electronic devices active filters · of active filters are: low pass filter, high pass filter, band...

ET Training

Electronic Devices Active Filters Instructor: H.Pham Email: [email protected]

0

f = 1/T

T

distance 0

time

Wavelength =V/F

Where V=3x108 m/s

And F=hz

ET Training


ET Training


+

Time domain (Oscilloscope )

Pure of 1 kHz square-wave

Frequency domain (Spectrum Analyzer)

Picture of 1 kHz square-wave

ET Training


A 1 kHz square wave

500 milliV/Div

200 usec/Div

On the oscilloscope

A 1 kHz square wave

100 milliV/Div

That shows on the spectrum analyzer

ET Training


10/25/2018 5

• Filters are circuits that are capable of passing signals with certain selected frequencies while rejecting signals with other frequencies. This property is called selectivity.

• Active filters use transistors or op-amps combined with passive RC, RL or RCL circuits. The active devices provide voltage gain , and the passive circuits provide frequency selectivity. The four basic categories of active filters are: low pass filter, high pass filter, band pass filter, band stop filter (Notch filter )

1. LOW PASS FILTER

0 dB

fc f

Av(dB)=Vo/Vi

BW=Band Width

Ideal Response

-3 dB

0 dB

fc

Slope=-20dB/decade

10fc f

BW=Band Width

Av(dB)=Vo/Vi

Practical Response

ET Training


10/25/2018 6

• One pole Low Pass Filter (LPF) ( one resistor, one capacitor )

• LPF only allows low frequency signals from 0 hertz to its cut-off frequency fc point to pass while blocking those any higher

• cut-off frequency or corner frequency or critical frequency is calculated by the following formula: fc = 1/ 2 RC

Where f= Hz, R=Ohm, C=farad.

• at cut-off frequency the Gain(AV) is down -3dB or the output going down to 70.7% ( same as 0.707 )of its maximum values.

• after fc output decreases at a constant rate a the frequency increases. That is, when the frequency is increased tenfold (one decade) the voltage gain is divided by 10, in other words, the gain decreased 20dB(=20log10) each time the frequency is increases by 10.

• One pole LPF (one R, one C) roll-off rate -20dB/Decade or 0.1

•Two pole LPF (two R, two C ) roll-off rate -40dB/decade or 0.01

• Three pole LPF (three R, three C ) roll-off rate -60dB/decade or 0.001

ET Training


10/25/2018 7

• The following shows the response for LPF with different poles

0 dB

fc

Av(dB)

-20dB/decade (1pole LPF)

10fc 10fc 10fc



-20dB

-40dB

-60dB

• EXAMPLE 01

For the LPF circuit shown, R=1k, C=0.1uF, Vin=10Vrms. Calculate:

1. Gain AV = _____NA

2. cut-off frequency fc=______ kHz

3. Vout at fc =_____________ Vrms

4. Vout at 10fc=____________ Vrms

5. Vout at 500hZ =_________ Vrms

ET Training


10/25/2018 8

• EXAMPLE 02

For the LPF circuit shown, Given Vin= 2.5Vrms. Calculate:

1. Gain AV = (1 + R2/R1): ____ 2. cut-off frequency fc=____ Hz

3. Vout max =______ Vrms 4. Vout at fc=______Vrms

5. Vout at 10fc =_____Vrms

fc

Vomax

0.707xVomax

0.1xVomax

0.01xVomax

Vo AV(db)

10fc 100fc 20log10=10dB 20log0.707=-3dB

20log0.1=-20dB 20log0.01=-40dB

___dB

___dB

___dB

___dB

Complete AV(dB )

ET Training


10/25/2018 9

2. ACTIVE HIGH PASS FILTER ( HPF )

• Pass any frequencies higher than cut-off frequency ( fc )

• Cut-off frequency formula is the same as LPF: fc = 1 / 2 R C

-3 dB

0 dB

fc fc/10 f

BW=Band Width

Or pass band

Av(dB)=Vo/Vi

HPF Response

-20 dB

-40 dB

• EXAMPLE 03

For the HPF circuit shown, Given Vin= 2.0Vrms, R=910,C=0.1uF. Calculate:

1. Gain AV = ____ 2. cut-off frequency fc=_____ kHz

3. Vout at fc =_____ Vrms 4. Vout at fc/10=______Vrms

5. Vout at 100hZ =________Vrms

ET Training


10/25/2018 10

3. ACTIVE HIGH PASS FILTER ( HPF ) using Inverting Amplifier

• EXAMPLE 04

For the HPF circuit shown, Given Vin= 1.5Vrms, R1=1k, R2=2.7k,

C=10nF.Calculate:

1. Gain AV = ____ 2. cut-off frequency fc=_____ kHz

3. Vout max =______ Vrms 4. Vout at fc =_________Vrms

5. Vout at 1.59kHz =________Vrms

ET Training


10/25/2018 11

4. Two Poles ACTIVE HIGH PASS FILTER ( HPF )

• 2 resistors R3, R4 and two capacitors C1, C2

• If R3=R4 and C1=C2 then cut-off frequency is FC = 1 / 2 RC

ET Training


10/25/2018 12

5. BAND PASS FILTER (BPF ) 5. BAND PASS FILTER (BPF )

ET Training


10/25/2018 13

5. BAND PASS FILTER RESPONSE

• Pass frequencies higher Fl and less than fH

• BW (bandwidth) = fH – fL fcenter=squareRoot ( f1xf2 )

• The quality Q=fcenter/BW

• Quality Q controls the roll-off rate and the bandwidth of the filter

ET Training


10/25/2018 14

ET Training


10/25/2018 15

6. BAND STOP (Notch) FILTER RESPONSE

• Pass frequencies less than Fl and higher than fH

ET Training


10/25/2018 16

ET Training


10/25/2018 17

electronic devices active filters · of active filters are: low pass filter, high pass filter, band...

Documents