© 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.

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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

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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.

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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

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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.

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Figure 4.7: AC High Frequency Model

Figure 4.8: Generic Amplifier with feedback.

Figure 4.9: Capacitances in the generic amplifier model

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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)

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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)

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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

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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

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Find the Mid Band Gain

Find the Low Frequency Response

Find the High Frequency Response

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Solution:

Find the Bias Point

Find the Mid Band Gain

Find the Low Frequency Response

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Find the High Frequency Response

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