ELE2ANALOGUE

REVISION NOTES

OP-AMPs

• Differential amplifier• Large input resistance 109• Large open-loop voltage

gain, A, 106

• Low output resistance 100• Frequency compensated• Gain-bandwidth product

V+ V–

Vout

+Vs

-Vs

0V

+

)VV(AVout

Inverting Amplifier

• Input resistance = R1

• Output inverted• Virtual earth point• Saturation/clipping• Bandwidth/voltage gain• Minimum resistance 1k

R

R

1

f

0V

VoutVin

+

_

R

R

V

V

1

f

in

out

Summing Amplifier

• Recognise by multiple inputs• Input resistance of each input

= R• Behaves the same as the inverting amplifier

• Used in DAC & Mixer

R

R1

f

0V

Vout

+

_

V1

V2

V3

R2

R3

R

V

R

V

R

VRV

3

3

2

2

1

1fout

Non-inverting amplifier

• Very large input resistance 109

• Beware input resistor to 0V• Output in phase with input• Voltage gain/bandwidth• Minimum voltage gain = 1

• Voltage follower when Rf = 0 or R1=

R1

R f

0V

Vout

Vin

+

_

R

R1V

V

1

f

in

out

Source Follower

• Voltage gain 0.7 – 1• Current gain 109

• Output voltage in phase with input

• Very large input resistance 109

• Remember to subtract Vgs from Vin to obtain Vout

• N-channel and p-channel

0V

Vin VoutR1

+Vs

Push-pull amplifier 1

• Provides power gain• Needs op-amp to provide

voltage gain• Cross-over distortion at

low volume• Bias MOSFETs into

conduction• Saturation/clipping at high

volume• Increase supply voltage

or reduce input voltage

0Vload

+Vs

–Vs

Vin

0V

n-channel

p-channel

Push-pull amplifier 2

• Negative feedback• Heat sinks

– Metal• conduction

– Black• radiation

– Large surface area• Convection

• Radiation

– Fan• Convection

0V0V

IRF630

IRF9630

Vin+–

8

10k

47k

47k

–12V

+12V

10k

10k

R2VsP2

Capacitor reactance

• Reactance – the opposition to the flow of alternating current.

• Measured in ohms.• Capacitor reactance

decreases with frequency.

fCπ2

1Xc

frequency (f)

reactance, Xcapacitive

c

Low pass filter

• Allows low frequency signals to pass, attenuates high frequency signals

• Break point Xc = R

C

R

VoutVin

0.001

0.01

0.1

1

0.01f 0.1f 1f 10f 100f 1000fofrequency

Vout

Vin

o o o o o

RC2

1fo

High pass filter

• Allows high frequency signals to pass, attenuates low frequency signals

• Break point Xc = R

RC2

1fo

C

R VoutVin

0.001

0.01

0.1

1

0.01fofrequency

Vout

Vin

0.1fo 1fo 10fo 100fo0.001fo

ELE2 Questions

• Set in context – read the context – underline relevant details.

• Look carefully at circuit diagrams – circle and name sub-systems.

• Focus on relevant bookwork descriptions and calculations – answer these.

• Attempt all sections – no credit can be given if there is no attempt at an answer.

• Estimate, Calculate – show your working.• State and explain – do both!