MSEE20-003

Saadia Mansoor Kazmi

Assignment 1
Sensors & Systems

Derive the formulations for internal capacitance of IC designing and how they are
effective at high frequency response of the amplifier

As the transistors are the most commonly used component in IC’s we shall discuss the internal
capacitance effects on the high frequency response

Recall from basic circuit theory that XC = 1/(2 pi fC).

At high frequencies, the coupling and bypass capacitors become effective ac shorts and do not
affect an amplifier's response. Internal transistor junction capacitances, however, do come into
play, reducing an amplifier's gain and introducing phase shift as the signal frequency increases.

FIG. 1 shows the internal pn junction capacitances for both a bipolar junction transistor and a
JFET. In the case of the BJT, Cbe is the base-emitter junction capacitance and Cbc is the base-
collector junction capacitance. In the case of the JFET, Cgs is the capacitance between gate
and source and Cgd is the capacitance between gate and drain.

FIG. 1--Internal transistor capacitances.

FIG. 2--AC equivalent circuit for a BJT amplifier showing effects of the internal capacitances
Cbe and Cbc.
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

(a) Effect of Cbe, where Vb is reduced by the voltage-divider action of Rs and XCbe;

(b) Effect of Cbc, where part of Vout (Vfb) goes back through Cbc to the base and reduces the
input signal because it is approximately 180° out of phase with Vin.

Datasheets often refer to the BJT capacitance Cbc as the output capacitance, often designated
Cob. The capacitance Cbe is often designated as the input capacitance Cib.

Datasheets for FETs normally specify input capacitance Ciss and reverse transfer capacitance
Crss. From these, Cgs and Cgd can be calculated.

At lower frequencies, the internal capacitances have a very high reactance because of their low
capacitance value (usually only a few picofarads) and the low frequency value. Therefore, they
look like opens and have no effect on the transistor's performance. As the frequency goes up,
the internal capacitive reactances go down, and at some point they begin to have a significant
effect on the transistor's gain. When the reactance of Cbe (or Cgs) becomes small enough, a
significant amount of the signal voltage is lost due to a voltage-divider effect of the signal source
resistance and the reactance of Cbe, as illustrated in FIG. 2(a). When the reactance of Cbc (or
Cgd) becomes small enough, a significant amount of output signal voltage is fed back out of
phase with the input (negative feedback), thus effectively reducing the volt age gain. This is
illustrated in FIG. 2(b).

Miller's Theorem

Miller's theorem is used to simplify the analysis of inverting amplifiers at high frequencies where
the internal transistor capacitances are important. The capacitance Cbc in BJTs (Cgd in FETs)
between the input (base or gate) and the output (collector or drain) is shown in FIG. 3(a) in a
generalized form. Av is the absolute voltage gain of the inverting amplifier at midrange
frequencies, and C represents either Cbc or Cgd.

FIG. 3--General case of Miller input and output capacitances. C represents Cbc or Cgd.

Miller's theorem states that C effectively appears as a capacitance from input to ground, as
shown in FIG. 3(b), that can be expressed as follows:

EQN. 1 Cin(Miller) = C(Av + 1)

MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

This formula shows that Cbc (or Cgd) has a much greater impact on input capacitance than its
actual value. For example, if Cbc = 6 pF and the amplifier gain is 50, then Cin(Miller) =306 pF.
FIG. 4 shows how this effective input capacitance appears in the actual ac equivalent circuit in
parallel with Cbe (or Cgs).

FIG. 4--Amplifier ac equivalent circuits showing internal capacitances and effective Miller
capacitances.

Miller's theorem also states that C effectively appears as a capacitance from output to ground,
as shown in FIG. 4(b), that can be expressed as follows:

EQN. 2 Cout(Miller)=C((Av+1)/Av)

This formula indicates that if the voltage gain is 10 or greater, Cout(Miller) is approximately
equal to Cbc or Cgd because (Av = 1) Av is approximately equal to 1. FIG. 4 also shows how
this effective output capacitance appears in the ac equivalent circuit for BJTs and FETs.

If the frequency is increased sufficiently, a point is reached where the transistor's internal
capacitances begin to have a significant effect on the gain. The basic differences between BJTs
and FETs are the specifications of the internal capacitances and the input resistance.

BJT Amplifiers
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FIG. 5 Capacitively coupled amplifier and its high-frequency equivalent circuit.

FIG. 6 High-frequency equivalent circuit after applying Miller's theorem.

A high-frequency ac equivalent circuit for the BJT amplifier in FIG. 5(a) is shown in FIG. 5(b).
Notice that the coupling and bypass capacitors are treated as effective shorts and do not appear
in the equivalent circuit. The internal capacitances, Cbe and Cbc, which are significant only at
high frequencies, do appear in the diagram. As previously mentioned, Cbe is sometimes called
the input capacitance Cib, and Cbc is sometimes called the output capacitance Cob. Cbe is
specified on datasheets at a certain value of VBE. Often, a datasheet will list Cib as Cibo and
Cob as Cobo. The "o" as the last letter in the subscript indicates the capacitance is measured
with the base open. For example, a 2N2222A transistor has a Cbe of 25 pF at VBE=0.5 V dc, IC
= 0, and f = 1 MHz. Also, Cbc is specified at a certain value of VBC. The 2N2222A has a
maximum Cbc of 8 pF at VBC= 10 V dc.

Miller's Theorem in High-Frequency Analysis

By applying Miller's theorem to the inverting amplifier in FIG. 5(b) and using the midrange
voltage gain, you have a circuit that can be analyzed for high-frequency response. Looking in
from the signal source, the capacitance Cbc appears in the Miller input capacitance from base
to ground.
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

Cin(Miller) = Cbc (Av + 1)

Cbe simply appears as a capacitance to ac ground, as shown in FIG. 6, in parallel with

Cin(Miller). Looking in at the collector, Cbc appears in the Miller output capacitance from
collector to ground. As shown in FIG. 6, the Miller output capacitance appears in parallel with
Rc.

Cout(Miller) = Cbc ( Av + 1 / Av )

These two Miller capacitances create a high-frequency input RC circuit and a high frequency
output RC circuit. These two circuits differ from the low-frequency input and output circuits,
which act as high-pass filters, because the capacitances go to ground and therefore act as low-
pass filters. The equivalent circuit in FIG. 6 is an ideal model because stray capacitances that
are due to circuit interconnections are neglected.

The Input RC Circuit

At high frequencies, the input circuit is as shown in FIG. 7(a), where Beta_ac r’e is the input
resistance at the base of the transistor because the bypass capacitor effectively shorts the
emitter to ground. By combining Cbe and Cin (Miller) in parallel and repositioning, you get the
simplified circuit shown in FIG. 7(b). Next, by thevenizing the circuit to the left of the capacitor,
as indicated, the input RC circuit is reduced to the equivalent form shown in FIG. 7(c).

FIG. 7 Development of the equivalent high-frequency input RC circuit.

MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

As the frequency increases, the capacitive reactance becomes smaller. This causes the signal
voltage at the base to decrease, so the amplifier's voltage gain decreases. The reason for this is
that the capacitance and resistance act as a voltage divider and, as the frequency increases,
more voltage is dropped across the resistance and less across the capacitance. At the critical
frequency, the gain is 3 dB less than its midrange value. The upper critical high frequency of the
input circuit, fcu (input ), is the frequency at which the capacitive reactance is equal to the total
resistance.

EQN. 16

where Rs is the resistance of the signal source and Ctot = Cbe + Cin(Miller). As the frequency
goes above fcu(input), the input RC circuit causes the gain to roll off at a rate of -20 dB/decade
just as with the low-frequency response.

Phase Shift of the Input RC Circuit

Because the output voltage of a high-frequency input RC circuit is across the capacitor, the
output of the circuit lags the input. The phase angle is expressed as:

EQN. 17

At the critical frequency, the phase angle is 45° with the signal voltage at the base of the
transistor lagging the input signal. As the frequency increases above fc, the phase angle in
creases above 45° and approaches 90° when the frequency is sufficiently high.

The Output RC Circuit

MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FIG. 8 Development of the equivalent high-frequency output RC circuit.

The high-frequency output RC circuit is formed by the Miller output capacitance and the
resistance looking in at the collector, as shown in FIG. 8(a). In determining the out put
resistance, the transistor is treated as a current source (open) and one end of RC is effectively
ac ground, as shown in FIG. 8(b). By rearranging the position of the capacitance in the diagram
and thevenizing the circuit to the left, as shown in FIG. 8(c), you get the equivalent circuit in FIG.
8(d). The equivalent output RC circuit consists of a resistance equal to the parallel combination
of RC and RL in series with a capacitance that is determined by the following Miller formula:

If the voltage gain is at least 10, this formula can be approximated as

The upper critical frequency for the output circuit is determined with the following equation,
where Rc = RC || RL.

EQN. 18 Cout(millewr)=Cbc((AV+1)/Av)=Cbc and Rc=RcIIRl

Fcu(output)=1/2piRcCout(miller)

Theta= tan-1(Rc/Xcout(miller))

Just as in the input RC circuit, the output RC circuit reduces the gain by 3 dB at the critical
frequency. When the frequency goes above the critical value, the gain drops at a rate. The
phase angle introduced by the output RC circuit is -20 dB/dec
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FET Amplifiers

The approach to the high-frequency analysis of a FET amplifier is similar to that of a BJT
amplifier. The basic differences are the specifications of the internal FET capacitances and the
determination of the input resistance.

FIG. 9(a) shows a JFET common-source amplifier that will be used to illustrate high-frequency
analysis. A high-frequency equivalent circuit for the amplifier is shown in FIG. 9(b). Notice that
the coupling and bypass capacitors are assumed to have negligible reactances and are
considered to be shorts. The internal capacitances Cgs and Cgd appear in the equivalent circuit
because their reactances are significant at high frequencies.

FIG. 9 Example of a JFET amplifier and its high-frequency equivalent circuit.

Values of Cgs, Cgd, and Cds

FET datasheets do not normally provide values for Cgs, Cgd, or Cds. Instead, three other
values are usually specified because they are easier to measure.

These are Ciss, the input capacitance; Crss, the reverse transfer capacitance; and Coss, the
output capacitance. Because of the manufacturer's method of measurement, the following
relationships allow you to determine the capacitor values needed for analysis.

EQN. 20; EQN. 21; EQN. 22 shown here:

MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

Coss is not specified as often as the other values on datasheets. Sometimes, it is designated as
Cd (sub), the drain-to-substrate capacitance. In cases where a value is not available, you must
either assume a value or neglect Cds.

Using Miller's Theorem

Miller's theorem is applied the same way in FET inverting amplifier high-frequency analysis as
was done in BJT amplifiers. Looking in from the signal source in FIG. 9(b), Cgd effectively
appears in the Miller input capacitance, which was given in EQN. 1, as follows:

Cin (Miller) = Cgd (Av + 1)

Cgs simply appears as a capacitance to ac ground in parallel with Cin(Miller), as shown in FIG.
10. Looking in at the drain, Cgd effectively appears in the Miller output capacitance (from EQN.
2) from drain to ground in parallel with Rd, as shown in FIG. 10.

Cout(Miller) = Cgd ( Av + 1 / Av )

These two Miller capacitances contribute to a high-frequency input RC circuit and a high
frequency output RC circuit. Both are low-pass filters, which produce phase lag.

FIG. 10 High-frequency equivalent circuit after applying Miller's theorem.

MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FIG. 11 Input RC circuit.

The Input RC Circuit

The high-frequency input circuit forms a low-pass type of filter and is shown in FIG. 11(a).
Because both RG and the input resistance at the gate of FETs are extremely high, the
controlling resistance for the input circuit is the resistance of the input source as long as Rs <<
Rin.

This is because Rs appears in parallel with Rin when Thevenin's theorem is applied. The
simplified input RC circuit appears in FIG. 11(b). The upper critical frequency for the input circuit
is

where Ctot Cgs Cin(Miller ). The input RC circuit produces a phase angle of

The effect of the input RC circuit is to reduce the midrange gain of the amplifier by 3 dB at the
critical frequency and to cause the gain to decrease at -20 dB/decade above fc.

The Output RC Circuit

The high-frequency output RC circuit is formed by the Miller output capacitance and the output
resistance looking in at the drain, as shown in FIG. 12(a). As in the case of the BJT, the FET is
treated as a current source. When you apply Thevenin's theorem, you get an equivalent output
RC circuit consisting of RD in parallel with RL and an equivalent output capacitance.
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FIG. 12 Output RC circuit.

This equivalent output circuit is shown in FIG. 12(b). The critical frequency of the output RC lag
circuit is EQN. 25, EQN. 26, shown below:

Total High-Frequency Response of an Amplifier

As you have seen, the two RC circuits created by the internal transistor capacitances influence
the high-frequency response of both BJT and FET amplifiers. As the frequency increases and
reaches the high end of its midrange values, one of the RC circuits will cause the amplifier's
gain to begin dropping off. The frequency at which this occurs is the dominant upper critical
frequency; it is the lower of the two upper critical high frequencies. An ideal high-frequency
Bode plot is shown in FIG. 13(a). It shows the first break point at fcu(input) where the voltage
gain begins to roll off at -20 dB/decade. At fcu(output), the gain begins dropping at -10
dB/decade because each RC circuit is providing a -20 dB/decade roll-off. FIG. 13(b) shows a
nonideal Bode plot where the voltage gain is actually -3 dB/decade below midrange at
fcu(input). Other possibilities are that the output RC circuit is dominant or that both circuits have
the same critical frequency.
MSEE20-003
Saadia Mansoor Kazmi
Assignment 1
Sensors & Systems

FIG. 13 High-frequency Bode plots.

References:

1. https://www.industrial-electronics.com/electrnc-dvcs-9e_10.html

2. http://www.ittc.ku.edu/~jstiles/412/handouts/5.8%20BJT%20Internal%20Capacitances
%20and%20high%20frequency%20model/section%205_8%20BJT%20Internal
%20Capacitances%20lecture.pdf

3. https://staff-old.najah.edu/sites/default/files/Chapter%2010.pdf