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Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

8 Operational Amplifiers

Op-amp design is a specialty, peopled by a small group of engineers forever dedicated to the quest of finding the universal building block. None has ever been found (hence the large number of different op-amps) but still they toil. Year after year they come up with small improvements; and each new design has one imperative requirement; it must work in any application without creating smoke or - heaven forbid - oscillation.

When designing an op-amp for an ASIC the precise application is known, so the circuit does not need to be universal and the task is easier. Not exactly a cinch, but nothing compared to what a designer of commercial op-amps has to face.

The majority of op-amps have three stages. The first stage converts the differential signal into a single-ended one; the second one provides the bulk on the gain and the third one the required output power. There is no law that says it has to be this way, it just turned out to be an approach that works well.

Bipolar Op-Amps

In our first circuit Q1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and Q2 form the differential

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1

 

 

 

 

 

 

 

Q3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

pair and Q3, a split-collector

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q4

 

 

 

 

100u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

lateral PNP transistor, is

 

 

 

 

 

 

 

In-

 

 

 

 

 

 

 

In+

 

 

C 1

 

 

 

 

 

 

connected as a current mirror

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10p

 

 

 

 

Q5

or active load. At the

 

 

 

 

 

 

 

 

 

 

Q1 Q2

 

 

 

 

 

 

 

 

 

 

 

 

 

Out

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

collectors of Q2 and Q3 we

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

have a high impedance, limited

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q7

 

 

 

 

 

 

 

 

 

Q8

 

 

Q6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q9

only by the base current of Q4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and the Early effects in Q2 and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q3. The second stage, Q4, has

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

as a load the current sink Q8

Fig. 8-1: Simple 3-stage op-amp.

and the base current of Q5, thus

its gain is also limited only by those two. The output stage is a simple emitter follower with a pull-down current sink.

Preliminary Edition September 2004

8-1

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

The operating currents for all three stages are derived from I1 through a multiple current mirror. Q9, with two emitters, delivers two times I1, an arbitrary choice; more emitters could be used if more current is required at the output for negative-going signals.

Q7 and Q8 are identical, which is no arbitrary choice. By making the collector currents of Q4 and Q3 (both halves) equal, Q3 takes as much base current from Q1 as Q4 does from Q2. Thus there is no designed-in base current error and the offset at the input is zero (for ideal matching).

Common Mode Range: Describes the minimum to maximum DC level at which the two inputs are functional. Desirable is rail-to-rail, i.e. from the negative supply to the positive one, but many op-amps only work with the inputs a volt or two above -V (or ground for a single supply) to some voltage below +V.

Common-Mode Rejection: If you connect both inputs together, bias them at a functional DC level and superimpose on the bias a small AC voltage, no signal should ideally appear at the output. In reality a small signal leaks through and the measure describes how much smaller this signal is (in dB) compared to the input.

The common-mode range in this design has limitations. The two inputs need a dc level of at least one VBE (Q1, Q2) plus a saturation voltage (Q7) above -V. If they move below this level the input pair gets no operating current. Also, the inputs need to be about 200mV below +V, otherwise Q1 or Q2 saturate and Q3 or Q4 are without current. At the output Q5 can pull the output only to within a VBE of +V.

There is also another flaw: being bases of bipolar transistors, the inputs need a current. With a minimum hFE of 100 and 50uA flowing per transistor, this amounts to base current of 0.5uA worst case. With a 100kOhm input resistance for one input and zero for the other this could amount to as much as a 50mV error at the output.

Frequency compensation is achieved with a single capacitor from the output to the high-impedance node at the output of the first stage. It

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

could have been placed just from collector

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+V

 

 

to the base of Q4, but the simulation shows

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In+

 

 

 

 

5

 

 

 

 

a small advantage for the shown

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-1

O u t

 

 

 

 

 

 

 

 

configuration.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let's first double-check this

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In-

 

 

 

 

 

 

-V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C 1

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

frequency compensation. To do this (as

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

explained in more detail in chapter 6) we

 

 

 

 

 

 

 

 

1Meg

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

place a very large inductor in the feedback

 

 

 

1

 

 

 

 

 

 

L1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vac

 

 

 

 

 

 

 

 

 

 

 

loop and feed an AC signal into the input

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

through a very large capacitor. The

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-2: Simulating loop gain

 

 

inductor (1MH, i.e. 1 million Henry)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and phase.

 

 

provides the DC bias to the input but blocks

Preliminary Edition September 2004

8-2

All rights reserved

100 1k Frequency / Hertz
Fig. 8-3: For stable operation the gain of a loop must reach 0dB before the phase reaches 0 degrees.

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

the AC signal. You want to choose the inductor and capacitor large enough so they have no effect at even the lowest frequency of interest for the circuit. Neither 1 million Henrys nor 1 Farad are practical values; they don't need to be.

And here is what we

get: The open-loop gain is

 

Y 2

 

Y 1

 

 

 

 

180

 

about 85dB. The dominant

 

 

 

 

 

160

 

160

Phase

pole, given by the

 

 

 

 

 

140

 

140

 

 

 

 

 

compensation capacitor, is at

 

120

 

120

 

 

 

 

 

 

 

 

 

 

about 2kHz, after which the

deg

100

dB

100

 

 

 

 

 

 

gain decreases steadily and

/

80

/

80

 

Phase

Gain

Gain

60

60

 

reaches unity (0dB) at about

 

 

 

 

 

40

 

40

 

12MHz. At that point the

 

 

 

 

 

20

 

20

 

phase is still at about 50

 

 

 

 

 

0

 

0

 

degrees, marginal but probably

 

-20

 

 

10k 100k 1M 10M 50M

 

 

 

 

adequate.

But, as pointed out in chapter 6, a phase-margin analysis is not the real test of stability. AC analysis uses

infinitely small signals (even if it says the signal is 1 Volt) and the operating currents and voltages are not disturbed. So, to be certain, we would have to repeat this analysis for the DC conditions over the entire (large-signal)

 

 

 

 

 

 

 

 

 

 

 

 

range of the circuit, a tedious task at best.

 

 

 

 

 

 

 

 

+V

 

We get a more immediate picture by

 

 

 

 

 

 

 

 

 

observing a large-signal pulse.

 

 

+ V

 

 

 

 

5

 

 

 

 

 

In+

 

 

 

 

 

 

 

 

Vin

 

 

 

 

 

 

 

 

To do this we eliminate the

 

 

Fig. 8-1

Out

 

 

 

 

 

 

 

 

 

-V

 

 

 

 

 

 

 

 

inductor and capacitor in the feedback

 

 

 

 

 

 

 

 

 

 

 

 

In-

 

 

 

 

 

-V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

Fig. 8-4: Buffer connection.

 

0.8

 

 

 

 

 

 

 

 

 

 

0.6

 

 

 

 

 

 

Output

 

path and connect the signal to the

 

0.4

 

 

 

 

 

 

 

input. Without resistors in the

V

0.2

 

 

 

Input

 

 

 

 

 

-0

 

 

 

 

 

 

 

 

feedback path, this is a buffer

 

-0.2

 

 

 

 

 

 

 

 

 

connection, i.e. a closed loop

 

-0.4

 

 

 

 

 

 

 

 

 

gain of one; with the entire open-

 

-0.6

 

 

 

 

 

 

 

 

 

 

-0.8

 

 

 

 

 

 

 

 

 

loop gain (85dB) being judged,

 

 

 

 

 

 

 

 

 

 

 

-1

 

 

 

 

 

 

 

 

 

this is the most severe test for

 

0

100

200

300

400

500

600

700

800

900

 

Time/nSecs

 

 

 

 

 

100nSecs/div

stability.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The output waveform

Fig. 8-5: Input and output waveforms for the

shows the amplifier to be very

 

 

 

buffer connection.

 

 

Preliminary Edition September 2004

8-3

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

stable, there is no ringing, just a small overshoot.

But the curve shows something else: rise and fall-times are substantial and almost straight lines. This is the slew-rate, the time it takes to charge and discharge the 10pF capacitor over a 2-Volt span with the operating current. You can speed it up by increasing the operating current, at the cost of power consumption.

In doing this test we assume that the op-amp needs to be operated as a buffer. What if, in a specific application, the closed loop gain is never lower than 40dB? In such a case we only have an excess open-loop gain of about 45dB, which makes compensation considerably easier.

Let's examine this. In figure 8-6 we have the same circuit as in the loop gain analysis before, except that the feedback resistors are in the loop also.

 

 

 

 

 

 

Y 2

 

Y1

 

 

 

 

 

 

 

 

 

+V

 

 

 

 

 

 

 

 

 

 

In+

+V

 

5

 

160

 

160

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-1

Out

 

 

140

 

140

Phase

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-V

 

 

 

120

 

120

 

 

 

 

 

 

In-

 

 

-V

deg/

100

dB/

100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

Phase

80

Gain

80

 

 

 

 

 

R2

R1

 

1Meg

 

60

60

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1K

99k

 

L1

 

 

40

 

40

Gain

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

20

 

20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

0

 

 

 

 

 

 

 

 

 

 

 

-20

 

100

1k

10k

100k

1M

10M

 

 

 

 

 

 

 

 

 

 

Vac

 

 

 

 

Frequency / Hertz

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-6: Simulation of loop gain and phase with closed-loop gain of 100.

 

1.5

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

/ V

0.5

 

 

 

 

 

 

 

 

 

Voltage

0

 

 

 

 

 

 

 

 

 

Output

 

 

 

 

 

 

 

 

 

-0.5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

- 1

 

 

 

 

 

 

 

 

 

 

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

 

Time/µSecs

 

 

 

 

 

200nSecs/div

Fig. 8-8: Pulse response with 40dB loop gain and a 2pF compensation capacitor).

Fig. 8-7: With the lower gain the loop shows greater phase margin, even though the compensation capacitor (C1 in figure 8-1 is reduced to 2pF.

The gain now shows a 45dB maximum instead of 85dB and thus drops to 0dB at a lower frequency. We can now reduce the value of the compensation capacitor from 10pF to 2pF and still have a phase margin of over 60 degrees.

When we check the stability with a pulse (with only the resistors in the feedback loop and the signal connected to the positive input) we see that the amplifier is just stable enough (fewer than 4 peaks in the damped

Preliminary Edition September 2004

8-4

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

oscillation). And, because the compensation capacitor is smaller, the slewrate is substantially higher.

No consideration has been given so far to noise and noise performance is in fact not that great in this design. The majority of the noise is created in the input stage; in subsequent stages the signal is larger

and the influence of noise is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

correspondingly reduced. To

 

 

200n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

lower the noise in Q1 and Q2 the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

devices need to be larger and

 

 

100n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

their operating current higher.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/ V/rtHz

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 8-9 shows the

 

 

50n

 

 

 

 

 

 

 

I1 =

10uA

 

 

 

 

 

 

 

 

 

 

 

 

 

 

noise performance of this

 

 

Noise

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

amplifier in the buffer

 

 

Output

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1 =

100uA

 

 

 

 

 

 

 

 

configuration (since the gain is 1,

 

 

 

20n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

output noise and input noise are

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

equal). As you can see, lowering

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

100 200 400

 

 

1k 2k 4k 10k 20k 40k 100k

the operating current increases

 

 

 

 

 

Frequency / H

rtz

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-9: Noise of the amplifier in the buffer

noise, an unpleasant fact of life.

 

 

 

 

 

 

 

 

connection.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V+

 

 

 

 

 

 

Naturally you can invert

 

 

 

 

 

Q7

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the polarity of the transistors and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q8

 

 

Q 9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

design an op-amp whose input

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In-

 

 

 

 

 

 

 

 

 

 

 

In+

 

 

 

 

 

 

 

 

 

 

 

 

can operate close to the negative

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Out supply (but loses its ability

 

Q1

Q2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q6

 

 

within about 1 Volt of the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

positive supply. Figure 8-10 is

 

 

 

 

 

 

 

 

5p

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q 5

 

 

 

 

 

the PNP-input equivalent of

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

50u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q 3

 

 

 

 

Q 4

 

 

 

 

SUB

 

 

 

 

 

figure, still with a rather

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V-

 

 

primitive output stage (Q6),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which can pull the output to

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-10: PNP-input equivalent of figure 8-1.

 

 

within about 150mV of the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

positive supply (if the load does

not require more the 50uA), but only down to about 1 Volt above the negative rail.

Let's now consider a design in which the inputs can be operated all the way down to the level of the negative rail and the output swings (almost) rail-to-rail. The input stage in figure 8-11 is a configuration known as the folded cascode stage . With an operating current (I1) of 10uA the voltage drop across R1 and R2 is a mere 50mV, thus the inputs can go about 250mV below the negative supply rail without saturating Q1 or Q2.

Preliminary Edition September 2004

8-5

All rights reserved

the loop gain is 110dB.
Frequency / Hertz
Fig. 8-12: Phase margin of figure 8-11.

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

The collector currents of Q1 and Q2 upset the balance of the Wilson current mirror Q3, Q4 and Q5 and the difference signal is picked up by Q6.

The output

 

 

 

 

 

+V

Q15

Q16

 

 

 

 

stage has two

 

 

 

 

 

 

 

 

Q17 Q10

Q13

branches to it. The

 

 

 

 

 

 

first one is simply

 

 

 

 

Q11

 

 

 

 

 

 

 

Q14, a grounded

 

 

 

 

Q7

 

 

 

 

Q6

 

 

emitter amplifier.

 

 

 

C 1

 

 

 

 

 

 

Q1

Q2

 

 

 

Out

All other transistors

In+

In-

 

 

5p

 

 

 

Q4

 

 

 

 

 

Q8

 

in this block serve

 

 

 

 

 

 

 

 

 

 

 

its antipode, Q13.

I1

Q3

Q5

 

 

 

 

 

 

Q9

 

Note the

 

 

 

 

 

10u

 

 

 

 

 

 

 

 

 

Q12

 

three diode

 

 

 

 

Q14

 

R 1

R2

R 3

 

 

 

 

 

5k

5k

5 0 k

 

 

connected

 

 

 

 

 

 

 

SUB

-V

 

 

 

 

 

transistors Q7, Q8

Fig. 8-11: Op-amp with folded-cascode input stage and

and Q9. They set a

 

(almost) rail-to-rail output.

 

voltage for the base

of Q11. If you follow the emitter base of Q11, Q12 and Q14, you notice there are also three diodes in series to the V- rail. Thus, as the input signal to the output stage moves up and down (by a few millivolts), the current in Q11 fluctuates. It is this current, amplified by the size ratio of Q13 to Q10 (here about 6) that becomes the pull-up portion at the output. Q7 to Q9 are deliberately made larger than Q11, Q12 and Q14 so that the idle current in the output is small. This creates a small "dead-band" (see chapter 16) but, because of the large loop gain the distortion is very small (0.0004% for a ± 4.7Vp signal at 1kHz).

Phase / deg

Y 2

180

160

140

120

100

80

60

40

20

0

-20

Gain / dB

 

In this circuit we are

Y1

fortunate to find a node ideally

180

suited for the connection of a

160

140

compensation capacitor to the

120

output: at the base of Q6 the signal

Phase

100

has a phase opposite to that at the

80

Gain

output; the base of Q6 and the

60

40

collectors of Q4 and Q16 all

 

20

represent a high impedance; and

0

there is substantial voltage gain

-20 1 10 100 1k 10k 100k 1M 10M 100M

between it and the output. Which all says that this op-amp can be compensated (at unity gain) with a single 5pF capacitor, even though

Not all designs behave that well.

Preliminary Edition September 2004

8-6

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

Though the inputs can

work at the level of V- (or

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ground if you have only a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

single supply), the output

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cannot. It is the sad truth for a

/ V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

bipolar transistor that there is a

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Out

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

saturation voltage. Unlike an

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MOS transistor, which is

 

-2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

simply a voltage-controlled

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

resistor, the bipolar transistor is

 

-4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the interaction of two junctions

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

0.2

 

0.4

0.6

0.8

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

with different doping levels

 

 

 

Time/mSecs

 

 

 

 

 

 

200µSecs/div

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and sizes. Even when fully

 

Figure 8-13: Output swing is limited by the

 

turned on there is a minimum

saturation voltages of Q13 and Q14.

 

voltage drop of about 150mV

 

between emitter and collector in transistors Q13 and Q14. Thus the output can never be at the rails, only approach them.

The use of lateral PNP transistors at the input and the low operating current is not kind to noise: 27nV/rtHz at 10kHz and up (white noise). At 1Hz the flicker noise rises to 80nV/rtHz. If you have vertical PNP transistors at your disposal and can afford a higher operating current, these figures drop by a large factor (but you need to carefully re-simulate the entire circuit; frequency behavior is bound to be entirely different).

Another unsatisfactory parameter is the input current. Each input transistor runs at 5uA; with a minimum hFE of 100 (in a good process) the base current can be as high as 50nA. But there is a solution to this: more transistors.

 

 

 

 

 

 

 

 

 

 

 

V+

Q15

 

Q16

 

Q17

 

 

 

Q10

 

 

 

 

 

Q18

 

 

 

Q13

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q23

 

 

 

Q7

Q11

 

 

 

Q1

 

 

 

 

 

 

 

In+

 

In-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q6

 

C1

 

 

 

 

Q2

 

 

 

 

 

 

 

Out

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q22

 

Q4

 

Q8

5p

 

 

 

 

Q20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1

 

 

 

Q24

 

 

 

 

 

 

Q19

 

 

Q21

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10u

 

 

 

 

 

Q3

Q5

 

Q9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

R1

R2

R3

 

Q12

Q14

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q25

 

 

 

 

Q26

5k

5k

50k

SUB

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V-

Fig. 8-14: Op-amp of figure 8-11 with base-current compensation for the input stage.

Preliminary Edition September 2004

8-7

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

In figure 8-14 eight transistors are added. Their job is to pull as much current out of the bases of Q1 and Q2 as they naturally require, so that he external circuit does not have to do this.

The key is Q22. It is identical in size and design to Q1 and Q2, has the same operating current and very close to the same collector-base voltage (created by the base-emitter voltage of Q23 and the diode-connected transistor Q24). Thus its base current must be the same as those of the input transistor. This base current is mirrored by Q21, Q20 and Q19 and fed to the inputs, opposing the two base currents.

The cancellation of the base currents is never perfect of course, the current levels are too small to get precision matching. But the net input currents are down to 2nA, a 25:1 improvement.

In the last bipolar op-amp the goal is not an ultra-low input current, but low-noise performance with a reasonably low input current.

 

 

 

 

 

 

 

 

 

 

 

 

 

V+

Q18

 

 

 

1

Q3

Q4

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

Q5

 

 

 

 

 

 

200u

 

Q11

Q12

 

C1

 

 

 

 

 

Q6

I1

 

 

Q13

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10

Q15

5p

 

In-

Q1

Q2

In+

 

 

Out

10

10

 

 

 

 

Q16

 

 

 

 

 

 

 

 

 

 

Q8

Q14

 

 

Q9

Q7Q10

SUB

V-

Fig. 8-15: Bipolar op-amp optimized for low noise.

This circuit is almost identical to that of figure 8-1. The input transistors are now again NPN, but with the base-current canceling scheme added. The key transistor is Q13; since the hFE of an NPN transistor changes much less with current than that of an PNP device, we can afford to run it at twice the current and then divide the base current by two in the current mirrors Q12/Q11. This brings the input current down to 20nA.

The operating current is much higher than that of the previous circuit and the input transistors are large, which lowers the white noise to 5nV/rtHz.

Preliminary Edition September 2004

8-8

All rights reserved

Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

CMOS Op-Amps

CMOS devices have two advantages for op-amps over bipolar ones: there is no input current (at least not at DC) and, when the transistor is fully turned on, there is only a simple resistance between drain and source (not some complex cancellation of two junctions, resulting in an offset voltage and a resistance).

Let's again

 

 

M 1

 

 

M 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M 1 1

 

 

 

 

 

M 1 3

 

 

 

 

 

 

M 1 7

 

 

 

V +

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

first look at a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

simple design. As

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=30u

 

W=30u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=30

u

 

 

W=30

u

 

W=90

u

 

 

L=5u

 

 

L=5u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L=5u

 

 

 

 

L=5u

 

 

 

 

 

L=5u

 

 

 

 

in figure 8-13 a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M 2

 

 

M 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

"folded cascode"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M 1 2

 

 

 

 

 

 

M 1 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

input stage is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=100u

 

W=100

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

used, this time

 

 

L=2u

 

 

L=2u

 

 

 

 

 

 

 

 

 

 

 

 

 

W=100u

 

 

 

W=100u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M 5

 

 

 

M 6

 

 

 

L=2u

 

 

 

 

L=2u

 

 

 

C 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In+

 

 

 

 

 

In-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

using N-channel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Out

transistors (M5,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=40u

 

W=40u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 p

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 0 u

 

 

 

 

 

 

 

 

 

 

 

 

 

L = 2 u

 

 

L=2u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M6). The primary

 

 

 

 

 

 

 

M 7

 

 

M 9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

current, I1, is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=100u

 

W=100u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M 1 8

 

 

 

 

mirrored with M1-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L = 2 u

L=2u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M4 and than again

 

 

 

 

 

 

 

 

 

 

 

M 8

 

 

M10

 

 

 

 

 

 

 

 

 

 

 

 

 

M15

 

 

M 1 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

W=50

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

in M7-M10, using

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L=0.5

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the circuit of

 

 

 

 

 

 

 

 

 

 

W=10u

W=10u

 

 

 

 

 

 

 

 

 

 

 

W=20u W=20

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L = 5 u

 

L=5u

 

 

 

 

 

 

 

 

 

 

 

 

L=3u

 

L=3u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V-

figure 3-24, so the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 8-16: Op-amp with folded-cascode input stage and a

operating current

 

 

 

 

for the input pair is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

simple (and limited) output stage.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a constant 20uA.

The four transistors M1-M4 also steer M11/M12 and M13/14, producing two more accurate currents of 20uA each. The drains of the input transistors are connected to the sources of M12 and M14, which have a potential about 200mV below V+; thus the inputs can operate up to (and about 100mV above) the positive supply.

At balance (In+ = In-) the input pair diverts half of the 20uA current produced in M11 and M13, i.e. M12 and M14 are left with only half the current, about 10uA each.. The current out of M12 is mirrored in M15/M16 and opposed to that flowing out of M14. With a large input signal the two currents become unbalanced and each can vary between zero to 20uA. The voltage created by this unbalance is amplified by the output stage (M18 and a simple pull-up current source, M17). The idle current of the output stage is set by the ratio of the channel widths of M1 to M17, i.e. 60uA.

Preliminary Edition September 2004

8-9

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Camenzind: Designing Analog Chips

Chapter 9: Operational Amplifiers

 

 

The circuit is

 

 

Y 2

 

Y 1

 

 

 

 

 

 

 

compensated with a single

 

 

 

 

 

 

 

 

 

 

 

1pF capacitor, utilizing the

 

150

 

8 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Miller effect of M18. This

 

100

 

6 0

 

 

 

 

 

 

 

works well as long as the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

deg

5 0

dB

4 0

 

 

 

 

 

 

 

load capacitance is small.

 

 

 

 

 

 

 

/

 

/

 

 

 

 

 

 

 

 

At 10pF the stability is

 

Phase

 

Gain

2 0

 

 

 

 

 

 

 

marginal when connected

0

 

 

 

10pF Load

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

as a buffer; if the closed-

 

 

 

 

 

 

 

 

 

 

 

-50

 

 

 

 

 

 

 

 

 

loop gain is never lower

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

than 10 however, the op-

 

-100

 

10

100

1k

10k

100k

1M

10M

100M

amp is very stable, even

 

 

 

 

 

 

 

 

Frequency / Hertz

 

 

 

 

 

with a load capacitance as

 

 

 

 

 

 

 

 

 

Fig. 8-17: Phase margin becomes critical with a

high as 50pF.

 

 

 

 

 

 

capacitive load, unless the closed-loop gain is

 

 

With a load of greater

 

 

 

 

40dB or larger.

 

 

 

 

than 25kOhms (60uA) the

 

 

 

output can move rail-to-rail (or,

 

 

1.5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

more precisely, to within about

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

100mV of each rail).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

0.5

 

 

 

 

 

 

 

All CMOS examples in

 

/

 

 

 

 

 

 

 

 

 

Voltage

0

 

 

 

 

 

 

this chapter assume a split power

Output

 

 

 

 

 

 

 

supply of 3 Volts total, or ±1.5V;

-0.5

 

 

 

 

 

 

they are operational down to

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-1

 

 

 

 

 

 

±0.8V, though with reduced

 

 

-1.5 0

 

 

 

 

 

 

performance.

 

 

 

 

 

 

0.2

0.4

0.6

0.8

1

 

 

You need to be aware of

 

 

Time/mSecs

 

 

200µSecs/div

 

 

 

 

 

 

 

the changing open-loop gain. At

 

 

 

 

 

 

 

 

 

 

Fig. 8-18: Output swing.

 

 

 

ground level it amounts to 100dB.

 

 

 

 

 

 

 

 

As the output moves close to

 

either supply there is a marked drop (66dB at -1.4V, 60dB at +1.4V). This can be improved by making the output devices larger.

What cannot be improved is a fundamental dependence of loop gain on the load impedance. The lower the load, the lower the loop gain.

The input stage has a limited common-mode range. It will work up to about 100mV above the positive rail, but not below about -0.8V, i.e. about 0.7V above the negative rail.

Let's convert the wimpy output stage into a true rail-to-rail one, with some current capability:

Preliminary Edition September 2004

8-10

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