AJ Syllabus: Transistor Frequency Response

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

• At the high – frequency end, there are two factors that define the – 3dB cutoff point:

– The network capacitance ( parasitic and introduced) and

– the frequency dependence of hfe(b)

Network parameters

• In the high frequency region, the RC network of the amplifier has the configuration

shown below.

• At increasing frequencies, the reactance XC will decrease in magnitude, resulting

in a short effect across the output and a decreased gain.

Vo = Vi(-jXC) / R -jXC

Vo / Vi = 1/[ 1+j(R/XC)] ; XC = 1/2pfC

AV = 1/[ 1+j(2pfRC)];

AV = 1/[ 1+jf/f2]

o This results in a magnitude plot that drops off at 6dB / octave with increasing

frequency.

Network with the capacitors that affect the high frequency response

• Capacitances that will affect the high-frequency response:

· Cbe, Cbc, Cce – internal capacitances

· Cwi, Cwo – wiring capacitances

· CS, CC – coupling capacitors

· CE – bypass capacitor

The capacitors CS, CC, and CE are absent in the high frequency equivalent of the BJT

amplifier.The capacitance Ci includes the input wiring capacitance, the transition

capacitance Cbe, and the Miller capacitance CMi.The capacitance Co includes the

output wiring capacitance Cwo, the parasitic capacitance Cce, and the output Miller

capacitance CMo.In general, the capacitance Cbe is the largest of the parasitic

capacitances, with Cce the smallest.

As per the equivalent circuit,

fH = 1 / 2pRthiCi

Rthi = Rs|| R1||R2||Ri

Ci = Cwi+Cbe+CMi = CWi + Cbe+(1- AV) Cbe

At very high frequencies, the effect of Ci is to reduce the total impedance of the

parallel combination of R1, R2, Ri, and Ci.The result is a reduced level of voltage

across Ci, a reduction in Ib and the gain of the system.

For the output network,

fHo = 1/(2pRThoCo)

RTho = RC||RL||ro

Co = Cwo+Cce+CMo

At very high frequencies, the capacitive reactance of Co will decrease and

consequently reduce the total impedance of the output parallel branches.

The net result is that Vo will also decline toward zero as the reactance Xc becomes

smaller.The frequencies fHi and fHo will each define a -6dB/octave asymtote.

If the parasitic capacitors were the only elements to determine the high – cutoff

frequency, the lowest frequency would be the determining factor.However, the

decrease in hfe(or b) with frequency must also be considered as to whether its break

frequency is lower than fHi or fHo.

hfe (or ) variation

• The variation of hfe( or b) with frequency will approach the following relationship

hfe = hfe mid / [1+(f/fb)]

• fb is that frequency at which hfe of the transistor falls by 3dB with respect to its

mid band value.

• The quantity fb is determined by a set of parameters employed in the hybrid p

model.

• In the hybrid p model, rb includes the

• base contact resistance

• base bulk resistance

• base spreading resistance

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

• A voltage divider BJT bias configuration with load is considered for this analysis.

• For such a network of voltage divider bias, the capacitors CS, CC and CE will

determine the low frequency response.

CS:

• At mid or high frequencies, the reactance of the capacitor will be sufficiently

small to permit a short – circuit approximations for the element.

• The voltage Vi will then be related to Vs by

Vi |mid = VsRi / (Ri+Rs)

• At f = FLS, Vi = 70.7% of its mid band value.

• The voltage Vi applied to the input of the active device can be calculated using the

voltage divider rule:

Vi = RiVs / ( Ri+ Rs – jXCs)

Effect of CC:

• Since the coupling capacitor is normally connected between the output of the

active device and applied load, the RC configuration that determines the low

cutoff frequency due to CC appears as in the figure given below.

• Ro = Rc|| ro

Effect of CE:

• The effect of CE on the gain is best described in a quantitative manner by recalling

that the gain for the amplifier without bypassing the emitter resistor is given by:

AV = - RC / ( re + RE)

• Maximum gain is obviously available where RE is 0W.

• At low frequencies, with the bypass capacitor CE in its “open circuit” equivalent

state, all of RE appears in the gain equation above, resulting in minimum gain.

• As the frequency increases, the reactance of the capacitor CE will decrease,

reducing the parallel impedance of RE and CE until the resistor RE is effectively

shorted out by CE.

• The result is a maximum or midband gain determined by AV = - RC / re.

• The input and output coupling capacitors, emitter bypass capacitor will affect only

the low frequency response.

• At the mid band frequency level, the short circuit equivalents for these capacitors

can be inserted.

• Although each will affect the gain in a similar frequency range, the highest low

frequency cutoff determined by each of the three capacitors will have the greatest

impact.

Problem:

Determine the lower cutoff freq. for the network shown using the following

parameters:

Cs = 10μF, CE = 20μF, Cc = 1μF

Rs = 1kΩ, R1= 40kΩ, R2 = 10kΩ,

RE = 2kΩ, RC = 4kΩ, RL = 2.2kΩ,

β = 100, ro = ∞Ω, Vcc = 20V

• Solution:

a. To determine re for the dc conditions, let us check whether bRE > 10R2

Here, bRE = 200kW, 10R2 = 100kW. The condition is satisfied. Thus approximate

analysis can be carried out to find IE and thus re.

VB = R2VCC / ( R1+R2) = 4V

VE = VB – 0.7 = 3.3V

IE = 3.3V / 2kW = 1.65mA

re = 26mV / 1.65mA = 15.76 W

Mid band gain:

AV = Vo / Vi = -RC||RL / re = - 90

• Input impedance

Zi = R1 || R2|| bre = 1.32K

• Cut off frequency due to input coupling capacitor ( fLs)

fLs = 1/ [2p(Rs +Ri)CC1 = 6.86Hz.

fLc = 1 / [2p(RC + RL) CC

= 1 / [ 6.28 (4kW + 2.2kW)1uF]

= 25.68 Hz

Effect of CE:

R¢S = RS||R1||R2 = 0.889W

Re = RE || (R¢S/b + re) = 24.35 W

fLe = 1/2p ReCE = 327 Hz

fLe = 327 Hz

fLC = 25.68Hz

fLs = 6.86Hz

In this case, fLe is the lower cutoff frequency.

• In the high frequency region, the capacitive elements of importance are the interelectrode

( between terminals) capacitances internal to the active device and the

wiring capacitance between leads of the network.

• The large capacitors of the network that controlled the low frequency response are

all replaced by their short circuit equivalent due to their very low reactance level.

• For inverting amplifiers, the input and output capacitance is increased by a

capacitance level sensitive to the inter-electrode capacitance between the input

and output terminals of the device and the gain of the amplifier.

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

• Analysis of the above circuit indicates that,

XC = 1/2pfC @ 0 W

• Thus, Vo = Vi at high frequencies.

• At f = 0 Hz, XC = , Vo = 0V.

• Between the two extremes, the ratio, AV = Vo / Vi will vary.

As frequency increases, the capacitive reactance decreases and more of the input

voltage appears across the output terminals.

The output and input voltages are related by the voltage – divider rule:

Vo = RVi / ( R – jXC)

the magnitude of Vo = RVi / ÖR2 + XC2

• For the special case where XC = R,

Vo =RVi / RÖ2 = (1/Ö2) Vi

AV = Vo / Vi = (1/Ö2) = 0.707

• The frequency at which this occurs is determined from,

XC = 1/2pf1C = R

where, f1 = 1/ 2pRC

• Gain equation is written as,

AV = Vo / Vi

= R / (R – jXC) = 1/ ( 1 – j(1/wCR)

= 1 / [ 1 – j(f1 / f)]

• In the magnitude and phase form,

AV = Vo / Vi

= [1 /Ö 1 + (f1/f)2 ] Ð tan-1 (f1 / f)

• In the logarithmic form, the gain in dB is

AV = Vo / Vi = [1 /Ö 1 + (f1/f)2 ]

= 20 log 10 [1 /Ö 1 + (f1/f)2 ]

= - 20 log 10 Ö [ 1 + (f1/f)2]

= - 10 log10 [1 + (f1/f)2]

• For frequencies where f << f1 or (f1/ f)2 the equation can be approximated by

AV (dB) = - 10 log10 [ (f1 / f)2]

= - 20 log10 [ (f1 / f)] at f << f1

• At f = f1 ;

f1 / f = 1 and

– 20 log101 = 0 dB

• At f = ½ f1;

f1 / f = 2

– 20 log102 = - 6 dB

• At f = ¼ f1;

f1 / f = 4

– 20 log102 = - 12 dB

• At f = 1/10 f1;

f1 / f = 10

– 20 log1010 = - 20dB

• The above points can be plotted which forms the Bode – plot.

• Note that, these results in a straight line when plotted in a logarithmic scale.

Although the above calculation shows at f = f1, gain is 3dB, we know that f1 is

that frequency at which the gain falls by 3dB. Taking this point, the plot differs

from the straight line and gradually approaches to 0dB by f = 10f1.

Observations from the above calculations:

• When there is an octave change in frequency from f1 / 2 to f1, there exists

corresponding change in gain by 6dB.

• When there is an decade change in frequency from f1 / 10 to f1, there exists

corresponding change in gain by 20 dB.

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

• At low frequencies the coupling and bypass capacitors can no longer be replaced

by the short – circuit approximation because of the increase in reactance of these

elements.

• The frequency – dependent parameters of the small signal equivalent circuits and

the stray capacitive elements associated with the active device and the network

will limit the high frequency response of the system.

• An increase in the number of stages of a cascaded system will also limit both the

high and low frequency response.

• The horizontal scale of frequency response curve is a logarithmic scale to permit a

plot extending from the low to the high frequency

• For the RC coupled amplifier, the drop at low frequencies is due to the increasing

reactance of CC and CE, whereas its upper frequency limit is determined by either

the parasitic capacitive elements of the network or the frequency dependence of

the gain of the active device.

• In the frequency response, there is a band of frequencies in which the magnitude

of the gain is either equal or relatively close to the midband value.

• To fix the frequency boundaries of relatively high gain, 0.707AVmid is chosen to

be the gain at the cutoff levels.

• The corresponding frequencies f1 and f2 are generally called corner, cutoff, band,

break, or half – power frequencies.

• The multiplier 0.707 is chosen because at this level the output power is half the

midband power output, that is, at mid frequencies,

•

PO mid = | Vo2| / Ro = | AVmidVi|2 / RO

• And at the half – power frequencies,

POHPF = | 0.707 AVmidVi|2 / Ro

= 0.5| AVmid Vi|2 / Ro

• And, POHPF = 0.5 POmid

• The bandwidth of each system is determined by f2 – f1

• A decibel plot can be obtained by applying the equation,

(AV / AVmid )dB

= 20 log10 (AV / AVmid)

§ Most amplifiers introduce a 180° phase shift between input and output signals. At low

frequencies, there is a phase shift such that Vo lags Vi by an increased angle. At high

frequencies, the phase shift drops below 180°.

Low – frequency analysis – Bode plot

In the low frequency region of the single – stage BJT amplifier, it is the RC combinations

formed by the network capacitors CC and CE, the network resistive parameters that

determine the cutoff frequencies.

All Syllabus

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

Network with the capacitors that affect the high frequency response

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

ANALOG ELECTRONIC CIRCUITS × Transistor Frequency Response

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