JFET Frequency Response - Welcome to MATCecampus.matc.edu/lokkenr/electronics/elctec-121/lectures/JFET... · JFET Frequency Response ... Figure 4.6: FET internal Capacitances. ©

© 2009 Richard Lokken 4/11/2009 1

JFET Frequency Response

2 2c cBW f f (4.1)

Where:

BW =

fc1 =

fc2 =

What defines midband?

FrequencyHz

AmplitudedB

Ap(mid)

.5Ap(mid)

Bandwidth

fc1 fc2

High-frequencydrop in power gain

Low-frequency dropin power gain

Figure 4.1: A simplified frequency response curve

© 2009 Richard Lokken 4/11/2009 2

The geometric center frequency of the amplifier is _____________________________________.

0 1 2c cf f f (4.2)

Measuring fc1 and fc2

When using the oscilloscope the following procedure is used:

1.

2.

3.

4.

5.

6.

© 2009 Richard Lokken 4/11/2009 3

Graphing the frequency response

1.

2.

3.

Review dB Power Gain

Explain

( ) 10 log outp dB

in

PA

P (4.3)

Convert equation 4.3 to voltage gain.

( )v dBA (4.4)

Low Frequency Response

© 2009 Richard Lokken 4/11/2009 4

Figure 4.2: JFET CS Amplifier Circuit

First we will examine the circuit with only the coupling capacitors in place.

Figure 4.3: AC Model with only coupling capacitors.

© 2009 Richard Lokken 4/11/2009 5

1

bggen i genR R C

(4.5)

1

bLD L LR R C

(4.6)

Figure 4.4: AC Model with only CS

3

1b

S SR C (4.7)

4

1 m Sb

S S

g RR C

(4.8)

RG

RD

RS

RL

CS

iV

gsV

oV

m gsgV

© 2009 Richard Lokken 4/11/2009 6

Figure 4.5: Frequency Response due to CS

High Frequency Response

The high frequency response of the JFET is limited by values of ___________________________ just as

for the BJT.

Figure 4.6: FET internal Capacitances.

© 2009 Richard Lokken 4/11/2009 7

Figure 4.7: AC High Frequency Model

Figure 4.8: Generic Amplifier with feedback.

Figure 4.9: Capacitances in the generic amplifier model

© 2009 Richard Lokken 4/11/2009 8

Write a KCL at the input

inTI (4.9)

1I (4.10)

fI (4.11)

inTI (4.12)

Now Solve for total admittance

inTY (4.13)

inTY (4.14)

By direct comparison of equation 4.13 and equation 4.14

MiC (4.15)

Perform a similar analysis at the output

oTI (4.16)

oI (4.17)

© 2009 Richard Lokken 4/11/2009 9

fI (4.18)

oTI (4.19)

Solve for output admittance

oTY (4.20)

oTY (4.20)

By direct comparison of equation 4.20 and 4.21

MoC (4.21)

Figure 4.6: AC Model with Miller Capacitances.

gen ibi

gen i i

R RR R C

(4.22)

© 2009 Richard Lokken 4/11/2009 10

1bo

o oR C (4.23)

Where

i Mi gsC C C (4.24)

1Mi gd vC C A (4.25)

i GR R (4.26)

o MoC C (4.27)

11Mo gd

v

C CA

(4.28)

||o D LR R R (4.29)

© 2009 Richard Lokken 4/11/2009 11

Example 4.1

Figure 4.7: Circuit for example 4.1

Given:

RG = 250 kΩ

RS = 500 Ω

RD = 2 kΩ

RL = 5 kΩ

VDD = 20 V

IDSS = 6 mA

© 2009 Richard Lokken 4/11/2009 12

VP = ‐3 V

C1 = 10 μF

C2 = 10 μF

CS = 100 μF

Cgs = 4 pF

Cgd = 2 pF

Desired:

Mid band Gain

Low Frequency Response

High Frequency Response

Strategy:

Find the Bias Point

© 2009 Richard Lokken 4/11/2009 13

Find the Mid Band Gain

Find the Low Frequency Response

Find the High Frequency Response

© 2009 Richard Lokken 4/11/2009 14

Solution:

Find the Bias Point

Find the Mid Band Gain

Find the Low Frequency Response

© 2009 Richard Lokken 4/11/2009 15

Find the High Frequency Response

© 2009 Richard Lokken 4/11/2009 16