one and hence when log is taken over the ratio Rp

Rs+Rpit yields a

value less than 0dB. Since the first term is totally independent of frequency it assumesuniform value for all frequencies.

In section (b) of the figure, the second term is examined. Now, in this case theargument is directly proportional to frequency. When we plot it in log-log scale thisturns out to be a straight line of slope 20db per decade. This line intersects the 0dBline when f = 1

2πτs. This can be verified by substituting for f in the second term

which will yield log10 1 = 0.Section (c) is the interesting part. The third term appears to be different and

difficult. An engineer by the name Bode, came up with an innovative idea to plotthe curve as described by the third term. According to Bode’s suggestion ratherthan working with the equation for the curve we can work with the asymptotes. Anasymptote is defined as a straight line that approaches the curve at ∞ but it nevertouches for any finite value.

Bode’s idea is to consider the third term at two extremes, the first being whenf → 0 (low frequencies) and the second being when f → ∞ (high frequencies). Letus indentify the curve at these two extremes. At the left extreme,

limf→0

−20 log10√1 + (2πfτs)2 = 0 (14)

and hence at values close to 0Hz the curve will lie exactly on top of the frequencyaxis. We have got the first asymptote. Next we proceed to the second asymptote byconsidering,

limf→∞

−20 log10√

1 + (2πfτs)2 = −20 log10 (2πfτs) (15)

This equation looks similar to the curve obtained by using the second term. How-ever, now the slope is negative. For a moment we will use these approximate asymp-totes and extend them to finite frequencies. Being said that they are approximate wemust look how much approximation that we can afford. As engineers we should beopen minded to approximations but we should not overtly approximate as it could

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dB

0 f(log scale)

20.log(Rp/(Rp+Rs))

(a)

dB

20 dB/decade

0 f(log scale)f=1/2 s

(b)

dB f=1/2 s 3dB

0 f(log scale)

20 dB/decade

(c)

Figure 4: Series capacitor effects with each terms considered seperately

totally throw away the picture. Approximations are good as long as we are careful inusing them.

When we extend both the asymptotes they meet at f = 12πτs

. f is called as the

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break-point frequency of the system. In Fig.(4), the actual curve is also plotted alongwith the asymptotes. At the place where they meet it can be noted that the actualcurve is 3dB less than the asymptotic extension. This error is usually negligible.Remember, we are working with first order systems. For a system with nth order,the error can be as large as 3ndB. So for higher order systems Bode plot is a crudeapproximation. Besides, Bode’s approximation is effective only if the poles of thesystem are well separated. It is important to keep an eye on both these conditions toapply Bode’s approximation sucessfully. Sadly, very often this is forgotten but youshould’nt.

Figure 5: Plot obtained after combining all the terms

It is now time to combine all the graphs to obtain the final plot. We first combineplots (b) and (c) in Fig.(4), before the break-point frequency plot (b) is a straight linewith slope 20dB/decade and plot (c) assumes the value of 0dB uniformly. This resultsin a straight line of slope 20dB/decade. Beyond the break-point frequency we cansee that plot (b) is a line with positive slope and plot (c) is a line with negative slopeof equal magnitude. When added it results in 0dB throughout. Thus the resultantis a line of 20dB/decade before and 0dB beyond the break-point frequency. Plot (a)is independent of frequency, so when all three are added plot (a) pushes down thecombination of (b) and (c) by 20 log10

Rp

Rp+Rs. This is shown in Fig.(5).

1.1.3 Phase Response

We rewrite the transfer function to find the phase response,

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T (jf) =Rp

Rp +Rs

j2πfτs1 + j2πfτs

(16)

=Rp

Rp +Rs× j2πfτs × 1

1 + j2πfτs(17)

= T1 × T2 × T3 (18)

where T1 =Rp

Rp+Rs, T2 = j2πfτs and T3 =

11+j2πfτs

. Now, the phase response is,

� T (jf) = � T1 + � T2 + � T3 (19)

T1 has a phase of 0 as it is purely real, T2 contributes a phase of 90◦ as it ispurely imaginary and T3 is a complex number and its contibution is arc tan of ratioof imaginary and real part. That is,

� T (jf) = Θ(f) = 0 + 90◦ − tan−1 (2πfτs) (20)

= 90◦ − tan−1 (2πfτs) (21)

Here too, Bode has an interesting suggestion. Once again we shall begin to con-struct asymptotes at the low and high extremes of the frequency axis. For the lowfrequency conditions,

limf→0

Θ(f) = 90◦ − tan−1 0 = 90◦ (22)

and at high frequency conditions

limf→∞

Θ(f) = 90◦ − tan−1 (∞) (23)

= 90◦ − 90◦ = 0◦. (24)

So we end up with two asymptotes one having the value 90◦ at low frequenciesand 0◦ at very high frequencies. At the break-point frequency we have,

Θ

(f =

1

2πτs

)= 90◦ − tan−1 (1) = 90◦ − 45◦ = 45◦. (25)

Now, with these quantities we shall try to plot the phase response,which is shownin Fig.(6). It is important to realize that the phase assumes a value of 90◦ upto thefrequency 10 times less than the break-point frequency f

10. Then it assumes a straight

line and touches 45◦ at the break-point frequency with slope 45◦ per decade. Thiscontinues till 10 times the break-point frequency 10f where the phase reaches 0◦ andremains at this value for all higher frequencies.

Bode’s suggestion is to consider 10f as high frequency and f10

as low frequencywhen we are operating at a frequency f.

8

(f)

degree

90

45

0 f/10 f 10f

Figure 6: Phase response of Circuit with Series Capacitor

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