operational amplifier (op amp)

ideal op amp

circuit with two inputs and one output

• Rin = infinity (should not draw current from the source)

• Rout is zero

• A is infinite asVo/Vi and we want an extremely small Vi to produce large Vo

equivalent circuit of op amp

V- and V+ on left, Y on right

Vload

Vsource - IRo, if Ro is zero, then Vload = Vsource which is ideal

real op amp

numbers labelled anticlockwise from notch when looking from above

negative feedback

if output falls then V- will fall by the same amount so that V+ - V- will increase, this then causes Y to rise as Y = A(V+ - V-) = A(X-Y)

analyzing op amp circuits

1) check for negative feedback to ensure an increase in Y makes V+ - V- decrease, Y must be connected to V-

2) assume that V+ = V-, since it =Y xA, this is the same thing as assuming A = infinity

3) Assume zero input current

4) apply KCL at each node separately, with zero input current, do not apply at output as unknown

non-inverting amplifier

gain is positive, choose ratio R2/R1

voltage follower

R1= infinity and R2 = 0

• gain is (1+R2)/R1 = 1

• voltage follower makes Y=U

• op amp takes almost no current

• output provides lots of current to the load

• makes voltage easier to deliver

inverting amplifiers

1) V+ = 0, V- = 0

2) find that gain is negative so this is inverting amplifier

3) negative feedback holds V- close to V+

4) If V+ = 0, V- is called virtual ground

point of amplifier if gain is 1

isolate source from load

inverting summing amplifier

• connects several input signals to inverting amplifier

• as before, V- = 0 is virtual earth due to negative feedback (V+ is 0)

• KCL at V- node, weight of Xi = Rf/Ri

• Sum gain from final and each input resistor multiplied by input voltage to get final

differential amplifier

2 input circuit combining inverting and non-inverting amplifiers

• Linearity is Z = aX + bY

choosing resistor values

behaviour of an op-amp circuit depends on the ratio of resistor values

Gain = - R2/R1, how do you choose between giga, kilo and Mega?

• small resistors = large current

• large resistors make current in no longer negligble

• - so use 1-100k

Bandwidth of real op amp

infinite gain at low frequency but decreases rapidly as signal frequency increases

• op amps are characterised by frequency at which the gain becomes 1

• this is unity gain bandwidth

• 1MHz for MCP6002

benefits of negative feedback

• improves bandwidth

• gain is reduced, bandwidth is increased due to negative feedback

• gain x bandwidth = constant (gain-bandwidth product)